Properties of Scheme Morphisms

In the previous post, we explored several perspectives for understanding scheme morphisms. In this post, we formally define properties of scheme morphisms. First, we define the following property shared by them.

Definition 1 A property \(P\) of scheme morphisms is said to be local on target if the following two conditions hold:

1. If a scheme morphism \(\varphi:X \rightarrow Y\) satisfies \(P\), then on \(Y\), for any open subscheme \(V\), the scheme morphism \(\varphi\vert_{\varphi^{-1}(V)}: \varphi^{-1}(V) \rightarrow V\) also satisfies \(P\).
2. If, for a scheme morphism \(\varphi:X \rightarrow Y\), \(Y\) admits an open covering \(\{V_j\}\) such that the scheme morphisms \(\varphi\vert_{\varphi^{-1}(V_j)}: \varphi^{-1}(V_j) \rightarrow V_j\) all satisfy \(P\), then so does \(\varphi\).

Schemes are built from affine schemes. If a property \(P\) of scheme morphisms is local on target, for a scheme morphism \(\varphi:X \rightarrow Y\) we may assume that the target \(Y\) is \(\Spec B\), and then via the adjunction

\[\Hom_\Sch(X, \Spec B)\cong \Hom_\cRing(B, \Gamma(X, \mathcal{O}_X))\]

we can always reduce to the case where the target is affine.

Quasi-compact and Quasi-separated Morphisms

Definition 2 A scheme morphism \(\varphi: X \rightarrow Y\) is quasi-compact if for every affine open subset \(V\subseteq Y\), the preimage \(\varphi^{-1}(V)\) is quasi-compact.

Proposition 3 A scheme morphism \(\varphi: X \rightarrow Y\) is quasi-compact if and only if the preimage of every quasi-compact open subset of \(Y\) is quasi-compact.

Proof

Since every affine scheme is quasi-compact (§The Spectrum, ⁋Lemma 12), it is obvious that the given condition implies the condition of Definition 2.

Conversely, suppose a quasi-compact morphism \(\varphi: X \rightarrow Y\) is given. Now, if for \(Y\) an arbitrary quasi-compact open subset \(V\) is given, there exists a covering of \(V\) by finitely many affine open subsets \(\{V_j\}\), and their preimages \(\varphi^{-1}(V_j)\) are all quasi-compact. Now,

\[\varphi^{-1}(V)=\varphi^{-1}\left(\bigcup_{j\in J} V_j\right)=\bigcup_{j\in J}\varphi^{-1}(V_j)\]

and since a finite union of quasi-compact sets is again quasi-compact, we obtain the desired result.

Then from the equivalence in Proposition 3, we see that the composition of quasi-compact morphisms is again quasi-compact. Moreover, the following holds.

Proposition 4 For a Noetherian scheme \(X\), a scheme morphism \(\varphi: X \rightarrow Y\) is always quasi-compact.

Proof

Suppose an affine open subset \(V\subseteq Y\) is given; we must show that \(\varphi^{-1}(V)\) is quasi-compact. However, by [Topology] §Dimension, ⁋Proposition 12 and the first result of [Topology] §Dimension, ⁋Proposition 13, any subspace of a Noetherian topological space is quasi-compact.

Similarly, we define quasi-separated morphisms. To this end, we first define quasi-separated schemes.

Definition 5 A scheme \(X\) is quasi-separated if the intersection of any two quasi-compact open subsets of \(X\) is again quasi-compact. A scheme morphism \(\varphi: X \rightarrow Y\) is quasi-separated if for every affine open set \(V\subseteq Y\), \(\varphi^{-1}(V)\) is quasi-separated.

Then the following holds.

Proposition 6 A locally Noetherian scheme is always quasi-separated.

Proof

Let \(X\) be a locally Noetherian scheme, and let \(V_1=\Spec B_1, V_2=\Spec B_2\) be any two affine open subsets; we must show that \(V_1\cap V_2\) is quasi-compact.

