Chapter 4 the Algebra–Geometry Dictionary

Chapter 4 The Algebra–Geometry Dictionary

In this chapter, we will explore the correspondence between ideals and varieties. In §§1 and 2, we will prove the Nullstellensatz, a celebrated theorem which iden- tiﬁes exactly which ideals correspond to varieties. This will allow us to construct a “dictionary” between geometry and algebra, whereby any statement about varieties can be translated into a statement about ideals (and conversely). We will pursue this theme in §§3 and 4, where we will deﬁne a number of natural algebraic operations on ideals and study their geometric analogues. In keeping with the computational emphasis of the book, we will develop algorithms to carry out the algebraic operations. In §§5 and 6, we will study the more important algebraic and geometric concepts arising out of the Hilbert Basis Theorem: notably the possibility of decom- posing a variety into a union of simpler varieties and the corresponding algebraic notion of writing an ideal as an intersection of simpler ideals. In §7, we will prove the Closure Theorem from Chapter 3 using the tools developed in this chapter.

§1 Hilbert’s Nullstellensatz

In Chapter 1, we saw that a varietyV k n can be studied by passing to the ideal ⊆

I(V)= f k[x 1,...,x n] f(a)= 0 for alla V { ∈ | ∈ } of all polynomials vanishing onV. Hence, we have a map

afﬁne varieties ideals V −→ I(V).

Conversely, given an idealI k[x 1,...,x n], we can deﬁne the set ⊆ V(I)= a k n f(a)= 0 for allf I . { ∈ | ∈ }

© Springer International Publishing Switzerland 2015 175 D.A. Cox et al., Ideals, Varieties, and Algorithms, Undergraduate Texts in Mathematics, DOI 10.1007/978-3-319-16721-3_4 176 Chapter 4 The Algebra–Geometry Dictionary

The Hilbert Basis Theorem assures us thatV(I) is actually an afﬁne variety, for it tells us that there exists aﬁnite set of polynomialsf 1,...,f s I such thatI= ∈ f 1,...,f s , and we proved in Proposition 9 of Chapter 2, §5 thatV(I) is the set of common⟨ roots⟩ of these polynomials. Thus, we have a map

ideals affine varieties I −→ V(I). These two maps give us a correspondence between ideals and varieties. In this chapter, we will explore the nature of this correspondence. Thefirst thing to note is that this correspondence (more precisely, the mapV) is not one-to-one: different ideals can give the same variety. For example, x and x 2 are different ideals ink[x] which have the same varietyV(x) =V(x 2) =⟨ ⟩0 . More⟨ ⟩ serious problems can arise if thefieldk is not algebraically closed. For{ example,} consider the three polynomials 1,1+x 2, and 1+x 2 +x 4 inR[x]. These generate different ideals

2 2 4 I1 = 1 =R[x],I 2 = 1+x ,I 3 = 1+x +x , ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ but each polynomial has no real roots, so that the corresponding varieties are all empty: V(I1) =V(I 2) =V(I 3) = . ∅ Examples of polynomials in two variables without real roots include 1+x 2 +y 2 and 1+x 2 +y 4. These give different ideals inR[x,y] which correspond to the empty variety. Does this problem of having different ideals represent the empty variety go away if theﬁeldk is algebraically closed? It does in the one-variable case when the ring is k[x]. To see this, recall from §5 of Chapter 1 that any idealI ink[x] can be generated by a single polynomial becausek[x] is a principal ideal domain. So we can write I= f for some polynomialf k[x]. ThenV(I) is the set of roots off ; i.e., the⟨ set⟩ ofa k such thatf(a)∈ = 0. But sincek is algebraically closed, every nonconstant∈ polynomial ink[x] has a root. Hence, the only way that we could have V(I) = would be to havef be a nonzero constant. In this case, 1/f k. Thus, 1=(1/∅f) f I, which means thatg=g 1 I for allg k[x]. This shows ∈ thatI=k[x·] ∈is the only ideal ofk[x] that· ∈ represents∈ the empty variety whenk is algebraically closed. A wonderful thing now happens: the same property holds when there is more than one variable. In any polynomial ring, algebraic closure is enough to guarantee that the only ideal which represents the empty variety is the entire polynomial ring itself. This is the Weak Nullstellensatz, which is the basis of (and is equivalent to) one of the most celebrated mathematical results of the late nineteenth century, Hilbert’s Nullstellensatz. Such is its impact that, even today, one customarily uses the original German name Nullstellensatz: a word formed, in typical German fashion, from three simpler words: Null (= Zero), Stellen (= Places), Satz (= Theorem). §1 Hilbert’s Nullstellensatz 177

Theorem 1 (The Weak Nullstellensatz). Let k be an algebraically closedﬁeld and let I k[x 1,...,x n] be an ideal satisfyingV(I)= . Then I=k[x 1,...,x n]. ⊆ ∅ Proof. Our proof is inspired by GLEBSKY (2012). We will prove the theorem in contrapositive form: I!k[x 1,...,x n] = V(I)= . ⇒ ̸ ∅ We will make frequent use of the standard equivalenceI=k[x 1,...,x n] 1 I. This is part (a) of Exercise 16 from Chapter 1, §4. ⇔ ∈ Givena k andf k[x 1,...,x n], let ¯f=f(x 1,...,x n 1,a) k[x 1,...,x n 1]. Similar to∈ Exercise∈2 of Chapter 3, §5 and Exercise 15 of Chapter− ∈ 3, §6, the set −

Ix a = ¯f f I n= { | ∈ } is an ideal ofk[x 1,...,x n 1]. The key step in the proof is the following claim. −

Claim. Ifk is algebraically closed andI!k[x 1,...,x n] is a proper ideal, then there isa k such thatI xn=a !k[x 1,...,x n 1]. ∈ −

Once we prove the claim, an easy induction gives elementsa 1,...,a n k such ∈ thatI xn=an,...,x1=a1 !k. But the only ideals ofk are 0 andk (Exercise 3), so that { } Ixn=an,...,x1=a1 = 0 . This implies(a 1,...,a n) V(I). We conclude thatV(I)= , and the theorem{ will} follow. ∈ ̸ ∅ To prove the claim, there are two cases, depending on the size ofI k[x n]. ∩ Case 1.I k[x n]= 0 . Letf I k[x n] be nonzero,and note thatf is nonconstant, ∩ ̸ { } ∈ ∩ since otherwise 1 I k[x n] I, contradicting? I=k[x 1,...,x n]. ∈ ∩ ⊆ r ̸ mi Sincek is algebraically closed,f=c (xn b i) wherec,b 1,...,b r k and i=1 − ∈ c= 0. Suppose thatI xn=bi =k[x 1,...,x n 1] for alli. Then for alli there isB i I ̸ − ∈ withB i(x1,...,x n 1,b i) = 1. This implies that −

1=B i(x1,...,x n 1,b i) =B i(x1,...,x n 1,x n (x n b i)) =B i +A i(xn b i) − − − − − for someA i k[x 1,...,x n]. Since this holds fori=1,...,r, we obtain ∈ @r @r mi mi 1= (Ai(xn b i) +B i) =A (xn b i) +B, − − i=1 i=1 ? ? r mi r mi 1 whereA= A andB I. This and (xn b i) =c − f I imply that i=1 i ∈ i=1 − ∈ 1 I, which contradictsI=k[x 1,...,x n]. ThusI xn=bi =k[x 1,...,x n 1] for somei. ∈ ̸ ̸ − Thisb i is the desireda.

Case 2.I k[x n] = 0 . Let g 1,...,g t be a Gröbner basis ofI for lex order with ∩ { } { } x1 > >x n and write ···

αi αi (1)g i =c i(xn)x + terms

αi wherec i(xn) k[x n] is nonzero andx is a monomial inx 1,...,x n 1. ∈ − 178 Chapter 4 The Algebra–Geometry Dictionary

Now picka k such thatc i(a)= 0 for alli. This is possible since algebraically closedﬁelds are∈ inﬁnite by Exercise̸ 4. It is easy to see that the polynomials

¯gi =g i(x1,...,x n 1,a) − form a basis ofI xn=a (Exercise 5). Substitutingx n =a into equation ( 1), one easily αi αi sees that LT(¯gi) =c i(a)x sincec i(a)= 0. Also note thatx = 1, since otherwise ̸ ̸ gi =c i I k[x n] = 0 , yetc i = 0. This shows that LT(¯gi) is nonconstant for alli. ∈ ∩ { } ̸ We claim that the¯gi form a Gröbner basis ofI xn=a. Assuming the claim, it follows that 1/ I xn=a since no LT(¯gi) can divide 1. ThusI xn=a =k[x 1,...,x n 1], which is what we∈ want to show. ̸ − To prove the claim, takeg i,g j G and consider the polynomial ∈ xγ xγ S=c j(xn) gi c i(xn) gj, xαi − xαj wherex γ = lcm(xαi ,x αj ). By construction,x γ > LT("S) (be sure you understand t this). SinceS I, it has a standard representationS= Al gl. Then evaluating ∈ l=1 atx n =a gives γ γ " x x t ¯ cj(a) ¯gi c i(a) ¯gj = S= l Al ¯gl. xαi − xαj =1

αi Since LT(¯gi) =c i(a)x , we see that S is theS-polynomialS(¯g i,¯gj) up to the nonzero constantc i(a)cj(a). Then

γ x > LT(S) LT(Algl),A lgl =0 ≥ ̸ implies that γ x > LT(A¯l¯gl), A¯l¯gl =0 ̸ γ (Exercise 6). Sincex = lcm(LM(¯gi), LM(¯gj)), it follows thatS(¯gi,¯gj) has an lcm representation for alli,j and hence is a Gröbner basis by Theorem 6 of Chapter 2, §9. This proves the claim and completes the proof of the theorem.!

In the special case whenk=C, the Weak Nullstellensatz may be thought of as the “Fundamental Theorem of Algebra for multivariable polynomials”—every system of polynomials that generates an ideal strictly smaller thanC[x 1,...,x n] has a common zero inC n. The Weak Nullstellensatz also allows us to solve the consistency problem from §2 of Chapter 1. Recall that this problem asks whether a system

f1 =0,

f2 =0, . .

fs =0 §1 Hilbert’s Nullstellensatz 179 of polynomial equations has a common solution inC n. The polynomials fail to have a common solution if and only ifV(f 1,...,f s) = . By the Weak Nullstellensatz, the ∅ latter holds if and only if 1 f 1,...,f s . Thus, to solve the consistency problem, we need to be able to determine∈⟨ whether⟩ 1 belongs to an ideal. This is made easy by the observation that for any monomial ordering, 1 is the only reduced Gröbner { } basis of the ideal 1 =k[x 1,...,x n]. ⟨ ⟩ To see this, let g 1,...,g t be a Gröbner basis ofI= 1 . Thus, 1 LT(I) = { } ⟨ ⟩ ∈⟨ ⟩ LT(g1),..., LT(gt) , and then Lemma 2 of Chapter 2, §4 implies that 1 is divisible ⟨ ⟩ by some LT(gi), say LT(g1). This forces LT(g1) to be constant. Then every other LT(gi) is a multiple of that constant, so thatg 2,...,g t can be removed from the Gröbner basis by Lemma 3 of Chapter 2, §7. Finally, since LT(g1) is constant,g 1 itself is constant since every nonconstant monomial is> 1 (see Corollary 6 of Chapter 2, §4). We can multiply by an appropriate constant to makeg 1 = 1. Our reduced Gröbner basis is thus 1 . To summarize, we have the{ following} consistency algorithm: if we have polynomialsf 1,...,f s C[x 1,...,x n], we compute a reduced Gröbner basis of the ideal they generate with∈ respect to any ordering. If this basis is 1 , the polynomials have { } no common zero inC n; if the basis is not 1 , they must have a common zero. Note that this algorithm works over any algebraically{ } closedfield. If we are working over afieldk which is not algebraically closed, then the consistency algorithm still works in one direction: if {1} is a reduced Gröbner basis of f 1,...,f s , then the equationsf 1 = =f s = 0 have no common solution. The ⟨converse is⟩ not true, as shown by the··· examples preceding the statement of the Weak Nullstellensatz. Inspired by the Weak Nullstellensatz, one might hope that the correspondence between ideals and varieties is one-to-one provided only that one restricts to algebraically closedfields. Unfortunately, our earlier exampleV(x) =V(x 2) = 0 works over anyfield. Similarly, the ideals x 2,y and x,y (and, for that matter,{ } xn,y m wheren andm are integers greater⟨ than one)⟩ are⟨ different⟩ but define the same⟨ variety:⟩ namely, the single point (0,0) k 2. These examples illustrate a basic reason why different ideals can de{fine the}⊆ same variety (equivalently, that the mapV can fail to be one-to-one): namely, a power of a polynomial vanishes on the same set as the original polynomial. The Hilbert Nullstellensatz states that over an algebraically closedfield, this is the only reason that different ideals can give the same variety: if a polynomialf vanishes at all points of some varietyV(I), then some power off must belong toI.

Theorem 2 (Hilbert’s Nullstellensatz). Let k be an algebraically closedﬁeld. If f,f 1,...,f s k[x 1,...,x n], then f I(V(f 1,...,f s)) if and only if ∈ ∈ m f f 1,...,f s ∈ ⟨ ⟩ for some integer m 1. ≥ Proof. Given a nonzero polynomialf which vanishes at every common zero of the polynomialsf 1,...,f s, we must show that there exists an integerm 1 and ≥ 180 Chapter 4 The Algebra–Geometry Dictionary polynomialsA 1,...,A s such that

!s m f = Ai fi. i=1 The most direct proof is based on an ingenious trick. Consider the ideal

I˜= f 1,...,f s ,1 yf k[x 1,...,x n,y], ⟨ − ⟩⊆ wheref,f 1,...,f s are as above. We claim that

V(I˜)= . ∅ n+1 To see this, let(a 1,...,a n,a n 1) k . Either + ∈ (a 1,...,a n) is a common zero off 1,...,f s, or • (a 1,...,a n) is not a common zero off 1,...,f s. • In theﬁrst casef(a 1,...,a n) = 0 sincef vanishes at any common zero off 1,...,f s. Thus, the polynomial 1 yf takes the value 1 a n 1f(a 1,...,a n) =1= 0 at − − + ̸ the point(a 1,...,a n,a n 1). In particular,(a 1,...,a n,a n 1)/ V( I˜). In the second + + ∈ case, for somei,1 i s, we must havef i(a1,...,a n) = 0. Thinking off i as a function ofn+ 1≤ variables≤ which does not depend on the last variable, we have fi(a1,...,a n,a n+1)= 0. In particular, we again conclude that(a 1,...,a n,a n+1)/ ̸ n+1 ∈ V(I˜). Since(a 1,...,a n,a n+1) k was arbitrary, we obtainV( I˜)= , as claimed. Now apply the Weak Nullstellensatz∈ to conclude that 1 I˜. Hence∅ ∈ !s (2) 1= pi(x1,...,x n,y)f i +q(x 1,...,x n,y)(1 yf) − i=1 for some polynomialsp i,q k[x 1,...,x n,y]. Now sety=1/f(x 1,...,x n). Then relation (2) above implies that∈

!s (3) 1= pi(x1,...,x n,1/f)f i. i=1 Multiply both sides of this equation by a powerf m, wherem is chosen sufﬁciently large to clear all the denominators. This yields

!s m (4) f = Ai fi, i=1 for some polynomialsA i k[x 1,...,x n], which is what we had to show. ∈ ! §2 Radical Ideals and the Ideal–Variety Correspondence 181

EXERCISES FOR §1

2 3 3 1. Recall thatV(y x ,z x ) is the twisted cubic inR . − − a. Show thatV((y x 2)2 + (z x 3)2) is also the twisted cubic. − − n b. Show that any varietyV(I) R ,I R[x 1,...,x n], can be defined by a single equation (and hence by a principal⊆ ideal).⊆ 2. LetJ= x 2 +y 2 1,y 1 . Findf I(V(J)) such thatf/ J. 3. Prove that⟨ 0 and−k are− the⟩ only ideals∈ of afieldk. ∈ 4. Prove that{ an} algebraically closedfieldk must be infinite. Hint: Givenn elements a1,...,a n of afieldk, can you write down a nonconstant polynomialf k[x] with ∈ the property thatf(a i) = 1 for alli?

5. In the proof of Theorem 1, prove thatI xn=a = ¯g1,...,¯gt . δ ⟨ ⟩ δ 6. In the proof of Theorem 1, letx be a monomial inx 1,...,x n 1 satisfyingx > LT(f) δ − for somef k[x 1,...,x n]. Prove thatx > LT(¯f), where ¯f=f(x 1,...,x n 1,a). − 7. In deducing∈ Hilbert’s Nullstellensatz from the Weak Nullstellensatz, we made the sub- stitutiony=1/f(x 1,...,x n) to deduce relations (3) and (4) from (2). Justify this rigor- ously. Hint: In what set is 1/f contained? 8. The purpose of this exercise is to show that ifk is anyfield that is not algebraically closed, then any varietyV k n can be defined by a single equation. n ⊆ n 1 a. Ifg=a 0x +a 1x − + +a n 1x+a n is a polynomial of degreen inx, define the homogenization gh of···g with respect− to some variabley to be the polynomial h n n 1 n 1 n g =a 0x +a 1x − y+ +a n 1xy − +a ny . Show thatg has a root ink if and only if there is(a,b) k ···2 such that− (a,b)= (0,0) andg h(a,b) = 0. Hint: Show that h n h ∈ ̸ g (a,b) =b g (a/b,1) whenb= 0. b. Ifk is not algebraically closed,̸ show that there existsf k[x,y] such that the variety ∈2 defined byf= 0 consists of just the origin(0,0) k . Hint: Choose a polynomial in k[x] with no root ink and consider its homogenization.∈ c. Ifk is not algebraically closed, show that for each integerl> 0 there existsf ∈l k[x1,...,x l] such that the only solution off= 0 is the origin(0,...,0) k . Hint: Use induction onl and part (b) above. ∈ n d. IfW=V(g 1,...,g s) is any variety ink , wherek is not algebraically closed, then show thatW can be defined by a single equation. Hint: Consider the polynomial f(g 1,...,g s) wheref is as in part (c). 9. Letk be an arbitraryfield and letS be the subset of all polynomials ink[x 1,...,x n] that n have no zeros ink . IfI is any ideal ink[x 1,...,x n] such thatI S= , show that V(I)= . Hint: Whenk is not algebraically closed, use the previous∩ exercise.∅ ̸ ∅ 10. In Exercise 1, we encountered two ideals inR[x,y] that give the same nonempty variety. Show that one of these ideals is contained in the other. Can youfind two ideals inR[x,y], neither contained in the other, which give the same nonempty variety? Can you do the same forR[x]?

§2 Radical Ideals and the Ideal–Variety Correspondence

To further explore the relation between ideals and varieties, it is natural to recast Hilbert’s Nullstellensatz in terms of ideals. Can we characterize the kinds of ideals that appear as the ideal of a variety? In other words, can we identify those ideals that consist of all polynomials which vanish on some varietyV? The key observation is contained in the following simple lemma. 182 Chapter 4 The Algebra–Geometry Dictionary

Lemma 1. Let V be a variety. If f m I(V), then f I(V). ∈ ∈ Proof. Leta V. Iff m I(V), then(f(a)) m = 0. But this can happen only if f(a)= 0. Since∈ a V was arbitrary,∈ we must havef I(V). ∈ ∈ ! Thus, an ideal consisting of all polynomials which vanish on a varietyV has the property that if some power of a polynomial belongs to the ideal, then the polynomial itself must belong to the ideal. This leads to the following deﬁnition.

Deﬁnition 2. An idealI is radical iff m I for some integerm 1 implies that f I. ∈ ≥ ∈ Rephrasing Lemma 1 in terms of radical ideals gives the following statement.

Corollary 3. I(V) is a radical ideal.

On the other hand, Hilbert’s Nullstellensatz tells us that the only way that an arbitrary idealI can fail to be the ideal of all polynomials vanishing onV(I) is for I to contain powersf m of polynomialsf which are not inI—in other words, forI to fail to be a radical ideal. This suggests that there is a one-to-one correspondence between afﬁne varieties and radical ideals. To clarify this and get a sharp statement, it is useful to introduce the operation of taking the radical of an ideal.

