Basic Algebraic Geometry

Home , Irreducible component, Zariski topology

1.1 Basic deﬁnitions. 1.1.1 Zariski and standard topology.

Let F be a ﬁeld (almost always R or C). Then we say that a subset X F n ⊂ is Zariski closed if there is a set of polynomials S F [x1, ..., xn] such that X = F n(S) = x F n s(x) = 0, s S . One checks⊂ that the set of Zariski closed sets, Z,{ in ∈F n satisﬁes| the axioms∈ } for the closed sets in a topology. That is

F n and Z. • ∅ ∈

If T Z then Y T Y Z. • ⊂ ∩ ∈ ∈

If Y1, ..., Ym Z then jYj Z. • ∈ ∪ ∈ The Hilbert basis theorem (e.g. A.1.2 [GW]) implies that we can take the sets S to be finite. If X is a subset of F n then the Z-topology on X is the subspace topology corresponding to the Zariski topology. If F = R or C then we can also endow F n with the standard metric 2 n 2 topology corresponding to d(x, y) = x y where u = ui for k − k k k i=1 | | u = (u1, ..., un). We will call this topology the standard topology (in the literature it is also called the classical, Hausdorff or Euclidean topology).P Our first task will be to study relations be tween these very different topologies. Examples. n = 1. Then if Y Z then Y = F or Y is finite. n = 2.A Zariski closed subset∈ is a finite union of plane curves or all of F 2. Thus the Zariski topology on F 2 is not the product topology.

1.1.2 Noetherian topologies. The Zariski topology on a closed subset, X, of F n is an example of a Noetherian topology. That is if Y1 Y2 ... Ym ... is a decreasing sequence of closed subsets of X then⊃ there⊃ exists⊃ N such⊃ that if i, j N ≥ then Yi = Yj (this is a direct interpretation of the Hilbert basis theorem). If X is a topological space that be cannot written X = Y Z with Y and Z closed and both are proper the X is said to be irreducible.∪ Clearly an irreducible space is connected but the converse is not true.

1 Exercise. What are the irreducible Hausdorﬀ topological spaces?

Lemma 1 Let X be a Noetherian topological space. Then X is a ﬁnite union of irreducible closed subspaces. If X = X1 X2 Xm with Xi closed ∪ ∪ · · · ∪ and irreducible and if Xi * Xj for i = j then the Xi are unique up to order. 6

For a proof see A.1.12 [GW].The decomposition X = X1 X2 Xm ∪ ∪ · · · ∪ with Xi closed and irreducible and Xi * Xj for i = j is called the irredundant 6 decomposition of X into irreducible components. Each of the Xi is called an irreducible component.

1.1.3 Nullstellensatz.

n If X is a subset of F then we set X = f F [x1, ..., xn] f X = 0 . If I n { ∈ n | | } I F [x1, ..., xn] is an ideal then we set F ( ) = x F f(x) = 0, f . ⊂ I { ∈ | ∈ I} n Lemma 2 A closed subset, X, of F is irreducible if and only if X is a I prime ideal (i.e. F [x1, ..., xn] X is an integral domain). | We now recall the Hilbert Nullstellensatz.

Theorem 3 Let F be algebraically closed and let be an ideal in F [x1, ..., xn]. Then I n 1. F ( ) = if and only if = F [x1, ..., xn]. I ∅ I 2. IX(I) = √ . I We will give a proof of 1. in this theorem for the case of F = C since the method of proof will be consistent with the techniques of the later material to be presented. That 1. implies 2. is the trick of Rabinowitz which we defer to the references. We prove that if I is a proper ideal then Cn( ) = . Let be a maximal I 6 ∅ M proper ideal with . Then k = C[x1, ..., xn]/ is a ﬁeld containing C. I ⊂ M m M Since the image of the monomials x 1 xmn span k as a vector space over 1 ··· n C we see that dimC k is at most countable. Since C is algebraically closed, if k = C there must be t k that is transcendental over C. We assert that the 6 1 ∈ elements t a a C are independent over C. Indeed, if { − | ∈ } m c i = 0 t a i=1 i X − 2 with a1, ..., am C and distinct and c1, ..., cm in C then multiplying by ∈ m

