ADDITIONAL NOTES in ALGEBRAIC GEOMETRY, II 1. Basic Properties

ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II

ANDREAS LEOPOLD KNUTSEN

Abstract. These notes supplement [Hart, Chapter II] and form part of the syl- labus in the course MAT321-Introduction to sheaves and schemes, taught at the University of Bergen, Spring 2016.

1. Basic properties of coverings and morphisms of schemes This section supplements [Hart, II, §2]. The following summarizes basic properties of an aﬃne scheme that in particular are used in the proof of [Hart, II, Prop. 3.2]: Lemma 1.1. Let X = Spec A be an aﬃne scheme. Then the following holds: (a) The open sets of the form D(α), for α ∈ A, form a basis for the topology and

(D(α), OX |D(α)) ' Spec Aα. (b) For any aﬃne subscheme Spec B ⊆ X, we have (1) D(α) ∩ Spec B = D(α), where α = ϕ(α), with ϕ : A → B corresponding to the inclusion morphism Spec B → Spec A as in the proof of [Hart, II,Prop. 2.3] (see also below). In particular, if D(α) ⊆ Spec B, then Aα ' Bα. (c) We have

(2) Spec A = ∪i∈I D(αi) ⇐⇒ ({αi | i ∈ I}) = A. (d) Given (2), we have

Spec A = D(α1) ∪ · · · ∪ D(αr),

for ﬁnitely many αi. In particular, sp(X) is quasi-compact ([Hart, II, Exc. 2.13(b)]). (e) sp(X) is not necessarly noetherian, but is noetherian if A is noetherian ([Hart, II, Exc. 2.13(b,c)]). Proof. (a) The ﬁrst part is proved in [Hart, p. 70-71] (bottom and start of pages) and the second is [Hart, II, Exc. 2.1] and is proved in the same way as [Hart, II, Prop. 2.5(b)] (and is partially proved in [Hart, II, Prop. 2.2(b,c)]). (b) We have D(α) ∩ Spec B = {p ∈ Spec A | α 6∈ p} ∩ Spec B = {q ∈ Spec B | α 6∈ ϕ−1q} = {q ∈ Spec B | ϕ(α) 6∈ q} = D(α),

Date: April 18th, 2016. 1 2 ANDREAS LEOPOLD KNUTSEN as desired. (This will be generalized in Remark 2.4 below). In particular, if D(α) ⊆ Spec B, then D(α) = D(α), so that, by (a), we have Spec Aα ' Spec Bα, so that Aα ' Bα by (the proof of) [Hart, II, Prop. 2.3]. (c) We have

∪i∈I D(αi) = X − V (({αi | i ∈ I})) by [Hart, II, Lemma 2.1(b)] and V (({αi | i ∈ I})) = ∅ if and only if ({αi | i ∈ I}) = A. (d) ({αi | i ∈ I}) = A means that we can write X 1A = tiαi, for some ti ∈ A. i∈I

But this must hold for finitely many i’s, so that (α1, . . . , αr) = A, which means that Spec A = D(α1) ∪ · · · ∪ D(αr). By (a), this implies that sp(X) is quasi-compact. (e) One easily sees that Spec k[x1, x2,...] is not noetherian. However, if A is noetherian and V (a1) % V (a2) % ··· is a descending chain of closed subsets, then the corresponding chain √ √ a1 ⊆ a2 ··· √ must eventually stabilize, as A is noetherian. Now the rest follows as V (ai) = V ( ai). An almost immediate consequence of Lemma 1.1 is that any open subset U ⊂ X is itself a scheme, with structure sheaf OU = OX |U , cf. [Hart, II, Exc. 2.2]. (Indeed, X can be covered by open affine subschemes Spec A, each of which has a basis for the topology consisting of open sets that are spectra of rings, whence also U is a union of such spectra of rings.) We next study morphisms between affine schemes. Let X = Spec A and Y = Spec B be affine schemes and

(f,f ]) Y / X

p / ϕ−1p be a morphism, corresponding to a ring homomorphism

ϕ A / B, as given in the proof of [Hart, II,Prop. 2.3] (so that ϕ is given by

f ](X) A ' Γ(X, OX ) / Γ(Y, OY ) ' B, (using [Hart, II, Prop. 2.2(c)]). In particular, we have that for any y ∈ Y , the induced morphism on the stalks

] fy (3) OX,f(y) / OY,y ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 3 is just the natural localization of ϕ:

ϕpy −1 (4) Aϕ py = Af(py) / Bpy , where py ∈ Spec B is the prime ideal corresponding to y. It was noted in the proof of [Hart, II, Prop. 2.3] that (5) f −1(V (a)) = V (ϕ(a)) for any ideal a ⊆ A. Conversely, it is easy to check that (6) f(V (b)) = V (ϕ−1(b)) for any ideal b ⊆ B. The following properties of morphisms is for instance used in [Hart, II, Example 3.2.3]: Lemma 1.2. ([Hart, II, Exc. 2.18(b,c)]) Let X = Spec A and Y = Spec B be aﬃne schemes and f, f ] and ϕ as in the beginning of the section. Then ] (i) ϕ is injective if and only if f : OX → f∗OY is injective. Moreover, f is dominant in this case. (ii) If ϕ is surjective, then f is a closed immersion 1.

Proof. (i) If ϕ is injective, then also (4) is injective for every py ∈ Spec B, whence also (3), and therefore also f ] is injective. Conversely, if f ] is injective, then also ϕ = f ](X) is. The fact that f is dominant in this case follows from (6):

(∗) f(Y ) = f(Spec B) = f(V ((0))) = V (ϕ−1((0))) = V ((0)) = Spec A where the injectivity of ϕ is used in (∗). (ii) If ϕ is surjective, then B ∼= A/ ker ϕ, and, as is well-known from commutative algebra, there is a one-to-one correspondence (given by ϕ−1) between prime ideals in B and prime ideals in A containing ker ϕ. This turns (6) into (7) f(V (b)) = V (ϕ−1(b)) for any ideal b ⊆ B and yields that f(Y ) = f(Spec B) = f(V ((0))) = V (ker ϕ), and f induces a bijection between Y = Spec B and the closed subset V (ker ϕ) ⊆ A. It is continuous as f is, and it is bicontinuous since f maps closed sets into closed sets by (7). The surjectivity of f ] can be checked on stalks, where it follows since (4) and (3) are the same map. There are similar results concerning the schemes Proj S associated to graded rings, as we will now see. We use the same notation as in [Hart, p. 76-77]. Proposition 1.3. ([Hart, II, Exc. 2.14(b-c) and Exc. 3.12(a)]) Let ϕ : S → T be a graded homomorphism of graded rings. Let U := {p ∈ Proj T | ϕ(S+) 6⊆ p}.

1By deﬁnition (see [Hart, Def. p. 85]) this means that f is a homeomorphism onto a closed subset of X (necessarily equal to V (ker ϕ), as seen in the proof) and f ] is surjective. 4 ANDREAS LEOPOLD KNUTSEN

(a) U is an open subset of Proj T and ϕ determines a natural morphism f : U → Proj S (given on the topological spaces by f(p) = ϕ−1p). (b) If ϕ is surjective, then U = Proj T and f : Proj T → Proj S is a closed immersion. (c) If ϕd : Sd → Td is an isomorphism for all d ≥ d0, for some d0 ∈ Z, then U = Proj T and f : Proj T → Proj S is an isomorphism.

Proof. (a) The ideal ϕ(S+)T in T is homogeneous, as it is generated by the homogeneous elements ϕ(α) for homogeneous α ∈ S+. Hence, as

(8) Proj T − U = V (ϕ(S+)T ), the set U is an open subset of Proj T . −1 −1 We note that if p ∈ U, then S+ 6⊆ ϕ p, whence ϕ p ∈ Proj S and there is a map f : U → Proj S on the topological spaces by f(p) = ϕ−1p. If V (a) ⊆ Proj S is a closed set, for a homogeneous ideal a ⊂ S, then f −1V (a) = {p ∈ U | ϕ−1p ⊇ a} = V (ϕ(a)T ), whence f is continuous. ] We have left to deﬁne the morphism of sheaves f : OProj S → f∗OU. We proceed as in the proof of [Hart, II, Prop. 2.3]. For each p ∈ U, we obtain a local homomorphism of local rings ϕp : S(ϕ−1p) → T(p). For any open set V ⊆ Proj S we thus obtain a homomorphism of rings

f ](V ) −1 OProj S(V ) / OU(f V )