First, since \(X\) is locally Noetherian, we can cover \(X\) by spectra \(U_i=\Spec A_i\) of Noetherian rings. Now for each \(i\), by §The Topology of Schemes, ⁋Lemma 11 (Nike), we can cover \(U_i\cap V_1\) by spectra \(\Spec (A_i)_g\) of Noetherian rings. Collecting all of these covers \(V_1\) by spectra of Noetherian rings, and by §The Spectrum, ⁋Lemma 12, \(V_1=\Spec B_1\) is covered by finitely many spectra of Noetherian rings. Therefore, by §The Topology of Schemes, ⁋Lemma 13, \(B_1\) is a Noetherian ring, and hence \(V_1=\Spec B_1\) is Noetherian. Furthermore, since any subspace of a Noetherian topological space is quasi-compact by [Topology] §Dimension, ⁋Proposition 12 and the first result of [Topology] §Dimension, ⁋Proposition 13, in particular \(V_1\cap V_2\) is also quasi-compact. By the same reasoning, any affine open of \(X\) is Noetherian, and since a quasi-compact open is a union of finitely many affine opens, it is also Noetherian. Since a subspace of a Noetherian topological space is quasi-compact, the intersection of any two quasi-compact opens is also quasi-compact, and thus by Definition 5, \(X\) is quasi-separated.

Then quasi-compactness and quasi-separatedness not only satisfy the properties of Definition 1, but are also affine-local on target, as can be seen in the following proposition. (§The Topology of Schemes, ⁋Definition 9)

Proposition 7 For a scheme morphism \(\varphi: X \rightarrow Y\), the following hold.

1. If \(Y\) has an affine open covering \(\{V_j\}\) such that each \(\varphi^{-1}(V_j)\) is quasi-compact, then \(\varphi\) is quasi-compact.
2. If \(Y\) has an affine open covering \(\{V_j\}\) such that each \(\varphi^{-1}(V_j)\) is quasi-separated, then \(\varphi\) is quasi-separated.

Proof

1. Let an arbitrary affine open subset \(V\) of \(Y\) be given. Then by §The Topology of Schemes, ⁋Lemma 11 (Nike), we can cover \(V\cap V_j\) by open sets that are principal open sets in both \(V\) and \(V_j\), and considering this for all \(j\) and using the quasi-compactness of \(V\), we can choose only finitely many such sets. Let this be \(V=\bigcup W_l\).
   On the other hand, for each \(l\), choosing \(V_{j(l)}\) that has \(W_l\) as a principal open subset, since \(\varphi^{-1}(V_{j(l)})\) is quasi-compact, we can cover it by finitely many affine open subsets \(U_{j(l)k}\). Now since \(\varphi^{-1}(W_l)\cap U_{j(l)k}\) is a principal open set of \(U_{j(l)k}\) by §The Spectrum, ⁋Proposition 8, each \(\varphi^{-1}(W_l)\) can be expressed as a finite union of affine open sets, and hence \(\varphi^{-1}(V)\) can also be expressed as a finite union of affine open sets. Since a finite union of quasi-compact spaces is quasi-compact, we obtain the desired result.

2. First, a scheme \(Z\) is quasi-separated if and only if the intersection of any two affine open subsets of \(Z\) is quasi-compact. Since affine schemes are quasi-compact, one direction is trivial; conversely, since any quasi-compact open subset of \(Z\) is a union of finitely many affine open subsets, the intersection of two quasi-compact open subsets is the union of finitely many intersections of affine open sets, and is therefore quasi-compact.
   Now, as in the proof of the first statement, choose a finite covering \(V=\bigcup_{l=1}^n W_l\) of \(V\) such that each \(W_l\) is a principal open subset in both \(V\) and a suitable \(V_{j(l)}\). Given two affine open subsets \(U_1,U_2\) of \(\varphi^{-1}(V)\), we have

   \[U_1\cap U_2=\bigcup_{l=1}^n\left(U_1\cap \varphi^{-1}(W_l)\right)\cap\left(U_2\cap \varphi^{-1}(W_l)\right)\]

   Here, since \(W_l\) is a principal open subset of \(V\), \(U_1\cap \varphi^{-1}(W_l)\) is the principal open set defined on the affine scheme \(U_1\) by the pullback of the function defining \(W_l\) (§The Spectrum, ⁋Proposition 8), and hence is affine; the same holds for \(U_2\). But these are all affine open subsets of \(\varphi^{-1}(W_l)\subseteq \varphi^{-1}(V_{j(l)})\), and since \(\varphi^{-1}(V_{j(l)})\) is quasi-separated, each intersection is quasi-compact by the above criterion. From the above, \(U_1\cap U_2\) is quasi-compact as it is a union of finitely many quasi-compact sets, and again by the above criterion, \(\varphi^{-1}(V)\) is quasi-separated.