Deﬁnition 4. LetI k[x 1,...,x n] be an ideal. The radicalofI, denoted √I, is the set ⊆ f f m I for some integerm 1 . { | ∈ ≥ } Note that we always haveI √I sincef I impliesf 1 I and, hence,f √I by deﬁnition. It is an easy exercise⊆ to show∈ that an idealI is∈ radical if and∈ only if I= √I. A somewhat more surprising fact is that the radical of an ideal is always an ideal. To see what is at stake here, consider, for example, the idealJ= x 2,y 3 k[x,y]. Although neitherx nory belongs toJ, it is clear thatx √J and⟨ y √⟩ ⊆J. Note that(x y) 2 =x 2y2 J sincex 2 J; thus,x y √∈J. It is less obvious∈ that x+y √J·. To see this, observe∈ that ∈ · ∈ ∈ (x+y) 4 =x 4 +4x 3y+6x 2y2 +4xy 3 +y 4 J ∈ becausex 4,4x 3y,6x 2y2 J (they are all multiples ofx 2) and 4xy3,y 4 J (because they are multiples ofy 3).∈ Thus,x+y √J. By way of contrast, neither∈ xy norx+y belong toJ. ∈

Lemma 5. If I is an ideal in k[x1,...,x n], then √I is an ideal in k[x1,...,x n] containing I. Furthermore, √I is a radical ideal.

Proof. We have already shown thatI √I. To show √I is an ideal, supposef,g √I. Then there are positive integers⊆m andl such thatf m,g l I. In the binomial∈ m+l 1 i j ∈ expansion of(f+g) − every term has a factorf g withi+j=m+l 1. Since eitheri morj l, eitherf i org j is inI, whencef igj I and every− term in the ≥ ≥ ∈ §2 Radical Ideals and the Ideal–Variety Correspondence 183

m+l 1 binomial expansion is inI. Hence,(f+g) − I and, therefore,f+g √I. ∈ m ∈ Finally, supposef √I andh k[x 1,...,x n]. Thenf I for some integerm 1. SinceI is an ideal,∈ we have(h∈f) m =h mf m I. Hence,∈hf √I. This shows≥ that · ∈ ∈ √I is an ideal. In Exercise 4, you will show that √I is a radical ideal.! We are now ready to state the ideal-theoretic form of the Nullstellensatz.

Theorem 6 (The Strong Nullstellensatz). Let k be an algebraically closedﬁeld. If I is an ideal in k[x1,...,x n], then

I(V(I)) = √I.

Proof. We certainly have √I I(V(I)) becausef √I implies thatf m I for somem. Hence,f m vanishes on⊆V(I), which implies∈ thatf vanishes onV(I). Thus,∈ f I(V(I)). ∈ Conversely, takef I(V(I)). Then, by deﬁnition,f vanishes onV(I). By Hilbert’s Nullstellensatz,∈ there exists an integerm 1 such thatf m I. But this means thatf √I. Sincef was arbitrary,I(V(I)) ≥ √I, and we are done.∈ ∈ ⊆ ! It has become a custom, to which we shall adhere, to refer to Theorem 6 as the Nullstellensatz with no further qualiﬁcation. The most important consequence of the Nullstellensatz is that it allows us to set up a “dictionary” between geometry and algebra. The basis of the dictionary is contained in the following theorem.

Theorem 7 (The Ideal–Variety Correspondence). Let k be an arbitraryfield. (i) The maps affine varieties I ideals −→ and ideals V affine varieties −→ are inclusion-reversing, i.e., if I1 I 2 are ideals, thenV(I 1) V(I 2) and, ⊆ ⊇ similarly, if V1 V 2 are varieties, thenI(V 1) I(V 2). (ii) For any variety⊆ V, ⊇ V(I(V)) =V, so thatI is always one-to-one. On the other hand, any ideal I satisfies

V(√I)=V(I).

(iii) If k is algebraically closed, and if we restrict to radical ideals, then the maps

afﬁne varieties I radical ideals −→ and radical ideals V afﬁne varieties −→ are inclusion-reversing bijections which are inverses of each other. 184 Chapter 4 The Algebra–Geometry Dictionary

Proof. (i) The proof will be covered in the exercises. n (ii) LetV=V(f 1,...,f s) be an affine variety ink . Since everyf I(V) vanishes onV, the inclusionV V(I(V)) follows directly from the definition∈ ⊆ ofV. Going the other way, note thatf 1,...,f s I(V) by the definition ofI, ∈ and, thus, f 1,...,f s I(V). SinceV is inclusion-reversing, it follows that ⟨ ⟩ ⊆ V(I(V)) V( f 1,...,f s ) =V. This proves thatV(I(V)) =V, and, consequently, I is one-to-one⊆ ⟨ since it has⟩ a left inverse. Thefinal assertion of part (ii) is left as an exercise. (iii) SinceI(V) is radical by Corollary 3, we can think ofI as a function which takes varieties to radical ideals. Furthermore, we already knowV(I(V)) =V for any varietyV. It remains to proveI(V(I)) =I wheneverI is a radical ideal. This is easy: the Nullstellensatz tells usI(V(I)) = √I, andI being radical implies √I=I (see Exercise 4). This gives the desired equality. Hence,V andI are inverses of each other and, thus, define bijections between the set of radical ideals and affine varieties. The theorem is proved. ! As a consequence of this theorem, any question about varieties can be rephrased as an algebraic question about radical ideals (and conversely), provided that we are working over an algebraically closedfield. This ability to pass between algebra and geometry will give us considerable power. In view of the Nullstellensatz and the importance it assigns to radical ideals, it is natural to ask whether one can compute generators for the radical from generators of the original ideal. In fact, there are three questions to ask concerning an ideal I= f 1,...,f s : ⟨ ⟩ (Radical Generators) Is there an algorithm which produces a set g 1,...,g m of • { } polynomials such that √I= g 1,...,g m ? (Radical Ideal) Is there an algorithm⟨ which⟩ will determine whetherI is radical? • (Radical Membership) Givenf k[x 1,...,x n], is there an algorithm which will • determine whetherf √I? ∈ ∈ The existence of these algorithms follows from the work of HERMANN (1926) [see also MINES,RICHMAN, and RUITENBERG (1988) and SEIDENBERG (1974, 1984) for more modern expositions]. More practical algorithms forfinding radicals follow from the work of GIANNI,TRAGER and ZACHARIAS (1988), KRICK and LOGAR (1991), and EISENBUD,HUNEKE and VASCONCELOS (1992). These algorithms have been implemented in CoCoA, Singular, and Macaulay2, among others. See, for example, Section 4.5 of GREUEL and PFISTER (2008). For now, we will settle for solving the more modest radical membership problem. To test whetherf √I, we could use the ideal membership algorithm to check whetherf m I for∈ all integersm> 0. Thisis notsatisfactory becausewe mighthave to go to very∈ large powers ofm, and it will never tell us iff/ √I (at least, not until ∈ we work out a priori bounds onm). Fortunately, we can adapt theA proof of Hilbert’s Nullstellensatz to give an algorithm for determining whetherf f 1,...,f s . ∈ ⟨ ⟩ Proposition 8 (Radical Membership). Let k be an arbitraryfield and let I= f 1,...,f s k[x 1,...,x n] be an ideal. Then f √I if and only if the constant ⟨ ⟩ ⊆ ∈ §2 Radical Ideals and the Ideal–Variety Correspondence 185 polynomial1 belongs to the ideal I˜= f 1,...,f s,1 yf k[x 1,...,x n,y], in ⟨ − ⟩ ⊆ which case I˜=k[x 1,...,x n,y]. Proof. From equations (2), (3), and (4) in the proof of Hilbert’s Nullstellensatz in §1, we see that 1 I˜ impliesf m I for somem, which, in turn, impliesf √I. Going the other way,∈ suppose that∈f √I. Thenf m I I˜ for somem∈. But we also have 1 yf I˜, and, consequently,∈ ∈ ⊆ − ∈ m m m m m m m 1 m 1 1=y f + (1 y f ) =y f + (1 yf) (1+yf+ +y − f − ) I˜, − · − · ··· ∈ as desired. ! Proposition 8, together with our earlier remarks on determining whether 1 belongs to an ideal (see the discussion of the consistency problem in §1), imme- diatelyA leads to the following radical membership algorithm: to determine if f f 1,...,f s k[x 1,...,x n], we compute a reduced Gröbner basis of the ideal ∈ ⟨ ⟩ ⊆ f 1,...,f s,1 yf k[x 1,...,x n,y] with respect to some ordering. If the result is ⟨1 , thenf −√⟩⊆I. Otherwise,f/ √I. { }As an example,∈ consider the∈ idealI= xy 2 +2y 2,x 4 2x 2 +1 ink[x,y]. Let us test iff=y x 2 + 1 lies in √I. Using⟨ lex order onk[x−,y,z], one⟩ checks that the ideal − I˜= xy 2 +2y 2,x 4 2x 2 +1,1 z(y x 2 +1) k[x,y,z] ⟨ − − − ⟩ ⊆ has reduced Gröbner basis 1 . It follows thaty x 2 +1 √I by Proposition 8. Using the division algorithm,{ } we can check what− power of∈y x 2 + 1 lies inI: − G y x 2 +1 =y x 2 +1, − G − (y x 2 +1) 2 = 2x 2y+2y, − G − (y x 2 +1) 3 =0, − whereG= x 4 2x 2 +1,y 2 is a Gröbner basis ofI with respect to lex order and pG is the remainder{ − ofp on division} byG. As a consequence, we see that(y x 2 +1) 3 I, but no lower power ofy x 2 + 1 is inI (in particular,y x 2 +1/ −I). ∈ We can also see what− is happening in this example− geometrically.∈ As a set, V(I) = ( 1,0) , but (speaking somewhat imprecisely) every polynomial inI vanishes{ to± order} at least 2 at each of the two points inV(I). This is visible from the form of the generators ofI if we factor them:

xy2 +2y 2 =y 2(x+2) andx 4 2x 2 +1=(x 2 1) 2. − − Even thoughf=y x 2 + 1 also vanishes at( 1,0),f only vanishes to order 1 there. We must take− a higher power off to obtain± an element ofI. We will end this section with a discussion of the one case where we can compute the radical of an ideal, which is when we are dealing with a principal idealI= f . A nonconstant polynomialf is said to be irreducible if it has the property that ⟨whenever⟩ f=g h for some polynomialsg andh, then eithergorh is a constant. As noted in §2· of Appendix A, any nonconstant polynomialf can always be written 186 Chapter 4 The Algebra–Geometry Dictionary as a product of irreducible polynomials. By collecting the irreducible polynomials which differ by constant multiples of one another, we can writef in the form

f=cf a1 f ar ,c k, 1 ··· r ∈ where thef i’s, 1 i r, are distinct irreducible polynomials, meaning thatf i andf j are not constant≤ multiples≤ of one another wheneveri=j. Moreover, this expression ̸ forf is unique up to reordering thef i’s and up to multiplying thef i’s by constant multiples. (This unique factorization is Theorem 2 from Appendix A, §2.) If we havef expressed as a product of irreducible polynomials, then it is easy to write down the radical of the principal ideal generated byf.

Proposition 9. Let f k[x 1,...,x n]andI= f be the principal ideal generated a ∈ ⟨ ⟩ by f . If f=cf 1 f ar is the factorization of f into a product of distinct irreducible 1 ··· r polynomials, then A √I= f = f 1 f2 f r . ⟨ ⟩ ⟨ ··· ⟩ Proof. Weﬁrst show thatf 1 f2 f r belongs to √I. LetN be an integer strictly ··· greater than the maximum ofa 1,...,a r. Then

N N a1 N a2 N ar (f 1 f2 f r) =f − f − f − f ··· 1 2 ··· r N is a polynomial multiple off . This shows that(f 1 f2 f r) I, which implies that ··· ∈ f1 f2 f r √I. Thus f 1 f2 f r √I. Conversely,··· ∈ suppose⟨ that···g ⟩ ⊆√I. Then there exists a positive integerM such thatg M I= f , so thatg ∈ M is a multiple off and hence a multiple of each ∈ ⟨ ⟩ M irreducible factorf i off. Thus,f i is an irreducible factor ofg . However, the unique factorization ofg M into distinct irreducible polynomials is theMth power of the factorization ofg. It follows that eachf i is an irreducible factor ofg. This implies thatg is a polynomial multiple off 1 f2 f r and, therefore,g is contained in the ideal ··· f 1 f2 f r . The proposition is proved. ⟨ ··· ⟩ ! In view of Proposition 9, we make the following deﬁnition:

Deﬁnition 10.Iff k[x 1,...,x n] is a polynomial,A we deﬁne the reduction off, ∈ denotedf red, to be the polynomial such that f red = f . A polynomial is said to ⟨ ⟩ ⟨ ⟩ be reduced (or square-free) iff=f red.

Thus,f red is the polynomialf with repeated factors “stripped away.” So, for ex- 2 3 2 ample, iff=(x+y ) (x y), thenf red = (x+y )(x y). Note thatf red is only unique up to a constant factor− ink. − The usefulness of Proposition 9 is mitigated by the requirement thatf be factored into irreducible factors. We might ask if there is an algorithm to computef red fromf without factoringfﬁrst. It turns out that such an algorithm exists. To state the algorithm, we will need the notion of a greatest common divisor of two polynomials.

Deﬁnition 11. Letf,g k[x 1,...,x n]. Thenh k[x 1,...,x n] is called a greatest common divisoroff and∈ g, and denotedh= gcd∈(f,g), if §2 Radical Ideals and the Ideal–Variety Correspondence 187

(i)h dividesf andg. (ii) Ifp is any polynomial that divides bothf andg, thenp dividesh. It is easy to show that gcd(f,g) exists and is unique up to multiplication by a nonzero constant ink (see Exercise 9). Unfortunately, the one-variable algorithm forﬁnding the gcd (i.e., the Euclidean Algorithm) does not work in the case of several variables. To see this, consider the polynomials xyand xz ink[x,y,z]. Clearly, gcd(xy, xz) =x. However, no matter what term ordering we use, dividing xyby xz gives 0 plus remainder xy and dividing xz by xy gives 0 plus remainder xz. As a result, neither polynomial “reduces” with respect to the other and there is no next step to which to apply the analogue of the Euclidean Algorithm. Nevertheless, there is an algorithm for calculating the gcd of two polynomials in several variables. We defer a discussion of it until the next section after we have studied intersections of ideals. For the purposes of our discussion here, let us assume that we have such an algorithm. We also remark that given polynomialsf 1,...,f s ∈ k[x1,...,x n], one can deﬁne gcd(f 1,f 2,...,f s) exactly as in the one-variable case. There is also an algorithm for computing gcd(f 1,f 2,...,f s). Using this notion of gcd, we can now give a formula for computing the radical of a principal ideal.

Proposition 12. Suppose that k is aﬁeld containing the rational numbersQ and let I= f be a principal ideal in k[x 1,...,x n] . Then √I= f red , where ⟨ ⟩ ⟨ ⟩ f fred = 6 (. gcd f, ∂f , ∂f ,..., ∂f ∂x1 ∂x2 ∂xn

Proof. Writingf as in Proposition 9, we know that √I= f 1 f2 f r . Thus, it sufﬁces to show that ⟨ ··· ⟩ 0 / f f ∂ ∂ a1 1 a2 1 ar 1 (1) gcd f, ,..., =f 1 − f2 − f r − . ∂x1 ∂xn ··· Weﬁrst use the product rule to note that 0 / ∂f a1 1 a2 1 ar 1 ∂f1 ∂fr =f 1 − f2 − f r − a1 f2 f r + +a r f1 f r 1 . ∂xj ··· ∂xj ··· ··· ··· − ∂xj

a1 1 a2 1 ar 1 This proves thatf − f − f − divides the gcd. It remains to show that for 1 2 ··· r eachi, there is some ∂f which is not divisible byf ai . ∂xj i ai Writef=f i hi, whereh i is not divisible byf i. Sincef i is nonconstant, some variablex j must appear inf i. The product rule gives us 0 / ∂f ai 1 ∂fi ∂hi =f i − a1 hi +f i . ∂xj ∂xj ∂xj

ai ∂fi If this expression is divisible byf , then hi must be divisible byf i. Sincef i is i ∂xj irreducible and does not divideh , this forcesf to divide ∂fi . In Exercise 13, you i i ∂xj 188 Chapter 4 The Algebra–Geometry Dictionary

∂fi ∂fi will show that is nonzero sinceQ k andx j appears inf i. As also has ∂xj ⊆ ∂xj smaller total degree thanf , it follows thatf cannot divide ∂fi . Consequently, ∂f is i i ∂xj ∂xj ai not divisible byf i , which proves (1), and the proposition follows.! It is worth remarking that forﬁelds which do not containQ, the above formula forf red may fail (see Exercise 13).

EXERCISES FOR §2 @ 1. Given aﬁeldk (not necessarily algebraically closed), show that x2,y 2 = x,y and, @ ⟨ ⟩ ⟨ ⟩ more generally, show that xn,y m = x,y for any positive integersn andm. ⟨ ⟩ ⟨ ⟩ 2. Letf andg be distinct nonconstant polynomials ink[x,y] and letI= f 2,g 3 . Is it ⟨ ⟩ necessarily true that √I= f,g ? Explain. 2 ⟨ ⟩ 2 3. Show that x +1 R[x] is a radical ideal, but thatV(x +1) is the empty variety. ⟨ ⟩ ⊆ 4. LetI be an ideal ink[x 1,...,x n], wherek is an arbitraryﬁeld. a. Show that √I is a radical ideal. b. Show thatI@ is radical if and only ifI= √I. c. Show that √I= √I. 5. Prove thatI andV are inclusion-reversing and thatV( √I)=V(I) for any idealI. 6. LetI be an ideal ink[x 1,...,x n].

mi m1+m2 1 a. In the special case when √I= f 1,f 2 , withf I, prove thatf − I for all ⟨ ⟩ i ∈ ∈ f √I. b. Now∈ prove that for anyI, there exists a single integerm such thatf m I for all ∈ f √I. Hint: Write √I= f 1,...,f s . ∈ ⟨ ⟩ 7. Determine whether the following polynomials lie in the following radicals. If the answer is yes, what is@ the smallest power of the polynomial that lies in the ideal? a. Isx+y x3,y 3, xy(x+y) ? ∈ ⟨@ ⟩ b. Isx 2 +3xz x+z,x 2y,x z 2 ? ∈ ⟨ − ⟩ 2 2 8. Letf 1 =y +2xy 1 andf 2 =x + 1. Prove that f 1,f 2 is not a radical ideal. Hint: − ⟨ ⟩ What isf 1 +f 2? 9. Givenf,g k[x 1,...,x n], use unique factorization to prove that gcd(f,g) exists. Also prove that∈ gcd(f,g) is unique up to multiplication by a nonzero constant ofk. 10. Prove the following ideal-theoretic characterization of gcd(f,g): given polynomials f,g,h ink[x 1,...,x n], thenh= gcd(f,g) if and only ifh is a generator of the smallest principal ideal containing f,g (i.e., if h J, wheneverJ is a principal ideal such thatJ f,g ). ⟨ ⟩ ⟨ ⟩ ⊆ 11. Find⊇⟨ a basis⟩ for the ideal @ x5 2x 4 +2x 2 x,x 5 x 4 2x 3 +2x 2 +x 1 . ⟨ − − − − − ⟩ Compare with Exercise 17 of Chapter 1, §5. 5 4 3 2 4 2 2 3 3 3 2 4 3 4 5 2 5 6 7 12. Letf=x +3x y+@3x y 2x y +x y 6x y 6x y +x y 2xy +3x y +3xy +y − − − − ∈ Q[x,y]. Compute f . ⟨ ⟩ 13. Afieldk has characteristic zero if it contains the rational numbersQ; otherwise,k has positive characteristic. 2 2 a. Letk be thefieldF 2 from Exercise 1 of Chapter 1, §1. Iff=x 1 + +x n ∂f ··· ∈ 2[x1,...,x n], then show that = 0 for alli. Conclude that the formula given in F ∂xi Proposition 12 may fail when thefield isF 2. §3 Sums, Products, and Intersections of Ideals 189

b. Letk be aﬁeld of characteristic zero and letf k[x 1,...,x n] be nonconstant. If the ∂f ∈ ∂f variablex j appears inf , then prove that = 0. Also explain why has smaller ∂xj ̸ ∂xj total degree thanf. 14. LetJ= xy,(x y)x . DescribeV(J) and show that √J= x . 15. Prove that⟨ I=−xy, xz⟩ , yz is a radical ideal. Hint: If you divide⟨ f⟩ k[x,y,z] by xy, xz, yz, what does the⟨ remainder⟩ look like? What doesf m look like? ∈ 16. LetI k[x 1,...,x n] be an ideal. Assume thatI has a Gröbner basisG= g 1,...,g t ⊆ { } such that for alli, LT(gi) is square-free in the sense of Deﬁnition 10. m a. Iff √I, prove that LT(f) is divisible by LT(gi) for somei. Hint:f I. b. Prove∈ thatI is radical. Hint: Use part (a) to show thatG is a Gröbner basis∈ of √I. 17. This exercise continues the line of thought begun in Exercise 16. a. Prove that a monomial ideal ink[x 1,...,x n] is radical if and only if its minimal generators are square-free. b. Given an idealI k[x 1,...,x n], prove that if LT(I) is radical, thenI is radical. c. Give an example⊆ to show that the converse of⟨ part (b)⟩ can fail.

