Fiber Product of Schemes

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition and Existence of Fiber Products

In §Morphisms of Schemes, ⁋Definition 3, we agreed to call a scheme morphism \(X \rightarrow S\) an \(S\)-scheme. In this post, we define the product in the category \(\Sch_{/S}\).

Definition 1 The fiber product of two scheme morphisms \(\varphi_X:X \rightarrow S\), \(\varphi_Y:Y \rightarrow S\) is denoted by \(X\times_SY\). (§Limits, ⁋Example 8 (Fiber product))

That is, \(X\times_SY\) satisfies the following property.

The following diagram

commutes. Moreover, whenever arbitrary \(\psi_X:Z \rightarrow X\), \(\psi_Y:Z \rightarrow Y\) satisfying the equation \(\varphi_Y\circ\psi_Y=\varphi_X\circ\psi_X\) are given, there exists a unique \(\psi:Z \rightarrow X\times_SY\) such that \(\psi_X=\rho_X\circ\psi\) and \(\psi_Y=\rho_Y\circ\psi\).

Therefore, there exists a canonical morphism from \(X\times_SY\) to \(S\), and from this we may view \(X\times_SY\) as an \(S\)-scheme. Moreover, from this perspective it is immediate from the definition that \(X\times_SY\) is also the product in \(\Sch_{/S}\).

After §Morphisms of Schemes, ⁋Example 4, we saw that an arbitrary scheme \(X\) can always be regarded as a \(\mathbb{Z}\)-scheme in a unique way. Thus, assuming that a fiber product \(X\times_SY\) satisfying Definition 1 always exists, we know that for any two schemes \(X, Y\), the product \(X\times_{\Spec \mathbb{Z}}Y\) gives the product of \(X\) and \(Y\).

Since Definition 1 guarantees nothing about the existence of the fiber product \(X\times_SY\), for this to be a genuine definition we must separately prove the existence of \(X\times_SY\). (Theorem 8) However, the existence of fiber products in \(\AffSch\) is almost obvious, and this will be the starting point of our proof.

Lemma 2 Given morphisms \(\Spec A \rightarrow \Spec C\), \(\Spec B \rightarrow\Spec C\) between affine schemes, we have

\[\Spec A\times_{\Spec C}\Spec B\cong\Spec (A\otimes_C B).\]

Proof

Using the equivalence \(\AffSch\cong\cRing^\op\), convert \(\Spec A \rightarrow \Spec C\), \(\Spec B \rightarrow \Spec C\) into \(C \rightarrow A\), \(C \rightarrow B\), and compare the universal property of [Algebraic Structures] §Direct Products, Direct Sums, Tensor Products of Algebras, ⁋Theorem 8 with the universal property of the fiber product.

To show that fiber products exist for general schemes, it now suffices to show that we can glue together the results for affine schemes examined in Lemma 2.

First, when an open subscheme \(U\) of \(Z\) is given, writing it in the form \(\iota:U \rightarrow Z\) using the inclusion morphism, the following lemma is almost a tautology.

Lemma 3 Given a scheme morphism \(\varphi: Y \rightarrow Z\) and an open subscheme \(\iota: U \rightarrow Z\) of \(Z\), the following diagram

is a fiber diagram.

Proof

\(\varphi^{-1}(U)\) satisfies the universal property of the fiber product.

Now, using this slightly, we can prove the following lemma.

Lemma 4 Given affine schemes \(X, Y, Z\) and an open subscheme \(Y'\hookrightarrow Y\) of \(Y\), the fiber product \(X\times_ZY'\) of \(X\rightarrow Z\) and \(Y'\hookrightarrow Y \rightarrow Z\) exists.

Proof

First, from Lemma 2 we know that the following fiber diagram

exists. Now, considering the following data

we can check from Lemma 3 that the open subscheme \(\rho_Y^{-1}(Y')\) of \(X\times_SY\) becomes a fiber product. Now, in general, if the two small squares in the following diagram

are fiber diagrams, then the outer large square is also a fiber diagram, so we obtain the desired result.

Now, using this, we can show that the fiber product of an affine scheme and an arbitrary scheme exists.

Lemma 5 For affine schemes \(X, Z\) and an arbitrary scheme \(Y\), the fiber product \(X\times_ZY\) of \(X\rightarrow Z\) and \(Y \rightarrow Z\) exists.

Proof

For this, cover \(Y\) by affine open subsets \(Y_i\). Then we know from Lemma 2 that the \(X\times_ZY_i\) exist. Also, since \(Y_{ij}=Y_i\cap Y_j\) is an open subscheme of the affine scheme \(Y_i\), the \(X\times_Z Y_{ij}\) also exist by Lemma 4.