Affine Morphisms

From the adjunction

\[\Hom_\Sch(X, \Spec B)\cong\Hom_\cRing (B, \Gamma(X, \mathcal{O}_X))\]

we know that in particular when \(X=\Spec A\),

\[\Hom_\Sch(\Spec A,\Spec B)\cong\Hom_\cRing (B, A)\]

holds. (§Affine Scheme, ⁋Proposition 11) Therefore, when examining properties of scheme morphisms that are affine-local on target as above, it would be desirable if for any affine open subset \(V\cong\Spec B\) of \(Y\), \(U=\varphi^{-1}(V)\) were also an open subscheme \(U\cong \Spec A\) of \(X\), so that \(\varphi\vert_U: U \rightarrow V\) becomes a morphism between affine schemes and we could obtain this property from the ring homomorphism

\[(\varphi\vert_U)^\sharp(V): \mathcal{O}_V(V) \rightarrow \mathcal{O}_U(U)\]

However, of course, for an arbitrary scheme morphism \(\varphi: X \rightarrow Y\), the preimage of an affine open subset of \(Y\) need not be affine. (§Schemes, ⁋Example 8)

Definition 8 A scheme morphism \(\varphi: X \rightarrow Y\) is an affine morphism if for any affine open subset \(V\) of \(Y\), \(\varphi^{-1}(V)\) is an affine open subset of \(X\).

Then it is clear that the composition of affine morphisms is affine. Furthermore, this property also satisfies the conditions of Definition 1, the proof of which is given in the next proposition.

Proposition 9 For a scheme morphism \(\varphi:X \rightarrow Y\), if there exists an affine open covering \(\{V_j\}\) of \(Y\) such that each \(\varphi^{-1}(V_j)\) is affine, then \(\varphi\) is affine.

Proof

For an affine open subset \(\Spec B\) of \(Y\), define the property \(P\) by “\(\varphi^{-1}(\Spec B)\) is an affine open subset of \(X\)”. Then \(\varphi\) being affine means that every affine open subset of \(Y\) satisfies \(P\), and the given assumption is that some affine open covering of \(Y\) satisfies \(P\). Thus, by the equivalence between the second and third conditions of §The Topology of Schemes, ⁋Lemma 12, it suffices to show that \(P\) is an affine-local property in the sense of §The Topology of Schemes, ⁋Definition 9. In what follows, we write the component of the sheaf morphism \(\varphi^\sharp: \mathcal{O}_Y \rightarrow \varphi_\ast \mathcal{O}_X\) on an open set \(V\) as \(\varphi^\sharp(V): \mathcal{O}_Y(V) \rightarrow \mathcal{O}_X(\varphi^{-1}(V))\).

First, let us verify the first condition of §The Topology of Schemes, ⁋Definition 9. If \(\varphi^{-1}(\Spec B)\) is an affine open subset \(\Spec A\), then by §Affine Scheme, ⁋Proposition 11, \(\Spec A \rightarrow \Spec B\) is of the form \(\Spec\phi\) for some ring homomorphism \(\phi: B \rightarrow A\), and from the formula

\[(\Spec\phi)^{-1}(D(f))=D(\phi(f))\]

obtained in the proof of §The Spectrum, ⁋Proposition 8, for any \(f\in B\) we have \(\varphi^{-1}(\Spec B_f)=D(\phi(f))\cong\Spec A_{\phi(f)}\), so \(\Spec B_f\) also satisfies \(P\).