§3 Sums, Products, and Intersections of Ideals

Ideals are algebraic objects and, as a result, there are natural algebraic operations we can define on them. In this section, we consider three such operations: sum, intersection, and product. These are binary operations: to each pair of ideals, they associate a new ideal. We shall be particularly interested in two general questions which arise in connection with each of these operations. Thefirst asks how, given generators of a pair of ideals, one can compute generators of the new ideals which result on applying these operations. The second asks for the geometric significance of these algebraic operations. Thus, thefirst questionfits the general computational theme of this book; the second, the general thrust of this chapter. We consider each of the operations in turn.

Sums of Ideals

Deﬁnition 1.IfI andJ are ideals of the ringk[x 1,...,x n], then the sumofI andJ, denotedI+J, is the set

I+J= f+g f I andg J . { | ∈ ∈ }

Proposition 2. If I and J are ideals in k[x1,...,x n], then I+ J is also an ideal in k[x1,...,x n]. In fact, I+ J is the smallest ideal containing I and J. Furthermore, if I= f 1,...,f r andJ= g 1,...,g s , then I+J= f 1,...,f r,g 1,...,g s . ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ Proof. Noteﬁrst that 0=0+0 I+J. Supposeh 1,h 2 I+J. By the deﬁnition of ∈ ∈ I+J, there existf 1,f 2 I andg 1,g 2 J such thath 1 =f 1 +g 1,h 2 =f 2 +g 2. Then, ∈ ∈ after rearranging terms slightly,h 1 +h 2 = (f 1 +f 2) + (g1 +g 2). Butf 1 +f 2 I ∈ becauseI is an ideal and, similarly,g 1 +g 2 J, whenceh 1 +h 2 I+J. To ∈ ∈ check closure under multiplication, leth I+J andp k[x 1,...,x n] be any ∈ ∈ 190 Chapter 4 The Algebra–Geometry Dictionary polynomial. Then, as above, there existf I andg J such thath=f+g. But thenp h=p (f+g)=p f+p g. Nowp f I and∈ p g J∈ becauseI andJ are ideals.· · Consequently,· p· h I+J.· This∈ shows· ∈ thatI+J is an ideal. IfH is an ideal which contains· ∈ I andJ, thenH must contain all elementsf I andg J. SinceH is an ideal,H must contain allf+g, wheref I,g J. In ∈ particular,∈ H I+J. Therefore, every ideal containingI andJ contains∈ ∈ I+J and, thus,I+J must⊇ be the smallest such ideal. Finally, ifI= f 1,...,f r andJ= g 1,...,g s , then f 1,...,f r,g 1,...,g s is ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ an ideal containingI andJ, so thatI+J f 1,...,f r,g 1,...,g s . The reverse ⊆⟨ ⟩ inclusion is obvious, so thatI+J= f 1,...,f r,g 1,...,g s . ⟨ ⟩ ! The following corollary is an immediate consequence of Proposition 2.

Corollary 3.Iff 1,...,f r k[x 1,...,x n], then ∈

f 1,...,f r = f 1 + + f r . ⟨ ⟩ ⟨ ⟩ ··· ⟨ ⟩ To see what happens geometrically, letI= x 2 +y andJ= z be ideals in ⟨ ⟩ ⟨ ⟩ R[x,y,z]. We have sketchedV(I)andV(J) on the next page.ThenI+J= x 2 +y,z contains bothx 2 +y andz. Thus, the varietyV(I+J) must consist of those⟨ points ⟩ where bothx 2 +y andz vanish, i.e., it must be the intersection ofV(I) andV(J). z

←V(x2 +y)

V(z) ↓

←V(x 2 +y,z) x

The same line of reasoning generalizes to show that addition of ideals corresponds geometrically to taking intersections of varieties.

Theorem 4. If I and J are ideals in k[x1,...,x n], thenV(I+J)=V(I) V(J). ∩ Proof. Ifa V(I+J), thena V(I) becauseI I+J; similarly,a V(J). Thus, a V(I) V∈(J) and we conclude∈ thatV(I+J⊆) V(I) V(J). ∈ ∈To get∩ the opposite inclusion, supposea ⊆V(I) ∩V(J). Leth be any polynomial inI+J. Then there existf I andg J such that∈ h=f∩+g. We havef(a) =0 becausea V(I) andg(a)∈ = 0 because∈ a V(J). Thus,h(a) =f(a)+g(a) = 0+0= 0.∈ Sinceh was arbitrary, we conclude∈ thata V(I+J). Hence,V(I+J) V(I) V(J). ∈ ⊇ ∩ ! §3 Sums, Products, and Intersections of Ideals 191

An analogue of Theorem 4 stated in terms of generators was given in Lemma 2 of Chapter 1, §2.

Products of Ideals

In Lemma 2 of Chapter 1, §2, we encountered the fact that an ideal generated by the products of the generators of two other ideals corresponds to the union of varieties:

V(f 1,...,f r) V(g 1,...,g s) =V(f igj,1 i r,1 j s). ∪ ≤ ≤ ≤ ≤ Thus, for example, the varietyV(xz, yz) corresponding to an ideal generated by the product of the generators of the ideals, x,y and z ink[x,y,z] is the union of V(x,y) (thez-axis) andV(z) [the(x,y)-plane].⟨ ⟩ This⟨ suggests⟩ the following deﬁni- tion.

Deﬁnition 5.IfI andJ are two ideals ink[x 1,...,x n], then their product, denoted I J, is deﬁned to be the ideal generated by all polynomialsf g wheref I and g· J. · ∈ ∈ Thus, the productI JofI andJ is the set · I J= f 1g1 + +f rgr f 1,...,f r I,g 1,...,g r J,r a positive integer . · { ··· | ∈ ∈ } To see that this is an ideal, note that 0=0 0 I J. Moreover, it is clear that · ∈ · h1,h 2 I J implies thath 1 +h 2 I J. Finally, ifh=f 1g1 + +f rgr I J and p is any∈ polynomial,· then ∈ · ··· ∈ ·

ph=(pf 1)g1 + +(pf r)gr I J ··· ∈ · sincepf i I for alli, 1 i r. Note that the set of products would not be an ideal because∈ it would not≤ ≤ be closed under addition. The following easy proposition shows that computing a set of generators forI J given sets of generators forI and J is completely straightforward. ·

Proposition 6. Let I= f 1,...,f r andJ= g 1,...,g s . Then I J is generated by the set of all products⟨ of generators⟩ of I and⟨ J: ⟩ ·

I J= f igj 1 i r,1 j s . · ⟨ | ≤ ≤ ≤ ≤ ⟩

Proof. It is clear that the ideal generated by productsf igj of the generators is contained inI J. To establish the opposite inclusion, note that any polynomial inI J is a sum of· polynomials of the form fg withf I andg J. But we can writef and· ∈ ∈ g in terms of the generatorsf 1,...,f r andg 1,...,g s, respectively, as

f=a 1f1 + +a rfr,g=b 1g1 + +b sgs ··· ··· for appropriate polynomialsa 1,...,a r,b 1,...,b s. Thus, fg," and consequently any sum of polynomials of this form, can be written as a sum cij figj, wherec ij ij ∈ k[x1,...,x n]. ! 192 Chapter 4 The Algebra–Geometry Dictionary

The following proposition guarantees that the product of ideals does indeed correspond geometrically to the operation of taking the union of varieties.

Theorem 7. If I and J are ideals in k[x1,...,x n], thenV(I J)=V(I) V(J). · ∪ Proof. Leta V(I J). Theng(a)h(a)= 0 for allg I and allh J. Ifg(a)=0 for allg I, then∈ a ·V(I). Ifg(a)= 0forsomeg I, then∈ we must∈ haveh(a)=0 for allh∈J. In either∈ event,a V̸ (I) V(J). ∈ Conversely,∈ supposea V∈(I) V∪(J). Eitherg(a)= 0 for allg I orh(a)=0 for allh J. Thus,g(a)h(a∈) = 0∪ for allg I andh J. Thus,f(a)∈ = 0 for all f I J and,∈ hence,a V(I J). ∈ ∈ ∈ · ∈ · ! In what follows, we will often write the product of ideals asIJ rather thanI J. · Intersections of Ideals

The operation of forming the intersection of two ideals is, in some ways, even more primitive than the operations of addition and multiplication.

Deﬁnition 8. The intersectionI J of two idealsI andJ ink[x 1,...,x n] is the set of polynomials which belong to∩ bothI andJ.

As in the case of sums, the set of ideals is closed under intersections.

Proposition 9. If I and J are ideals in k[x1,...,x n], then I J is also an ideal. ∩ Proof. Note that 0 I J since 0 I and0 J. Iff,g I J, thenf+g I becausef,g I. Similarly,∈ ∩ f+g J and,∈ hence,∈f+g I∈J∩. Finally, to∈ check ∈ ∈ ∈ ∩ closure under multiplication, letf I J andh be any polynomial ink[x 1,...,x n]. Sincef I andI is an ideal, we have∈ ∩h f I. Similarly,h f J and, hence, h f I ∈J. · ∈ · ∈ · ∈ ∩ ! Note that we always haveIJ I J since elements ofIJ are sums of polynomials of the form fg withf I andg J⊆. But∩ the latter belongs to bothI (sincef I) andJ (sinceg J). However,∈ IJ∈ can be strictly contained inI J. For example,∈ ifI=J= x,y ,∈ thenIJ= x 2, xy,y 2 is strictly contained∩ inI J=I= x,y (x I J,⟨ but⟩x/ IJ). ⟨ ⟩ ∩ ⟨ ⟩ ∈Given∩ two∈ ideals and a set of generators for each, we would like to be able to compute a set of generators for the intersection. This is much more delicate than the analogous problems for sums and products of ideals, which were entirely straightforward. To see what is involved, supposeI is the ideal inQ[x,y] generated by the polynomialf=(x+y) 4(x2 +y) 2(x 5y) and letJ be the ideal generated by the polynomialg=(x+y)(x 2 +y) 3(x+3−y). We leave it as an (easy) exercise to check that I J= (x+y) 4(x2 +y) 3(x 5y)(x+3y) . ∩ ⟨ − ⟩ §3 Sums, Products, and Intersections of Ideals 193

This computation is easy precisely because we were given factorizations off andg into irreducible polynomials. In general, such factorizations may not be available. So any algorithm which allows one to compute intersections will have to be powerful enough to circumvent this difﬁculty. Nevertheless, there is a nice trick that reduces the computation of intersections to computing the intersection of an ideal with a subring (i.e., eliminating variables), a problem which we have already solved. To state the theorem, we need a little notation: ifI is an ideal ink[x 1,...,x n] andf(t) k[t] a polynomial in the single ∈ variablet, thenf(t)I denotes the ideal ink[x 1,...,x n,t] generated by the set of polynomials f(t) h h I . This is a little different from our usual notion of product in that the ideal{ ·I and| ∈ the} ideal generated byf(t) ink[t] lie in different rings: in fact, the idealI k[x 1,...,x n] is not an ideal ink[x 1,...,x n,t] because it is not closed un- ⊆ der multiplication byt. When we want to stress that a polynomialh k[x 1,...,x n] ∈ involves only the variablesx 1,...,x n, we writeh=h(x). Along the same lines, if we are considering a polynomialg ink[x 1,...,x n,t] and we want to emphasize that it can involve the variablesx 1,...,x n as well ast, we will writeg=g(x,t). In terms of this notation,f(t)I= f(t)h(x) h(x) I . So, for example, iff(t)=t 2 t and I= x,y , then the ideal⟨f(t)I ink[|x,y,t]∈ contains⟩ (t 2 t)x and(t 2 t)y. In fact,− it is not⟨ difﬁ⟩ cult to see thatf(t)I is generated as an ideal by−(t 2 t)x and−(t 2 t)y. This is a special case of the following assertion. − −

Lemma 10.

(i) If I is generated as an ideal in k[x1,...,x n]byp 1(x),...,p r(x), then f(t)I is generated as an ideal in k[x1,...,x n,t]byf(t) p 1(x),...,f(t) p r(x). (ii)Ifg(x,t) f(t)I and a is any element of theﬁeld· k, then g(x,a) I.· ∈ ∈ Proof. To prove theﬁrst assertion, note that any polynomialg(x,t) f(t)I canbe ∈ expressed as a sum of terms of the formh(x,t) f(t) p(x) forh k[x 1,...,x n,t] · · ∈ andp I. But becauseI is generated byp 1,...,p r the polynomialp(x) can be ∈ expressed as a sum of terms of the formq i(x)pi(x), 1 i r. In other words, ≤ ≤ !r p(x) = qi(x)pi(x). i=1 Hence, !r h(x,t) f(t) p(x) = h(x,t)q i(x)f(t)p i(x). · · i=1

Now, for eachi,1 i r,h(x,t) q i(x) k[x 1,...,x n,t]. Thus,h(x,t) f(t) p(x) ≤ ≤ · ∈ · · belongs to the ideal ink[x 1,...,x n,t] generated byf(t) p 1(x),...,f(t) p r(x). Since g(x,t) is a sum of such terms, · ·

g(x,t) f(t) p 1(x),...,f(t) p r(x) , ∈⟨ · · ⟩ which establishes (i). The second assertion follows immediately upon substituting a k fort. ∈ ! 194 Chapter 4 The Algebra–Geometry Dictionary

Theorem 11. Let I, J be ideals in k[x1,...,x n]. Then

I J=(tI+(1 t)J) k[x 1,...,x n]. ∩ − ∩

Proof. Note thattI+(1 t)J is an ideal ink[x 1,...,x n,t]. To establish the desired equality, we use the usual− strategy of proving containment in both directions. Supposef I J. Sincef I, we havet f tI. Similarly,f J implies (1 t) f (1 ∈t)∩J. Thus,f=∈t f+(1 t) f ·tI∈+(1 t)J. Since∈I,J − · ∈ − · − · ∈ − ⊆ k[x1,...,x n], we havef (tI+(1 t)J) k[x 1,...,x n]. This shows thatI J ∈ − ∩ ∩ ⊆ (tI+(1 t)J) k[x 1,...,x n]. − ∩ To establish the opposite containment, takef (tI+(1 t)J) k[x 1,...,x n]. Thenf(x) =g(x,t) +h(x,t), whereg(x,t) tI and∈h(x,t) −(1 t)∩J. First sett= 0. Since every element oftI is a multiple∈ oft, we∈ have−g(x,0) = 0. Thus, f(x) =h(x,0) and hence,f(x) J byLemma 10. On the other hand, sett= 1 in the relationf(x) =g(x,t)+h(x,t)∈. Since every element of(1 t)J is a multiple of 1 t, we haveh(x,1)= 0. Thus,f(x) =g(x,1) and, hence,f(x)−I by Lemma 10−. Since ∈ f belongs to bothI andJ, we havef I J. Thus,I J (tI+(1 t)J) k[x 1,...,x n] ∈ ∩ ∩ ⊇ − ∩ and this completes the proof. ! The above result and the Elimination Theorem (Theorem 2 of Chapter 3, §1) lead to the following algorithm for computing intersections of ideals: ifI= f 1,...,f r andJ= g 1,...,g s are ideals ink[x 1,...,x n], we consider the ideal ⟨ ⟩ ⟨ ⟩

tf1,...,tf r,(1 t)g 1,...,(1 t)g s k[x 1,...,x n,t] ⟨ − − ⟩ ⊆ and compute a Gröbner basis with respect to lex order in whicht is greater than thex i. The elements of this basis which do not contain the variablet will form a basis (in fact, a Gröbner basis) ofI J. For more efﬁcient calculations, one could also use one of the orders described∩ in Exercises 5 and 6 of Chapter 3, §1. An algorithm for intersecting three or more ideals is described in Proposition 6.19 of BECKER and WEISPFENNING (1993). As a simple example of the above procedure, suppose we want to compute the intersection of the idealsI= x 2y andJ= xy 2 inQ[x,y]. We consider the ideal ⟨ ⟩ ⟨ ⟩ tI+(1 t)J= tx 2y,(1 t)xy 2 = tx 2y, txy2 xy2 − ⟨ − ⟩ ⟨ − ⟩ inQ[t,x,y]. Computing theS-polynomial of the generators, we obtain tx 2y2 (tx2y2 x 2y2) =x 2y2. It is easily checked that tx 2y, txy2 xy2,x 2y2 is a Gröbner− basis of−tI+(1 t)J with respect to lex order with{ t>x>y. By− the Elimination} − Theorem, x 2y2 is a (Gröbner) basis of(tI+(1 t)J) Q[x,y]. Thus, { } − ∩ I J= x 2y2 . ∩ ⟨ ⟩ As another example, we invite the reader to apply the algorithm for computing intersections of ideals to give an alternate proof that the intersectionI J of the ideals ∩ I= (x+y) 4(x2 +y) 2(x 5y) andJ= (x+y)(x 2 +y) 3(x+3y) ⟨ − ⟩ ⟨ ⟩ §3 Sums, Products, and Intersections of Ideals 195 inQ[x,y] is I J= (x+y) 4(x2 +y) 3(x 5y)(x+3y) . ∩ ⟨ − ⟩ These examples above are rather simple in that our algorithm applies to ideals which are not necessarily principal, whereas the examples given here involve intersections of principal ideals. We shall see a somewhat more complicated example in the exercises. We can generalize both of the examples above by introducing the following definition.

Deﬁnition 12. A polynomialh k[x 1,...,x n] is called a least common multiple ∈ off,g k[x 1,...,x n] and denotedh= lcm(f,g) if ∈ (i)f dividesh andg dividesh. (ii) Iff andg both divide a polynomialp, thenh dividesp. For example, lcm(x2y, xy2) =x 2y2 and

lcm((x+y) 4(x2 +y) 2(x 5y),(x+y)(x 2 +y) 3(x+3y)) − = (x+y) 4(x2 +y) 3(x 5y)(x+3y). −

a1 ar More generally, supposef,g k[x 1,...,x n] and letf= cf 1 ...f r andg= b1 bs ∈ c′g1 ...g s be their factorizations into distinct irreducible polynomials. It may happen that some of the irreducible factors off are constant multiples of those ofg. In this case, let us suppose that we have rearranged the order of the irreducible polynomials in the expressions forf andg so that for somel, 1 l min(r,s),f i is a ≤ ≤ constant (nonzero) multiple ofg i for 1 i l and for alli,j>l,f i is not a constant ≤ ≤ multiple ofg j. Then it follows from unique factorization that

max a b max a b b a (1) lcm(f,g)=f ( 1, 1) f ( l, l) g l+1 g bs f l+1 f ar . 1 ··· l · l+1 ··· s · l+1 ··· r [In the case thatf andg share no common factors, we have lcm(f,g)=f g.] This, in turn, implies the following result. · Proposition 13.

(i) The intersection I J of two principal ideals I,J k[x 1,...,x n] is a principal ideal. ∩ ⊆ (ii)IfI= f ,J= g andI J= h , then ⟨ ⟩ ⟨ ⟩ ∩ ⟨ ⟩ h= lcm(f,g).

Proof. The proof will be left as an exercise. ! This result, together with our algorithm for computing the intersection of two ideals immediately gives an algorithm for computing the least common multiple of two polynomials: to compute the least common multiple of two polynomials 196 Chapter 4 The Algebra–Geometry Dictionary f,g k[x 1,...,x n], we compute the intersection f g using our algorithm for computing∈ the intersection of ideals. Proposition⟨13⟩∩⟨assures⟩ us that this intersection is a principal ideal (in the exercises, we ask you to prove that the intersection of principal ideals is principal) and that any generator of it is a least common multiple off andg. This algorithm for computing least common multiples allows us to clear up a point which we left unﬁnished in §2: namely, the computation of the greatest common divisor of two polynomialsf andg. The crucial observation is the following.

Proposition 14. Let f,g k[x 1,...,x n]. Then ∈ lcm(f,g) gcd(f,g)= fg. · Proof. This follows by expressingf andg as products of distinct irreducibles and then using the remarks preceding Proposition 13, especially equation (1). You will provide the details in Exercise 5. ! It follows immediately from Proposition 14 that f g (2) gcd(f,g)= · . lcm(f,g)

This gives an algorithm for computing the greatest common divisor of two polynomialsf andg. Namely, we compute lcm(f,g) using our algorithm for the least common multiple and divide it into the product off andg using the division algorithm. We should point out that the gcd algorithm just described is rather cumber- some. In practice, more efﬁcient algorithms are used [see DAVENPORT,SIRET and TOURNIER (1993)]. Having dealt with the computation of intersections, we now ask what operation on varieties corresponds to the operation of intersection on ideals. The following result answers this question.