On the other hand, looking at the proof of Lemma 4, we see that \(X\times_ZY_{ij}\) is an open subscheme of each of \(X\times_ZY_i\) and \(X\times_ZY_{ij}\). We can easily check that these data satisfy the conditions of §Schemes, ⁋Lemma 9, so we can glue them together to form a scheme \(X\times_ZY\). That this satisfies the universal property of the fiber product can be checked by restricting the codomain of a scheme morphism \(W \rightarrow Y\) to the \(Y_i\), using the universal property of each \(X\times_ZY_i\), and then gluing the scheme morphisms together as in §Morphisms of Schemes, ⁋Proposition 1.

In this lemma, the assumption that \(X\) is an affine scheme was used only to show that \(X\times_ZY_i\) exists. Therefore, given arbitrary schemes \(X,Y\), an affine scheme \(Z\), and scheme morphisms \(X \rightarrow Z\) and \(Y \rightarrow Z\), we choose an affine open cover \(\{Y_i\}\) of \(Y\), then know that \(X\times_ZY_i\) exists by Lemma 5, and thus can glue them together to form \(X\times_ZY\). That is, the following holds.

Lemma 6 For an affine scheme \(Z\), arbitrary schemes \(X,Y\), and scheme morphisms \(X \rightarrow Z\), \(Y \rightarrow Z\), the fiber product \(X\times_ZY\) exists.

Finally, we must extend \(Z\) to an arbitrary scheme. First, the following holds.

Lemma 7 Given arbitrary schemes \(X,Y,Z\), scheme morphisms \(\varphi_X:X \rightarrow Z\), \(\varphi_Y:Y \rightarrow Z\), and a morphism \(\iota: Z \rightarrow Z'\) to an affine scheme \(Z'\), the fiber product \(X\times_{Z'}Y\) of \(\iota\circ\varphi_X\) and \(\iota\circ\varphi_Y\) satisfies the universal property of \(X\times_ZY\), and therefore \(X\times_ZY\) exists.

Now, using the above lemma, for arbitrary \(X,Y,Z\) and scheme morphisms \(\varphi_X:X \rightarrow Z\), \(\varphi_Y: Y \rightarrow Z\), if we cover \(Z\) by an affine open cover \(\{Z_i\}\), we know that fiber products \(X_i\times_{Z_i}Y_i\) exist for \(\varphi_X\vert^{Z_i}:\varphi_X^{-1}(Z_i) \rightarrow Z_i\) and \(\varphi_Y\vert^{Z_i}:\varphi_Y^{-1}(Z_i) \rightarrow Z_i\). Now, the intersection \(Z_{ij}=Z_i\cap Z_j\) is an open subset of \(Z_i\), so by Lemma 7 the fiber products of \(\varphi_X\vert^{Z_{ij}}\) and \(\varphi_Y\vert^{Z_{ij}}\) also exist and are open subschemes of \(X_i\times_{Z_i}Y_i\) and \(X_j\times_{Z_j}Y_j\). Therefore, just as in the proof of Lemma 5, if we show that these data satisfy the conditions of §Schemes, ⁋Lemma 9, we obtain the following theorem.

Theorem 8 For arbitrary schemes \(X,Y,Z\) and scheme morphisms \(X \rightarrow Z\), \(Y \rightarrow Z\), the fiber product \(X\times_ZY\) exists.

Interpretations of the Fiber Product

Just as there are various ways to interpret a scheme morphism, there are various ways to understand the fiber product.

Earlier, we agreed to think of a scheme morphism \(X \rightarrow S\) as a family parametrized by \(S\) (§Morphisms of Schemes, ⁋Example 10), and from this perspective \(S\) can be thought of as the base of the family \(X\). Now, given an arbitrary \(S\)-family \(X \rightarrow S\) and a scheme morphism \(S' \rightarrow S\), through the fiber product we obtain a new \(S'\)-family \(X\times_SS' \rightarrow S'\). From this perspective, we often call the fiber product a base change.

Example 9 Narrowing our scope to affine schemes, the fact that \(\Spec B\) is a \(C\)-scheme means that a scheme morphism \(\Spec B \rightarrow \Spec C\) is given, which in turn means that a ring homomorphism \(C \rightarrow B\) is given, which is the same as saying that \(B\) is a \(C\)-algebra.

Now, additionally given a scheme morphism \(\Spec A \rightarrow \Spec C\), let us see what the above base change gives. By Lemma 2, we know that what is obtained in this way is

\[\Spec A\times_{\Spec C}\Spec B=\Spec(A\otimes_CB) \rightarrow \Spec A\]

that is, the ring homomorphism \(A \rightarrow A\otimes_CB\). In other words, base change is (in the case of affine schemes) nothing other than §Change of Base Ring, ⁋Definition 3.

