Algebraic Geometry Homework II.3 #’s 6,7,8,12,14 II.4 #’s 2, 4

William A. Stein

April 4, 1996

Problem 1 (3.6) Let X be an integral scheme. Show that the local ring O_{ξ} of the generic point ξ of X is a field. Call it K(X). Show also that if U = Spec A is any open affine subset of X, then K(X) is isomorphic to the quotient field of A.

Proof. Let U = Spec A be any nonempty open affine subset of X. Then since the closure of a generic point of X is all of X, every open set must contain a generic point. Thus if ξ is a generic point, then ξ ∈ U. But A is an integral domain so (0) is the unique generic point of U, whence ξ = (0). This shows the generic point is unique if it exists. Since X is integral it is irreducible so every open set intersects U. Thus every open set contains (0) ∈ Spec A = U, so X actually contains a generic point ξ = (0). Furthermore, O_{ξ} ≅ A_{(0)} is the quotient field of A.

Problem 2 (3.7) Let f : X → Y be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset U ⊆ Y such that the induced morphism f^{−1}(U) → U is finite.

Proof. Let U = Spec B be an open affine subset of Y which contains the generic point of Y. Let V = Spec A be an open affine subset of f^{−1}(U). Since f is of finite type A is a finitely generated B-algebra. The generic point of X is in V since every open set contains the generic point. Let φ : B → A be the homomorphism corresponding to the induced morphism of affine schemes f : V → U. Since f is dominant, we know that φ is injective. The induced map on stalks then gives an inclusion of function fields K(Y) = B_{(0)} ↪ A_{(0)} = K(X). Since A is a finitely generated Balgebra, K(X) is a finitely generated field extension of K(Y). If this field extension is not of finite degree then K(X) must contain an element which is transcendental over K(Y). Thus A must contain an element t which is transcendental over B. But then infinitely many primes of A lie over (0). Indeed, since K(X) is finitely generated over K(Y), K(X) is a finite algebraic extension of k(t) for some field k. Then (since the algebraic closure of a field is infinite), there are infinitely many irreducible polynomials in k[t]. Since K(X) is finite algebraic over k(t) infinitely many of these must remain irreducible in A. Multiplying through denominators this gives infinitely many irreducible elements of A which generate prime ideals which lie over (0). This would contradict the fact that f is generically finite. Thus K(X) is finite over K(Y).

Let x_{1}, …, x_{n} generate A as a B-algebra. Then, since K(X) is finite over K(Y) each x_{i} satisfies some polynomial f_{i} with coefficients in B. Let b be the product of all of the leading coefficients of the polynomials f_{i}. If b is a unit in B then so are all of the leading coefficients of the f_{i} so we can divide by them and hence assume the f_{i} are monic polynomials. If not, replace B by the localization B_{b} and repeat the whole argument with U = Spec B_{b}. In either case we may assume the f_{i} are monic from which we conclude that A is a finitely generated integral extension of B, thus A is a finite module over B.

Now let U = Spec B be an open affine subset of Y which contains the generic point. Since f is of finite type we may write f^{−1}(U) = ∪_{i=1,...,n}V_{i} where each V_{i} = Spec A_{i} is finitely generated B-algebra. By the work above we may shrink U so that we can assume each A_{i} is actually a finitely generated B-module. To complete the proof we need to show that there is a distinguised open subset of U (which necessarily contains the generic point) whose inverse image under f is an open affine which is the spectrum of a finitely generated B-module. Let φ_{i} : B → A_{i} be the homomorphism which induces f|_{V_{i}}. Since f is dominant, each φ_{i} is an injection. Thus we may, for notational convenience, identify B with it’s images in the various A_{i}. The morphism f is then induced by the inclusion map B ↪ A_{i}.