Now we must check the second condition. That is, assuming \(B=(f_1,\ldots, f_r)\) and that each \(U_i=\varphi^{-1}(D(f_i))\) is affine, we must show that \(U=\varphi^{-1}(\Spec B)\) is affine. For convenience, let \(R=\Gamma(U, \mathcal{O}_X)\), and let the image of \(f_i\) under \(\varphi^\sharp(\Spec B): B \rightarrow R\) be \(g_i\in R\). Also, for \(g\in \Gamma(U, \mathcal{O}_X)\), for points at which the stalk of \(g\) does not belong to the maximal ideal of \(\mathcal{O}_{X,x}\), we write the set of such points \(x\) as \(U_g\). Then for any affine open subset of \(U\), say \(\Spec A\), it is clear by definition that \(U_g\cap \Spec A=D(g\vert_{\Spec A})\), and in particular \(U_g\) is open.

We make the following three observations. First, since \(B=(f_1,\ldots, f_r)\), we have \(\Spec B=\bigcup_{i=1}^rD(f_i)\), and hence \(\{U_i\}_{i=1}^r\) is a finite affine open covering of \(U\). Furthermore, since \(1=\sum_{i=1}^rb_if_i\) for some \(b_i\in B\), applying \(\varphi^\sharp(\Spec B)\) shows that \(g_1,\ldots, g_r\) generate the unit ideal of \(R\). Second, since \(\varphi\) is a morphism of locally ringed spaces, at each \(x\in U\) the map \(\varphi^\sharp_x:\mathcal{O}_{Y,\varphi(x)} \rightarrow \mathcal{O}_{X,x}\) is a local homomorphism, and hence \(\varphi(x)\in D(f_i)\) is equivalent to \(x\in U_{g_i}\), so \(U_i=U_{g_i}\). Third, writing \(U_i\cong \Spec A_i\) and letting the ring homomorphism corresponding to \(\Spec A_i \rightarrow \Spec B_{f_i}\) be \(\phi_i\), since \(D(f_j)\cap D(f_i)\) is a principal open set of \(\Spec B_{f_i}\), applying §The Spectrum, ⁋Proposition 8 as above shows that \(U_i\cap U_j\) is a principal open set of \(\Spec A_i\), and in particular affine.

Now, for each \(i\), we show that the canonical map \(R_{g_i} \rightarrow \Gamma(U_{g_i}, \mathcal{O}_X)\) is an isomorphism. For the open covering of \(U\) given by \(\{U_j\}_{j=1}^r\), the two conditions of [Topology] §Sheaves, ⁋Definition 1 are equivalent to the existence of the following exact sequence:

\[0 \rightarrow R \rightarrow \bigoplus_{j=1}^r \mathcal{O}_X(U_j) \rightarrow \bigoplus_{j,k=1}^r \mathcal{O}_X(U_j\cap U_k)\]

Here, the second map is \((s_j)_j\mapsto (s_j\vert_{U_j\cap U_k}-s_k\vert_{U_j\cap U_k})_{j,k}\). Since this is an exact sequence of \(R\)-modules and the direct sums are finite, by [Commutative Algebra] §Properties of Localization, ⁋Proposition 2, exactness is preserved upon localizing at \(g_i\), so we obtain the exact sequence

\[0 \rightarrow R_{g_i} \rightarrow \bigoplus_{j=1}^r \mathcal{O}_X(U_j)_{g_i} \rightarrow \bigoplus_{j,k=1}^r \mathcal{O}_X(U_j\cap U_k)_{g_i}\]

(Here, the localization of \(\mathcal{O}_X(U_j)\) means localization at the restriction of \(g_i\) to \(U_j\).) On the other hand, both \(U_j\) and \(U_j\cap U_k\) are affine, and for an affine scheme \(\Spec A\) and \(a\in A\) we have \(\mathcal{O}_{\Spec A}(D(a))=A_a\) (§Schemes, ⁋Lemma 2), so from \(U_g\cap \Spec A=D(g\vert_{\Spec A})\) observed above, we obtain

\[\mathcal{O}_X(U_j)_{g_i}=\mathcal{O}_X(U_j\cap U_{g_i}),\qquad \mathcal{O}_X(U_j\cap U_k)_{g_i}=\mathcal{O}_X(U_j\cap U_k\cap U_{g_i})\]

However, this exact sequence is precisely the exact sequence given by the sheaf condition for the open covering of \(U_{g_i}\) by \(\{U_j\cap U_{g_i}\}_{j=1}^r\), so \(R_{g_i}\) and \(\Gamma(U_{g_i}, \mathcal{O}_X)\) are both kernels of the same map, and thus the canonical map \(R_{g_i} \rightarrow \Gamma(U_{g_i}, \mathcal{O}_X)\) is an isomorphism.