Theorem 15. If I and J are ideals in k[x1,...,x n], thenV(I J)=V(I) V(J). ∩ ∪ Proof. Leta V(I) V(J). Thena V(I)ora V(J). This means that either f(a) = 0 for∈ allf I∪ orf(a) = 0 for∈ allf J.∈ Thus, certainly,f(a) = 0 for all f I J. Hence,a ∈V(I J). Hence,V(I) ∈V(J) V(I J). ∈ On∩ the other∈ hand,∩ note that since∪IJ I ⊆J, we∩ haveV(I J) V(IJ). ButV(IJ) =V(I) V(J) by Theorem ⊆7,∩ and we immediately∩ ⊆ obtain the reverse ∪ inequality. ! Thus, the intersection of two ideals corresponds to the same variety as the product. In view of this and the fact that the intersection is much more difﬁcult to compute than the product, one might legitimately question the wisdom of bothering with the intersection at all. The reason is that intersection behaves much better with respect to the operation of taking radicals: the product of radical ideals need not be a radical ideal (considerIJ whereI=J), but the intersection of radical ideals is always a radical ideal. The latter fact is a consequence of the next proposition. §3 Sums, Products, and Intersections of Ideals 197

Proposition 16.IfI, J are any ideals, then

√I J= √I √J. ∩ ∩ Proof. Iff √I J, thenf m I J for some integerm> 0. Sincef m I, we havef √∈I. Similarly,∩ f √J.∈ Thus,∩ √I J √I √J. ∈ For∈ the reverse inclusion,∈ takef √I ∩√⊆J. Then,∩ there exist integersm,p>0 such thatf m I andf p J. Thusf ∈m+p =∩f mf p I J, sof √I J. ∈ ∈ ∈ ∩ ∈ ∩ ! EXERCISES FOR §3

1. Show that inQ[x,y], we have

(x+y)4(x2+y)2(x 5y) (x+y)(x 2+y)3(x+3y) = (x+y) 4(x2+y)3(x 5y)(x+3y) . ⟨ − ⟩∩⟨ ⟩ ⟨ − ⟩ 2. Prove formula (1) for the least common multiple of two polynomialsf andg. 3. Prove assertion (i) of Proposition 13. In other words, show that the intersection of two principal ideals is principal. 4. Prove assertion (ii) of Proposition 13. In other words, show that the least common multiple of two polynomialsf andgink[x 1,...,x n] is the generator of the ideal f g . 5. Prove Proposition 14. In other words, show that the least common multiple⟨ of⟩∩⟨ two⟩ polynomials times the greatest common divisor of the same two polynomials is the product of the polynomials. Hint: Use the remarks following the statement of Proposition 14. 6. LetI 1,...,I r andJ be ideals ink[x 1,...,x n]. Show the following:

a.(I 1 +I 2)J=I 1J+I 2J. m m m b.(I 1 I r) =I I . ··· 1 ··· r 7. LetI andJ be ideals ink[x 1,...,x n], wherek is an arbitraryfield. Prove the following: ℓ a. IfI J for@ some integerℓ> 0, then √I √J. ⊆ ⊆ b. √I+J= √I+ √J. 8. Let f=x 4 +x 3y+x 3z2 x 2y2 +x 2yz2 xy3 xy2z2 y 3z2 − − − − and g=x 4 +2x 3z2 x 2y2 +x 2z4 2xy 2z2 y 2z4. − − − @ a. Use a computer algebra program to compute generators for f g and f g . b. Use a computer algebra program to compute gcd(f,g). ⟨ ⟩∩⟨ ⟩ ⟨ ⟩⟨ ⟩ c. Letp=x 2 + xy+ xz+ yz andq=x 2 xy xz+ yz. Use a computer algebra program to calculate f,g p,q . − − ⟨ ⟩ ∩ ⟨ ⟩ 9. For an arbitraryfield, show that √IJ= √I J. Give an example to show that the product ∩ of radical ideals need not be radical. Also give an example to show that √IJ can differ from √I√J. 10. IfI is an ideal ink[x 1,...,x n] and f(t) is an ideal ink[t], show that the idealf(t)I ⟨ ⟩ defined in the text is the product of the ideal I˜ generated by all elements ofI in k[x1,...,x n,t] and the ideal f(t) generated byf(t)ink[x 1,...,x n,t]. ⟨ ⟩ 11. Two idealsI andJ ofk[x 1,...,x n] are said to be comaximal if and only ifI+J= k[x1,...,x n]. a. Show that ifk=C, thenI andJ are comaximal if and only ifV(I) V(J)= . Give an example to show that this is false in general. ∩ ∅ b. Show that ifI andJ are comaximal, thenIJ=I J. ∩ 198 Chapter 4 The Algebra–Geometry Dictionary

c. Is the converse to part (b) true? That is, ifIJ=I J, does it necessarily follow thatI andJ are comaximal? Proof or counterexample?∩ d. IfI andJ are comaximal, show thatI andJ 2 are comaximal. In fact, show thatI r and Js are comaximal for all positive integersr ands. A e. LetI 1,...,I r be ideals ink[x 1,...,x n] and suppose thatI i andJ i = j=i Ij are comaximal for alli. Show that ̸

m m m m I I = (I1 I r) = (I1 I r) 1 ∩···∩ r ··· ∩···∩ for all positive integersm. m 12. LetI,J be ideals ink[x 1,...,x n] and suppose thatI √J. Show thatI J for some integerm> 0. Hint: You will need to use the Hilbert⊆ Basis Theorem. ⊆ 13. LetA be anm n constant matrix and suppose thatx= Ay, where we are thinking of x k m andy ×k n as column vectors of variables. Deﬁne a map ∈ ∈

αA :k[x 1,...,x m] k[y 1,...,y n] −→

by sendingf k[x 1,...,x m]toα A(f) k[y 1,...,y n], whereα A(f) is the polynomial ∈ ∈ deﬁned byα A(f)(y) =f(Ay).

a. Show thatα A isk-linear, i.e., show thatα A(rf+sg) =rα A(f)+sα A(g) for allr,s k ∈ and allf,g k[x 1,...,x n]. ∈ b. Show thatα A(f g) =α A(f) α A(g) for allf,g k[x 1,...,x n]. (As we will see in Deﬁnition 8 ·of Chapter 5, §·2, a map between∈ rings which preserves addition and multiplication and also preserves the multiplicative identity is called a ring homomorphism. Since it is clear thatα A(1) = 1, this shows thatα A is a ring homomorphism.) c. Show that the set f k[x 1,...,x m] α A(f)=0 is an ideal ink[x 1,...,x m]. [This { ∈ | } set is called the kernel ofα A and denoted ker(αA).] d. IfI is an ideal ink[x 1,...,x n], show that the setα A(I) = α A(f) f I need { | ∈ } not be an ideal ink[y 1,...,y n]. [We will often write α A(I) to denote the ideal in ⟨ ⟩ k[y1,...,y n] generated by the elements ofα A(I)—it is called the extension ofI to k[y1,...,y n].] 1 e. IfI ′ is an ideal ink[y 1,...,y n], setα A− (I′) = f k[x 1,...,x m] α A(f) I ′ . 1 { ∈ | ∈ } Show thatα A− (I′) is an ideal ink[x 1,...,x m] (often called the contraction ofI ′). 14. LetA andα A be as above and letK= ker(α A). LetI andJ be ideals ink[x 1,...,x m]. Show that:

a.I J implies α A(I) αA(J) . ⊆ ⟨ ⟩ ⊆ ⟨ ⟩ b. α A(I+J) = α A(I) + α A(J) . ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ c. α A (IJ) = α A(I) αA(J) . ⟨ ⟩ ⟨ ⟩⟨ ⟩ d. α A(I J) @α A(I) αA(J) , with equality ifI K orJ K andα A is onto. ⟨ ∩ ⟩⊆⟨ ⟩ ∩ ⟨ ⟩ ⊇ ⊇ e. α A(√I) αA(I) with equality ifI K andα A is onto. ⟨ ⟩ ⊆ ⟨ ⟩ ⊇ 15. LetA,α A, andK= ker(α A) be as above. LetI ′ andJ ′ be ideals ink[y 1,...,y n]. Show that: 1 1 a.I ′ J ′ impliesα A− (I′) α A− (J′). ⊆1 1 ⊆ 1 b.α A− (I′ +J ′) α A− (I′) +α A− (J′), with equality ifα A is onto. 1 ⊇ 1 1 c.α A− (I′J′) (α A− (I′))(αA− (J′)), with equality ifα A is onto and the right-hand side containsK⊇. 1 1 1 d.α − (I′ J ′) =α − (I′) α − (J′). A ∩ 9 A ∩ A 1 1 e.α A− (√I′) = αA− (I′). §4 Zariski Closures, Ideal Quotients, and Saturations 199 §4 Zariski Closures, Ideal Quotients, and Saturations

We have already encountered a number of examples of sets which are not varieties. Such sets arose very naturally in Chapter 3, where we saw that the projection of a variety need not be a variety, and in the exercises in Chapter 1, where we saw that the (set-theoretic) difference of varieties can fail to be a variety. Whether or not a setS k n is an afﬁne variety, the set ⊆

I(S)= f k[x 1,...,x n] f(a)= 0 for alla S { ∈ | ∈ } is an ideal ink[x 1,...,x n] (check this!). In fact, it is radical. By the ideal–variety correspondence,V(I(S)) is a variety. The following proposition states that this variety is the smallest variety that contains the setS.

Proposition 1.If S k n, the affine varietyV(I(S)) is the smallest variety that contains S [in the⊆ sense that if W k n is any affine variety containing S, then V(I(S)) W]. ⊆ ⊆ Proof. IfW S, thenI(W) I(S) becauseI is inclusion-reversing. But then V(I(W)) V⊇(I(S)) because⊆V also reverses inclusions. SinceW is an affine variety, ⊇ V(I(W)) =W by Theorem 7 from §2, and the result follows.! This proposition leads to the following definition.

Definition 2. The Zariski closure of a subset of affine space is the smallest affine algebraic variety containing the set. IfS k n, the Zariski closure ofS is denoted S and is equal toV(I(S)). ⊆

We note the following properties of Zariski closure.

Lemma 3. Let S and T be subsets of kn. Then: (i)I( S)=I(S). (ii)IfS T, then S T. (iii) S ⊆T= S T. ⊆ ∪ ∪ Proof. For (i), the inclusionI( S) I(S) follows fromS S. Going the other way, f I(S) impliesS V(f). ThenS ⊆ S V(f) by Deﬁnition⊆ 2, so thatf I( S). ∈ ⊆ ⊆ ⊆ ∈ The proofs of (ii) and (iii) will be covered in the exercises.! A natural example of Zariski closure is given by elimination ideals. We can now prove theﬁrst assertion of the Closure Theorem (Theorem 3 of Chapter 3, §2).

Theorem 4 (The Closure Theorem,ﬁrst part). Assume k is algebraically closed. n n n l Let V=V(f 1,...,f s) k , and letπ l :k k − be projection onto the last n l ⊆ → − coordinates. If Il is the l-th elimination ideal Il = f 1,...,f s k[x l 1,...,x n], then ⟨ ⟩ ∩ + V(Il) is the Zariski closure ofπ l(V). 200 Chapter 4 The Algebra–Geometry Dictionary

Proof. In view of Proposition 1, we must show thatV(I l) =V(I(π l(V))). By Lemma 1 of Chapter 3, §2, we haveπ l(V) V(I l). SinceV(I(π l(V))) is the small- ⊆ est variety containingπ l(V), it follows immediately thatV(I(π l(V))) V(I l). ⊆ To get the opposite inclusion, supposef I(π l(V)), i.e.,f(a l 1,...,a n) =0for ∈ + all(a l 1,...,a n) π l(V). Then, considered as an element ofk[x 1,x 2,...,x n], we + ∈ certainly havef(a 1,a 2,...,a n) = 0 for all(a 1,...,a n) V. By Hilbert’s Nullstel- N ∈ lensatz,f f 1,...,f s for some integerN. Sincef does not depend onx 1,...,x l, ∈ ⟨N ⟩ N neither doesf , and we havef f 1,...,f s k[x l 1,...,x n] =I l. Thus,f √I , ∈ ⟨ ⟩ ∩ + ∈ l which impliesI(π l(V)) √I . It follows thatV(I l) =V( √I ) V(I(π l(V))), and ⊆ l l ⊆ the theorem is proved. !

The conclusion of Theorem 4 can be stated asV(I l) = πl(V). In general, ifV is a variety, then we say that a subsetS V is Zariski dense in V ifV= S, i.e.,V is the ⊆ Zariski closure ofS. is Thus Theorem 4 tells us thatπ l(V) is Zariski dense inV(I l) when thefield is algebraically closed. One context in which we encountered sets that were not varieties was in taking the difference of varieties. For example, letV=V(I) whereI k[x,y,z] is the ideal xz, yz andW=V(J) whereJ= z . Then we have already⊆ seen thatV is the union⟨ of⟩ the(x,y)-plane and thez⟨-axis.⟩ SinceW is the(x,y)-plane,V W is the z-axis with the origin removed [because the origin also belongs to the\(x,y)-plane]. We have seen in Chapter 1 that this is not a variety. Thez-axis [i.e.,V(x,y)] is the Zariski closure ofV W. We could ask if\ there is a general way to compute the ideal corresponding to the Zariski closure V W of the difference of two varietiesV andW. The answer is affirmative, but it involves\ two new algebraic constructions on ideals called ideal quotients and saturations. We begin with thefirst construction.

Deﬁnition 5.IfI,J are ideals ink[x 1,...,x n], thenI:J is the set

f k[x 1,...,x n] fg I for allg J { ∈ | ∈ ∈ } and is called the ideal quotient (or colon ideal) ofIbyJ.

So, for example, ink[x,y,z] we have

xz, yz : z = f k[x,y,z] f z xz, yz ⟨ ⟩ ⟨ ⟩ { ∈ | · ∈⟨ ⟩} = f k[x,y,z] f z= Axz+ Byz { ∈ | · } = f k[x,y,z] f=Ax+By { ∈ | } = x,y . ⟨ ⟩

Proposition 6. If I, J are ideals in k[x1,...,x n], then the ideal quotient I: J is an ideal in k[x1,...,x n]andI: J contains I.

Proof. To showI:J containsI, note that becauseI is an ideal, iff I, then fg I ∈ ∈ for allg k[x 1,...,x n] and, hence, certainly fg I for allg J. To show thatI:J ∈ ∈ ∈ §4 Zariski Closures, Ideal Quotients, and Saturations 201 is an ideal,ﬁrst note that 0 I:J because 0 I. Letf 1,f 2 I:J. Thenf 1g andf 2g ∈ ∈ ∈ are inI for allg J. SinceI is an ideal(f 1 +f 2)g=f 1g+f 2g I for allg J. Thus, ∈ ∈ ∈ f1 +f 2 I:J. To check closure under multiplication is equally straightforward: if ∈ f I:J andh k[x 1,...,x n], then fg I and, sinceI is an ideal, hfg I for all g∈ J, which∈ means thathf I:J. ∈ ∈ ∈ ∈ ! The algebraic properties of ideal quotients and methods for computing them will be discussed later in the section. For now, we want to explore the relation between ideal quotients and the Zariski closure of a difference of varieties. Proposition 7.

(i) If I and J are ideals in k[x1,...,x n], then

V(I)=V(I+J) V(I:J). ∪ (ii) If V and W are varieties kn, then

V=(V W) ( V W). ∩ ∪ \ (iii) In the situation of (i), we have

V(I) V(J) V(I:J). \ ⊆ Proof. We begin with (ii). SinceV containsV W andV is a variety, the smallest variety containingV W must be contained inV\. Hence, V W V. SinceV W V, we have(V W) \( V W) V. \ ⊆ ∩ ⊆ To get the reverse∩ ∪ containment,\ ⊆ note thatV=(V W) (V W). SinceV W V W, the desired inclusionV (V W) V W∩ follows∪ \ immediately.\ ⊆ \For (iii), weﬁrst claim that⊆I:J ∩I(V∪(I) V(J))\. For suppose thatf I:J and a V(I) V(J). Then fg I for allg⊆ J. Since\ a V(I), we havef(a)g(a∈)=0 for∈ allg\ J. Sincea/ V∈(J), there∈ is someg ∈J such thatg(a)= 0. Hence, f(a)= 0∈ for alla V(∈I) V(J). Thus,f I(V(I∈) V(J)), which proves̸ the claim. SinceV reverses∈ inclusions,\ we have∈V(I:J) \V(I(V(I) V(J))) = V(I) V(J). Finally, for (i), note thatV(I+J) =V(I)⊇V(J) by Theorem\ 4 of §\3. Then applying (ii) withV=V(I) andW=V(J) gives∩

V(I)=V(I+J) V(I) V(J) V(I+J) V(I:J), ∪ \ ⊆ ∪ where the inclusion follows from (iii). ButI I+J andI I:J imply that ⊆ ⊆ V(I+J) V(I) andV(I:J) V(I). ⊆ ⊆ These inclusions giveV(I+J) V(I:J) V(I), and then we are done. ∪ ⊆ ! In Proposition 7, note thatV(I+J) from part (i) matches up withV W in part (ii) sinceV(I+J) =V(I) V(J). So it is natural to ask ifV(I:J) in∩ part (i) matches up with V W in part∩ (ii). This is equivalent to asking if the inclusion V(I) V(J) V(I:J) in\ part (iii) is an equality. \ ⊆ 202 Chapter 4 The Algebra–Geometry Dictionary

Unfortunately, this can fail, even when theﬁeld is algebraically closed. To see what can go wrong, letI= x 2(y 1) andJ= x in the polynomial ringC[x,y]. Then one easily checks that⟨ − ⟩ ⟨ ⟩

2 V(I)=V(x) V(y 1)=V(J) V(y 1) C , ∪ − ∪ − ⊆ which is the union of they-axis and the liney= 1. It follows without difﬁculty that V(I) V(J)=V(y 1). However, the ideal quotient is \ − 2 2 I:J= x (y 1) : x = f C[x,y] f x=Ax (y 1) ⟨ − ⟩ ⟨ ⟩ { ∈ | · − } = f C[x,y] f=Ax(y 1) = x(y 1) . { ∈ | − } ⟨ − ⟩ ThenV(I:J) =V(x(y 1)) =V(x) V(y 1), which is strictly bigger than V(I) V(J)=V(y 1)−. In other words,∪ the− inclusion in part (iii) of Proposition 7 can be\ strict, even− over an algebraically closedﬁeld. However, if we replaceJ withJ 2, then a computation similar to the above gives I:J 2 = y 1 , so thatV(I:J 2) = V(I) V(J). In general, higher powers may be required,⟨ − which⟩ leads to our second algebraic\ construction on ideals.

Deﬁnition 8.IfI,J are ideals ink[x 1,...,x n], thenI:J ∞ is the set

N f k[x 1,...,x n] for allg J, there isN 0 such that fg I { ∈ | ∈ ≥ ∈ } and is called the saturationofI with respect toJ.

Proposition 9. If I, J are ideals in k[x1,...,x n], then the saturation I:J ∞ is an ideal in k[x1,...,x n]. Furthermore:

(i)I I:J I:J ∞. ⊆ ⊆ N (ii)I:J ∞ =I:J for all sufﬁciently large N. (iii) √I:J ∞ = √I:J.

