Fiber Products

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. 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 \(X\times_SY\). ([Category Theory] §Limits, ⁋Example 8)

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 is a canonical morphism from \(X\times_SY\) to \(S\), and from this we can regard \(X\times_SY\) as an \(S\)-scheme. Moreover, from this viewpoint 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 any scheme \(X\) can always be thought of 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 object \(X\times_{\Spec \mathbb{Z}}Y\) gives the product of \(X\) and \(Y\).

Since Definition 1 does not guarantee anything about the existence of the fiber product \(X\times_SY\), for this to become a genuine definition we must prove the existence of \(X\times_SY\) separately. (Theorem 6) However, the existence of fiber products in \(\AffSch\) is almost obvious, and this will be the starting point of our proof.

Lemma 2 Let morphisms of affine schemes \(\Spec A \rightarrow \Spec C\), \(\Spec B \rightarrow\Spec C\) be given. Then

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

holds.

Proof

Via the equivalence \(\AffSch\cong\cRing^\op\), replace \(\Spec A \rightarrow \Spec C\), \(\Spec B \rightarrow \Spec C\) with \(C \rightarrow A\), \(C \rightarrow B\) and compare the universal property of ##ref## with the universal property of the fiber product.

Now, the fact that fiber products exist for general schemes follows from the result for affine schemes in Lemma 2: we need only show that these can be glued together well.

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 Let a scheme morphism \(\varphi: Y \rightarrow Z\) and an open subscheme \(\iota: U \rightarrow Z\) of \(Z\) be given. Then the following diagram

is a fiber diagram.

Proof

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

Now, by exploiting this slightly we can prove the following lemma.

Lemma 4 Let affine schemes \(X, Y, Z\) be given, and let an open subscheme \(Y'\hookrightarrow Y\) of \(Y\) be given. Then 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 \(X\times_SY\) is the 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 fiber product \(X\times_Z Y_{ij}\) also exists by Lemma 4.

On the other hand, looking at the proof of Lemma 4 we can see that \(X\times_ZY_{ij}\) is an open subscheme of each of \(X\times_ZY_i\) and \(X\times_ZY_j\). We can easily check that these data satisfy the conditions of §Schemes, ⁋Lemma 9, so we can glue them to construct 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 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. Hence, if any two schemes \(X,Y\) and an affine scheme \(Z\) are given, together with scheme morphisms \(X \rightarrow Z\) and \(Y \rightarrow Z\), we can choose an affine open cover \(\{Y_i\}\) of \(Y\), then know that \(X\times_ZY_i\) exists by Lemma 5, and thus glue them to construct \(X\times_ZY\). 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.

Now finally we must extend to the case where \(Z\) is an arbitrary scheme. First, the following holds.

Lemma 7 Let arbitrary schemes \(X,Y,Z\) be given, and let scheme morphisms \(\varphi_X:X \rightarrow Z\), \(\varphi_Y:Y \rightarrow Z\) and a morphism \(\iota: Z \rightarrow Z'\) to an affine scheme \(Z'\) be given. Then 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 \(\{Z_i\}\) then 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, in the same way 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.

Interpretation of Fiber Products

Just as there are several ways to interpret a scheme morphism, there are also several ways to understand fiber products.

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 viewpoint \(S\) 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 point of view we often call the fiber product a base change.

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

Now, furthermore, let a scheme morphism \(\Spec A \rightarrow \Spec C\) be given and let us see what the above base change yields; 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 [Algebraic Structures] §Change of Base Ring, ⁋Definition 3.

In particular, for a \(B\)-algebra \(B[\x_1,\ldots,\x_n]\) and an arbitrary ring homomorphism \(B \rightarrow A\), from the fact that 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 point of view is important, but for now the geometric intuition here is not easy to see. For this, let us consider the case where \(S' \rightarrow S\) is an embedding in particular.

First, for an arbitrarily given \(S\)-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 \(S'\). Furthermore, assuming that \(X \rightarrow S\) is also an open embedding, we know that \(X\times_SS'\) is the intersection (inside \(S\)) of \(X\) and \(S'\).

The above argument also works for closed embeddings. For this we need to 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

From the ideal \(\mathfrak{b}\) we obtain the following exact sequence

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

and taking \(\otimes_BA\) gives the following 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, since any closed embedding locally always comes from \(B \rightarrow B/\mathfrak{b}\), the above discussion applies equally to closed embeddings. 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

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

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

Now let us replace \(Y\) in the above computation by 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 point of view we can also see how the fiber \(\varphi^{-1}(y_0)\) of a scheme morphism \(\varphi:X \rightarrow Y\) at a point \(y_0\in Y\) should be defined. Whether \(y_0\) is a closed point or not, viewing it as \(\iota:\{y_0\}\hookrightarrow Y\) and taking the fiber product of \(\iota\) and \(\varphi\) 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 \(\mathfrak{q}_y\), through the canonical morphism

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

the morphism \(\Spec\kappa(y)\rightarrow \Spec B\) is defined 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 by

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

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 resulting scheme morphism \(\varphi: \Spec \mathbb{K}[\y] \rightarrow \Spec \mathbb{K}[\x]\). 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 \(\mathbb{K}[\x]\) we have \(\kappa((0))=\mathbb{K}(\x)\), so

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

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 \(B\), the tensor product \(A\otimes_B\kappa(\mathfrak{q})\) has only finitely many prime ideals. But 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. (##ref##)

From the above example 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 not a coincidence; in fact, most properties we are interested in 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 we proved this in [Commutative Algebra] §Integral Extensions, ⁋Proposition 14, and for the other properties the above proposition can be shown without difficulty.