Chapter 4 the Algebra–Geometry Dictionary

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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 op- erations. 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 chap- ter, 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<x , 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.
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