(t ai) i=1 − Y we ﬁnd that m

cj (t ai) = 0. − j=1 i=j X Y6 This defines a polynomial satisfied by t that is trivial if and only if all of the cj are 0 (evaluate at ai and get ci (ai aj)) Since C is not countable this − j=i Y6 is a contradiction. We have proved that k is one dimensional over C with basis 1 + . Thus xi + = zi1 + with zi C. We have shown that M M M ∈ xi zi for all i = 1, ..., n. Since the ideal z is maximal we see that − ∈ M n I{ } = z and since we have z C ( ). M I{ } I ⊂ M ∈ I 1.2 Affi ne varieties and dimension. 1.2.1 Affi ne varieties. Let X,Y be sets and let f : X Y be a mapping. If g is a function from Y → to F then we set f ∗g = g f which is a function on X. In this section we will◦ assume that F is algebraically closed (you can assume it is C). An affi ne variety over F is a pair (X,R) with X a topological space and R and algebra over F of continuous functions from X to F (F is endowed with the Zariski topology) such that there exists a Z-closed subset n Z in F for some n and a homeomorphism, f, of X onto Z such that f ∗ : F [x1, ..., xn] Z R is an algebra isomorphism. A morphism affi ne varieties || → (X,R) and (Y,S) is a map f : X Y such that f ∗S R. An isomorphism → ⊂ is a morphism that is one to one and onto and f ∗ is an isomorphism of algebras. We note that if (X,R) is an affi ne variety then R must be a finitely generated algebra over F and if a R is such that ak = 0 for some k > 0 then a = 0. That is, the only nilpotent∈ element in R is 0. The converse is also true. Let R be a finitely generated algebra over F without nilpotents. Let r1, ..., rm generate R as an algebra over F . Then we have an algebra homomorphism µ : F [x1, ..., xm] R defined by µ(xi) = ri. Let = ker µ and set Z = n → I F ( ). Then since R = F [x1, ..., xm]/ and R has no nilpotents we must I ∼ I 3 have √ = . Thus = Z (by the nullstellensatz). Thus R is isomorphic I I I I with (Z) = F [x1, ..., xm] Z . Actually more is true. The topological space X isO also completely determined| by R. Indeed, the nullstellensatz implies that the maximal ideals in (Z) are the deals z + for z Z. Thus the set Z can be identifiedO with the set of maximalI{ } idealsI in ∈(Z). The topology is determined as follows. If Y is a closed subset of Z andO is the ideal of elements in (Z) such that Y = z Z g(z) = 0, g J then O { ∈ | ∈ J } Y + z Z z for an ideal of R. Returning to R we may define { ∈ |J ⊂ I{ }} I R to be the set of maximal proper ideals endowed with the topology that M has as its closed sets the sets R( ) = maximal with . We have just seen that if we use theM factI that{M|R is isomorphic withI ⊂ M}(Z) and O the maximal ideals in (Z) are the images of the ideals z for z Z. This O I{ } ∈ implies that if R then R/ = F 1 + . Thus if r R then we can define r( ) =Mc if ∈r M= c1 + . WeM thereforeM have made R∈into an algebra M M of F valued functions on R. The bottom line is that all of the information is in the algebra R. M Recall that if F = C and if Y Cn is Z-closed then we also have the standard (metric) topology on Y . If⊂(X,R) is an affi ne variety and if (X,R) is isomorphic with (Y, C[x1, ..., xn] Y ) and (V, C[x1, ..., xm] V ) with Y Z-closed | | in Cn and V Z-closed in Cm. Then we have f : X Y and g : X V → → homeomorphisms in the Z-topology such that f ∗ : C[x1, ..., xn] Y R and 1 | → g∗ : C[x1, ..., xm] V R algebra isomorphisms. This g f − : Y V is | → 1 ◦ → given by the map Φ = (φ1, ..., φm) with φj = (g f − )(xj V ) C[x1, ..., xn] Y . 1 ◦ | ∈ | Thus g f − is continuous in the standard metric topology. This implies that endowed◦ with the standard topology Y and V are homeomorphic. We therefore see that this implies that (X,R) also has a natural metric topology which we will also call standard. We will say that an irreducible affi ne variety (X,R) is normal if R is integrally closed in K(X) the quotient field of R.

Theorem 4 If (X,R) is an irreducible affi ne variety then if S is the integral closure of R in K(X) then S is finitely generated over F and ( S,S) is a normal affi ne variety. M

For a proof see [S] Theorem II.5.4. We will call the variety ( S,S) with M the map M z with z X such that M R = Mz the normalization of (X,R) one can→ show that∈ this map is ﬁnite∩ (in particular ﬁnite to one).

4 1.2.2 Dimension. Let (X,R) be an irreducible affi ne variety over an algebraically closed field F . Then R is an integral domain. We denote by K(X) the quotient field of R. Then K(X) is a field extension of F.If (X,R) is isomorphic with a Z-closed subset of F n then the field K(X) is isomorphic is generated as a field by at most n generators. We define the dimension of X to be the transcendence degree of K(X) over F . n n Example. (F ,F [x1, ..., xn]) has dimension n since K(F ) = F (x1, ..., xn) the field of rational functions in n variables. If X F n is Z-closed then dim X n. ⊂ ≤ This is a precise definition but is hard to compute. We will now discuss an effective way to calculate the dimension of an affi ne variety. Let R be an algebra over F then a filtration of R is an increasing sequence 0R 1R ... mR ... of subspaces of R such that iR jR F ⊂ F ⊂ ⊂ F ⊂ F ·F ⊂ i+jR If is a filtration of R then we define Gr R = j 0 jR/ j 1R F F F ≥ F F − ( 1R = 0 ) with multiplication F− { } L (a + i 1R)(b + j 1R) = ab + i+j 1R F − F − F − j for a iR and b jR. If we set Gr R = jR/ j 1R then Gr R is a graded∈ algebra F over∈F F. Here, a gradedF algebraF over F −is an algebraFS such j i j i+j that as a vector space over F , S = j 0 S such that S S S . ≥ ⊂ We will say that a filtration, , of R is good if dim Gr 1R < and it L generates Gr R as an algebra overF F . F ∞ F n Example. Let X F be Z-closed with R = F [x1, ..., xn] X . Then we set ⊂ a | n jR equal to the span of the restrictions x X with a = (a1, ..., an) N and Fa a1 an | ∈ 1 x = x x . Then Gr R is generated by the images of the xi in Gr R. 1 ··· n F F Theorem 5 Let R be an algebra over F with a good filtration then there F exists, N N and a polynomial h (q) such that dim jR = h (j) for j N. Furthermore,∈ if and are goodF filtrations of R thenF deg hF = deg h ≥. F G F G Examples. n 1. Let X = F with R = F [x1, ..., xn] with the filtration, FT, given by n+j jn degree as in the example above. Then dim FjR = n = n! + lower in j. 2. Let f F [x1, ..., xn] be a non-constant polynomial. Let ∈