In particular, from the fact that for a \(B\)-algebra \(B[\x_1,\ldots,\x_n]\) and an arbitrary ring homomorphism \(B \rightarrow A\), the following equation

\[A\otimes_BB[\x_1,\ldots,\x_n]\cong A[\x_1,\ldots, \x_n]\]

holds, we know that the following diagram

is a fiber diagram.

This perspective is important, but for now the geometric intuition here is not very visible. For this, let us think specifically of the case where \(S' \rightarrow S\) is an embedding.

First, given an arbitrary \(S\)-family \(X \rightarrow S\) and an open embedding \(S' \rightarrow S\), Lemma 3 shows that the \(S'\)-family \(X\times_SS' \rightarrow S'\) is simply obtained by restricting the base of \(X \rightarrow S\) to \(S'\). Moreover, if we also assume that \(X \rightarrow S\) is an open embedding, we know that \(X\times_SS'\) is the intersection of \(X\) and \(S'\) (inside \(S\)).

The above argument also holds for closed embeddings. For this, we must show the following lemma corresponding to Lemma 3.

Lemma 10 For a ring homomorphism \(\phi: B \rightarrow A\) and an arbitrary ideal \(\mathfrak{b}\) of \(B\), there exists an isomorphism

\[A/\phi(\mathfrak{b})A\cong A \otimes_B(B/\mathfrak{b}).\]

Proof

Taking \(\otimes_BA\) of the following exact sequence obtained from the ideal \(\mathfrak{b}\)

\[\mathfrak{b} \rightarrow B \rightarrow B/\mathfrak{b} \rightarrow 0\]

gives the exact sequence

\[A\otimes_B \mathfrak{b} \rightarrow A\otimes_BB \rightarrow A\otimes_B (B/\mathfrak{b}) \rightarrow 0\]

and since the image of \(A\otimes_B \mathfrak{b}\) in \(A\otimes_BB\cong A\) is \(\phi(\mathfrak{b})A\), we obtain the desired result.

Now, an arbitrary closed embedding locally always comes from some \(B \rightarrow B/\mathfrak{b}\), so the above discussion can be applied identically to closed embeddings as well. In particular, the intersection of two closed embeddings is well-defined.

Example 11 Consider the two closed subschemes of \(Z=\Spec\mathbb{K}[\x,\y]\)

\[X=\Spec \mathbb{K}[\x,\y]/(\y)=\Spec \mathbb{K}[\x],\qquad Y=\Spec \mathbb{K}[\x,\y]/(\x)=\Spec \mathbb{K}[\y].\]

Then \(X\) and \(Y\) correspond to the \(\x\)-axis and \(\y\)-axis of \(Z=\mathbb{A}^2_\mathbb{K}\) respectively, and their closed embeddings are given by the projections

\[\mathbb{K}[\x,\y] \rightarrow \mathbb{K}[\x],\qquad \mathbb{K}[\x,\y] \rightarrow \mathbb{K}[\y].\]

Now, \(X\times_ZY\) is, by Lemma 2, given by

\[\Spec\left(\frac{\mathbb{K}[\x,\y]}{(\x)}\otimes_{\mathbb{K}[\x,\y]} \frac{\mathbb{K}[\x,\y]}{(\y)}\right)\cong \Spec \mathbb{K}[\x,\y]/(\x,\y)\cong\Spec \mathbb{K}\]

which can be checked to correspond exactly to the origin, the intersection point of the \(\x\)-axis and the \(\y\)-axis.

Now, let us replace \(Y\) in the above computation with the following closed subscheme

\[Y=\Spec \mathbb{K}[\x,\y]/(\y-\x^2).\]

The intersection of \(\y=\x^2\) and the \(\x\)-axis is again the origin, but this time a double root exists, so the scheme structure must be given differently from above. Indeed, repeating the computation, \(X\times_ZY\) becomes

\[\Spec\left(\frac{\mathbb{K}[\x,\y]}{(\y)}\otimes_{\mathbb{K}[\x,\y]}\frac{\mathbb{K}[\x,\y]}{(\y-\x^2)}\right)\cong\Spec \mathbb{K}[\x,\y]/(\y,\y-\x^2)\cong\Spec \mathbb{K}[\x]/(\x^2).\]

From this perspective, we can also see how to define the fiber \(\varphi^{-1}(y_0)\) of a scheme morphism \(\varphi:X \rightarrow Y\) at a point \(y_0\in Y\). Whether \(y_0\) is a closed point or not, viewing it as \(\iota:\{y_0\}\hookrightarrow Y\) and taking the fiber product with \(\varphi\) suffices. For this, we must describe \(\iota\) as a scheme morphism.