Since ∩_{i=1,...,n}V_{i} is open we can, for each i, 1 ≤ i ≤ n − 1, find α_{i} ∈ A_{i} such that Spec (A_{i})_{α_{i}} ⊆ ∩_{i=1,...,n}V_{i}. Since A_{i} is a finite module over B there is an integral equation

α_{i}^{n} + b_{n−1}α_{i}^{n−1} + ⋯ + b_{0} = 0

where each b_{j} ∈ B and b_{0} ≠ 0. Let b = ∏_{i=1,...,n−1} b_{i}. Then any prime of A_{i} which contains α_{i} must also contain b_{i} and hence b. Therefore Spec (A_{i})_{b} ⊆ Spec (A_{i})_{α}. We then have that g^{−1}(Spec B_{b}) = ∪_{i=1,...,n−1}Spec (A_{i})_{b}∪Spec (A_{n})_{b} = Spec (A_{n})_{b}. The latter equality follows since, for 1 ≤ i ≤ n − 1, Spec (A_{i})_{b} ⊆ ∩_{i=1,...,n}V_{i} ∩ f^{−1}(U_{b}) ⊆ f^{−1}(U_{b}) ∩ V_{n} = Spec (A_{n})_{b}. We thus see that U_{b} is a dense open subset of Y such that the morphism f : f^{−1}(U_{b}) → U_{b} is finite (since f^{−1}(U_{b}) is the affine scheme Spec (A_{n})_{b} which a finite B_{b}-module). This completes the proof.

Problem 3 (3.8) Normalization. Let X be an integral scheme. For each open affine subset U = Spec A, let Ã be the integral closure of A in its quotient field, and let Ũ = Spec Ã. Show that one can glue the schemes Ũ to obtain a normal integral scheme X̃, called the normalization of X. Show also that there is a morphism X̃ → X, having the following universal property: for every normal integral scheme Z, and for every dominant morphism f : Z → X, f factors uniquely through X̃. If X is of finite type over a field k, then the morphism X̃ → X is a finite morphism.

Proof.

We first verify the universal property for affine schemes where it is clear what the normalization is.

Proposition 1 Suppose X = Spec A, X̃ = Spec Ã its normalization and Z = Spec B is a normal integral scheme. Then every dominant morphism f : Z → X factors uniquely through X̃.

Proof. Let φ : A → B be the homomorphism corresponding to f. Then, since f is dominant, φ is injective. Indeed, f(Z) ⊆ V (ker(φ)) so if ker(φ) ≠ 0 then f(Z) doesn’t meet the nonempty open set X − V (ker(φ)) (nonempty since A is a domain so (0) is prime so (0) ∈ X − V (ker(φ)). So there is a unique extension of φ to a homomorphism from Ã → B. Indeed, define \overline{φ}(a/b) = φ(a)/φ(b). Then since φ is injective this is well-defined since b ≠ 0 implies φ(b) ≠ 0. Furthermore, if ψ is another possible extension of φ to Ã, then ψ(b)ψ(a/b) = ψ(a) = φ(a) = φ(b)\overline{φ}(a/b) = ψ(b)\overline{φ}(a/b) so cancelling shows that ψ(a/b) = \overline{φ}(a/b). Thus there is a unique morphism f′ : Z → X̃ whose composition with the natural map X̃ → X equals f. (The natural map is induced by the inclusion A ↪ Ã.)

Now we prove the universal property holds when Z is an arbitrary normal integral scheme but X is still affine.

Proposition 2 Let X = Spec A, X̃ = Spec Ã its normalization and Z be any normal integral scheme. Then every dominant morphism f : Z → X factors uniquely through X̃.

Proof. Let U_{i} be a cover of Z by open affines. If U = U_{i} is any U_{i} then U is a normal integral affine scheme and f|_{U} is a dominant morphism. Indeed, U is dense in Z since Z is irreducible (Proposition 3.1). Thus f^{−1}(\overline{f(U)}) ⊇ \overline{U} = Z so f^{−1}(\overline{f(U)}) = Z so \overline{f(U)} ⊇ f(Z) so \overline{f(U)} = \overline{f(Z)} = X. We can thus apply the above proposition to find a unique morphism g_{i} : U_{i} → X̃ such that ψ ∘ g_{i} = f|_{U} where ψ : X̃ → X. By uniqueness on a cover of U_{i} ∩ U_{j} by open affines, g_{i}|_{U_{i}∩U_{j}} = g_{j}|_{U_{i}∩U_{j}}. We can thus glue the morphism g_{i} to obtain a morphism g : Z → X̃ such that ψ ∘ g = f. The morphism g is evidently unique.