s / t[p∈f −1V ]ϕp ◦ s ◦ f,

] ] ] such that ff(p) = ϕp on the stalks. This deﬁnes f and proves that (f, f ) is a morphism. (b) Surjectivity of ϕ implies that ϕ(S+) = T+, proving that U = Proj T . As T ' S/ ker ϕ, we have that f(Proj T ) = V (ker ϕ), which is closed in Proj S. Finally, the surjectivity of f ] can be proved on the stalks, where we get the local homomorphisms ϕϕ−1p : S(ϕ−1p) → T(p), which are surjective by hypothesis. Hence f is a closed immersion. (c) Assume that ϕd : Sd → Td is an isomorphism for all d ≥ d0. If p ∈ Proj T − U, d0 then ϕ(S+) ⊆ p, whence ⊕d≥d0 Td ⊂ p. Therefore (T+) ⊆ p. As p is prime, we must have T+ ⊂ p, a contradiction. Thus U = Proj T . We now prove that f is injective. If f(p) = f(q), then ϕ−1p = ϕ−1q, whence p ∩ Td = q ∩ Td for all d ≥ d0. Thus, if a ∈ p is a homogeneous element, then n n a ∈ p ∩ Td for some n and d ≥ d0. Hence a ∈ q, which implies a ∈ q, as q is prime. This proves that p ⊆ q. By symmetry, we have p = q, which proves that f is injective. Now Proj S is covered by the basic open sets D+(s) for s ∈ S+. The morphism f −1 satisﬁes f D+(s) = D+(ϕ(s)). By the construction in (a) and [Hart, II, Prop. 2.5], ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 5

the morphism f|D+(ϕ(s)) is simply the morphism

f|D+(ϕ(s)) : Spec T(ϕ(s)) → Spec S(s) induced by the localization homomorphisms

ϕs : S(s) → T(ϕ(s)). But the latter are isomorphisms by the hypotheses, whence so are also the morphisms fD+(s). Thus f is an isomorphism on all members of an open cover. As f is injective, it is an isomorphism. Example 1.4. Part (c) in the previous proposition shows that the morphism f can be an isomorphism even if ϕ is not, as opposed to the case of affine schemes, where clearly ϕ is an isomomorphism if f is. For a concrete such example, let T be the graded ring k[x, y] and S = T0 + T2 + T3 + ··· . Let ϕ : S → T be the inclusion. Then ϕd is an isomorhism for all d ≥ 2, but ϕ is not. Example 1.5. Any homogeneous ideal I ⊂ S gives rise to a closed subscheme of S. Indeed, let T = S/I. Then Proposition 1.3(b) yields a natural closed immersion Proj T → Proj S. However, as opposed to the case of affine schemes, cf. [Hart, II, Cor. 5.10], different ideals can give rise to the same subschemes. For instance, for 0 any integer d0, the ideal I := ⊕Id≥d0 determines the same closed subscheme as I, by Proposition 1.3(c).

2. The residue field, and the distinguished open subsets and zeros of a section This section supplements [Hart, II, §2] and the beginning of [Hart, II, §3]. In particular, the result proved in Lemma 2.3 below is used in the proof of [Hart, II, Prop. 3.1]. Let X be a scheme and x ∈ X a point (closed or not). Let mx ⊂ Ox denote the maximal ideal. Definition 2.1. ([Hart, II, Exc. 2.7]) The residue field of x (on X) is the field

k(x) := Ox/mx.

Note that this only depends on the local ring Ox, so the residue field is a local property of the point x, that is, it does not change whether we consider the point x ∈ X, or whether we consider x ∈ U, a neighborhood. By the definition of scheme, we may find an affine neighbourhood U = Spec A of x. Denoting by p ∈ Spec A the point corresponding to x, we see from [Hart, II, Prop. 2.2] that

(9) k(x) ' Ap/pAp.

Let s ∈ Γ(X, OX ) be a global section. Look at the natural morphisms

(10) Γ(X, OX ) / Ox / k(x) = Ox/mx

s / sx / s(x). 6 ANDREAS LEOPOLD KNUTSEN

We call the images of s in Ox and k(x) the stalk sx of s at x and the evaluation s(x) of s at x, respectively. The ﬁelds k(x) may vary from point to point, as will be discussed in Section 5 below, but the following set is well deﬁned: Z(s) = {x ∈ X | s(x) = 0}, and its complement is the set

Xs := {x ∈ X | sx 6∈ mx ⊂ Ox}.

Lemma 2.3 below shows that Xs is open, equivalently that Z(s) is closed, a fact that is used in the proof of [Hart, II, Prop. 3.1].