N+1 N Proof. First observe thatJ 1 J 2 impliesI:J 2 I:J 1. SinceJ J for allN, we obtain the ascending chain⊆ of ideals ⊆ ⊆

(1)I I:J I:J 2 I:J 3 . ⊆ ⊆ ⊆ ⊆··· N N+1 By the ACC, there isN such thatI:J =I:J = . We claim thatI:J ∞ = I:J N. One inclusion is easy, for iff I:J N andg ···J, theng N J N. Hence, N ∈ ∈ ∈ fg I, proving thatf I:J ∞. For the other inclusion, takef I:J ∞ and let ∈ ∈ ∈ J= g 1,...,g s . By Deﬁnition 8,f times a power ofg i lies inI. IfM is the largest such⟨ power, then⟩ fgM I fori=1,...,s. In the exercises, you will show that i ∈ J sM gM,...,g M . ⊆ ⟨ 1 s ⟩ This impliesfJ sM I, so thatf I:J sM. Thenf I:J N since (1) stabilizes atN. ⊆ ∈ ∈ N Part (ii) follows from the claim just proved, andI:J ∞ =I:J implies thatI:J ∞ is an ideal by Proposition 6. Note also that part (i) follows from (1) and part (ii) . §4 Zariski Closures, Ideal Quotients, and Saturations 203

For part (iii), weﬁrst show √I:J ∞ √I:J. This is easy, forf √I:J ∞ m ⊆ m N ∈ impliesf I:J ∞ for somem. Giveng J, it follows thatf g I for someN. Then(fg) M∈ I forM= max(m,N), so that∈ fg √I. Since this holds∈ for allg J, we conclude∈ thatf √I:J. ∈ ∈ ∈ For the opposite inclusion, takef √I:J and writeJ= g 1,...,g s . Then M ∈M ⟨ ⟩ fgi √I, so we canﬁndM withf gi I for alli. The argument from (ii) implies f MJ∈sM I, so ∈ ⊆ M sM f I:J I:J ∞. ∈ ⊆ It follows thatf √I:J , and the proof is complete. ∈ ∞ ! Later in the section we will discuss further algebraic properties of saturations and how to compute them. For now, we focus on their relation to geometry.

Theorem 10. Let I and J be ideals in k[x1,...,x n]. Then:

(i)V(I)=V(I+J) V(I:J ∞). ∪ (ii) V(I) V(J) V(I:J ∞). \ ⊆ (iii) If k is algebraically closed, thenV(I:J ∞) = V(I) V(J). \ Proof. In the exercises, you will show that (i) and (ii) follow by easy modiﬁcations of the proofs of parts (i) and (iii) of Proposition 7. For (iii), suppose thatk is algebraically closed. Weﬁrst show that

(2) I(V(I) V(J)) √I:J. \ ⊆ Letf I(V(I) V(J)). Ifg J, then fg vanishes onV(I) becausef vanishes on V(I)∈V(J) and\ gonV(J).∈ Thus, fg I(V(I)), so fg √I by the Nullstellensatz. Since\ this holds for allg J, we have∈ f √I:J,∈ as claimed. SinceV is inclusion-reversing,∈ ( 2)∈ implies

V(√I:J) V(I(V(I) V(J))) = V(I) V(J). ⊆ \ \ However, we also have

V(I:J ∞) =V( √I:J ∞) =V( √I:J), where the second equality follows from part (iii) of Proposition 9. Combining the last two displays, we obtain

V(I:J ∞) V(I) V(J). ⊆ \ Then (iii) follows immediately from this inclusion and (ii).! Whenk is algebraically closed, Theorem 10 and Theorem 4 of §3 imply that the decomposition V(I)=V(I+J) V(I:J ∞) ∪ is precisely the decomposition 204 Chapter 4 The Algebra–Geometry Dictionary

V(I)=(V(I) V(J)) ( V(I) V(J)) ∩ ∪ \ from part (ii) of Proposition 7. This shows that the saturationI:J ∞ is the ideal- theoretic analog of the Zariski closure V(I) V(J). In some situations, saturations can be replaced\ with ideal quotients. For example, the proof of Theorem 10 yields the following corollary when the idealI is radical.

Corollary 11. Let I and J be ideals in k[x1,...,x n]. If k is algebraically closed and I is radical, then V(I:J)= V(I) V(J). \ You will prove this in the exercises. Another nice fact (also covered in the exercises) is that ifk is arbitrary andV andW are varieties ink n, then

I(V):I(W)=I(V W). \ The following proposition takes care of some simple properties of ideal quotients and saturations.

Proposition 12. Let I and J be ideals in k[x1,...,x n]. Then:

(i)I:k[x 1,...,x n] =I:k[x 1,...,x n]∞ =I. (ii)J I if and only if I:J=k[x 1,...,x n]. ⊆ (iii)J √I if and only if I:J ∞ =k[x 1,...,x n]. ⊆ Proof. The proof is left as an exercise. ! When theﬁeld is algebraically closed, the reader is urged to translate parts (i) and (iii) of the proposition into terms of varieties (upon which they become clear). The following proposition will help us compute ideal quotients and saturations.

Proposition 13. Let I and J1,...,J r be ideals in k[x1,...,x n]. Then: 0 !r / Br 6 ( (3) I: Ji = I:J i , i=1 i=1 0 !r / Br 6 ( ∞ (4) I: Ji = I:J i∞ . i=1 i=1

Proof. We again leave the (straightforward) proofs to the reader.! Iff is a polynomial andI an ideal, we will often writeI:f instead ofI: f , and ⟨ ⟩ similarlyI:f ∞ instead ofI: f ∞. Note that (3) and (4) imply that ⟨ ⟩ Br Br (5)I: f 1,f 2,...,f r = (I:f i) andI: f 1,f 2,...,f r ∞ = (I:f ∞). ⟨ ⟩ ⟨ ⟩ i i=1 i=1 We now turn to the question of how to compute generators of the ideal quotient I:J and saturationI:J ∞, given generators ofI andJ. Inspiredby ( 5), we begin with the case whenJ is generated by a single polynomial. §4 Zariski Closures, Ideal Quotients, and Saturations 205

Theorem 14. Let I be an ideal and g an element of k[x1,...,x n]. Then: (i)If h 1,...,h p is a basis of the ideal I g , then h 1/g,...,h p/g is a basis of I{:g. } ∩⟨ ⟩ { } (ii)If f 1,...,f s is a basis of I and I˜= f 1,...,f s,1 yg k[x 1,...,x n,y], where y{ is a new variable,} then ⟨ − ⟩⊆

I:g ∞ = I˜ k[x 1,...,x n]. ∩

Furthermore, if G is a lex Gröbner basis of I˜ for y>x 1 > >x n, then ··· G k[x 1,...,x n] is a basis of I:g ∞. ∩ Proof. For (i), observe that ifh g , thenh= bg for some polynomialb ∈⟨ ⟩ ∈ k[x1,...,x n]. Thus, iff h 1/g,...,h p/g , then ∈⟨ ⟩

hf= bgf h 1,...,h p =I g I. ∈⟨ ⟩ ∩⟨ ⟩⊆ Thus,f I:g. Conversely, supposef I:g. Then fg I. Since fg g ," we ∈ ∈ ∈ ∈⟨ ⟩ have fg I g . IfI g = h 1,...,h p , this means fg= rihi for some ∈ ∩⟨ ⟩ ∩⟨ ⟩ ⟨ ⟩ polynomials" r i. Since eachh i g , eachh i/g is a polynomial, and we conclude ∈ ⟨ ⟩ thatf= ri(hi/g), whencef h 1/g,...,h p/g . Thefirst assertion of (ii) is∈⟨ left as an exercise.⟩ Then the Elimination Theorem from Chapter 3, §1 implies thatG k[x 1,...,x n] is a Gröbner basis ofI:g ∞. ∩ ! This theorem, together with our procedure for computing intersections of ideals and equation (5), immediately leads to an algorithm for computing a basis of an ideal quotient: givenI= f 1,...,f r andJ= g 1,...,g s , to compute a basis ⟨ ⟩ ⟨ ⟩ ofI:J, wefirst compute a basis forI:g i for eachi. In view of Theorem 14, this means computing a basis h 1,...,h p of f 1,...,f r gi . Recall that we do this via the algorithm for computing{ intersections} ⟨ of ideals⟩ ∩ from ⟨ ⟩ §3. Using the division algorithm, we divide each of basis elementh j byg i to get a basis forI:g i by part (i) of Theorem 14. Finally, we compute a basis forI:J by applying the intersection algorithms 1 times, computingfirst a basis forI: g 1,g 2 =(I:g 1) (I:g 2), then − ⟨ ⟩ ∩ a basis forI: g 1,g 2,g 3 =(I: g 1,g 2 ) (I:g 3), and so on. Similarly,⟨ we have an⟩ algorithm⟨ for⟩ ∩ computing a basis of a saturation: given I= f 1,...,f r andJ= g 1,...,g s , to compute a basis ofI:J ∞, wefirst compute ⟨ ⟩ ⟨ ⟩ a basis forI:g i∞ for eachi using the method described in part (ii) of Theorem 14. Then by (5), we need to intersect the idealsI:g i∞, which we do as above by applying the intersection algorithms 1 times. − EXERCISES FOR §4

1. Find the Zariski closure of the following sets: 2 a. The projection of the hyperbolaV(xy 1)inR onto thex-axis. − 2 b. The boundary of theﬁrst quadrant inR . 2 2 2 c. The set (x,y) R x +y 4 . 2. Complete the{ proof∈ of Lemma| 3. Hint:≤ } For part (iii), use Lemma 2 from Chapter 1, §2. 3. Letf=(x+y) 2(x y)(x+z 2) andg=(x+z 2)3(x y)(z+y). Compute generators for f : g . − − ⟨ ⟩ ⟨ ⟩ 206 Chapter 4 The Algebra–Geometry Dictionary

4. LetI andJ be ideals ink[x 1,...,x n]. Show that ifI is radical, thenI:J is radical and I:J=I: √J=I:J ∞. sM M M 5. As in the proof of Proposition 9, assumeJ= g 1,...,g s . Prove thatJ g1 ,...,g s . Hint: See the proof of Lemma 5 of §2. ⟨ ⟩ ⊆⟨ ⟩ 6. Prove parts (i) and (ii) of Theorem 10. Hint: Adapt the proofs of parts (i) and (iii) of Proposition 7. 7. Prove Corollary 11. Hint: Combine Theorem 10 and the Exercise 4. Another approach would be look closely at the proof of Theorem 10 whenI is radical. 8. LetV,W k n be varieties. Prove thatI(V):I(W)=I(V W). 9. Prove Proposition⊆ 12 andﬁnd geometric interpretations\ of parts (i) and (iii) 10. Prove Proposition 13 andﬁnd a geometric interpretation of ( 4).

11. ProveI:g ∞ = I˜ k[x 1,...,x n] from part (ii) of Theorem 14. Hint: See the proof of Proposition 8 of §∩2. 12. Show that Proposition 8 of §2 is a corollary of Proposition 12 and Theorem 14. 2 13. An example mentioned in the text usedI= x (y 1) andJ= x . ComputeI:J ∞ and explain how your answer relates to the⟨ discussion− in⟩ the text.⟨ ⟩ N N 14. LetI,J k[x 1,...,x n] be ideals. Prove thatI:J ∞ =I:J if and only ifI:J = N+⊆1 I:J . Then use this to describe an algorithm for computing the saturationI:J ∞ based on the algorithm for computing ideal quotients. N N 15. Show thatN can be arbitrarily large inI:J ∞ =I:J . Hint: Look atI= x (y 1) . ⟨ − ⟩ 16. LetI,J,K k[x 1,...,x n] be ideals. Prove the following: ⊆ a.IJ K if and only ifI K:J. ⊆ ⊆ b.(I:J):K=I:JK. - A . A r r 17. Given idealsI 1,...,I r,J k[x 1,...,x n], prove that i=1 Ii :J= i=1(Ii :J). Then prove a similar result for⊆ saturations and give a geometric interpretation. 18. LetA be anm n constant matrix and suppose thatx= Ay. where we are thinking of x k m andy ×k n as column vectors of variables. As in Exercise 13 of §3, deﬁne a map ∈ ∈

αA :k[x 1,...,x m] k[y 1,...,y n] −→

by sendingf k[x 1,...,x m]toα A(f) k[y 1,...,y n], whereα A(f) is the polynomial ∈ ∈ deﬁned byα A(f)(y) =f(Ay). a. Show thatα A(I:J) α A(I):α A(J) with equality ifI ker(α A) andα A is onto. 1 ⊆ 1 1 ⊇ b. Show thatα A− (I′ :J ′) =α A− (I′):α A− (J′) whenα A is onto.

§5 Irreducible Varieties and Prime Ideals

We have already seen that the union of two varieties is a variety. For example, in Chapter 1 and in the last section, we consideredV(xz, yz), which is the union of a line and a plane. Intuitively, it is natural to think of the line and the plane as “more fundamental” thanV(xz, yz). Intuition also tells us that a line or a plane are “irreducible” or “indecomposable” in some sense: they do not obviously seem to be a union ofﬁnitely many simpler varieties. We formalize this notion as follows.

Definition 1. An affine varietyV k n is irreducible if wheneverV is written in ⊆ the formV=V 1 V 2, whereV 1 andV 2 are affine varieties, then eitherV 1 =V or ∪ V2 =V. §5 Irreducible Varieties and Prime Ideals 207

Thus,V(xz, yz) is not an irreducible variety. On the other hand, it is not completely clear when a variety is irreducible. If this deﬁnition is to correspond to our geometric intuition, it is clear that a point, a line, and a plane ought to be irreducible. For that matter, the twisted cubicV(y x 2,z x 3) inR 3 appears to be irreducible. But how do we prove this? The key is− to capture− this notion algebraically: if we can characterize ideals which correspond to irreducible varieties, then perhaps we stand a chance of establishing whether a variety is irreducible. The following notion turns out to be the right one.

Deﬁnition 2. An idealI k[x 1,...,x n] is prime if wheneverf,g k[x 1,...,x n] and fg I, then eitherf I⊆org I. ∈ ∈ ∈ ∈ If we have set things up right, an irreducible variety will correspond to a prime ideal and conversely. The following theorem assures us that this is indeed the case.

Proposition 3. Let V k n be an afﬁne variety. Then V is irreducible if and only if I(V) is a prime ideal.⊆

Proof. First, assume thatV is irreducible and let fg I(V). SetV 1 =V V(f) and ∈ ∩ V2 =V V(g); these are affine varieties because an intersection of affine varieties is ∩ a variety. Then fg I(V) easily implies thatV=V 1 V 2. SinceV is irreducible, we ∈ ∪ have eitherV=V 1 orV=V 2. Say the former holds, so thatV=V 1 =V V(f). This implies thatf vanishes onV, so thatf I(V). Thus,I(V) is prime. ∩ ∈ Next, assume thatI(V) is prime and letV=V 1 V 2. Suppose thatV=V 1. We ∪ ̸ claim thatI(V)=I(V 2). To prove this, note thatI(V) I(V 2) sinceV 2 V. For ⊆ ⊆ the opposite inclusion,first note thatI(V)!I(V 1) sinceV 1 !V. Thus, we can pick f I(V 1) I(V). Now take anyg I(V 2). SinceV=V 1 V 2, it follows that fg vanishes∈ on\ V, and, hence, fg I(V∈). ButI(V) is prime, so that∪f org lies inI(V). ∈ We know thatf/ I(V) and, thus,g I(V). This provesI(V) =I(V 2), whence ∈ ∈ V=V 2 becauseI is one-to-one. Thus,V is an irreducible variety.! It is an easy exercise to show that every prime ideal is radical. Then, using the ideal-variety correspondence between radical ideals and varieties, we get the following corollary of Proposition 3.

Corollary 4. When k is algebraically closed, the functionsI andV induce a one- to-one correspondence between irreducible varieties in kn and prime ideals in k[x1,...,x n].

As an example of how to use Proposition 3, let us prove that the idealI(V) of the twisted cubic is prime. Suppose that fg I(V). Since the curve is parametrized by (t,t 2,t 3), it follows that, for allt, ∈

f(t,t 2,t 3)g(t,t 2,t 3) =0.

This implies thatf(t,t 2,t 3)org(t,t 2,t 3) must be the zero polynomial, so thatf or g vanishes onV. Hence,f org lies inI(V), proving thatI(V) is a prime ideal. 208 Chapter 4 The Algebra–Geometry Dictionary

By the proposition, the twisted cubic is an irreducible variety inR 3. One proves that a straight line is irreducible in the same way:first parametrize it, then apply the above argument. In fact, the above argument holds much more generally. Proposition 5. If k is an infinitefield and V k n is a variety defined parametrically ⊆ x1 =f 1(t1,...,t m), . . xn =f n(t1,...,t m), where f1,...,f n are polynomials in k[t1,...,t m], then V is irreducible. Proof. As in §3 of Chapter 3, we letF:k m k n be defined by → F(t1,...,t m) = (f 1(t1,...,t m),...,f n(t1,...,t m)). Saying thatV is defined parametrically by the above equations means thatV is the Zariski closure ofF(k m). In particular,I(V)=I(F(k m)). For any polynomialg k[x 1,...,x n], the functiong F is a polynomial in ∈ ◦ k[t1,...,t m]. In fact,g F is the polynomial obtained by “plugging the polynomials ◦ f1,...,f n intog”:

g F=g(f 1(t1,...,t m),...,f n(t1,...,t m)). ◦ m Becausek is inﬁnite,I(V) =I(F(k )) is the set of polynomials ink[x 1,...,x n] whose composition withF is the zero polynomial ink[t 1,...,t m]:

I(V) = g k[x 1,...,x n] g F=0 . { ∈ | ◦ } Now suppose thatgh I(V). Then(gh) F=(g F)(h F)= 0. (Make sure you ∈ ◦ ◦ ◦ understand this.) But, if the product of two polynomials ink[t 1,...,t m] is the zero polynomial, one of them must be the zero polynomial. Hence, eitherg F=0or h F= 0. This means that eitherg I(V)orh I(V). This shows thatI(V) ◦is a ◦ ∈ ∈ prime ideal and, therefore, thatV is irreducible. ! With a little care, the above argument extends still further to show that any variety defined by a rational parametrization is irreducible. Proposition 6. If k is an infinitefield and V is a variety defined by the rational parametrization

f1(t1,...,t m) x1 = , g1(t1,...,t m) . . fn(t1,...,t m) xn = , gn(t1,...,t m) where f1,...,f n,g 1,...,g n k[t 1,...,t m], then V is irreducible. ∈ §5 Irreducible Varieties and Prime Ideals 209

m n Proof. SetW=V(g 1g2 g n) and letF:k W k deﬁned by ··· & \ → ' f1(t1,...,t m) fn(tn,...,t m) F(t1,...,t m) = ,..., . g1(t1,...,t m) gn(t1,...,t m)

ThenV is the Zariski closure ofF(k m W), which implies thatI(V) is the set of \ m h k[x 1,...,x n] such that the functionh F is zero for all(t 1,...,t m) k W. The∈ difﬁculty is thath F need not be a polynomial,◦ and we, thus, cannot∈ directly\ apply the argument in◦ the latter part of the proof of Proposition 5. We can get around this difﬁculty as follows. Leth k[x 1,...,x n]. Since ∈

g1(t1,...,t m)g2(t1,...,t m) g n(t1,...,t m)=0 ··· ̸ m N for any(t 1,...,t m) k W, the function(g 1g2 g n) (h F) is equal to zero ∈ \ m ··· ◦ at precisely those values of(t 1,...,t m) k W for whichh F is equal to zero. ∈ \ ◦ Moreover, if we letN be the total degree ofh k[x 1,...,x n], then we leave it as N ∈ an exercise to show that(g 1g2 g n) (h F) is a polynomial ink[t 1,...,t m]. We ··· ◦ N m deduce thath I(V) if and only if(g 1g2 g n) (h F) is zero for allt k W. ∈ ··· ◦ ∈ N \ But, by Exercise 11 of Chapter 3, §3, this happens if and only if(g 1g2 g n) (h F) ··· ◦ is the zero polynomial ink[t 1,...,t m]. Thus, we have shown that

N h I(V) if and only if(g 1g2 g n) (h F)=0 k[t 1,...,t m]. ∈ ··· ◦ ∈

Now, we continue our proof thatI(V) is prime. Supposep,q k[x 1,...,x n] satisfyp q I(V). If the total degrees ofp andq areM andN, respectively,∈ then · ∈ M+N the total degree ofp q isM+N. Thus,(g 1g2 g n) (p F) (q F)= 0. But the · ··· M ◦ · ◦ N former is a product of the polynomials(g 1g2 g n) (p F)and(g 1g2 g n) (q F) ··· ◦ ··· ◦ ink[t 1,...,t m]. Hence one of them must be the zero polynomial. In particular, either p I(V)orq I(V). This shows thatI(V) is a prime ideal and, therefore, thatV is ∈ ∈ an irreducible variety. ! The simplest variety ink n given by a parametrization consists of a single point, (a1,...,a n) . In the notation of Proposition 5, it is given by the parametrization in { } which eachf i is the constant polynomialf i(t1,...,t m) =a i,1 i n. It is clearly ≤ ≤ irreducible and it is easy to check thatI( (a 1,...,a n) )= x 1 a1,...,x n an (see { } ⟨ − − ⟩ Exercise 7), which implies that the latter is prime. The ideal x 1 a 1,...,x n a n has another distinctive property: it is maximal in the sense that⟨ the− only ideal which− ⟩ strictly contains it is the whole ringk[x 1,...,x n]. Such ideals are important enough to merit special attention.