Finally, in the adjunction of §Affine Scheme, ⁋Theorem 13,

\[\Hom_\Sch(U, \Spec R)\cong \Hom_\cRing(R, \Gamma(U, \mathcal{O}_X))\]

consider the scheme morphism \(\psi: U \rightarrow \Spec R\) corresponding to \(\id_R\). Since \(\psi\) is also a morphism of locally ringed spaces, at each \(x\in U\) the map \(\psi^\sharp_x\) is a local homomorphism, and since the composition \(R \rightarrow \mathcal{O}_{\Spec R, \psi(x)} \rightarrow \mathcal{O}_{X,x}\) is the canonical map \(R=\Gamma(U, \mathcal{O}_X) \rightarrow \mathcal{O}_{X,x}\), the point \(\psi(x)\) is the preimage under this canonical map of the maximal ideal of \(\mathcal{O}_{X,x}\). Therefore,

\[\psi^{-1}(D(g_i))=U_{g_i}=U_i\]

and since \(\psi\vert_{U_i}: U_i \rightarrow D(g_i)\cong \Spec R_{g_i}\) is a morphism between affine schemes corresponding to the isomorphism \(R_{g_i}\cong \Gamma(U_i, \mathcal{O}_X)\) just obtained, it is an isomorphism by §Affine Scheme, ⁋Proposition 11. Now, since the \(g_i\) generate the unit ideal of \(R\), the open sets \(\{D(g_i)\}\) cover \(\Spec R\) and \(\{U_i\}\) covers \(U\), so \(\psi\) is an isomorphism. That is, \(U\cong\Spec R\) is affine.

Finite Morphisms, Integral Morphisms, and Morphisms of Finite Type

Definition 10 A scheme morphism \(\varphi:X \rightarrow Y\) is finite if \(\varphi\) is affine, and for every affine open subset \(V\) of \(Y\), the ring homomorphism

\[(\varphi\vert_{\varphi^{-1}(V)})^\sharp(V): \mathcal{O}_V(V) \rightarrow \mathcal{O}_{\varphi^{-1}(V)}(\varphi^{-1}(V))\]

is a finite ring homomorphism. ([Commutative Algebra] §Integral Extensions, ⁋Definition 3)

To help understand this, let us write an affine open subset \(V\subseteq Y\) as \(\Spec B\). Then, from the assumption that \(\varphi\) is affine, \(U=\varphi^{-1}(V)\) is an affine open subset of \(X\), and thus there exists \(A\) such that \(U\cong\Spec A\). Through this identification, the scheme morphism \(\varphi\vert_U: U \rightarrow V\) is the same as a morphism between spectra \(\Spec A \rightarrow \Spec B\), and \(\varphi\) being finite now means that the ring homomorphism \(B \rightarrow A\) corresponding to this morphism is finite. Similarly, we define the following.

Definition 11 A scheme morphism \(\varphi:X \rightarrow Y\) is integral if \(\varphi\) is affine, and for every affine open subset \(V\) of \(Y\), the ring homomorphism

\[(\varphi\vert_{\varphi^{-1}(V)})^\sharp(V): \mathcal{O}_V(V) \rightarrow \mathcal{O}_{\varphi^{-1}(V)}(\varphi^{-1}(V))\]

is an integral ring homomorphism. ([Commutative Algebra] §Integral Extensions, ⁋Definition 3)

Now from the definitions, we see that finite morphisms and integral morphisms are closed under composition. In addition, since we can see that they satisfy the affine-local property condition of §The Topology of Schemes, ⁋Definition 9 from Proposition 9 and [Commutative Algebra] §Integral Extensions, ⁋Proposition 14, [Commutative Algebra] §Integral Extensions, ⁋Proposition 15, they are both affine-local on target.