Deﬁnition 2.2. We often call the open sets Xs, for s ∈ Γ(X, OX ), distinguished open sets. The closed set Z(s), the complement of Xs, is called the set of zeros of s.

Lemma 2.3. If U = Spec A ⊂ X is an aﬃne subscheme and s|U = ρXU (s) ∈ Γ(U, OX |U ) = A is the restriction of s, then

Xs ∩ U = D(s|U ), whence Z(s) ∩ U = V ((s|U )).

Proof. For any x ∈ Spec A ⊂ X, denote by px the corresponding prime ideal. By

[Hart, II, Prop. 2.2] we have that Ox ' Apx and (10) factorizes as

Γ(X, OX ) / Γ(U, OX |U ) = A / Ox ' Apx / k(x) = Ox/mx ' Apx /pxApx

s / s|U / sx / s(x). Therefore,

D(s|U ) = {p ∈ U | s|U 6∈ px} = {x ∈ U | s|U 6∈ mx} = Xs ∩ U.

Remark 2.4. In particular, if X = Spec A is aﬃne and f ∈ Γ(X, OX ) = A (by [Hart, II, Prop. 2.2]), then

Xf = D(f) ⊆ Spec A, and the lemma generalizes Lemma 1.1(b).

2 Example 2.5. Let X = Ak = Spec k[x1, x2] be the aﬃne plane over an algebraically closed ﬁeld k. Then Γ(X, OX ) = k[x1, x2] by [Hart, II, Prop. 2.2]. Let x be a closed 2 point in Ak. Then x = (a, b) for some a, b ∈ k, equivalently, x corresponds to the maximal ideal mx = (x1 − a, x2 − b). Then (10) looks like

(11) Γ(X, OX ) = k[x1, x2] / Ox ' k[x1, x2](x1−a,x2−b) / Ox/mx ' k

s = s(x1, x2) / sx / s(x) = s(a, b), the evaluation of the polynomial s(x1, x2) at the point (a, b) in the usual sense. ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 7

The notions above behave nicely when it comes to morphisms, as the two next lemmas show. (The first one plays an important role in the proof of Corollary 5.2 below, and the second is useful in [Hart, II, Exc. 3.2].) Lemma 2.6. Let (f, f ]): X → Y be a morphism of schemes and x ∈ X a point. Then k(x) is a field extension of k(f(x)). In particular, if X is a scheme over k, then the residue field at any point of X is a field extension of k. Proof. By definition, the induced homomorphism on the local rings (or stalks)

] fx : OY,f(x) −→ OX,x

]−1 is local, that is, fx mx = mf(x). (In particular, it is nonzero.) Hence, we can mod out by the maximal ideals on both sides and obtain an injective homomorphism k(f(x)) → k(x), proving that k(x) is a field extension of k(f(x)). The last assertion follows setting Y = Spec k and using that the residue field of the only point of Spec k is k itself. ] Lemma 2.7. Let (f, f ): X → Y be any morphism of schemes and σ ∈ Γ(Y, OY ). Then −1 f Yσ = Xf ](Y )(σ) ] (where f (Y ) : Γ(Y, OY ) → Γ(X, OX ) denotes the map induced on global sections). Proof. By definition of a morphism of schemes, we have, for any x ∈ X, a commutative diagram

f ](Y ) Γ(Y, OY ) / Γ(X, OX )

] fx OY,f(x) / OX,x,

]−1 such that fx mx = mf(x). Hence ] ] f(x) ∈ Yσ ⇔ σf(x) 6∈ mf(x) ⇔ fx(σf(x)) 6∈ mx ⇔ (f (Y )(σ))x 6∈ mx ⇔ x ∈ Xf ](Y )(σ) ] ] and the result follows (we have used that fx(σf(x)) = (f (Y )(σ))x by the commuta- tivity of the diagram).

3. Reduced and integral schemes This section supplements [Hart, II, 3.0.1 and Prop. 3.1]. Proposition 3.1. ([Hart, II, Exc. 2.3]) Let X be a scheme. The following conditions are equivalent:

(i) OX (U) has no nilpotent elements for all open sets U ⊆ X; (ii) OX,P has no nilpotent elements for all points P ∈ X. 8 ANDREAS LEOPOLD KNUTSEN

m Proof. Assume that there is a nonzero f ∈ OX,P such that f = 0 for some m > 0. Then f has a representative hU, gi, where U ⊆ X is open and g ∈ OX (U), and g 6= 0, m g = 0 in OX (U), proving that there is a U such that OX (U) has nilpotent elements. This proves that (i) implies (ii). Conversely, assume that there is an open set U ⊂ X and a nonzero g ∈ OX (U) such m that g = 0 for some m > 0. Localizing at any point P ∈ U, we obtain gP ∈ OX,P . Now gP cannot be zero for all P , as otherwise, by the sheaf properties, g would be m zero also on U. Hence, there is a P ∈ U such that 0 6= gP ∈ OX,P and gP = 0, whence OX,P has nilpotent elements. This proves that (ii) implies (i). If the equivalent conditions above are satisﬁed, then we say that X is reduced, cf. [Hart, Def. p. 82]. It is easy to see that it is enough to check conditions (i) on any base for the topology. We also recall the deﬁnition of X being integral, cf. [Hart, Def. p. 82], as well as [Hart, II, Prop. 3.1], which states that a scheme X is integral if and only if it is both reduced and irreducible. Example 3.2. ([Hart, II, Example 3.0.1]) If X = Spec A, then (a) X is irreducible if and only if the nilradical nil A is prime; (b) X is reduced if and only if nil A = 0; (c) X integral if and only if A is an integral domain. Proof. (a) X is reducible if and only if there are two nonempty open subsets of X with empty intersection. These can be taken as basic open subsets, that is, we have a, b ∈ A such that D(a),D(b) 6= ∅,D(a) ∩ D(b) = D(ab) = ∅ Recalling that D(f) = ∅ ⇔ f ∈ p for all p ∈ Spec A ⇔ f ∈ nil A, we get X is reducible ⇔ there exist a, b ∈ A such that a, b 6∈ nil A, ab ∈ nil A ⇔ nil A is not prime.

(b) By Proposition 3.1, X is reduced if and only if Ap has no nilpotents for all p ∈ X = Spec A. The latter is equivalent to A having no nilpotents, by standard algebra, which means precisely that nil A = 0. (c) By [Hart, II, Prop. 3.1], X is integral if and only if it is both reduced and irreducible, which by (a) and (b) occurs if and only if (0) is a prime ideal in A. The latter is equivalent to A being an integral domain. Example 3.3. (i) Let A = k[x, y]/(x2). Let x denote the residue class of x in A. Then nil A = (x), which is prime but nonzero. Hence, Spec A is irreducible, nonreduced, and nonintegral. (ii) Let A = k[x, y]/(xy). Let x and y denote the residue class of x and y, respectively, in A. Then nil A = (0) is not prime, as x, y 6= 0 but x · y = 0. Hence, Spec A is reduced, reducible, and nonintegral. ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 9

4. Generic point and function field This section supplements [Hart, II, Example 2.3.4 and beginning of §3]. Deﬁnition 4.1. ([Hart, II, Exc. 2.9]) Let X be a topological space and Z ⊂ X an irreducible closed subset. A point ξ ∈ Z is a generic point for Z if {ξ} = Z. Note that a generic point has the property that it lies in every nonempty open subset of Z. Therefore, for any sheaf F on X, and any point z ∈ Z we have natural maps on stalks

(12) Fz −→ Fξ. Lemma 4.2. ([Hart, II, Exc. 2.9]) If X is a scheme, then every (nonempty) irreducible closed subset has a unique generic point. Proof. Let Z ⊂ X be a (nonempty) irreducible closed subset. Choose any open, affine V = Spec A ⊂ X such that V ∩ Z 6= ∅. Then V ∩ Z is open, irreducible and dense in Z, by [Hart, I, 1.1.3]. At the same time, V ∩ Z is closed in V = Spec A, so that V ∩ Z = V (a) for some ideal a in A, that is, V ∩ Z = Spec(A/a). This is irreducible if and only if the nilradical of A/a is prime, by [Hart, II, Example 3.0.1] or by Example 3.2 above. The nilradical is the√ intersection of all primes in√A/a, that is, of all primes in A containing a, which is a, the radical of a. Hence a = p, a prime, so that √ V (a) = V ( a) = V (p) the closure of {p} in Spec A. Letting x ∈ X be the point corresponding to p, we have Z = V ∩ Z = {x}, proving the existence of a generic point. If also Z = {x0}, then x0 ∈ V by the property of a generic point, so that x = p0 ∈ Spec A. But then p0 ∈ V (p), whence p0 ⊇ p. Likewise, p ⊇ p0, so that p = p0, proving the uniqueness. In particular, if X is an integral scheme, then it is reduced and irreducible by [Hart, II, Prop. 3.1], so it has a unique generic point ξ. For any x ∈ X, pick an open affine neighborhood V = Spec A of x, and let p be the prime ideal corresponding to x. Then Ox ' Ap by [Hart, II, Prop. 2.2] and since A = OX (V ) is an integral domain by definition, the point ξ ∈ V ⊆ X must be the zero ideal. Hence Oξ ' A((0)) = K(A), the quotient field of A, again by [Hart, II, Prop. 2.2] so that Oξ is a field. Also note that we have a natural injective homomorphism