R = F [x1, ..., xn]/fF [x1, ..., xn]

5 on R we ﬁlter by degree in the xi. Then if deg f = m > 0 we have

0 k jF [x1, ..., xn] kF [x1, ..., xn] kR 0 → F − → F → F → is exact with the map from ﬁltered degree k j to k given by multiplication by f. Thus −

n 1 n + k n + k j k − dim kR = − = j + lower degree in k. F n − n (n 1)! − Theorem 6 Let (X,R) be an aﬃ ne variety over F then if is a good ﬁl- tration of R then deg h is the maximum of the dimensions ofF the irreducible components of X. F

We will denote by DimR the degree of h for a good ﬁltration of R. More important for computations is F F

Theorem 7 Let I be an ideal in F [x1, ..., xn] then

DimF [x1, ..., xn]/I = DimF [x1, ..., xn]/√I.

This implies that we can compute the dimension of a variety from a deﬁning set of equations. We also note that Theorem 6 tells us that we should (and do) deﬁne dim X = DimR = max dim Z for Z an irreducible component of X. We record here two results that we will need later.

Theorem 8 If X is an irreducible aﬃ ne variety and if Y is a closed subvariety then dim Y dim X with equality if an only if Y = X. ≤ Theorem 9 Let (X,R) be an irreducible aﬃ ne variety of dimension n. Then if f is a non-constant element of R then every irreducible component of X(f) = x X f(x) = 0 is of dimension n 1. { ∈ | } − For a proof see Theorem I.6.5 p. 74 in [S].

6 1.2.3 Tangent space.

If f F [x1, ..., xn] then we deﬁne for ∈ n n p = (p1, ..., pn) F , dfp : F F ∈ → by ∂f dfp(z1, ..., zn) = (p)zi. ∂xi X Here ∂f (p) is the formal derivative. We have the Taylor formula ∂xi

α f(p + x) = f(p) + dfp(x) + cαx = α 2 |X|≥

f(p) + dfp(x) + xigi(x) i X α α1 αn here for α = (α1, ..., αn), x = x x and gi is an appropriate choice of 1 ··· n polynomial which can choose so that gi(0) = 0. If F = R or C this coincides n with the usual derivative. We set Tp(X) = v F dfp(v) = 0, f X . { ∈ | ∈ I } We will now deﬁne this space in terms of the algebra R = F [x1, ..., xn] X . If | p X then we set p = f R f(p) = 0 . Then the Zariski cotangent ∈ M { ∈ | 2 } space to X at p is the vector space p/ . We will now deﬁne a natural M Mp pairing between this space and Tp(X). If f R and f(p) = 0 then we ∈ choose g F [x1, ..., xn] such that g X = f. If z Tp(X) then we assert that ∈ | ∈ dgp(z) depends only on f and not on the choice of g. Indeed, if h X = f 2 | then g h IX . Thus d(g h)p(z) = 0. Now if f then f is a sum of − ∈ − ∈ Mp elements of the form u1u2 with ui p. If hi X = ui then using ∈ M |

d(h1h2)p = h1(p)d(h2)p + h2(p)d(h1)p = 0 we see that we do indeed have a natural (independent of choices) pairing of 2 2 p/ p and Tp(X). Furthermore, if f(p) = 0 and dfp = 0 then f + X p M(see Taylor’sformulaM above). I ∈ M 2 We thus have an intrinsic deﬁnition of Tp(X) = ( p/ )∗. M Mp

Theorem 10 Let X be an irreducible aﬃ ne variety. If p X then dim Tp(X) ∈ ≥ dim X. Furthermore, dim X = minp X dim Tp(X). ∈ For this c.f. [GW] A.3.2. We have

7 Theorem 11 Let X be an irreducible aﬃ ne variety then the set of p X ∈ such that dim Tp(X) = dim X is Z-open.

We can see this as follows. We assume that X is Z-closed in F n. Let n d d = dim X. Then if f = (f1, ..., fn d) X− then we set Uf equal to the subset of X consisting of those points,− ∈p, I such that some (n d) (n d) minor of − × − ∂f i (p) ∂x j is non-zero. One sees that each Uf is Z-open in X (perhaps empty) and the union of the Uf is the set of smooth elements. Let X be an irreducible aﬃ ne variety. We will call a point smooth if dim Tp(X) = dim X. In our application of these ideas we will use (the case n = 1) of the following theorem.

Theorem 12 If (X,R) is a normal aﬃ ne variety then the set of singular points (i.e. not smooth points) is Z-closed and of dimension at most n 2. − For a proof of this see [S], Theorem II.5.3 p.117. This result implies that since dim = 1 we have the normalization of a curve is smooth (all points are smooth).∅ − We will need to extend the concept of tangent space to non-irreducible varieties, For this we will need some new concepts.