We know that every finite morphism is integral by [Commutative Algebra] §Integral Extensions, ⁋Lemma 4. Now, in order to state this lemma completely in the language of algebraic geometry, we must define morphisms of finite type.

Definition 12 A scheme morphism \(\varphi:X \rightarrow Y\) is locally of finite type if for every affine open subset \(V\) of \(Y\) and every affine open subset \(U\) of \(\varphi^{-1}(V)\),

\[(\varphi\vert_{U})^\sharp(V): \mathcal{O}_V(V) \rightarrow \mathcal{O}_U(U)\]

is of finite type. ([Commutative Algebra] §Integral Extensions, ⁋Definition 3)

Just as above, let \(V\cong \Spec B\) and let \(U\cong\Spec A\subseteq \varphi^{-1}(V)\). Then we can view the scheme morphism \(\varphi\vert_U: U \rightarrow V\) as \(\Spec A \rightarrow \Spec B\), and this requires that the corresponding ring homomorphism \(B \rightarrow A\) be of finite type.

Since Definition 12 quantifies over all affine open subsets of \(\varphi^{-1}(V)\), we must show separately that a single affine open covering is sufficient to check this. This is the content of the following lemma.

Lemma 13 Let a scheme morphism \(\varphi: W \rightarrow \Spec B\) be given, and suppose \(W\) has an affine open covering \(\{\Spec A_i\}\) such that each \(B \rightarrow A_i\) is of finite type. Then on \(W\), for any affine open subset \(U\), the map \(B \rightarrow \mathcal{O}_W(U)\) is also of finite type.

Proof

For an affine open subset of \(W\), \(\Spec R\), define a property \(Q\) by

\(B \rightarrow R\) is of finite type,

and let us show that \(Q\) is an affine-local property in the sense of §The Topology of Schemes, ⁋Definition 9. Then, since the given covering \(\{\Spec A_i\}\) satisfies \(Q\), the desired result follows from the second condition of §The Topology of Schemes, ⁋Lemma 12.

The first condition is clear because when \(B \rightarrow R\) is of finite type, adjoining to the generators of \(R\) the element \(1/h\) makes \(R_h\) finitely generated as a \(B\)-algebra. For the second condition, suppose that \(R=(h_1,\ldots, h_m)\) and that each \(B \rightarrow R_{h_t}\) is of finite type. For each \(t\), choosing a finite generating set of \(R_{h_t}\) as a \(B\)-algebra and clearing denominators, there exist elements in \(R\), \(x_{t1},\ldots, x_{tn_t}\), such that \(R_{h_t}\) is generated by the \(x_{tk}/1\) and \(1/h_t\) as a \(B\)-algebra. Also, write \(1=\sum_{t=1}^ma_th_t\) with \(a_t\in R\). Now, let the finite set \(\{h_t\}\cup\{a_t\}\cup\{x_{tk}\}\) generate in \(R\) a \(B\)-subalgebra \(R'\); then since \(R'\) is a finite type \(B\)-algebra, it suffices to show that \(R'=R\). For any \(x\in R\), in \(R_{h_t}\) the element \(x/1\) is a polynomial in the \(x_{tk}/1\) and \(1/h_t\) with coefficients in \(B\), so for some \(r_t\in R'\) and \(n_t\geq 0\) we have \(x/1=r_t/h_t^{n_t}\), and thus for some \(N_t\), in \(R\) we have \(h_t^{N_t}(h_t^{n_t}x-r_t)=0\), that is, \(h_t^{N_t+n_t}x=h_t^{N_t}r_t\in R'\). Since there are only finitely many \(t\), we can choose a common \(M\) such that for all \(t\), we have \(h_t^Mx\in R'\). Meanwhile, from \(1=\sum_ta_th_t\), since \(a_t,h_t\in R'\), the elements \(h_1,\ldots, h_m\) generate the unit ideal of \(R'\), and by raising both sides of this equation to a sufficiently high power, we see that \(h_1^M,\ldots, h_m^M\) also generate the unit ideal of \(R'\). That is, we can write \(1=\sum_tc_th_t^M\) for some \(c_t\in R'\), and therefore

\[x=\sum_{t=1}^mc_t(h_t^Mx)\in R'\]

as desired.