Now we can define the identification maps φ_{ij}. Let {U_{i} = Spec A_{i}} be the open affine subsets of X. Let {Ũ_{i} = Spec Ã_{i}} be the associated normalizations. Let ψ_{i} : Ũ_{i} → U_{i} be the morphism induced by the inclusion A_{i} ↪ Ã_{i}. Let W_{ij} = ψ_{i}^{−1}(U_{i} ∩ U_{j}). Then W_{ij} is an open subset of a normal scheme hence normal. ψ_{i} : W_{ij} → U_{i} ∩ U_{j} ⊆ U_{j} so there is a unique morphism which we call φ_{ij} : W_{ij} → Ũ_{j} such that ψ_{j}|_{W_{ji}} ∘ φ_{ij} = ψ_{i}|_{W_{ij}}. By uniqueness we see that φ_{ij} ∘φ_{ji} = id so φ_{ij} = φ_{ji}^{−1}. Furthermore, for each i, j, k, φ_{ij}(W_{ij} ∩W_{ik}) = ψ_{j}^{−1}(ψ_{i}(W_{ij} ∩ W_{ik})) = ψ_{j}^{−1}(ψ_{i}(W_{ij})) ∩ ψ_{j}^{−1}(ψ_{i}(W_{ik})) = W_{ji} ∩ W_{jk}. By uniqueness, φ_{jk} ∘ φ_{ij} = φ_{ik} on W_{ij} ∩ W_{ik}. So by the glueing lemma (Exercise 2.12) we may glue to obtain a scheme X̃. We can also glue the morphisms ψ_{i} to obtain a morphism ψ : X̃ → X.

Next, we must verify that the universal property holds in general. Let Z be an arbitrary normal integral scheme, and let X and X̃ be as above and suppose f : Z → X is a morphism. Cover X be open affines U_{i}. Then for each morphism f|_{f^{−1}(U_{i})} we can apply the above proposition to find a morphism g_{i} such that ψ ∘ g_{i} = f|_{f^{−1}(U_{i})}. By uniqueness we can glue these morphism to obtain the required morphism g : Z → X̃.

Now we check that X̃ is a normal integral scheme. Note first that each Ũ_{i} is the spectrum of an integrally closed domain and is hence a normal integral scheme (since the localization of an integrally closed domain is integrally closed). Let x ∈ X̃. Then x is contained in some Ũ_{i}. But the local ring of x in X̃ is the same as the local ring of x in Ũ_{i} which is integrally closed. This shows that X̃ is normal.

Since X is irreducible, every U_{i} intersects every U_{j}. Thus every Ũ_{i} intersects every Ũ_{j} after glueing. Since each Ũ_{j} is irreducible and they all overlap this implies X̃ is irreducible. Indeed, if X̃ = A ∪ B with A and B closed, then every Ũ_{i} is either completely contained in A or in B. If they are not all contained in one of A or B then we can find an open subset U contained in A and an open subset V contained in B but not contained in A. Then V = (U ∩ V) ∪ (U^{c} ∩ V) = (A ∩ V) ∪ (U^{c} ∩ V) which expresses V as a union of two proper closed subsets of V. A ∩ V is a proper subset of V since V is not contained in A and U^{c} ∩ V is a proper subset of V since U ∩ V ≠ ∅. This contradicts the fact that V is irreducible. Thus X̃ = A or X̃ = B whence X̃ is irreducible.

Now we check that the structure sheaf has no nilpotents. Let U be an open subset of X̃ and suppose f ∈ OX̃(U) is nilpotent. Then since f is nonzero, there is some point x ∈ X̃ so that the stalk f_{x} of f at x in the local ring OX̃ is nonzero and nilpotent (use sheaf axiom (iii) and the definition of OX̃.) Let Ũ_{i} be some Ũ_{i} which contains x. Then the local ring at x is a localization of the integral domain Ã_{i} so it can’t contain any nilpotents. Thus the scheme X̃ is reduced.

Now we check that if X is of finite type over a field k, then the morphism f : X̃ → X is a finite morphism. The U_{i} form an open cover of X and f^{−1}(U_{i}) = Spec Ã_{i} is affine for each i, so we just need to check that Ã_{i} is a finite module over A_{i}. This follows from Theorem 3.9A of chapter I.

Problem 4 (3.12) Closed Subschemes of Proj S.

(a) Let φ : S → T be a surjective homomorphism of graded rings, preserving degrees. Show that the open set U of (Ex. 2.14) is equal to Proj T, and the morphism f : Proj T → Proj S is a closed immersion.