Ox ' Ap −→ A((0)) 'Oξ, which is nothing but (12). We summarize this in the following result, which corresponds to the same facts for varieties mentioned in [Hart, p. 16] Lemma 4.3. ([Hart, II, Exc. 3.6]) If X is an integral scheme and ξ its generic point, then for any point x ∈ X and any neighborhood U of x, the natural homomorphisms

OX (U) −→ Ox −→ Oξ 10 ANDREAS LEOPOLD KNUTSEN

are both injective. Moreover, Oξ is a field, and for any open, affine U = Spec A ⊆ X, we have that Oξ is isomorphic to the quotient field of A. Proof. We only have left to prove the injectivity of the left map. For this, it suffices to show the injectivity of the composed map to Oξ. Assume therefore that s ∈ OX (U) is mapped to zero in Oξ. Let V = Spec A be any affine open subset of U. (Recall that U, being open in X, has a natural structure of scheme, see the lines following Lemma 1.1.) Then ξ ∈ V , as ξ is a generic point. Since there is a base for the topology on V = Spec A consisting of open subsets of the form D(f) for f ∈ A (see [Hart, p. 70-71] or Lemma 1.1(a)), we can find a neighborhood of the form D(f) ⊆ U ∩ V of ξ such that s|D(f) = 0 in OX (D(f)) = Af (by [Hart, II, Prop. 2.2(b)]: here we also use that OX |V = OV by definition of a scheme, so that OX (D(f)) = OV (D(f)). Now A injects into Af , since A is an integral domain as X is integral. This means that s|V = 0 in OX (V ) = A (again by [Hart, II, Prop. 2.2]). Hence s = 0 by the sheaf properties. Definition 4.4. ([Hart, II, Exc. 3.6]) Let X be an integral scheme and ξ its generic point. The field Oξ (in Lemma 4.3) is called the function field of X and is denoted by K(X). Its elements are called rational functions. We say that a rational function f ∈ K(X) is regular at a point x ∈ X if f ∈ Ox ⊆ K(X) (by the right inclusion in Lemma 4.3). n ∼ Example 4.5. We have K(Ak ) = k(x1, . . . , xn), the field of rational functions in n n n variables. As Pk can be covered by open affine subschemes isomorphic to Ak , we also n ∼ have K(Pk ) = k(x1, . . . , xn) by Lemma 4.3. 5. The residue field of points in a scheme of finite type over a field This section supplements [Hart, II, §3] by collecting some important result on the residue field of points in a scheme of finite type over a field k (see [Hart, Def. p. 84]). As above, let X be a scheme, x ∈ X a point (closed or not) and mx ⊂ Ox denote the maximal ideal. We recall the definition of the residue field k(x) := Ox/mx from Definition 2.1. Proposition 5.1. If X is a scheme of finite type over k, then k(x) is a finitely generated field extension of k. Moreover x ∈ X is a closed point if and only if k(x) is a finite algebraic extension of k. In particular, if k is algebraically closed, then x is a closed point if and only if k(x) ' k. Proof. If X is of finite type over a field k, then for any open affine U = Spec A ⊆ X, we have that A is a finitely generated k-algebra, by [Hart, II, Exc. 3.3(c)]). We therefore have that A ' k[x1, . . . , xn]/I for some ideal I ⊂ A. If x ∈ U, and p ∈ Spec A is the corresponding prime ideal, then k(x) ' Ap/pAp (cf. (9)), so that k(x) is a finitely generated field extension of k. Now x is closed if and only if x is closed in any affine neighborhood U = Spec A. This happens if and only if the corresponding point p ∈ Spec A is a maximal ideal. If p is maximal, then k(x) ' A/p by (9), which is a finite algebraic extension of k, as A is a finitely generated k-algebra, by Hilbert’s Nullstellensatz [A-M, Cor. 5.24]. ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 11