Then morphisms of finite type are defined as follows.

Definition 14 A scheme morphism \(\varphi:X \rightarrow Y\) is a morphism of finite type if \(\varphi\) is a quasi-compact morphism locally of finite type.

From the definition, it is clear that morphisms locally of finite type are affine-local on the target. In addition, since quasi-compact morphisms are affine-local on the target by Proposition 7, morphisms of finite type are also affine-local on the target.

Then by [Commutative Algebra] §Integral Extensions, ⁋Lemma 4, the following holds.

Proposition 15 A scheme morphism \(\varphi:X \rightarrow Y\) is finite if and only if \(\varphi\) is an integral morphism (locally) of finite type.

Proof

First, suppose that \(\varphi\) is finite. For any affine open subset \(V=\Spec B\subseteq Y\), we have \(\varphi^{-1}(V)=\Spec A\) and \(B \rightarrow A\) is finite, so by [Commutative Algebra] §Integral Extensions, ⁋Lemma 4, this is integral and in particular of finite type. That is, \(\varphi\) is integral, and taking \(\{\varphi^{-1}(V)\}\) itself as an affine open covering and applying Lemma 13, for \(\varphi^{-1}(V)\), every affine open subset \(U\) also satisfies that \(B \rightarrow \mathcal{O}_X(U)\) is of finite type, so \(\varphi\) is locally of finite type. For the converse, from the assumption that \(\varphi\) is integral, we first know that for any affine open subset \(V\subseteq Y\), the preimage \(\varphi^{-1}(V)\) is an affine open subset of \(X\), and it suffices to apply [Commutative Algebra] §Integral Extensions, ⁋Lemma 4 to the ring map obtained in this way.

In the proposition above, since \(\varphi\) is an integral morphism, it is an affine morphism, and hence a quasi-compact morphism (§The Spectrum, ⁋Lemma 12); thus whether \(\varphi\) is of finite type or locally of finite type amounts to the same hypothesis.

Example 16 Let us look at examples of the morphisms examined in this section. In the world of affine schemes, this is merely looking at examples from [Commutative Algebra] §Integral Extensions, ⁋Definition 3. The purpose of this example is to give them geometric intuition.

First, for an algebraically closed field \(\mathbb{K}\), if we consider the ring map \(\iota:\mathbb{K}[\x] \rightarrow \mathbb{K}[\x,\y]\), then \(\mathbb{K}[\x,\y]\) is generated by a single element \(\y\) as a \(\mathbb{K}[\x]\)-algebra, so it is a ring homomorphism of finite type; however, as a \(\mathbb{K}[\x]\)-module, it is not finitely generated, and thus is not a finite ring homomorphism.

Now consider the corresponding morphism of schemes \(\Spec\iota: \Spec \mathbb{K}[\x,\y] \rightarrow\Spec \mathbb{K}[\x]\). This is the map that takes any prime ideal \(\mathfrak{p}\subseteq \mathbb{K}[\x,\y]\) and sends it to the prime ideal \(\mathfrak{p}\cap \mathbb{K}[\x]\) of \(\mathbb{K}[\x]\). Geometrically, this is the map sending a point \((x,y)\) on the affine plane \(\mathbb{A}^2_\mathbb{K}\) to the point \(x\) on the affine line \(\mathbb{A}^1_\mathbb{K}\).

An example of a related finite morphism is obtained by composing the ring homomorphism \(\iota:\mathbb{K}[\x]\rightarrow \mathbb{K}[\x,\y]\) above with the projection map \(\pi:\mathbb{K}[\x,\y] \rightarrow \mathbb{K}[\x,\y]/(\x-\y^2)\). Then \(\mathbb{K}[\x,\y]/(\x-\y^2)\) is generated by \(1\) and \(\y\) as a \(\mathbb{K}[\x]\)-module, so \(\phi:\mathbb{K}[\x] \rightarrow \mathbb{K}[\x,\y]/(\x-\y^2)\) is a finite ring homomorphism.