(b) If I ⊆ S is a homogeneous ideal, take T = S/I and let Y be the closed subscheme of X = Proj S defined as the image of the closed immersion Proj S/I → X. Show that different homogeneous ideals can give rise to the same closed subscheme. For example, let d_{0} be an integer, and let I′ = ⨁_{d≥d_{0}} I_{d}. Show that I and I′ determine the same closed subscheme.

Proof. (a) Since φ is graded and surjective, φ(S_{+}) = T_{+} from which it is immediate that U = Proj T. By the first isomorphism theorem T ≅ S/ ker(φ) so f(Proj T) = f(Proj S/ ker(φ)) = V (ker(φ)) which is a closed subset of Proj S. (This is just the fact that there is a one to one correspondence between homogeneous ideals of S/ ker(φ) and homogeneous ideals of S which contain ker(φ).) The map on the stalk corresponding to a point x ∈ Proj T is the map S_{(φ^{−1}(x))} → T_{(x)} induced by φ. This map is surjective since φ is surjective. Thus the induced map on sheaves is surjective.

(b) Let φ : S/I′ → S/I be the natural projection homomorphism. (This makes sense because S/I is a quotient of S/I′. Indeed, S/I = (S/I′)/ ⨁_{0≤d<d_{0}} I_{d}.) Then φ is a graded homomorphism of graded rings such that φ_{d} is the identity for d ≥ d_{0}. So by (Exercise 2.14c) φ induces an isomorphism f : Proj S/I → Proj S/I′. Since this is a morphism over Proj S (the corresponding triangle of homomorphisms commutes) it follows that I and I′ give rise to the same closed subscheme.

Problem 5 (3.14) If X is a scheme of finite type over a field, show that the closed points of X are dense. Give an example to show that this is not true for arbitrary schemes.

Proof. Since X is of finite type over k we can cover X with affine open sets U_{i} = Spec A_{i} where each A_{i} is a finitely generated k-algebra. Let U be an open subset of X. We must show that U contains a closed point. Since the U_{i} cover X, U must intersect some U_{i}. Then U ∩ U_{i} contains a distinguised open subset of U_{i}. So, to show that every open set contains a closed point, it suffices to show that every nonempty distinguised open subset of each U_{i} contains a closed points of X. Since a distinguished open subset (U_{i})_{x} of a U_{i} is also the spectrum of a finitely generated k-algebra Spec (A_{i})_{x} we can just add it to our collection {U_{i}}. The problem thus reduces to showing that each U_{i} contains a closed point.

Proposition 3 With the notation as above, if x ∈ U_{i} is closed in U_{i} (here U_{i} has the subspace topology) then x is closed in X.

Proof. Suppose x ∈ U_{j}. There is a natural injection U_{i} ∩ U_{j} ↪ U_{j}. Let Spec (B_{i})_{f} be a distinguished open subset of U_{i} contained in U_{i} ∩ U_{j} which contains x. Then we have a morphism Spec (B_{i})_{f} ↪ U_{j} = Spec B_{j}. We thus get a ring homomorphism φ : B_{j} → (B_{i})_{f} of Jacobson rings. Since it is induced by a restriction of the identity map X → X which is a morphism over k, φ is a k-algebra homomorphism. Since (B_{i})_{f} is a finitely generated k-algebra, (B_{i})_{f} is also a finitely generated B_{j}-algebra. Since x is closed in Spec (B_{i})_{f}, x is a maximal ideal of (B_{i})_{f}. Thus by page 132 of Eisenbud’s Commutative Algebra φ^{−1}(x) is a maximal ideal of B_{j}. Thus x is also a closed point of U_{j} in the subspace topology on U_{j}. Thus X − x = ∪_{i}(U_{i} − x) is a union of open subsets of X, hence open in X, so x is closed.

To finish we just need to know that U_{i} has a closed point.

Proposition 4 Let X = Spec A be an affine scheme with A a finitely generated k-algebra. Then any nonempty distinguished open subset of X contains a closed point.

Proof. The key observation is that A is a Jacobson algebra since it finitely generated over a field, so by page 131 of Eisenbud’s Commutative Algebra the Jacobson radical of A equals the nilradical of A. Let D(f) be a nonempty distinguised open subset of X. Then some prime omits f so f is not in the nilradical of A. Thus f is not in the Jacobson radical of A so there is some maximal ideal m so that f ∉ m. Then m ∈ D(f) and m is a closed point of X since m is maximal.

Strangely enough I never used the hypothesis that X is of finite type over k but just the weeker hypothesis that X is locally of finite type over k. Did I miss something?