1.2.4 Affi ne varieties revisited. We take F to be algebraically closed. Let (X,R) be an affi ne variety. Let γ R be non-constant. Let X γ = x X γ(x) = 0 . If (X,R) is isomorphic∈ with the affi ne algebraic{ } set {Y =∈ F n|(S) then6 R} is isomorphic with (Y ) = F [x1, ..., xn] Y under the map f : X Y (i.e. f is a O | → Z-homeomorphism and f ∗ : (Y ) R is an algebra isomorphism). Let O → φ F [x1, ..., xn] be such that if φ = φ Y then f ∗φ = γ. We consider the subset∈ | 1 Z = (x, ) x Y φ { φ(x) | ∈ { }} n+1 of F . We look upon F [x1, ..., xn] as the subalgebra of F [x1, ..., xn+1] spanned by the monomials that don’tinvolve xn+1. We note that if h(x1, ..., xn+1) =

8 xn+1φ(x1, ..., xn) 1 then − Z = F n+1(S h ). ∪ { } One checks that (Z) is the localization of (Y ) by the multiplicative set 2 O O 1, φ, φ , ... . Thus if R(γ) is the corresponding localization of R then we have { } (X γ ,R(γ)) is an aﬃ ne variety. { } Now let U be Z-open in Y.˙ Then U = Y W Y with W Z-closed in F n. That is W = F n(S) with Sa ﬁnite set of polynomials.− ∩ This implies that

U = φ SY φ . This implies that the affi ne subvarieties (X γ ,R(γ)) define a basis∪ for∈ the{ } open sets in the Z-topology. { } If U X is open then we define X (U) to be the functions α from U to ⊂ O F such that for each p U there is a γ R such that γ(p) = 0, X γ U ∈ ∈ 6 { } ⊂ and α Xγ R(γ). To make sure that this makes sense we need to know | } ∈

Theorem 13 Let (X,R) be an aﬃ ne variety. Then X (X) = R. O 1.2.5 The tangent space revisited.

Let (X,R) be an affi ne variety and let for each Z-open subset U X, X (U) be as in the previous section. If p X and if U and V are Z-open⊂ inOX and ∈ p U V then we say that α X (U) and β X (V )are equivalent if ∈ ∩ ∈ O ∈ O there exists W a Z-open subset with p W U V such that α W = β W . ∈ ⊂ ∩ | | We define X,p to be the set of equivalence classes under this relation. O If α X (U) for some Z-open U with p U let [α]p be the corresponding equivalence∈ O class. Then one checks that we∈ can define an algebra structure on X,p by defining [α]p [β]p = [α W ]p[β W ]p for α X (U) and β X (V ) andOp W U V . | | ∈ O ∈ O This∈ algebra⊂ ∩ is called the local ring of X at p. One can define this ring to be the localization of R at p (see section 1.4). If [α] X,p then the M p ∈ O value of α(p) depends only on [α]p and we use the notation τ(p) for the value at p of τ X,p. This defines a homomorphism of X,p onto F . We will ∈ O O use the notation mX,p for the kernel of this homomorphism. Then it is easily seen that X,p is a local ring with maximal ideal mX,p(see section 1.4). We O 2 2 2 consider the map ιp : p/ mX,p/m by α [α]p + m . M Mp → X,p → X,p

Theorem 14 If X is irreducible and if p X then ιp is a bijection. ∈

9 2 This leads to a deﬁnition of Zariski cotangent space. Tp∗(X) = mX,p/mX,p. As before we say that p X is smooth if dim Tp∗(X) = dim X. If X = m ∈ j=1Xjis its decomposition into irreducible components and if X0 = p Xj Xj ∪ ∪ ∈ then X,p = X0,pand the cotangent space depends only on this closed subvariety.O O

1.2.6 The formal algebra at a smooth point. Let (X,R) be an affi ne variety over F and algebraically closed field. We assume that p X is a smooth point. Let u1, ..., ud mX,p be such that if 2 ∈ 2 ∈ ui = ui + mX,p form a basis of mX,p/mX,p. If α = (a1, ..., αd) is a multiindex α α1 αd k then we set (as usual) u = u u . Let pk : R R/m . 1 ··· d → X,p Lemma 15 u1, .., ud are algebraically independent over F . α Lemma 16 In the notation above the set plu 0 α < l is a basis of l { | ≤ | | } X,p/m over F . O X.p Corollary 17 In the notation above k lim OX,p/mX,p ∼= F [[x1, ..., xd]] ←− the formal power series in the indeterminates x1, ..., xd. Theorem 18 Let (X,R) be an affi ne variety over F of dimension d and let p be a smooth point of X. Let X = X1 Xm be the decomposition ∪ · · · ∪ of X into irreducible components and let X0 = p Xj Xj (the union of the ∪ ∈ irreducible components containing X). Then X0 is irreducible.

Proof. We may assume that X = X0. Thus we need only show that if p is a smooth point of X and if every irreducible component of X contains p then k X is irreducible. We ﬁrst note that the map OX,p lim OX,p/mX,p given k → ←− k by a a + mX,p is injective. Indeed, if a 0 then a mX,p for all k and thus7−→ Corollary { 29} in the appendix (section7→ 1.4) implies that∈ a = 0. This implies that X,p is an integral domain. We assert that this implies that R is an integral domain.O Indeed, we assert that the map a a/1 injects R into 7−→ X,p. To see this we note that a/1 = 0 implies that there exists φ R such thatO φ(p) = 0 and φa = 0. This implies that if Y is an irreducible component∈ 6 of X then φa Y = 0. Since φ(p) = 0 the set y Y φ(y) = 0 is Z-dense in | 6 { ∈ | 6 } Y and thus a Y = 0. We have assumed that every irreducible component of X contains p |sot a = 0.This implies that Rinjects into an integral domain so Xis irreducible.