Conversely, if k(x) ⊇ k is ﬁnite and algebraic, then, as

k ⊆ A/p ⊆ (A/p)p ' Ap/pAp ' k(x) (using (9) again), we see in particular that A/p is integral over k, whence a field by [A-M, Prop. 5.7], so that p is maximal. We collect a few corollaries of this result: Corollary 5.2. Let f : X → Y be a morphism of schemes of finite type over k. Then f maps closed points to closed points. Proof. By definition of a morphism of schemes over k, we have a commutative diagram

f X / Y GG w GG ww GG ww GG ww G# {ww Spec k. For any x ∈ X, this induces, by Lemma 2.6, a commutative diagram of ﬁeld exten- sions k(x) o ? _k(f(x)) aB ; BB xx BB xx BB xx B0 P - xx k x If x ∈ X is closed, then k(x) is a ﬁnite algebraic extension of k, by Proposition 5.1. The latter diagram implies that also k(f(x)) is, whence also f(x) ∈ Y is closed by Proposition 5.1. A consequence of this is a proof of the last part of [Hart, II, Prop. 2.6] left as an exercise ([Hart, II, Exc. 2.15]): Corollary 5.3. With notation as in [Hart, II, Prop. 2.6], the natural map

HomVar(k)(V,W ) → HomSch(k)(t(V ), t(W )) is bijective. Proof. Injectivity is immediate. Let us prove that any morphism of schemes f : t(V ) → t(W ) over k is induced from a unique morphism of varieties V → W . We recall from the proof of [Hart, II, Prop. 2.6] that t(V ) = {(nonempty) closed irreducible subsets of V } and that the natural map αV V / t(V )

P / {P } induces a homeomorphism between V and the subspace t(V )cl of closed points of t(V ). By [Hart, II, Example 3.2.1], the schemes t(V ) and t(W ) are schemes of ﬁnite type over k. Hence, by Corollary 5.2, the morphism f induces a natural continuous 12 ANDREAS LEOPOLD KNUTSEN map f cl : t(V )cl → t(W )cl, whence also a continuous map between the homeomorphic −1 cl spaces αW ◦ f ◦ αV : V → W . To be precise, we have a commutative diagram: f (13) t(V ) / t(W ) O O iV iW ? f cl ? t(V )cl / t(W )cl O O αV =∼ αW =∼

V / W The unicity of V → W is clear, as it has to make the latter diagram commutative. The obtained map is continuous. The fact that it takes regular functions on W to regular functions on V follows since it is proved in [Hart, II, Prop. 2.6] that the sheaves of regular functions on V and W are obtained by restricting the structure sheaves on t(V ) (resp. t(W )) via the inclusion iV (resp. iW ) and the homeomorphism αV (resp. αW ). Hence, it is a morphism, as desired. Finally, the following consequence of Proposition 5.1 is used in the proof of [Hart, II, Prop. 4.10]: Corollary 5.4. ([Hart, II, Exc. 3.14]) Let X be a scheme of finite type over a field k. Then the set of closed points in X is dense. Proof. Assume, to get a contradiction, that Y := {x ∈ X | x is closed} $ X. Then we could find an open, affine subset Spec A ⊆ X − Y . Since X is of finite type, A must be a finitely generated k-algebra [Hart, II, Exc. 3.3(c)]. Pick a closed point x ∈ Spec A. Then k(x) (which depends only on a neighborhood of x) is a finite, algebraic extension of k by Proposition 5.1. But then, again by the same proposition, x is also closed in X, a contradiction, as x ∈ X − Y . Remark 5.5. ([Hart, II, Exc. 3.14]) The same is not true for general schemes. For instance, Spec Z(2) contains precisely one closed point, namely the ideal (2). In general, look at Spec of any DVR (see for instance [Hart, II, Example 2.3.2]).