Finally, we present a counterexample in the more general situation. Let X = Spec Z_{(2)}. Then X contains precisely one closed point, the ideal (2). So the set of closed points in X is not dense in X. In fact, if X is the spectrum of any DVR we also get a counterexample.

Problem 6 (4.2) Let S be a scheme, let X be a reduced scheme over S, and let Y be a seperated scheme over S. Let f_{1} and f_{2} be two S-morphisms of X to Y which agree on an open dense subset U of X. Show that f_{1} = f_{2}. Give examples to show that this result fails if either (a) X is nonreduced, or (b) Y is nonseparated.

Proof. Let g = (f_{1}, f_{2})_{S} : X → Y ×_{S} Y be the product of f_{1} and f_{2} over S. By hypothesis the diagonal T = Δ_{Y} (Y) is a closed subscheme of Y ×_{S} Y. Thus Z = g^{−1}(T) is a closed subscheme of X. If h : U → Y is the common restriction of f_{1} and f_{2} to U, then, since g|_{U} makes the correct diagram commute, the restriction of g to U is g′ = (h, h)_{S} and g′ = Δ_{Y} ∘ h since Δ_{Y} ∘ h makes the correct diagram commute.

Thus Δ_{Y}^{−1}(T) = Y implies

g^{−1}(T) ⊇ (g′)^{−1}(T) = h^{−1}(Δ_{Y}^{−1}(T)) = h^{−1}(Y) = U.

Thus g^{−1}(T) is a closed set which contains the dense set U. Thus g^{−1}(T) = X so g(X) ⊆ T. So, by the proposition below, since X is reduced, g factors as g = Δ_{Y} ∘ f where f : X → Y. From the definition of Δ_{Y}, we see that π_{1} ∘ Δ_{Y} = id_{Y} = π_{2} ∘ Δ_{Y}. Thus f_{1} = π_{1} ∘ g = π_{1} ∘ Δ_{Y} ∘ f = f and f_{2} = π_{2} ∘ g = π_{2} ∘ Δ_{Y} ∘ f = f so f = f_{1} = f_{2}, as desired.

Proposition 5 Let X be a reduced scheme, f : X → Y a morphism, Z a closed subscheme of Y, j : Z ↪ Y, such that f(X) ⊆ j(Z). Then f factors uniquely as

X →^{g} Z ↪^{j} Y.

Proof. First assume X and Y are affine, X = Spec A, Y = Spec B, Z = Spec B/I. (Use exercise 3.11 to see that every closed subscheme Z of Y is of the form Spec B/I.) Let φ : B → A be the homomorphism which induces f. Since f(X) ⊆ Z, the inverse image of any prime of A contains I. Since A is reduced the intersection of all primes of A equals {0}. Thus

ker(φ) = φ^{−1}({0}) = φ^{−1}(∩_{primesp}p) ⊆ I

so φ factors uniquely through B/I.

B →^{j^{#}} B/I ↪^{g^{#}} A

This proves the proposition when X and Y are affine.

Now suppose X is an arbitrary reduced scheme. Cover X by open affines X = ∪_{i}U_{i}. For each i let g_{i} be the unique map which factors f|_{U_{i}} through Z. By uniqueness g_{i}|_{U_{i}∩U_{j}} = g_{j}|_{U_{i}∩U_{j}}, so we can glue the g_{i} to obtain a morphism g : X → Z such that j ∘ g = f. Now suppose both X and Y are arbitrary. Cover Y by open affines, take their inverse images in X, perform the construction locally for each one, use uniqueness and glue. Counterexamples.

(a) Let A = k[x, y]/(x^{2}, xy), let X = Y = Spec A and let S = Spec k. Then Y is affine hence seperable over S, but X is not reduced. Let f : X → Y be the morphism induced by the identity homomorphism id : A → A. Let g : X → Y be the morphism induced by the homomorphism φ : A → A : x ↦ 0, y ↦ y. Let U = D(y) = Spec A_{y}. Then since A_{y} ≅ Spec k[y, y^{−1}], the localized homomorphisms agree, id_{y} = φ_{y}. Thus f|_{U} = g|_{U}. Now X is irreducible since A has just one minimal prime, namely (x), so U is dense in X. But, f ≠ g. since f^{#} = id ≠ φ = g^{#}.