Deﬁnition 7. An idealI k[x 1,...,x n] is said to be maximal ifI=k[x 1,...,x n] ⊆ ̸ and any idealJ containingI is such that eitherJ=IorJ=k[x 1,...,x n].

In order to streamline statements, we make the following deﬁnition.

Deﬁnition 8. An idealI k[x 1,...,x n] is said to be proper ifI is not equal to ⊆ k[x1,...,x n]. 210 Chapter 4 The Algebra–Geometry Dictionary

Thus, an ideal is maximal if it is proper and no other proper ideal strictly contains it. We now show that any ideal of the form x 1 a 1,...,x n a n is maximal. ⟨ − − ⟩

Proposition 9. If k is anyﬁeld, an ideal I k[x 1,...,x n] of the form ⊆

I= x 1 a 1,...,x n a n , ⟨ − − ⟩ where a1,...,a n k, is maximal. ∈ Proof. Suppose thatJ is some ideal strictly containingI. Then there must exist f J such thatf/ I. We can use the division algorithm to writef asA 1(x1 a 1) + ∈ ∈ − +A n(xn a n) +b for someb k. SinceA 1(x1 a 1) + +A n(xn a n) I ··· − ∈ − ··· − ∈ andf/ I, we must haveb= 0. However, sincef J and sinceA 1(x1 a 1) + + ∈ ̸ ∈ − ··· An(xn a n) I J, we also have − ∈ ⊆

b=f (A 1(x1 a 1) + +A n(xn a n)) J. − − ··· − ∈

Sinceb is nonzero, 1=1/b b J, soJ=k[x 1,...,x n]. · ∈ ! Since V(x1 a 1,...,x n a n) = (a 1,...,a n) , − − { } n every point(a 1,...,a n) k corresponds to a maximal ideal ofk[x 1,...,x n], ∈ namely x 1 a 1,...,x n a n . The converse does not hold ifk is not algebraically ⟨ − − ⟩ closed. In the exercises, we ask you to show that x 2 +1 is maximal inR[x]. The ⟨ ⟩ latter does not correspond to a point ofR. The following result, however, holds in any polynomial ring.

Proposition 10. If k is anyﬁeld, a maximal ideal in k[x 1,...,x n] is prime.

Proof. Suppose thatI is a proper ideal which is not prime and let fg I, where f/ I andg/ I. Consider the ideal f +I. This ideal strictly containsI because∈ ∈ ∈ ⟨ ⟩ f/ I. Moreover, if we were to have f +I=k[x 1,...,x n], then 1= cf+h forsome polynomial∈ c andsomeh I. Multiplying⟨ ⟩ through byg would giveg= cfg+hg I which would contradict∈ our choice ofg. Thus,I+ f is a proper ideal containing∈ ⟨ ⟩ I, so thatI is not maximal. !

Note that Propositions 9 and 10 together imply that x 1 a1,...,x n an is prime ⟨ − − ⟩ ink[x 1,...,x n] even ifk is not inﬁnite. Over an algebraically closedﬁeld, it turns out that every maximal ideal corresponds to some point ofk n.

Theorem 11. If k is an algebraically closedﬁeld, then every maximal ideal of k[x1,...,x n] is of the form x 1 a 1,...,x n a n for some a1,...,a n k. ⟨ − − ⟩ ∈

Proof. LetI k[x 1,...,x n] be maximal. SinceI=k[x 1,...,x n], we have V(I)= by⊆ the Weak Nullstellensatz (Theorem 1 ̸ of §1). Hence, there is some ̸ ∅ §5 Irreducible Varieties and Prime Ideals 211 point(a 1,...,a n) V(I). This means that everyf I vanishes at(a 1,...,a n), so ∈ ∈ thatf I( (a 1,...,a n) ). Thus, we can write ∈ { }

I I( (a 1,...,a n) ). ⊆ { }

We have already observed thatI( (a 1,...,a n) ) = x 1 a 1,...,x n a n (see Exercise 7), and, thus, the above inclusion{ becomes} ⟨ − − ⟩

I x 1 a 1,...,x n a n !k[x 1,...,x n]. ⊆⟨ − − ⟩

SinceI is maximal, it follows thatI= x 1 a 1,...,x n a n . ⟨ − − ⟩ ! Note the proof of Theorem 11 uses the Weak Nullstellensatz. It is not difﬁcult to see that it is, in fact, equivalent to the Weak Nullstellensatz. We have the following easy corollary of Theorem 11.

Corollary 12. If k is an algebraically closedﬁeld, then there is a one-to-one corre- n spondence between points of k and maximal ideals of k[x1,...,x n]. Thus, we have extended our algebra–geometry dictionary. Over an algebraically closedﬁeld, every nonempty irreducible variety corresponds to a proper prime ideal, and conversely. Every point corresponds to a maximal ideal, and conversely. We can use Zariski closure to characterize when a variety is irreducible.

Proposition 13. A variety V is irreducible if and only if for every variety W!V, the difference V W is Zariski dense in V. \ Proof. First assume thatV is irreducible and takeW!V. Then Proposition 7 of §4 gives the decompositionV=W V W. SinceV is irreducible andV=W, this forcesV= V W. ∪ \ ̸ \ For the converse, suppose thatV=V 1 V 2. IfV 1 !V, then V V 1 =V. But ∪ \ V V 1 V 2, so that V V 1 V 2. This impliesV V 2, andV=V 2 follows. \ ⊆ \ ⊆ ⊆ ! Let us make aﬁnal comment about terminology. Some references, such as HARTSHORNE (1977), use the term “variety” for what we call an irreducible variety and say “algebraic set” instead of variety. When reading other books on algebraic geometry, be sure to check the deﬁnitions!

EXERCISES FOR §5

1. Ifh k[x 1,...,x n] has total degreeN and ifF is as in Proposition 6, show that ∈ N (g1g2 ...g n) (h F) is a polynomial ink[t 1,...,t m]. 2. Show that a prime◦ ideal is radical. 3. Show that an idealI is prime if and only if for any idealsJ andK such thatJK I, either ⊆ J I orK I. A ⊆ ⊆ n 4. LetI 1,...,I n be ideals andA P a prime ideal containing i Ii. Then prove thatP I i for n =1 ⊇ somei. Further, ifP= i=1 Ii, show thatP=I i for somei. 2 4 3 5. Expressf=x z 6y +2xy z in the formf=f 1(x,y,z)(x+3) +f 2(x,y,z)(y 1) + − − f3(x,y,z)(z 2) for somef 1,f 2,f 3 k[x,y,z]. − ∈ 212 Chapter 4 The Algebra–Geometry Dictionary

6. Letk be an infinitefield. a. Show that any straight line ink n is irreducible. b. Prove that any linear subspace ofk n is irreducible. Hint: Parametrize and use Propo- sition 5. 7. Show that I( (a1,...,a n) ) = x 1 a 1,...,x n a n . { } ⟨ − − ⟩ 8. Show the following: 2 a. x +1 is maximal inR[x]. ⟨ ⟩ n b. IfI R[x 1,...,x n] is maximal, show thatV(I) is either empty or a point inR . Hint: Examine⊆ the proof of Theorem 11. c. Give an example of a maximal idealI inR[x 1,...,x n] for whichV(I) = . Hint: 2 ∅ Consider the ideal x 1 +1,x 2,...,x n . 9. Suppose thatkisafield⟨ which is not algebraically⟩ closed. n a. Show that ifI k[x 1,...,x n] is maximal, thenV(I) is either empty or a point ink . Hint: Examine⊆ the proof of Theorem 11. b. Show that there exists a maximal idealI ink[x 1,...,x n] for whichV(I) = . Hint: See the previous exercise. ∅ c. Conclude that ifk is not algebraically closed, there is always a maximal ideal of k[x1,...,x n] which is not of the form x 1 a 1,...,x n a n . 10. Prove that Theorem 11 implies the Weak⟨ Nullstellensatz.− − ⟩ 11. Iff C[x 1,...,x n] is irreducible, thenV(f) is irreducible. Hint: Show that f is prime. ∈ ⟨ ⟩ 12. Prove that ifI is any proper ideal inC[x 1,...,x n], then √I is the intersection of all maximal ideals containingI. Hint: Use Theorem 11. 13. Letf 1,...,f n k[x 1] be polynomials of one variable and consider the ideal ∈

I= f 1(x1),x 2 f 2(x1),...,x n f n(x1) k[x 1,...,x n]. ⟨ − − ⟩⊆

We also assume that deg(f 1) =m> 0. a. Show that everyf k[x 1,...,x n] can be written uniquely asf=q+r whereq I ∈ ∈ andr k[x 1] with eitherr= 0 or deg(r)

§6 Decomposition of a Variety into Irreducibles

In the last section, we saw that irreducible varieties arise naturally in many contexts. It is natural to ask whether an arbitrary variety can be built up out of irreducibles. In this section, we explore this and related questions. We begin by translating the ascending chain condition (ACC) for ideals (see §5 of Chapter 2) into the language of varieties. Proposition 1 (The Descending Chain Condition). Any descending chain of varieties V1 V 2 V 3 ⊇ ⊇ ⊇··· n in k must stabilize, meaning that there exists a positive integer N such that VN = VN 1 = . + ··· §6 Decomposition of a Variety into Irreducibles 213

Proof. Passing to the corresponding ideals gives an ascending chain of ideals

I(V1) I(V 2) I(V 3) . ⊆ ⊆ ⊆··· By the ascending chain condition for ideals (see Theorem 7 of Chapter 2, §5), there existsN such thatI(V N) =I(V N 1) = . SinceV(I(V)) =V for any varietyV, + ··· we haveV N =V N 1 = . + ··· ! We can use Proposition 1 to prove the following basic result about the structure of afﬁne varieties.

Theorem 2. Let V k n be an afﬁne variety. Then V can be written as aﬁnite union ⊆

V=V 1 V m, ∪···∪ where each Vi is an irreducible variety.

Proof. Assume thatV is an affine variety which cannot be written as afinite union of irreducibles. ThenV is not irreducible, so thatV=V 1 V ′ , whereV=V 1 and ∪ 1 ̸ V=V ′ . Further, one ofV 1 andV ′ must not be afinite union of irreducibles, for ̸ 1 1 otherwiseV would be of the same form. SayV 1 is not afinite union of irreducibles. Repeating the argument just given, we can writeV 1 =V 2 V ′ , whereV 1 =V 2,V 1 = ∪ 2 ̸ ̸ V2′ , andV 2 is not afinite union of irreducibles. Continuing in this way gives us an infinite sequence of affine varieties

V V 1 V 2 ⊇ ⊇ ⊇··· with V=V 1 =V 2 = . ̸ ̸ ̸ ··· This contradicts Proposition 1. ! As a simple example of Theorem 2, consider the varietyV(xz, yz) which is a union of a line (thez-axis) and a plane [the(x,y)-plane], both of which are irreducible by Exercise 6 of §5. For a more complicated example of the decomposition of a variety into irreducibles, consider the variety

V=V(xz y 2,x 3 yz). − − A sketch of this variety appears at the top of the next page. The picture suggests that this variety is not irreducible. It appears to be a union of two curves. Indeed, since both xz y 2 andx 3 yz vanish on thez-axis, it is clear that thez-axisV(x,y) is contained− inV. What− about the other curveV V(x,y)? \ 214 Chapter 4 The Algebra–Geometry Dictionary

By Corollary 11 of §4, this suggests looking at the ideal quotient

xz y 2,x 3 yz : x,y . ⟨ − − ⟩ ⟨ ⟩ (At the end of the section we will see that xz y 2,x 3 yz is a radical ideal.) We can compute this quotient using our algorithm⟨ − for computing− ⟩ ideal quotients (make sure you review this algorithm). By equation (5) of §4, the above is equal to

(I:x) (I:y), ∩ whereI= xz y 2,x 3 yz . To computeI:x, weﬁrst computeI x using our algorithm⟨ for− computing− intersections⟩ of ideals. Using lex order∩⟨ with⟩z>y>x, we obtain I x = x 2z xy 2,x 4 xyz,x 3y xz 2,x 5 xy3 . ∩⟨ ⟩ ⟨ − − − − ⟩ We can omitx 5 xy3 since it is a combination of theﬁrst and second elements in the basis. Hence− 9 : x2z xy 2 x4 xyz x3y xz 2 I:x= − , − , − x x x (1) = xz y 2,x 3 yz,x 2y z 2 ⟨ − − − ⟩ =I+ x 2y z 2 . ⟨ − ⟩ Similarly, to computeI: y , we compute ⟨ ⟩ I y = xyz y 3,x 3y y 2z,x 2y2 yz2 , ∩⟨ ⟩ ⟨ − − − ⟩ §6 Decomposition of a Variety into Irreducibles 215 which gives 9 : xyz y 3 x3y y 2z x2y2 yz2 I:y= − , − , − y y y = xz y 2,x 3 yz,x 2y z 2 ⟨ − − − ⟩ =I+ x 2y z 2 ⟨ − ⟩ =I:x.

(Do the computations using a computer algebra system.) SinceI:x=I:y, we have

I: x,y = xz y 2,x 3 yz,x 2y z 2 . ⟨ ⟩ ⟨ − − − ⟩ The varietyW=V(xz y 2,x 3 yz,x 2y z 2) turns out to be an irreducible curve. To see this, note that it can− be parametrized− − as(t 3,t 4,t 5) [it is clear that(t 3,t 4,t 5) W for anyt—we leave it as an exercise to show every point ofW is of this form],∈ so thatW is irreducible by Proposition 5 of the last section. It then follows easily that

V=V(I)=V(x,y) V(I: x,y ) =V(x,y) W ∪ ⟨ ⟩ ∪ (see Proposition 7 of §4), which gives decomposition ofV into irreducibles. Both in the above example and the case ofV(xz, yz), it appears that the decomposition of a variety into irreducible pieces is unique. It is natural to ask whether this is true in general. It is clear that, to avoid trivialities, we must rule out decompositions in which the same irreducible piece appears more than once, or in which one irreducible piece contains another. This is the aim of the following deﬁnition.

Deﬁnition 3. LetV k n be an afﬁne variety. A decomposition ⊆

V=V 1 V m, ∪···∪ where eachV i is an irreducible variety, is called a minimal decomposition (or, sometimes, an irredundant union) ifV i V j fori=j. Also, we call theV i the irreducible componentsofV. ̸⊆ ̸

With this deﬁnition, we can now prove the following uniqueness result.

Theorem 4. Let V k n be an afﬁne variety. Then V has a minimal decomposition ⊆

V=V 1 V m ∪···∪

(so each Vi is an irreducible variety and Vi V j for i= j). Furthermore, this ̸⊆ ̸ minimal decomposition is unique up to the order in which V1,...,V m are written.

Proof. By Theorem 2,V can be written in the formV=V 1 V m, where each ∪···∪ Vi is irreducible. Further, if aV i lies in someV j fori=j, we can dropV i, andV ̸ will be the union of the remainingV j’s forj=i. Repeating this process leads to a minimal decomposition ofV. ̸ 216 Chapter 4 The Algebra–Geometry Dictionary

To show uniqueness, suppose thatV=V ′ V ′ is another minimal decom- 1 ∪···∪ l position ofV. Then, for eachV i in theﬁrst decomposition, we have

Vi =V i V=V i (V ′ V ′) = (Vi V ′ ) (V i V ′). ∩ ∩ 1 ∪···∪ l ∩ 1 ∪···∪ ∩ l

SinceV i is irreducible, it follows thatV i =V i V ′ for somej, i.e.,V i V ′. Applying ∩ j ⊆ j the same argument toV j′ (using theV i’s to decomposeV) shows thatV j′ V k for somek, and, thus, ⊆ Vi V ′ V k. ⊆ j ⊆ By minimality,i=k, and it follows thatV i =V j′. Hence, everyV i appears in V=V ′ V ′, which impliesm l. A similar argument provesl m, and 1 ∪···∪ l ≤ ≤ m=l follows. Thus, theV i′’s are just a permutation of theV i’s, and uniqueness is proved. ! The uniqueness part of Theorem 4 guarantees that the irreducible components ofV are well-defined. We remark that the uniqueness is false if one does not insist that the union befinite. (A planeP is the union of all the points on it. It is also the union of some line inP and all the points not on the line—there are infinitely many lines inP with which one could start.) This should alert the reader to the fact that although the proof of Theorem 4 is easy, it is far from vacuous: one makes subtle use offiniteness (which follows, in turn, from the Hilbert Basis Theorem). Here is a result that relates irreducible components to Zariski closure.

Proposition 5. Let V, W be varieties with W! V. Then V W is Zariski dense in V if and only if W contains no irreducible component of V.\

Proof. Suppose thatV=V 1 V m as in Theorem 4 and thatV i W for alli. ∪···∪ ̸⊆ This impliesV i W!V i, and sinceV i is irreducible, we deduce Vi (V i W)=V i by Proposition 13∩ of §5. Then \ ∩

V W= (V1 V m) W= (V1 (V 1 W)) (V m (V m W)) \ ∪···∪ \ \ ∩ ∪···∪ \ ∩ = V1 (V 1 W) Vm (V m W) \ ∩ ∪···∪ \ ∩ =V 1 V m =V, ∪···∪ where the second line uses Lemma 3 of §4. The other direction of the proof will be covered in the exercises. ! Theorems 2 and 4 can also be expressed purely algebraically using the one-to-one correspondence between radical ideals and varieties.

Theorem 6. If k is algebraically closed, then every radical ideal in k[x1,...,x n] can be written uniquely as aﬁnite intersection of prime ideals P 1 P r, where ∩···∩ Pi P j for i= j. (As in the case of varieties, we often call such a presentation of a radical̸⊆ ideal̸ a minimal decompositionoran irredundant intersection). Proof. Theorem 6 follows immediately from Theorems 2 and 4 and the ideal– variety correspondence. ! §6 Decomposition of a Variety into Irreducibles 217

We can also use ideal quotients from §4 to describe the prime ideals that appear in the minimal representation of a radical ideal.

Theorem 7. If k is algebraically closed and I is a proper radical ideal such that

Br I= Pi i=1 is its minimal decomposition as an intersection of prime ideals, then the Pi’s are precisely the proper prime ideals that occur in the set I:f f k[x 1,...,x n] . { | ∈ }

Proof. First, note that sinceI is proper, eachP i is also a proper ideal (this follows from minimality). For anyf k[x 1,...,x n], we have ∈ 0 Br / Br I:f= Pi :f= (Pi :f) i=1 i=1 by Exercise 17 of §4. Note also that for any prime idealP, eitherf P, in which caseP:f= 1 , orf/ P, in which caseP:f=P (see Exercise 3).∈ Now suppose⟨ ⟩ that∈ I:f is a proper prime ideal. By Exercise 4 of §5, the above formula forI:f implies thatI:f=P i :f for somei. SinceP i :f=P i or 1 by the ⟨ ⟩ above observation, it follows thatI:f=P i. 6 C ( r To see that everyCP i can arise in this way,ﬁxi and pickf j=i Pj P i; such r ∈ ̸ \ anf exists because Pi is minimal. ThenP i :f=P i andP j :f= 1 forj=i. If i=1 ⟨ ⟩ ̸ we combine this with the above formula forI:f , then it follows thatI:f=P i.! We should mention that Theorems 6 and 7 hold for anyﬁeldk, although the proofs in the general case are different (see Corollary 10 of §8). For an example of these theorems, consider the idealI= xz y 2,x 3 yz . Recall that the varietyV=V(I) was discussed earlier in this section.⟨ − For the time− being,⟩ let us assume thatI is radical (eventually we will see that this is true). Can we write I as an intersection of prime ideals? We start with the geometric decomposition

V=V(x,y) W ∪ proved earlier, whereW=V(xz y 2,x 3 yz,x 2y z 2). This suggests that − − − I= x,y xz y 2,x 3 yz,x 2y z 2 , ⟨ ⟩ ∩ ⟨ − − − ⟩ which is straightforward to prove by the techniques we have learned so far (see Exercise 4). Also, from equation (1), we know thatI:x= xz y 2,x 3 yz,x 2y z 2 . Thus, ⟨ − − − ⟩ I= x,y (I:x). ⟨ ⟩ ∩ 218 Chapter 4 The Algebra–Geometry Dictionary

To represent x,y as an ideal quotient ofI, let us think geometrically. The idea is to removeW ⟨from⟩ V. Of the three equations definingW, thefirst two giveV. So it makes sense to use the third one,x 2y z 2, and one can check thatI:(x 2y z 2) = x,y (see Exercise 4). Thus, − − ⟨ ⟩ (2)I=(I:(x 2y z 2)) (I:x). − ∩ It remains to show thatI:(x 2y z 2) andI:x are prime ideals. Thefirst is easy sinceI:(x 2y z 2) = x,y is obviously− prime. As for the second, we have already seen thatW=−V(xz y⟨ ⟩2,x 3 yz,x 2y z 2) is irreducible and, in the exercises, you will show that−I(W) = xz−y 2,x−3 yz,x 2y z 2 =I:x. It follows from Proposition 3 of §5 thatI:x is⟨ a− prime ideal.− This completes− ⟩ the proof that ( 2) is the minimal representation ofI as an intersection of prime ideals. Finally, sinceI is an intersection of prime ideals, we see thatI is a radical ideal (see Exercise 1). The arguments used in this example are special to the caseI= xz y 2,x 3 yz . It would be nice to have more general methods that could be applied⟨ − to any− ideal.⟩ Theorems 2, 4, 6, and 7 tell us that certain decompositions exist, but the proofs give no indication of how tofind them. The problem is that the proofs rely on the Hilbert Basis Theorem, which is intrinsically nonconstructive. Based on what we have seen in §§5 and 6, the following questions arise naturally: (Primality) Is there an algorithm for deciding if a given ideal is prime? • (Irreducibility) Is there an algorithm for deciding if a given affine variety is irre- • ducible? (Decomposition) Is there an algorithm forfinding the minimal decomposition of • a given variety or radical ideal? The answer to all three questions is yes, and descriptions of the algorithms can be found in the works of HERMANN (1926), MINES,RICHMAN, and RUITEN- BERG (1988), and SEIDENBERG (1974, 1984). As in §2, the algorithms in these ar- ticles are very inefficient. However, the work of GIANNI,TRAGER and ZACHARIAS (1988) and EISENBUD,HUNEKE and VASCONCELOS (1992) has led to more effi- cient algorithms. See also Chapter 8 of BECKER and WEISPFENNING (1993) and §4.4 of ADAMS and LOUSTAUNAU (1994).