10 1.2.7 Relations between the standard and Zariski topologies.

We now look at the case when F = C. We look upon C with the standard topology as R2 and Cn as R2n. If f function from a set S to C we will write f(s) = fR(s) + ifI (s) with fR and fI n n real valued functions on S. If δ > 0 we set Bδ = x C x < δ thought { ∈ | k k } of as the δ ball in R2n. We will now prove

Theorem 19 Let X be Z-closed in Cn and irreducible be of dimension d. Let d n p X be a smooth point then there exists δ > 0 and a C∞ map Φ: Bδ C ∈ d → such that Φ(0) = p, Φ(Bδ ) X is an open neighborhood of p in the standard topology and Φ: Bd Φ(B⊂d) is a homeomorphism in the standard topology. δ → δ

Proof. Let f1, ..., fm generate X . After possible relabeling we may as- I n sume that (df1)p, ..., (dfn d)p are linearly independent elements of (C )∗. Let − ui = (fi)R and un d+i = (fi)I then we note that (dui)p = ((dfi)p)R and − 2n (dun d+i)p = ((dfi)p)I where duj is the standard calculus diﬀerential on R . − This implies that (du1)p, ..., (du2n 2d)pare linearly independent at p. Now af- − ter relabeling we may assume that the restrictions of x1 p1, ..., xd pd 2 − − to X form a basis of p modulo p. Set u2n 2d+i = (xi pi)R and M M − −2n u2n d+i = (xi pi)I . Then (du1)p, ..., (du2n)p form a basis of (R )∗. The real− analytic inverse− function theorem implies that there exists an open stan- 2n dard neighborhood U of p in R such that if Ψ = (u1, ..., u2n) then Ψ(U) is open and Ψ: U Ψ(U) is a (real analytic) diﬀeomorphism. Let δ > 0 be n→d d so small that B − B Ψ(U). Then δ × δ ⊂ 1 d n 1 n d d Ψ− (0 Bδ ) = C (f1, ..., fn d) Ψ− (Bδ − Bδ ). × − ∩ × 1 d We set Φ(z) = Ψ− (0, z) for z Bδ . n ∈ r Let Z = C (f1, ..., fn d) and let Z = i=1Zi be the decomposition of Z into irreducible components.− Since X is∪ irreducible we may assume that

X Z1. On the other hand since f1, ..., fn d Z1 we see that dim Tp(Z1) ⊂ − ∈ I ≤ d = dim X. Thus Z1 = X. d Let W be the Z-closure of Φ(Bδ ). Then W C(f1, ..., fn d). We also note d ⊂ − that xi Φ defines an analytic function on B for i = 1, ..., n. This implies that ◦ δ Φ∗ defines an injective homomorphism of (W ) into the analytic functions d O on Bδ . Since this algebra is an integral domain this implies that (W ) is an integral domain and hence that W is irreducible. Furthermore theO functions d xi pi, i = 1, .., d are algebraically independent on Φ(B )and hence on W . − δ 11 thus the dimension of W is at least d. Since p W we see that dim W = d (the tangent space at p is at most of dimension ∈d). Thus W is an irreducible component of Y containing p. Since p is smooth we must have Z = X the unique irreducible component of Y containing p. In our application of these results we will use (the case n = 1) of the following theorem. In general the standard metric topology (which we will denote as the S- topology) is much finer than the Z-topology on an affi ne variety (they are equal if and only if the variety is finite). However, we will show that the S-closure of a Z-open set in an affi ne variety is the Z-closure.

Theorem 20 Let X be Z-closed in Cn and irreducible and let U = be Z-open in X. Then the S-closure of U is X. 6 ∅ Proof. We prove this result by induction on dim X. If dim X = 1 then X U is ﬁnite. Let p X U. Let f : Y X be the normalization of −X then f is surjective∈ and− continuous in the→ S-topology. Theorem 13 1 implies that Y is smooth. Let q f − (p) as in Theorem 12 choose δ > 0 1 1 ∈ and Φ: Bδ Φ(Bδ ) be a homomorphism in the S-topology onto an S-open subset of Y→with Φ(0) = q. By taking δ possibly smaller we may assume 1 1 1 that Φ(B ) f − (X U) = q . Choose a sequence zj B 0 such that δ ∩ − { } ∈ δ − { } limj 0 zj = 0. Then xj = f(Φ(zj)) U and in the S-topology limj xj = p. Now→ assume that the result has∈ been proved for all X of dimension→∞ n 1. ≥ Assume dim X = n + 1. Let p X U. Let Y = X U = Y1 Yr ∈ − − ∪ · · · ∪ be its decomposition into irreducible components and assume that Y1, ..., Ys are exactly the ones that are not equal to p (usually s = r). Let pi Yi { } ∈ be such that pi = p. Let φ (X) be such that φ(p) = 0 and φ(pj) = 0 6 ⊂ O 6 for j = 1, ..., s. Set Z = x X φ(x) = 0 . Let Z = Z1 Zt be the decomposition of Z into{ irreducible∈ | components.} Assume∪ that · · · some ∪ Zj X U. Then since X is irreducible and U is non-empty we see that the ⊂ − Yi are all of dimension at most n. Since Zj is irreducible it must be contained in one of the Yj. On the other hand dim Zk = n for all k. Thus we must have Zj = Yi for some i.But pi / Z. This contradiction implies Zi * X U. Let ∈ − Zj be such that p Zj. Then Zj has dimension n and U Zj = . Thus the ∈ ∩ 6 ∅ inductive hypothesis implies that p is in the S-closure of U Zj hence of U. ∩ Exercises. 1. Why didn’twe start the induction from dimension 0(where the result is trivial)?