2 Example 5.6. Consider the affine plane Ak = Spec k[x, y] over an algebraically 2 closed field k. If p ∈ Ak is a closed point, then p = (a, b) for some a, b ∈ k, and the corresponding prime ideal in Spec k[x, y] is the maximal ideal mp = (x − a, y − b). Then k(p) = k[x, y](x−a,y−b)/(x − a, y − b)k[x, y](x−a,y−b) ' k, as anticipated by Proposition 5.1. The prime ideals that are neither maximal nor the zero ideal, are those of height one, whence principal (see e.g. [Hart, I, Prop. 1.12A]), that is, of the form p = (f), for an irreducible polynomial f ∈ k[x, y]. In this case, one easily sees that k(p) is 2 isomorphic to the function field over k in one variable, where p ∈ Ak denotes the corresponding point. ADDITIONAL NOTES IN ALGEBRAIC GEOMETRY, II 13

6. Fibers of a morphism This section supplements [Hart, II, §3] by supplying some results concerning fibers of a morphism (cf. [Hart, Def. p. 89]) that are left as exercises. We start with the following result: Lemma 6.1. ([Hart, II, Exc. 2.7]) Let K be any field and X a scheme. Then, giving a morphism Spec K → X is equivalent to giving a point x ∈ X and an extension k(x) ⊆ K. In particular, for any x ∈ X, we have a natural morphism (14) Spec k(x) −→ X Proof. Given a morphism f : Spec K → X, we obtain a point x = f((0)) ∈ X. By Lemma 2.6, the residue field k((0)) = K is a field extension of k(f((0)) = k(x). Conversely, if we are given a point x ∈ X and k(x) ⊆ K, we get an induced local homomorphism, by composing:

Ox −→ k(x) := Ox/mx −→ K. This induces a morphism of sheaves

] f : OX −→ Sx, where Sx is the skyscraper sheaf located at x, which can be described as f∗OSpec K , where f : {x} = Spec K −→ X is the inclusion (cf. [Hart, II, Exc. 1.17]). Therefore ] f and f define a morphism Spec K → X. The natural morphism (14) is used to define the fiber over a point of a morphism, cf. [Hart, Def. p. 89]: If f : X −→ Y is a morphism of schemes and y ∈ Y is any point, then we define the fiber of f over y to be Xy := X ×Y Spec k(y) given by the morphisms f : X −→ Y and k(y) −→ Y from (14). That is, we have a commutative diagram

(15) X ×Y Spec k(y) qq QQQ qq QQQ qqq QQQ qq QQQ xqqq ( X N Spec k(y) NNN m NN f mmm NNN mmm NN mmm NNN mmm & Y vmm

Therefore, Xy is a scheme over k(y). Furthermore it is the scheme-theoretic ﬁber over the point y, that is, it gives the set f −1y := {x ∈ X | f(x) = y} a structure of scheme, by the following result:

Lemma 6.2. ([Hart, II, Exc. 3.10]) The underlying topological space of Xy is homeomorphic to f −1y. 14 ANDREAS LEOPOLD KNUTSEN

Proof. By taking aﬃne coverings of X and Y we reduce to the aﬃne case. So let X = Spec B and Y = Spec A and ϕ : A −→ B be the ring homomorphism inducing f, as in [Hart, II, Prop. 2.3]. This means that

f (16) Spec B / Spec A

q / ϕ−1q Let p ∈ Spec A be the prime ideal corresponding to y. We have k(y) = Ap/pAp by (9) and Xy = X ×Y Spec k(y) = Spec(B ⊗A Ap/pAp) by Step 1 in the proof of [Hart, II, Thm. 3.3] (the aﬃne case of the theorem). Moreover, denoting by S is the multiplicative set ϕ(A − p), one can check that there is a natural isomorphism −1 B ⊗A Ap/pAp ' S (B/ϕ(p)B), where we use the notation ϕ(p)B for the ideal generated by ϕ(p) in B (sometimes called the extension of p by ϕ and denoted by pe, cf. [A-M]). We therefore have that −1 Xy ' Spec S (B/ϕ(p)B). Finally, recalling (16), we have f −1(y) = f −1(p) = {q ∈ Spec B | f(q) = p} = {q ∈ Spec B | ϕ−1(q) = p} = {q ∈ Spec B | ϕ(p)B ⊆ q and q ∩ S = ∅} = Spec S−1(B/ϕ(p)B), where the last equality is a homeomorphism of topological spaces. This proves the lemma.

Acknowledgements. I would like to thank Mirjam Solberg, Mauricio Godoy, Mar- tin Stolz and Tommy Lundemo for useful comments on and discovering mistakes and inaccuracies in previous versions of the notes.

References [A-M] M. F. Atiyah, I. G. Macdonald, Introduction to commutative algebra. Addison-Wesley Pub- lishing Co., Reading, Mass.-London-Don Mills, Ont. [Hart] R. Hartshorne, Algebraic Geometry. Graduate Texts in Mathematics, 52. Springer-Verlag, New York-Heidelberg, 1977.