EXERCISES FOR §6

1. Show that the intersection of any collection of prime ideals is radical. 2. Show that an irredundant intersection of at least two prime ideals is never prime. 3. IfP k[x 1,...,x n] is a prime ideal, then prove thatP:f=Piff/ P andP:f= 1 if f P⊆. ∈ ⟨ ⟩ ∈ 2 3 4. LetI= xz y ,x yz . ⟨ − − ⟩ a. Show thatI:(x 2y z 2) = x,y . 2 − 2 ⟨ ⟩ b. Show thatI:(x y z ) is prime. − c. Show thatI= x,y xz y 2,x 3 yz,z 2 x 2y . ⟨ ⟩ ∩ ⟨ − − − ⟩ 5. LetJ= xz y 2,x 3 yz,z 2 x 2y k[x,y,z], wherekisinﬁnite. ⟨ − − − ⟩ ⊆ a. Show that every point ofW=V(J)isof the form(t 3,t 4,t 5) for somet k. ∈ §7 Proof of the Closure Theorem 219

b. Show thatJ=I(W). Hint: Compute a Gröbner basis forJ using lex order with z>y>x and show that everyf k[x,y,z] can be written in the form ∈ f=g+a+ bz+ xA(x) + yB(x) +y 2C(x),

whereg J,a,b k andA,B,C k[x]. ∈ ∈ ∈ 6. Complete the proof of Proposition 5. Hint:V i W impliesV W V V i. 7. Translate Theorem 7 and its proof into geometry.⊆ \ ⊆ \ 2 3 5 8. LetI= xz y ,z x Q[x,y,z]. ⟨ − − ⟩ ⊆ a. ExpressV(I) as aﬁnite union of irreducible varieties. Hint: The parametrizations (t3,t 4,t 5) and(t 3, t 4,t 5) will be useful. b. ExpressI as an intersection− of prime ideals which are ideal quotients ofI and conclude thatI is radical. 9. LetV,W be varieties ink n withV W. Show that each irreducible component ofV is contained in some irreducible component⊆ ofW. a1 a2 ar 10. Letf C[x 1,...,x n] and letf=f 1 f2 f r be the decomposition off into irreducible ∈ ··· factors. Show thatV(f)=V(f 1) V(f r) is the decomposition ofV(f) into irre- ∪···∪ ducible components andI(V(f))= f 1 f2 f r . Hint: See Exercise 11 of §5. ⟨ ··· ⟩

§7 Proof of the Closure Theorem

This section will complete the proof of the Closure Theorem from Chapter 3, §2. We will use many of the tools introduced in this chapter, including the Nullstellensatz, Zariski closures, saturations, and irreducible components. We begin by recalling the basic situation. Letk be an algebraically closedfield, n n l and letπ l :k k − is projection onto the lastn l components. IfV=V(I) is → n − an affine variety ink , then we get thel-th elimination idealI l =I k[x l 1,...,x n]. ∩ + Thefirst part of the Closure Theorem, which asserts thatV(I l) is the Zariski closure n l ofπ l(V) ink − , was proved earlier in Theorem 4 of §4. The remaining part of the Closure Theorem tells us thatπ l(V)fills up “most” of V(Il). Here is the precise statement.

Theorem 1 (The Closure Theorem, second part). Let k be algebraically closed, n and let V=V(I) k . Then there is an afﬁne variety W V(I l) such that ⊆ ⊆

V(Il) W π l(V) and V(Il) W=V(I l). \ ⊆ \ This is slightly different from the Closure Theorem stated in §2 of Chapter 3. There, we assumedV= and asserted thatV(I l) W π l(V) for someW!V(I l). In Exercise 1 you will̸ ∅ prove that Theorem 1 implies\ ⊆ the version in Chapter 3. The proof of Theorem 1 will use the following notation. Renamex l+1,...,x n as yl+1,...,y n and writek[x 1,...,x l,y l+1,...,y n] ask[x,y] forx=(x 1,...,x l) and y=(y l+1,...,y n). Alsoﬁx a monomial order>onk[x,y] with the property that xα >x β impliesx α >x βyγ for allγ. The product order described in Exercise 9 of Chapter 2, §4 is an example of such a monomial order. Another example is given by lex order withx 1 > >x l >y l 1 > >y n. ··· + ··· 220 Chapter 4 The Algebra–Geometry Dictionary

An important tool in proving Theorem 1 is the following result.

Theorem 2. Fix aﬁeld k. Let I k[x,y] be an ideal and let G= g 1,...,g t bea Gröbner basis for I with respect⊆ to a monomial order as above.{ For1 i t with} ≤ ≤ gi / k[y], write gi in the form ∈

αi αi (1)g i =c i(y)x + terms

n l Finally, assume thatb=(a l 1,...,a n) V(I l) k − is a partial solution such + ∈ ⊆ that ci(b)=0 for all g i / k[y]. Then: ̸ ∈ (i) The set G= g i(x,b) g i / k[y] k[x] { | ∈ }⊆ is a Gröbner basis of the ideal f(x,b) f I . { | ∈ } l (ii) If k is algebraically closed, then there existsa=(a 1,...,a l) k such that (a,b) V=V(I). ∈ ∈ Proof. Givenf k[x,y], we set ∈ ¯f=f(x,b) k[x]. ∈

In this notation, G= ¯gi g i / k[y] . Also observe that¯ig= 0 wheng i k[y] since { | ∈ } ∈ b V(I l). If we set ¯I= ¯f f I , then it is an easy exercise to show that ∈ { | ∈ } ¯I= G k[x]. ⟨ ⟩ ⊆ In particular, ¯I is an ideal ofk[x]. To prove that G is a Gröbner basis of ¯I, takeg i,g j G k[y] and consider the polynomial ∈ \ xγ xγ S=c j(y) gi c i(y) gj, xαi − xαj wherex γ = lcm(xαi ,x αj ). Our chosen monomial order has the property that αi γ LT(gi) = LT(ci(y))x , and it" follows easily thatx > LT(S). SinceS I, it has a t ∈ standard representationS= k=1 Akgk. Then evaluating atb gives xγ xγ " c (b) ¯g c (b) ¯g= S= A ¯g j α i i α j ¯kg G k k x i − x j ∈ since¯gi =0forg i k[y]. ∈ Thenc i(b),c j(b)= 0 imply that S is theS-polynomialS(¯gi,¯gj) up to the nonzero ̸ constantc i(b)cj(b). Since

γ x > LT(S) LT(Akgk),A kgk =0, ≥ ̸ it follows that γ x > LT(A¯k¯gk), A¯k¯gk =0, ̸ §7 Proof of the Closure Theorem 221 by Exercise 3 of Chapter 2, §9. HenceS(¯gi,¯gj) has an lcm representation as deﬁned in Chapter 2, §9. By Theorem 6 of that section, we conclude that G is a Gröbner basis of ¯I, as claimed. For part (ii), note that by construction, every element of G has positive total degree in thex variables, so that¯gi is nonconstant for everyi. It follows that 1/ ¯I ∈ since G is a Gröbner basis of ¯I. Hence ¯I!k[x], so that by the Nullstellensatz, l there existsa k such that¯gi(a) = 0 for all¯gi G, i.e.,g i(a,b) = 0 for all ∈ ∈ gi G k[y]. Since¯g i = 0 wheng i k[y], it follows thatg i(a,b) = 0 for allg i G. Hence∈ \(a,b) V=V(I). ∈ ∈ ∈ ! Part (ii) of Theorem 2 is related to the Extension Theorem from Chapter 3. Com- pared to the Extension theorem, part (ii) is simultaneously stronger (the Extension Theorem assumesl= 1, i.e., just one variable is eliminated) and weaker [part (ii) requires the nonvanishing of all relevant leading coefﬁcients, while the Extension Theorem requires just one]. For our purposes, Theorem 2 has the following important corollary.

Corollary 3. With the same notation as Theorem 2, we have 6? ( V(I ) V c π (V). l gi G k[y] i l \ ∈ \ ⊆ 6? ( Proof. Takeb V(I ) V c . Thenb V(I ) andc (b)= 0 for all l gi G k[y] i l i ∈ \ ∈ \ l ∈ ̸ gi G k[y]. By Theorem 2, there isa k such that(a,b) V=V(I). In other ∈ \ ∈ ∈ words,b π l(V), and the corollary follows. ∈ ! SinceA B=A (A B), Corollary 3 implies that the intersection \ \ ∩ 6? ( W=V(I ) V c V(I ) l gi G k[y] i l ∩ ∈ \ ⊆ has the property thatV(I l) W π l(V). IfV(I l) W is also Zariski dense inV(I l), \ ⊆ \ thenW V(I l) satisﬁes the conclusion of the Closure Theorem. ⊆ Hence, to complete the proof of the6 Closure? Theorem,( we need to explore what happens when the differenceV(I l) V ci is not Zariski dense inV(I l). \ gi G k[y] The following proposition shows that in this∈ \ case, the original varietyV=V(I) decomposes into varieties coming from strictly bigger ideals.

Proposition 4. Assume6? that k is algebraically( closed and the Gröbner basis G is reduced. IfV(I ) V c is not Zariski dense inV(I ), then there is some l gi G k[y] i l \ ∈ \ gi G k[y] whose c i has the following two properties: ∈ \ (i)V=V(I+ c i ) V(I:c ∞). ⟨ ⟩ ∪ i (ii)I!I+ c i andI!I:c i∞. ⟨ ⟩

Proof. For (i), we haveV=V(I)=V(I+ c i ) V(I:c ∞) by Theorem 10 of §4. ⟨ ⟩ ∪ i For (ii), weﬁrst show thatI!I+ c i for allg i G k[y]. To see why, suppose ⟨ ⟩ ∈ \ thatc i I for somei. SinceG is a Gröbner basis ofI, LT(ci) is divisible by some ∈ LT(gj), and theng j k[y] since the monomial order eliminates thex variables. ∈ 222 Chapter 4 The Algebra–Geometry Dictionary

αi Henceg j =g i. But then (1) implies that LT(gj) divides LT(gi) = LT(ci)x , which ̸ contradicts our assumption thatG is reduced. Hencec i / I, andI!I+ c i follows. ∈ ⟨ ⟩ Now suppose thatI=I:c ∞ for alli withg i G k[y]. In Exercise 4, you will i ∈ \ show that this impliesI l :c i∞ =I l for alli. Hence

V(Il) =V(I l :c ∞) = V(Il) V(c i) = V(Il) (V(I l) V(c i)), i \ \ ∩ where the second equality uses Theorem 10 of §4. If follows thatV(I l) V(c i) ∩ contains no irreducible component ofV(I l) by Proposition 5 of §6. Since this holds for alli, thefinite union 1 1 6? ( V(I ) V(c ) =V(I ) V(c ) =V(I ) V c gi G k[y] l i l gi G k[y] i l gi G k[y] i ∈ \ ∩ ∩ ∈ \ ∩ ∈ \ also contains no irreducible component ofV(I l) (see Exercise 5). By the same proposition from §6, we conclude that the difference 6 6? (( 6? ( V(I ) V(I ) V c =V(I ) V c l l gi G k[y] i l gi G k[y] i \ ∩ ∈ \ \ ∈ \ is Zariski dense inV(I l). This contradiction shows thatI!I:c i∞ for somei and completes the proof of the proposition. ! In the situation of Proposition 4, we have a decomposition ofV into two pieces. The next step is to show that if we canfind aW that works for each piece, then we canfind aW what works forV. Here is the precise result.

Proposition 5. Let k be algebraically closed. Suppose that a variety V=V(I) can be written V=V(I (1)) V(I (2)) and that we have varieties ∪ (1) (2) W1 V(I ) and W2 V(I ) ⊆ l ⊆ l

(i) (i) (i) (i) such that V(I ) W i =V(I ) andV(I ) W i π l(V(I ) for i=1,2. Then l \ l l \ ⊆ W=W 1 W 2 is a variety contained in V that satisﬁes ∪

V(Il) W=V(I l) andV(I l) W π l(V). \ \ ⊆ (i) Proof. For simplicity, setV i =V(I ), so thatV=V 1 V 2. Theﬁrst part of the ∪ (i) Closure Theorem proved in §4 implies thatV(I l) = πl(V) andV(I l ) = πl(Vi). Hence

V(Il) = πl(V)= πl(V1 V 2) = πl(V1) π l(V2) = πl(V1) πl(V2) ∪ ∪ ∪ =V(I (1)) V(I (2)), l ∪ l where the last equality of theﬁrst line uses Lemma 3 of §4. (i) Now letW i V(I l ) be as in the statement of the proposition. By Proposition 5 ⊆ (i) of §6, we know thatW i contains no irreducible component ofV(I l ). As you will prove in Exercise 5, this implies that the unionW=W 1 W2 contains no irreducible ∪ §7 Proof of the Closure Theorem 223

(1) (2) component ofV(I l) =V(I ) V(I ). Using Proposition 5 of §6 again, we deduce l ∪ l thatV(I l) W is Zariski dense inV(I l). Since we also have \ 6 ( 6 ( 6 ( (1) (2) (1) (2) V(Il) W= V(I ) V(I ) (W 1 W 2) V(I ) W 1 V(I ) W 2 \ l ∪ l \ ∪ ⊆ l \ ∪ l \ π l(V1) π l(V2) =π l(V), ⊆ ∪ the proof of the proposition is complete. ! Theﬁnal ingredient we need for the proof of the Closure Theorem is the following maximum principle for ideals.

Proposition 6 (Maximum Principle for Ideals). Given a nonempty collection of ideals I α α in a polynomial ring k[x1,...,x n], there existsα 0 such that for allβ { , we} have∈A ∈A ∈A I I = I =I . α0 ⊆ β ⇒ α0 β In other words, I is maximal with respect to inclusion among the I forα . α0 α ∈A Proof. This is an easy consequence of the ascending chain condition (Theorem 7 of Chapter 2, §5). The proof will be left as an exercise. ! We are now ready to prove the second part of the Closure Theorem.

Proof of Theorem 1. Suppose the theorem fails for some idealI k[x 1,...,x n], ⊆ i.e., there is no afﬁne varietyW!V(I) such that

V(Il) W π l(V(I)) and V(Il) W=V(I l). \ ⊆ \ Our goal is to derive a contradiction. Among all ideals for which the theorem fails, the maximum principle of Propo- sition 6 guarantees that there is a maximal such ideal, i.e., there is an idealI such that the theorem fails forI but holds for every strictly larger idealI!J. Let us apply our results toI. By Corollary 3, we know that 6? ( V(I ) V c π (V). l gi G k[y] i l \ ∈ \ ⊆ 6? ( Since the theorem fails forI,V(I) V c cannot be Zariski dense in gi G k[y] i \ ∈ \ V(Il). Therefore, by Proposition 4, there is somei such that

(1) (2) I!I =I+ c i ,I!I =I:c i∞ ⟨ ⟩ and V(I)=V(I (1)) V(I (2)). ∪ Our choice ofI guarantees that the theorem holds for the strictly larger idealsI (1) (2) (i) andI . The resulting afﬁne varietiesW i V(I ),i=1, 2, satisfy the hypothesis ⊆ l of Proposition 5, and then the proposition implies thatW=W 1 W 2 V(I) ∪ ⊆ satisﬁes the theorem forI. This contradicts our choice ofI, and we are done.! 224 Chapter 4 The Algebra–Geometry Dictionary

The proof of the Closure Theorem just given is nonconstructive. Fortunately, in practice it is straightforward toﬁndW V(I l) with the required properties. We will give two examples and then describe⊆ a general procedure. Theﬁrst example is very simple. Consider the ideal

2 I= yx + yx+1 C[x,y]. ⟨ ⟩⊆ 2 We use lex order withx>y, andI 1 = 0 sinceg 1 = yx +yx+1 is a Gröbner basis { } forI. In the notation of Theorem 2, we havec 1 =y, and then Corollary 3 implies that V(I1) V(c 1) =C V(y) =C 0 π 1(V(I)) =C. \ \ \{ }⊆ Hence, we can takeW= 0 in Theorem 1 sinceC 0 is Zariski dense inC. { } \{ } The second example, taken from SCHAUENBURG (2007), uses the ideal

2 I= xz+y 1,w+y+z 2,z C[w,x,y,z]. ⟨ − − ⟩ ⊆ It is straightforward to check thatV=V(I) is the lineV=V(w 1,y 1,z) C 4, 2 − − ⊆ which projects to the pointπ 2(V)=V(y 1,z) C when we eliminatew andx. Thus,W= satisﬁes Theorem 1 in this− case.⊆ Here is∅ a systematic way to discover thatW= . A lex Gröbner basis ofI for w>x>y>z consists of ∅

2 2 g1 =w+y+z 2,g 2 = xz+y 1,g 3 =y 2y+1,g 4 = yz z,g 5 =z . − − − −

Eliminatingw andx givesI 2 = g 3,g 4,g 5 , and one sees easily that ⟨ ⟩

V(I2) =V(y 1,z). −

Sinceg 1 =1 w+y+z 2 andg 2 =z x+y 1, we havec 1 = 1 andc 2 =z. If we set · − · − J= c 1c2 = z , ⟨ ⟩ ⟨ ⟩ then Corollary 3 implies thatV(I 2) V(J) π 2(V). However,V(I 2) V(J)= , so \ ⊆ \ ∅ the difference is not Zariski dense inV(I 2). In this situation, we use the decomposition ofV(I) guaranteed to exist by Propo- sition 4. Note thatI=I:c 1∞ sincec 1 = 1. Hence we usec 2 =z in the proposition. This gives the two ideals

(1) 2 I =I+ c 2 = xz+y 1,w+y+z 2,z ,z = w 1,y 1,z , ⟨ ⟩ ⟨ − − ⟩ ⟨ − − ⟩ (2) 2 I =I:c ∞ =I:z ∞ = 1 sincez I. 2 ⟨ ⟩ ∈ Now we start over withI (1) andI (2). §7 Proof of the Closure Theorem 225

(1) (1) (1) ForI , observe that w 1,y 1,z is a Gröbner basis ofI , and onlyg 1 = { − − } (1) w 1/ C[y,z]. The coefﬁcient ofw isc 1 = 1, and then Corollary 3 applied to I(−1) gives∈ (1) (1) V(I ) V(1) π 2(V(I )). 2 \ ⊆ (1) SinceV(1)= , we can pickW 1 = forI in Theorem 1. ∅ ∅ (2) Applying the same systematic process toI = 1 , we see that there arenog i / ⟨ ⟩ ∈ C[y,z]. Thus Corollary 3 involves the product over the empty set. By convention (see Exercise 7) the empty product is 1. Then Corollary 3 tells us that we can pick (2) W2 = forI in Theorem 1. By Proposition 5, it follows that Theorem 1 holds for the ideal∅ I with W=W 1 W 2 = = . ∪ ∅∪∅ ∅ To do this in general, we use the following recursive algorithm to produce the desired subsetW:

Input: an idealI k[x,y] with varietyV=V(I) ⊆ Output: FindW(I)=W V(I l) withV(I l) W π l(V), V(Il) W=V(I l) ⊆ \ ⊆ \ G := reduced Gröbner basis forI for a monomial order as in Theorem 2

αi αi ci := coefﬁcient ing i =c i(y)x + terms

Selectg i G k[y] withI!I:c i∞ ∈ \ FindW(I) := FindW(I+ c i ) FindW(I:c ∞) ⟨ ⟩ ∪ i RETURN FindW(I)

The7? function8 FindW takes the input idealI and computes the idealsI l andJ= c . The IF statement asks whetherV(I ) V(J) is Zariski dense inV(I ). gi G k[y] i l l ∈ \ \ If the answer is yes, thenV(I l) V(J) has the desired property by Corollary 3, which ∩ is why FindW(I)=V(I l) V(J) in this case. In the exercises, you will describe an ∩ algorithm for determining whether V(Il) V(J)=V(I l). \ WhenV(I l) V(J) fails to be Zariski dense inV(I l), Proposition 4 guarantees \ that we canﬁndc i such that the ideals

(1) (2) I =I+ c i andI =I:c ∞ ⟨ ⟩ i are strictly larger thanI and satisfyV=V(I) =V(I (1)) V(I (2)). Then, as in the second example above, we repeat the process on the∪ two new ideals, which means computing FindW(I(1)) and FindW(I(2)). By Proposition 5, the union of these varieties works forI, which explains the last line of FindW. 226 Chapter 4 The Algebra–Geometry Dictionary

We say that FindW is recursive since it calls itself. We leave it as an exercise to show that the maximum principle from Proposition 6 implies that FindW always terminates infinitely many steps. When it does, correctness follows from the above discussion. We end this section by using the Closure Theorem to give a precise description n l n of the projectionπ l(V) k − of an affine varietyV k . ⊆ ⊆ Theorem 7. Let k be algebraically closed and let V k n be an affine variety. Then n l ⊆ there are affine varieties Zi W i k − for1 i p such that ⊆ ⊆ ≤ ≤ 5p πl(V)= (Wi Z i). \ i=1

Proof. We assumeV= . First letW 1 =V(I l). By the Closure Theorem, there is a ̸ ∅ n varietyZ 1 !W 1 such thatW 1 Z 1 π l(V). Then, back ink , consider the set \ ⊆ n V1 =V (a 1,...,a n) k (a l 1,...,a n) Z 1 . ∩{ ∈ | + ∈ }

One easily checks thatV 1 is an afﬁne variety (see Exercise 10), and furthermore, V1 !V since otherwise we would haveπ l(V) Z 1, which would implyW 1 Z 1 by Zariski closure. Moreover, you will check in⊆ Exercise 10 that ⊆

(2)π l(V)=(W1 Z 1) π l(V1). \ ∪

IfV 1 = , then we are done. IfV 1 is nonempty, letW 2 be the Zariski closure of ∅ πl(V1). Applying the Closure Theorem toV 1, we getZ 2 !W 2 withW 2 Z2 π l(V1). Then, repeating the above construction, we get the variety \ ⊂

n V2 =V 1 (a 1,...,a n) k (a l 1,...,a n) Z 2 ∩{ ∈ | + ∈ } such thatV 2 !V 1 and

πl(V)=(W1 Z 1) (W 2 Z 2) π l(V2). \ ∪ \ ∪

IfV 2 = , we are done, and if not, we repeat this process again to obtainW 3,Z 3 ∅ andV 3 !V 2. Continuing in this way, we must eventually haveV N = for someN, since otherwise we would get an inﬁnite descending chain of varieties∅

V"V 1 "V 2 " , ··· which would contradict Proposition 1 of §6. Once we haveV N = , the desired ∅ formula forπ l(V) follows easily. ! In general, a set of the form described in Theorem 7 is called constructible. As a simple example of Theorem 7, considerI= xy+z 1,y 2z2 C[x,y,z] ⟨ − ⟩ ⊆ and setV=V(I) C 3. We leave it as an exercise to show that ⊆

V(I1) =V(z) V(y,z 1)=V(z) (0,1) ∪ − ∪{ } §7 Proof of the Closure Theorem 227 and thatW=V(y,z) = (0,0) satisﬁesV(I 1) W π 1(V). However, we also { } \ ⊆ haveπ 1(V) V(I 1), and since xy+z 1 I, no point ofV has vanishingy andz ⊆ − ∈ coordinates. It follows thatπ 1(V) V(I 1) (0,0) . Hence ⊆ \{ }

π1(V)=V(I 1) (0,0) =(V(z) (0,0) ) (0,1) . \{ } \{ } ∪{ }

This gives an explicit representation ofπ 1(V) as a constructible set. You will work out another example of Theorem 7 in the exercises. More substantial examples can be found in SCHAUENBURG (2007), which also describes an algorithm for writing πl(V) as a constructible set. Another approach is described in ULLRICH (2006).

EXERCISES FOR §7

1. Prove that Theorem 3 of Chapter 3, §2 follows from Theorem 1 of this section. Hint: Show that theW from Theorem 1 satisfiesW!V(I l) whenV= . ̸ ∅ 2. In the notation of Theorem 2, prove that ¯I= G for ¯I= ¯f f I . 3. Given setsA andB, prove thatA B=A (A B⟨). ⟩ { | ∈ } \ \ ∩ 4. In the proof of Proposition 4, prove thatI=I:c i∞ implies thatI l =I l :c i∞. 5. This exercise will explore some properties of irreducible components needed in the proofs of Propositions 4 and 5. a. LetW 1,...,W r be affine varieties contained in a varietyV and assume that for each 1 i r,B no irreducible component ofV is contained inW i. Prove that the same is ≤ ≤ r true for i=1 Wi. (This fact is used in the proof of Proposition 4.) b. LetW i V i be affine varieties fori=1, 2 such thatW i contains no irreducible ⊆ component ofV i. Prove thatW=W 1 W 2 contains no irreducible component of ∪ V=V 1 V 2. (This fact is used in the proof of Proposition 5.) 6. Prove Proposition∪ 6. Hint: Assume that the proposition is false for some nonempty collection of ideals I α α and show that this leads to a contradiction of the ascending chain condition.{ } ∈A 7. In this exercise we will see why it is reasonable to make the convention that the empty product is 1. LetR be a commutative ring with 1 and let be afinite set such that for A everyα , we haver α R. Then we get the product ∈A ∈ 2 α rα. ∈A Although is unordered, the product is well-defined sinceR is commutative. A a. Assume isfinite and disjoint from such that for everyβ , we haver β R. Prove thatB A ∈B ∈ 2 -2 .-2 . γ rγ = α rα β rβ . ∈A∪B ∈A ∈B b. It is likely that2 the proof you gave in part (a) assumed that and are nonempty. A B Explain why α rα =& 1 makes the above formula& work in all cases. ∈∅ c. In a similar way, define α rα and explain why α rα = 0 is needed to make the analog of part (a) true for∈A sums. ∈∅ 8. The goal of this exercise is to describe an algorithm for deciding whether V(I) V(g) = V(I) when thefieldk is algebraically closed. \ a. Prove that V(I) V(g) =V(I) is equivalent toI:g ∞ √I. Hint: Use the Nullstel- \ ⊆ lensatz and Theorem 10 of §4. Also remember thatI I:g ∞. b. Use Theorem 14 of §4 and the Radical Membership⊆ Algorithm from §2 to describe an algorithm for deciding whetherI:g ∞ √I. ⊆ 228 Chapter 4 The Algebra–Geometry Dictionary

9. Give a proof of the termination of FindW that uses the maximum principle stated in Proposition 6. Hint: Consider the set of all ideals ink[x,y] for which FindW does not terminate. 10. This exercise is concerned with the proof of Theorem 7. n a. Verify thatV 1 =V (a 1,...,a n) k (a l+1,...,a n) Z 1 is an afﬁne variety. ∩{ ∈ | ∈ } b. Verify thatπ l(V)=(W1 Z 1) π l(V1). \ ∪ 2 2 11. As in the text, letV=V(I) forI= xy+z 1,y z C[x,y,z]. Show that ⟨ − ⟩ ⊆

V(I1) =V(z) V(y,z 1) =V(z) (0,1) ∪ − ∪{ }

and thatW=V(y,z) = (0,0) satisﬁesV(I 1) W π 1(V). { 3 } \ ⊆ 2 12. LetV=V(y xz) C . Theorem 7 tells us thatπ 1(V) C is a constructible set. Find − ⊆ ⊆ an explicit decomposition ofπ 1(V) of the form given by Theorem 7. Hint: Your answer will involveW 1,Z 1 andW 2. 13. Proposition 6 is the maximum principle for ideals. The geometric analog is the minimum principle for varieties, which states that among any nonempty collection of varieties in kn, there is a variety in the collection which is minimal with respect to inclusion. More precisely, this means that if we are given varietiesV α,α , where is a nonempty ∈A A set, then there is someα 0 with the property that for allβ ,V β V α implies ∈A ∈A ⊆ 0 Vβ =V α0 . Prove the minimum principle. Hint: Use Proposition 1 of §6. 14. Apply the minimum principle of Exercise 13 to give a different proof of Theorem 7. Hint: n Consider the collection of all varietiesV k for whichπ l(V) is not constructible. By ⊆ the minimum principle, there is a varietyV such thatπ l(V) is not constructible butπ l(W) is constructible for every varietyW!V. Show how the proof of Theorem 7 up to (2) can be used to obtain a contradiction and thereby prove the theorem.

§8 Primary Decomposition of Ideals

In view of the decomposition theorem proved in §6 for radical ideals, it is natural to ask whether an arbitrary idealI (not necessarily radical) can be represented as an intersection of simpler ideals. In this section, we will prove the Lasker-Noether decomposition theorem, which describes the structure ofI in detail. There is no hope of writing an arbitrary idealI as an intersection of prime ideals (since an intersection of prime ideals is always radical). The next thing that suggests itself is to writeI as an intersection of powers of prime ideals. This does not quite work either: consider the idealI= x,y 2 inC[x,y]. Any prime ideal containingI must containx andy and, hence, must⟨ equal⟩ x,y (since x,y is maximal). Thus, if I were to be an intersection of powers of prime⟨ ⟩ ideals,⟨ it would⟩ have to be a power of x,y (see Exercise 1 for the details). ⟨The⟩ concept we need is a bit more subtle.

Deﬁnition 1. An idealI ink[x 1,...,x n] is primary if fg I implies eitherf Ior some powerg m I for somem>0. ∈ ∈ ∈ It is easy to see that prime ideals are primary. Also, you can check that the ideal I= x,y 2 discussed above is primary (see Exercise 1). ⟨ ⟩ Lemma 2. If I is a primary ideal, then √I is prime and is the smallest prime ideal containing I. §8 Primary Decomposition of Ideals 229

Proof. See Exercise 2. ! In view of this lemma, we make the following deﬁnition.

Deﬁnition 3.IfI is primary and √I=P, then we say thatI isP-primary.

We can now prove that every ideal is an intersection of primary ideals.

Theorem 4. Every ideal I k[x 1,...,x n] can be written as aﬁnite intersection of primary ideals. ⊆

Proof. Wefirst define an idealI to be irreducible ifI=I 1 I 2 implies thatI=I 1 ∩ orI=I 2. We claim that every ideal is an intersection offinitely many irreducible ideals. The argument is an “ideal” version of the proof of Theorem 2 from §6. One uses the ACC rather than the DCC—we leave the details as an exercise. Next we claim that an irreducible ideal is primary. Note that this will prove the theorem. To see why the claim is true, suppose thatI is irreducible and that fg I withf/ I. We need to prove that some power ofg lies inI. Consider the saturation∈ ∈ N I:g ∞. By Proposition 9 of §4, we know thatI:g ∞ =I:g onceN is sufficiently large. We will leave it as an exercise to show that(I+ g N ) (I+ f ) =I. Since I is irreducible, it follows thatI=I+ g N orI=I+ f⟨. The⟩ latter∩ cannot⟨ ⟩ occur sincef/ I, so thatI=I+ g N . This⟨ proves⟩ thatg ⟨N ⟩ I. ∈ ⟨ ⟩ ∈ ! As in the case of varieties, we can define what it means for a decomposition to be minimal.

Deﬁnition 5.A primary decompositionC of an idealI is an expression ofI as an r intersection of primary ideals:I= C i=1 Qi. It is called minimalor irredundant if the √Qi are all distinct andQ i j=i Qj. ̸⊇ ̸ To prove the existence of a minimal decomposition, we will need the following lemma, the proof of which we leave as an exercise.

Lemma 6. If I, J are primary and √I= √J, then I J is primary. ∩ We can now prove theﬁrst part of the Lasker-Noether decomposition theorem.

Theorem 7 (Lasker-Noether). Every ideal I k[x 1,...,x n] has a minimal primary decomposition. ⊆ C r Proof. By Theorem 4, we know that there is a primary decompositionI= i=1 Qi. Suppose thatQ i andQ j have the same radical for somei=j. Then, by Lemma 6, ̸ Q=Q i Q j is primary, so that in the decomposition ofI, we can replaceQ i and ∩ Qj by the single idealQ. Continuing in this way, eventually all of theQ i’s will have distinct radicals. C Next, suppose that someQ i contains j=i Qj. Then we can omitQ i, andI will ̸ be the intersection of the remainingC Q j’s forj=i. Continuing in this way, we can ̸ reduce to the case whereQ i j=i Qj for alli. ! ̸⊇ ̸ 230 Chapter 4 The Algebra–Geometry Dictionary

Unlike the case of varieties (or radical ideals), a minimal primary decomposition need not be unique. In the exercises, you will verify that the ideal x 2, xy k[x,y] has the two distinct minimal decompositions ⟨ ⟩⊆

x2, xy = x x 2, xy,y 2 = x x 2,y . ⟨ ⟩ ⟨ ⟩ ∩ ⟨ ⟩ ⟨ ⟩ ∩ ⟨ ⟩ Although x 2, xy,y 2 and x 2,y are distinct, note that they have the same radical. To prove that⟨ this happens⟩ ⟨ in general,⟩ we will use ideal quotients from §4. We start by computing some ideal quotients of a primary ideal.

Lemma 8. If I is primary with √I=Pandf k[x 1,...,x n], then: ∈ (i)Iff I, then I:f= 1 . (ii)Iff/∈ I, then I: f⟨ is⟩ P-primary. (iii)Iff/∈ P, then I:f=I. ∈ Proof. See Exercise 7. ! The second part of the Lasker-Noether theorem tells us that the radicals of the ideals in a minimal decomposition are uniquely determined. C r Theorem 9 (Lasker-Noether). Let I= i=1 Qi be a minimal primary decomposition of a proper ideal I k[x 1,...,x n] and let Pi = √Qi. Then the Pi are precisely ⊆ the proper prime ideals occurring in the set √I:f f k[x 1,...,x n] . { | ∈ }

Remark. In particular, theP i are independent of the primary decomposition ofI. We say that theP i belong toI.

Proof. The proof is very similar to the proof of Theorem 7 from §6. The details are covered in Exercises 8–10. ! In §6, we proved a decomposition theorem for radical ideals over an algebraically closedﬁeld. Using the Lasker–Noether theorems, we can now show that these results hold over an arbitraryﬁeldk. C r Corollary 10. Let I= i=1 Qi be a minimal primary decomposition of a proper radical ideal I k[x 1,...,x n]. Then the Qi are prime and are precisely the proper ⊆ prime ideals occurring in the set I:f f k[x 1,...,x n] . { | ∈ } Proof. See Exercise 12. !

The two Lasker–NoetherC theorems do not tell the full story of a minimal primary r decompositionI= i=1 Qi. For example, ifP i is minimal in the sense that noP j is strictly contained inP i, then one can show thatQ i is uniquely determined. Thus there is a uniqueness theorem for some of theQ i’s [see Chapter 4 of ATIYAH and MACDONALD (1969) for the details]. We should also mention that the conclusion of Theorem 9 can be strengthened: one can show that theP i’s are precisely the proper prime ideals in the set I:f f k[x 1,...,x n] [see Chapter 7 of ATIYAH and { | ∈ } MACDONALD (1969)]. §8 Primary Decomposition of Ideals 231

Finally, it is natural to ask if a primary decomposition can be done constructively. More precisely, givenI= f 1,...,f s , we can ask the following: ⟨ ⟩ (Primary Decomposition) Is there an algorithm forﬁnding bases for the primary • idealsQ i in a minimal primary decomposition ofI? (Associated Primes) Can weﬁnd bases for the associated primesP i = √Qi? • If you look in the references given at the end of §6, you will see that the answer to these questions is yes. Primary decomposition has been implemented in CoCoA, Macaulay2, Singular, and Maple.

EXERCISES FOR §9

2 1. Consider the idealI= x,y C[x,y]. 2 ⟨ ⟩ ⊆ a. Prove that x,y !I! x,y , and conclude thatI is not a prime power. b. Prove that⟨I is primary.⟩ ⟨ ⟩ 2. Prove Lemma 2. 3. This exercise is concerned with the proof of Theorem 4. LetI k[x 1,...,x n] be an ideal. a. Using the hints given in the text, prove thatI is aﬁnite⊆ intersection of irreducible ideals. N N b. Suppose thatfg I andI:g ∞ =I:g . Then prove that(I+ g ) (I+ f ) =I. ∈ N ⟨ ⟩N∩ ⟨ ⟩ Hint: Elements of(I+ g ) (I+ f ) can be written asa+ bg =c+ df , where ⟨ ⟩ ∩ ⟨ ⟩ N N+1 a,c I andb,d k[x 1,...,x n]. Now multiply through byg and useI:g =I:g . 4. In the∈ proof of Theorem∈ 4, we showed that every irreducible ideal is primary. Surpris- ingly, the converse is false. LetI be the ideal x 2, xy,y 2 k[x,y]. a. Show thatI is primary. ⟨ ⟩ ⊆ b. Show thatI= x 2,y x,y 2 and conclude thatI is not irreducible. 5. Prove Lemma 6.⟨ Hint: Proposition⟩ ∩ ⟨ ⟩ 16 from §3 will be useful. 2 6. LetI be the ideal x , xy Q[x,y]. a. Prove that ⟨ ⟩ ⊆ I= x x 2, xy,y 2 = x x 2,y ⟨ ⟩ ∩ ⟨ ⟩ ⟨ ⟩ ∩ ⟨ ⟩ are two distinct minimal primary decompositions ofI. b. Prove that for anya Q, ∈ 2 I= x x ,y ax ⟨ ⟩ ∩ ⟨ − ⟩ is a minimal primary decomposition ofI. ThusI has inﬁnitely many distinct minimal primary decompositions. 7. Prove Lemma 8. 8. Prove that an ideal is proper if and only if its radical is. 9. Use Exercise 8 to show that the primes belonging to a proper ideal are also proper. 10. Prove Theorem 9. Hint: Adapt the proof of Theorem 7 from §6. The extra ingredient is that you will need to take radicals. Proposition 16 from §3 will be useful. You will also need to use Exercise 9 and Lemma 8. 11. LetP 1,...,P r be theA prime ideals belonging toI. √ r a. Prove that I= Ai=1 Pi. Hint: Use Proposition 16 from §3. √ r √ b. Show that I= i=1 Pi need not be a minimal decomposition of I. Hint: Exer- cise 4. 12. Prove Corollary 10. Hint: Use Proposition 9 of §4 to show thatI:f is radical. 232 Chapter 4 The Algebra–Geometry Dictionary §9 Summary

The table on the next page summarizes the results of this chapter. In the table, it is supposed that all ideals are radical and that theﬁeld is algebraically closed.

ALGEBRA GEOMETRY radical ideals varieties I V(I) −→ I(V) V ←− addition of ideals intersection of varieties A I+J V(I) V(J) −→ ∩ I(V)+I(W) V W ←− ∩ product of ideals union of varieties A IJ V(I) V(J) −→ ∪ I(V)I(W) V W ←− ∪ intersection of ideals union of varieties I J V(I) V(J) I(V)∩I( W) −→V W∪ ∩ ←− ∪ ideal quotients difference of varieties I:J V(I) V(J) −→ \ I(V):I(W) V W ←− \ elimination of variables projection of varieties I k[x l 1,...,x n] πl(V(I)) ∩ + ←→ prime ideal irreducible variety ←→ minimal decomposition minimal decomposition I=P 1 P m V(I)=V(P 1) V(P m) ∩···∩ −→ ∪···∪ I(V)=I(V 1) I(V m) V=V 1 V m ∩···∩ ←− ∪···∪ maximal ideal point of afﬁne space ←→ ascending chain condition descending chain condition ←→