12 2. Prove that if X is an aﬃ ne variety and U is a Z-open subset of X then the Z-closure of U is the same as the S-closure.

1.3 Projective and quasiprojective varieties. 1.3.1 Projective algebraic sets. We define projective n-space to be the set of all one dimensional subspaces of F n+1and denote it Pn(F ) or Pn if F is understood. If L Pn then L = F v ∈ with v L 0 . Thus we can look upon Pn as the set F n+1 0 modulo ∈ − { } − { } the equivalence relation v w if there exists c F × = F 0 such that ≡ n+1 ∈ − { } cw = v. If (x0, x1, ..., xn) F 0 then we denote the corresponding line ∈ −{ } by [x0, x1, ..., xn] and x0, x1, ..., xn are called the homogeneous coordinates of the point in Pn. Let f F [x0, x1, ..., xn] then f is said to be homogeneous of degree m if f(cx) =∈cmf(x) for x F n+1and c F . Assuming that F is infinite this condition is equivalent∈ to saying that∈ f is an F -linear combination of α α0 α1 αn monomials x = x0 x1 xn with α = α0 + ... + αn = m. We denote by m ··· | | F [x0, ..., xn] the space of polynomials that are homogeneous of degree m. n+1 m We note that if x F 0 and if f F [x0, ..., xn] then whether or not f(x) = 0 depends∈ only on−[x {].} Define the∈ closed sets in the Zariski topology on Pn to be the sets of the form

n n P (S) = [x] P f(x) = 0, f S { ∈ | ∈ } and S is a set of homogeneous polynomials. One checks that with these sets satisfy the conditions necessary to be the closed sets of a topology. n n n n Of particular interest are the Z-open sets Pj = P P (xj).Let [x] Pj − ∈ (that is xj = 0 ) then 6 1 x0 xj 1 xj+1 xn [x] = [ x] = [ , ..., − , 1, , ..., ]. xj xj xj xj xj

n n We therefore see that we have a bijective mapping from F to Pj , Ψj, given by Ψj :(a1, ..., an) [a1, ..., aj, 1, aj+1, ..., an]. 7→ Suppose that f F [x1, ..., xn] is of degree m. Then the j-th homogenization of f is the polynomial∈ obtained as follows:

13 α Let f(x1, ..., xn) = α m cαx . Write | |≤

P m α α1 αj αj +1 αn fj(xo, ..., xn) = cαxj −| |xo xj 1xj+1 xn . ··· − ··· α m | X|≤

Then fj is homogeneous of degree m in the variables x0, ..., xn and Ψj∗fj = f. n n n Thus if we endow Pj with the subspace Zariski topology then Ψj : F Pj defines a Z-homeomorphism. → m If f F [x0, ..., xn] then define Rif(x1, ..., xn) = f(x1, ..., xj, 1, xj+1, ..., xn). ∈ m m m Then Ri is a linear bijection between F [x0, ..., xn] and F [x1, ..., xn] = ⊕j=0 Fm[x1, ..., xn]. Let X be a Z-closed subset of Pn. Then we have an open covering X = n n n 1 n n i=0Pi X. Let Ψi be as above. If X = P (S) then Ψi− (Pi X) = F (RiS). ∪ ∩ ∩ 1.3.2 Sheaves of functions. Let X be a topological space then a sheaf of functions on X is an assignment : U (U) with (U) a subalgebra of the algebra of all functions from FU to F→satisfying F theF following properties: 1. If V U then (U) V (V ). ⊂ F | ⊂ F 2. If Vα α I is an open covering of U then if γα (Vα) is given such { } ∈ ∈ F that if x Vα Vβ then γ (x) = γ (x) then there exists γ (U) such that ∈ ∩ α β ∈ F γ Vα = γα for all α. Examples.| 1. It is easily seen that if (X,R) is an affi ne variety then X is a sheaf O of functions on X. We call X the structure sheaf of X. 2. If is a sheaf of functionsO on X and if U is an open subset of X then F we define U to be the assignment V (V ) for V open in U(hence in X). F| → F 3. If X is a topological space and if F = R or C then the assignment of the assignment U C(U, F ) (the continuous functions from U to F ) is a sheaf or functions. → n 4. If F = R then the assignment U open in R to C∞(U) (the functions from U to R such that all partial derivatives of all orders are continuous) is ω a sheaf of functions denoted C∞n . Similarly,U C (U) the assignment to R → U the real analytic functions from U to R. We consider the category of all pairs (X, ) with X a topological space and a sheaf of functions on X. We will callF this the category of spaces F

14 with structure. With morphisms between (X, ) and (Y, ) continuos maps 1 F G f : X Y such that f ∗ (U) (f − U). → G ⊂ F Example.4. The category of smooth manifolds is the full subcategory of spaces with structures (X, ) where X is a Hausdorff space such that for F each p X there is an open neighborhood of p, U, such that (U, U ) is ∈ n F| isomorphic with (V,C n ) for V an open set of for some fixed n. We will R∞ V R | denote the sheaf by C∞ and call n the dimension of X. F X We are now ready to give a definition of algebraic variety.