For this, consider the residue field \(\kappa(y)\) at \(y\). Then \(\Spec\kappa(y)\) is always a one-point set. Moreover, considering an affine open subset \(V=\Spec B\) of \(Y\) containing \(y\), and assuming that \(y\) corresponds to the prime ideal \(\mathfrak{q}_y\), the canonical morphism

\[B \rightarrow B_{\mathfrak{q}_y} \rightarrow B_{\mathfrak{q}_y}/\mathfrak{q}_y B_{\mathfrak{q}_y} =\kappa(\mathfrak{q}_y)=\kappa(y)\]

defines \(\Spec\kappa(y)\rightarrow \Spec B\), and the (unique) point \((0)\) of \(\Spec \kappa(y)\) is mapped to \(\mathfrak{q}_y\) via the above morphism. Therefore, we define the following.

Definition 12 For a scheme morphism \(\varphi: X \rightarrow Y\), the fiber at a point \(y\in Y\) is defined as

\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y).\]

If \(Y\) is irreducible, the fiber at the generic point of \(Y\) is called the generic fiber.

Example 13 For an algebraically closed field \(\mathbb{K}\), define the ring homomorphism \(\mathbb{K}[\x] \rightarrow \mathbb{K}[\y]\) by the formula \(\x \mapsto \y^2\), and consider the scheme morphism \(\varphi: \Spec \mathbb{K}[\y] \rightarrow \Spec \mathbb{K}[\x]\) obtained from it. Then the residue field at an arbitrary point \((\x-a)\) of \(\Spec\mathbb{K}[\x]\) is

\[\Frac(\mathbb{K}[\x]/(\x-a))=\mathbb{K}[\x]/(\x-a).\]

Now, for arbitrary \(a\in \mathbb{K}\),

\[\varphi^{-1}((\x-a))=\Spec \mathbb{K}[\y]\otimes_{\Spec \mathbb{K}[\x]}\Spec \mathbb{K}[\x]/(\x-a)\cong \Spec(\mathbb{K}[\y]\otimes_{\mathbb{K}[\x]}\mathbb{K}[\x]/(\x-a))=\Spec \mathbb{K}[\y]/(\y^2-a)\]

and therefore if \(a=0\) then \(\varphi^{-1}((\x))\cong\Spec \mathbb{K}[\y]/(\y^2)\), and if \(a\neq 0\) then from the assumption that \(\mathbb{K}\) is algebraically closed we know that

\[\Spec \mathbb{K}[\y]/(\y^2-a)\cong \Spec \mathbb{K}[\y]/(\y-\sqrt{a})\coprod \Spec \mathbb{K}[\y]/(\y+\sqrt{a}).\]

On the other hand, for the generic point \((0)\) of \(\mathbb{K}[\x]\), since \(\kappa((0))=\mathbb{K}(\x)\), we have

\[\varphi^{-1}((0))=\Spec \mathbb{K}[\y]\otimes_{\Spec \mathbb{K}[\x]}\Spec \mathbb{K}(\x)\cong \Spec\mathbb{K}(\y).\]

The above example is what we already examined in §Properties of Scheme Morphisms, ⁋Example 15. In that example, we claimed that a finite morphism is always quasi-finite, and now we can prove this.

Proposition 14 A finite morphism \(\varphi: X \rightarrow Y\) is a quasi-finite morphism.

Proof

It suffices to show the affine case. That is, it suffices to show that for an arbitrary finite ring homomorphism \(\phi: B \rightarrow A\) and a prime ideal \(\mathfrak{q}\) of \(B\), \(A\otimes_B\kappa(\mathfrak{q})\) has only finitely many prime ideals. Since \(\phi\) is finite, \(A\otimes_B\kappa(\mathfrak{q})\) is a finite \(\kappa(\mathfrak{q})\)-algebra and hence artinian, from which we obtain the desired result. (§The Jordan-Hölder Theorem, ⁋Theorem 4)

From the above examples and propositions, we can make an important observation: if \(X \rightarrow S\) satisfies some property \(P\) of scheme morphisms, then the base change \(X\times_SS' \rightarrow S'\) via an arbitrary \(S' \rightarrow S\) also satisfies it. This is no accident; in fact, most properties of interest to us are closed under base change.

Proposition 15 If a scheme morphism \(\varphi:X \rightarrow Z\) is quasicompact (resp. quasiseparated, affine, finite, integral, locally of finite type, finite type, locally of finite presentation, finite presentation, quasi-finite, surjective), then the base change \(X\times_ZY \rightarrow X\) of \(\varphi\) via an arbitrary scheme morphism \(Y \rightarrow Z\) is also such.

For instance, for integral morphisms and finite morphisms, this was proved in §Integral Extensions, ⁋Proposition 14, and for the other properties the above proposition can be shown without difficulty.