(b) Let X be the affine line and Y the affine line with a doubled origin both thought of as schemes over S = Spec k. Let f_{1} : X → Y be one of the inclusions of the affine line in Y and let f_{2} : X → Y be the other one. Then f_{1} and f_{2} agree on X minus the origin but not on X.

[Reference, EGA, I.8.2.2.1.]

Problem 7 (4.4) Let f : X → Y be a morphism of separated schemes of finite type over a noetherian scheme S. Let Z be a closed subscheme of X which is proper over S. Show that f(Z) is closed in Y, and that f(Z) with its image subscheme structure is proper over S.

Proof. First we show that since X, Y and Z are of finite type over S and S is noetherian, X, Y and Z are noetherian. Suppose g : X → S is the map from X to S. Cover S by finitely many Spec A_{i}, A_{i} noetherian. Then for each f^{−1}(Spec A_{i}) = ∪_{j}Spec B_{ij}, with B_{ij} a finitely generated A_{i}-algebra. Since A_{i} is noetherian each B_{ij} is noetherian (this is the Hilbert basis theorem). Since X = ∪_{ij}Spec B_{ij}, X is noetherian. One shows that Y and Z are noetherian in exactly the same way.

Since the following diagram commutes 4.8(e) implies f|_{Z} is proper.

Thus f|_{Z}(Z) is closed in Y. (I’m assuming Z is an S-subscheme of X so that the diagram must commute.)

We now have f(Z) ↪ Y → S. We must show the composition f(Z) → S is proper. By Corollary 4.8a the closed immersion f(Z) ↪ Y is proper. We are given that the map Y → S is seperated and of finite type. Since the composition of seperated morphisms is seperated and the composition of finite type morphisms is of finite type, the morphism f(Z) → S is seperated and of finite type. The hard part is to show that it is universally closed.

Since the morphism Z → S is proper it is closed and from above the morphism Z → f(Z) is closed so the morphism f(Z) → S is closed. Let W be an any scheme over S. We must show that the map f(Z) ×_{S} W → W is closed. We have the following diagram.

If we can show that g_{1} is surjective we will be done. For then if A is closed subset of f(Z) ×_{S} W,

g_{2}(A) = g_{3}(g_{1}^{−1}(A)),

which, since g_{3} is closed, is also closed. (We wouldn’t have equality in the above expression if g_{1} weren’t surjective.)

In order to establish the surjectivity of g_{1} we prove that the property of being a surjective morphism is preserved under base extension. It will then follow, since g_{1} is a base extension of the surjective morphism Z → f(Z), that g_{1} is surjective.

Proposition 6 Let X and Y be schemes over S. Suppose x ∈ X and y ∈ Y both lie over the same point s ∈ S. Then there exists α ∈ X ×_{S} Y such that p_{X}(α) = x and p_{Y} (α) = y.

Proof. Let g_{1} : Spec (k(x)) → X and g_{2} : Spec (k(y)) → Y be the natural maps with g_{1}((0)) = x and g_{2}((0)) = y. Let Z = Spec (k(x))×_{Spec (k(s))} Spec (k(y)) = Spec (k(x)⊗_{k(s)}k(y)), and let π_{1} be the projection to Spec (k(x)), π_{2} the projection to Spec (k(y)). Let g = g_{1} ×_{S} g_{2} be the product of g_{1} ∘ π_{1} with g_{2} ∘ π_{2}. So g : Z → X ×_{S} Y. See the following diagram.

Since Z is the spectrum of the tensor product of two fields over a common base field, Z ≠ ∅ so there is some z ∈ Z. By the definition of g, g_{1}∘π_{1} = p_{x}∘g and g_{2} ∘ π_{2} = p_{y} ∘ g so x = g_{1} ∘ π_{1}(z) = p_{x} ∘ g(z) and y = g_{2} ∘ π_{2}(z) = p_{y} ∘ g(z) so we may take α = g(z).

Proposition 7 If f : X → Y is a surjective S-morphism then f × 1 : X ×_{S} S′ → Y ×_{S} S′ is surjective.

Proof. We have the following diagram.

Let y′ ∈ Y ×_{S} S′. Then q(y′) ∈ Y = f(X) so there is x ∈ X such that f(x) = q(y′). Then by the above proposition there is some α ∈ X ×_{S} S′ such that p(α) = x and (f × 1)(α) = y′. Thus f × 1 is surjective.