1.3.3 Algebraic varieties. The example in the previous subsection indicates how one should define an algebraic variety. We define a pre-variety to be a topological space with a structure (X, X ) such that for every p X there exists an open neigh- O ∈ borhood, U, of p in X and an affi ne variety (Y, Y ) such that (U, X U ) is O O | isomorphic with (Y, Y ). O Examples. 1.We have seen that the structure that we attached to an affi ne variety (X,R) which we denoted as (X, X ) is a pre-variety. O n n 2. Let X be a Z-closed subset of P . Set Xi = X Pi and let Ψi be as 1 ∩ n in 1.3.1. Then in 1.3.1 we observed that Ψi− (Xi) is Z-closed in F . We pull 1 back the structure as in example on Yi = Ψi− (Xi). Then getting sheaves of functions Xi . If U Xi Xj then Xi U = Xj U . We define for U X, O ⊂ ∩ O | O | ⊂ Z-open X (U) to be the functions γ on U such that γ U Xi Xi (U Xi). O | ∩ ∈ O ∩ Then (X, X ) is a pre-variety. O n 3. If U is open in X a Z-closed subset in P then we take U to be the O restriction of X to U and (U, U ) is a pre-variety. O O We now come to the notion of algebraic variety. For this we need to define products of pre-varieties. If X F n and Y F m are Z-closed sets then we note that X Y is Z-closed in⊂F n F m =⊂F m+n (as a topological space—recall that this× is not the product topology).× To see this we take F n to be F n 0 and F m to be 0 F m then if X = F n(S) and Y = F m(T ) × × then we look at f S as a function of x1, ..., xn and g T as a function of ∈ n+m ∈ xn+1, ..., xn+m then X Y = F (S T ). × ∪

15 This defines a product of affi ne varieties. One can do this intrinsically by defining (X,R) (Y,S) to be (X Y,R F S). Here (γ δ)(x, y) = γ(x)δ(y). × × ⊗ ⊗ If (X, X ) and (Y, Y ) are pre-varieties then we can define the pre-variety O O (X Y, X Y ) as follows: If U X and V Y are open and (U, X U ) and × O × ⊂ ⊂ O | (V, Y V ) are isomorphic to affi ne varieties then take X Y U V to be U V as inO 1.3.2| Example 2. This defines a product (here theO × topology| × on UO ×V is the topology of the product of affi ne varieties defined as above). One× can prove that this is the categorical product in the category of pre-varieties. We can finally define an algebraic variety. Let (X, X ) be a pre-variety then it is an algebraic variety over F if ∆(X) = (x, xO) x X is a closed set in X. This condition is usually called separable{ . We| have∈ (c.f.} Lemma A.4.2 [GW])

Theorem 21 Let X be Z-open in the Z-closed subset, Y , in Pn. Then the corresponding pre-variety (as in example 3 above) is an algebraic variety.

A variety as in the above theorem will be called quasi-projective. An algebraic variety isomorphic with Z-closed subset of Pnwill be called a projective variety.

Theorem 22 If (X, X ) and (Y, Y ) are quasi-projective then so is (X O O × Y, X Y ). O × For this it is enough to show that Pn Pm is isomorphic with a Z-closed × subset of PN for some N. To prove this we consider the map (F n+1 0 ) (F m+1 0 ) F n+1 F m+1 0 given by − { } × − { } → ⊗ − { } (x, y) x y. 7−→ ⊗ n+1 m+1 If we take the standard bases ej of F and F then we can identify n+1 m+1 (m+1)(n+1) F F with F using the basis ei ej for 0 i n and 0 j⊗ m. One checks that this deﬁnes a morphism⊗ of (F≤n+1 ≤ 0 ) ≤m+1 ≤ n+1 m+1 − { } × (F 0 ) F F 0 .Also, [x y] = [x0 y0] with (x, y) n+1 − { } → m+1 ⊗ − { } ⊗ ⊗ ∈ (F 0 ) (F 0 ) if and only if [x] = [x0] and [y] = [y0]. Further, − { } × − { } n+1 m+1 one sees that if zij is the coordinate of z F F then the condition ∈ ⊗ that [z] is in the image of this map is that zijzkl = zkjzil = zilzkj. These are homogeneous equations of degree 2. Thus the image is closed. We have thus embedded Pn Pm as a closed subset of Pnm+n+m. This deﬁnes the topology on the corresponding× product of algebraic varieties.

16 Exercise. Prove the assertions in the argument above (hit if z = x y then n m ⊗ zij = xiyj). Also show that the images of the Pi Pj are open in the image and isomorphic to F n+m. × The following theorem is Theorem A.4.8 in [GW].

Theorem 23 If (X, X ) is an irreducible projective variety then X (X)consists of the constants. O O

1.3.4 Local properties of algebraic varieties. It is almost counter-intuitive that the assertion that for X a topological space the assertion that Y X is closed is a local property. That is if for each p X there exists U ⊂ X open with U Y closed in U then Y is closed in X∈. To see this we note⊂ that this local condition∩ implies that there is an open covering of X, Uα α I such that Uα Y is closed in Uα. This implies that { } ∈ ∩ Uα Uα Y is open in Uα hence in X. But α I (Uα Uα Y ) = X Y . −One notes∩ that the topology that we have∪ put∈ on− an algebraic∩ variety− is Noetherian so algebraic varieties have a unique (up to order) decomposition into irreducible algebraic varieties. Since any two non-empty open sets in an irreducible topological space must intersect we see that we can define the dimension of an irreducible algebraic set to be the dimension of any open non-empty affi ne subvariety. We define the dimension of an algebraic variety to be the maximum of the dimensions of the irreducible components. We will say that a point in an irreducible algebraic variety is smooth if it is smooth in an affi ne open set containing it. Thus the set of all smooth points is open in the variety. We note that if F = C then a variety has in addition a standard metric topology. To distinguish the two topologies we will use the terms Z-topology and S-topology. Theorem 19 now implies

Theorem 24 If (X, X ) is an irreducible d-dimensional algebraic variety O over C then the set of smooth points endowed with the S-topology is a C∞ manifold of dimension 2d. In addition using the fact that closedness is a local property we have

Theorem 25 Let (X, X ) be an algebraic variety over C. If U is Z-open in X then the S-closure ofOU is the same as the Z-closure.

17 In our study of algebraic groups we will be using the following theorem the local version is e.g. Theorem A.2.7 in [GW].

Theorem 26 Let (X, X ) and (Y, Y ) be non-empty irreducible algebraic varieties. Then if f : XO Y is aO morphism such that the image of X is → dense in Y (here we are using the Z-topology if F = C) and if U is open and non-empty in X then f(U) contains an open non-empty subset of Y.

Corollary 27 Let (X, X ) and (Y, Y ) be non-empty irreducible algebraic O O varieties over C. Then if f : X Y is a morphism such that the image of X is Z-dense in Y then it is S-dense.→

1.4 Appendix. Some local algebra. 1.4.1 The Artin Rees Lemma. Let R be a Noetherian algebra over a ﬁeld F . In this context we recall the Artin Rees lemma.

Lemma 28 Let I be an ideal in R. Let M be a ﬁnitely generated module over R and let N be an R-submodule of M. Then there exists l N such ∈ that Il+jM N = Ij(IlM N) for all j N. ∩ ∩ ∈ Proof. Let t be an indeterminate and consider the subalgebra

2 2 R∗ = R + tI + t I + ... of R[t],this algebra is sometimes called the Rees algebra of R with respect to I. If I is generated by r1, ..., rm (i.e. T = Rr1 + ... + Rim ) then R∗ is the algebra over R generated by tr1, ..., trm. Thus R∗ is Noetherian. We consider the R∗-module 2 2 M ∗ = M + tIM + t I M + ....

If M is generated as a module by m1, ..., mr as an R-module then so is M ∗ as an R∗ module. Let

2 r+1 r N1 = N + tIM N + t I(IM N) + ... + t I (IM N) + ... ∩ ∩ ∩ 2 2 r+2 r 2 N2 = N + tIM N + t (I M N) + ... + t I (I M N) + ... ∩ ∩ ∩

18 and in general 2 2 m m Nm = N + tIM N + t (I M N) + ... + t (I M N)+ ∩ ∩ ∩ tm+1I(ImM N) + ... + tm+jIj(ImM N) + ... ∩ ∩ Then N1 N2 ... and all are R∗-submodules. Thus there is an index l ⊂ ⊂ such that Nl+j = Nl for all j. Comparing coeﬃ cients of t yields the lemma.

Corollary 29 Let be a maximal proper ideal in R. If M is an R-module k M then k 1 M = 0. ∩ ≥ M k Proof. Let N = k 1 M. Let l be as in the Artin Rees lemma. Then ∩ ≥ M N = l+jM N = j( lM N) = jN. M ∩ M M ∩ M Thus N = N. Now Nakayama’slemma (below) implies that N = 0. We nowM recall Nakayama’sLemma. Lemma 30 Let R be a commutative ring with identity and let I be an ideal in R such that 1 + a is invertible for all a I (e.g. I is maximal). Then if M is a ﬁnitely generated R-module such that∈ IM = M then M = 0.

Proof. Let M = Rm1 + ... + Rmr. Then there exist aij I such that r ∈ j=1 aijmj = mi. Thus we have P r (δij aij)mj = 0. j=1 − X If we apply Cramer’s rule we have det(I [aij])mk = 0 for all k. But − det(I [aij]) = 1 + a with a I. Thus mk = 0 all k. − ∈ 1.4.2 Localization. Let R be a finitely generated algebra over F . Let S R be closed under multiplication containing 1 and doesn’t contain 0we⊂ will call S a multiplicative subset. Then we define R(S) to be the set R S modulo the equivalence relation (r, s) is equivalent with (t, u) if there exists× v S such ∈ that v(ru ts) = 0. We observe that if (r, s) is equivalent with (r0, s0) and − (t, u) is equivalent with (t0, u0) then (rt, su) is equivalent with ((r0t0, s0u0) and (ru + ts, su) is equivalent with (r0u0 + t0s0, s0u0) we can thus define an algebra structure on R(S). We write the equivalence class of (r, s) as r/s. We have an algebra homomorphism of R to R(S) given by r r/1. → 19 Lemma 31 If R and S are as above then R(S) is Noetherian.

Let be an ideal in R such that S = R is multiplicative. This M − M condition is equivalent to the primality of . Then we set R = R(S). M M Then R has the unique maximal ideal m = R(S). This is the deﬁnition of a localM ring. M l l Let R be a local ring with maximal ideal . Let jkl : R/ R/ .for l k be the natural projection. We deﬁne M M → M ≥ lim R/ k ←− M k to be the set of sequences ak with ak R/ and jklal = ak for l k. Then this ring with addition{ } and multiplication∈ M given by component≥ wise operations is called the completion of R and denoted R.