Meanwhile, we know that a ring homomorphism \(\pi:A \rightarrow A/\mathfrak{a}\) corresponds geometrically to the inclusion of the closed subset defined by \(\mathfrak{a}\). Therefore, the morphism of schemes defined by the composition

\[\phi: \mathbb{K}[\x] \rightarrow \mathbb{K}[\x,\y] \rightarrow \mathbb{K}[\x,\y]/(\x-\y^2)\]

which is

\[\Spec\phi: \Spec \frac{\mathbb{K}[\x,\y]}{(\x-\y^2)}\rightarrow \Spec \mathbb{K}[\x,\y] \rightarrow \Spec\mathbb{K}[\x]\]

can be viewed geometrically as the projection from the zero set \(Z(\x-\y^2)\) of \(\x=\y^2\) to the \(x\)-axis.

The geometric difference between these two examples is quite clear. In the first example, the fiber over a point of the target is an infinite set, whereas in the second example, the fiber over a point is a finite set. Algebraically, when we take any point \(\mathfrak{p}=(\x-a)\) of the target \(\mathbb{A}_\mathbb{K}^1\), any \(\mathfrak{q}_b=(\x-a, \y-b)\in \mathbb{A}_\mathbb{K}^2\) satisfies \((\Spec\iota)(\mathfrak{q}_b)=\mathfrak{p}\), whereas in the second example, since \(\y^2=\x\) must hold, only \(b\) such that \(b^2=a\) are possible, and thus there are at most two points satisfying \((\Spec\phi)(\mathfrak{q})=\mathfrak{p}\), namely \(\mathfrak{q}_+=(\x-a, \y-\sqrt{a})\) and \(\mathfrak{q}_-=(\x-a, \y+\sqrt{a})\). These two points are distinct when \(\operatorname{char}\mathbb{K}\neq 2\) and \(a\neq 0\); if \(a=0\), the two points coincide, and when \(\operatorname{char}\mathbb{K}=2\), we have \(\y^2-a=(\y-\sqrt{a})^2\), so the fiber is a single point for every \(a\).

In this way, a morphism of finite type is geometrically related to fibers being finite-dimensional, and a finite morphism is related to fibers being finite sets.

For now, to compute the fibers of a morphism of schemes in situations like Example 16 above, we have no choice but to compute them directly on a case-by-case basis, but once we compute fiber products later on, we will be able to use a more systematic approach. For that time, we define the following.

Definition 17 A morphism of schemes \(\varphi: X \rightarrow Y\) is quasi-finite if \(\varphi\) is a morphism of finite type and for every \(y\in Y\), the set \(\varphi^{-1}(y)\) is always a finite set.

Then the geometric intuition regarding finite morphisms in Example 16 is always true. That is, every finite morphism is always quasi-finite. While it is possible to prove this right now, we postpone it until after defining the fiber product.

Finally, we define the following.

Definition 18 A scheme morphism \(\varphi: X \rightarrow Y\) is locally of finite presentation if for every affine open subset \(V\cong \Spec B\) of \(Y\), there exists a covering \(\varphi^{-1}(V)=\bigcup \Spec A_i\) of \(\varphi^{-1}(V)\) such that each \(B \rightarrow A_i\) is finitely presented. If a scheme morphism \(\varphi:X \rightarrow Y\) is quasi-compact, quasi-separated, and locally of finite presentation, then \(\varphi\) is called a morphism of finite presentation.

In most cases, we consider the situation where all schemes are locally Noetherian, in which case this concept is not new. Indeed, if \(B\) is a Noetherian ring and \(B \rightarrow A\) is of finite type, then we can write \(A\cong B[\x_1,\ldots, \x_n]/\mathfrak{a}\), and since \(B[\x_1,\ldots, \x_n]\) is Noetherian by [Commutative Algebra] §Basic Notions, ⁋Theorem 12 (Hilbert basis theorem), \(\mathfrak{a}\) is finitely generated and thus \(B \rightarrow A\) is finitely presented. Furthermore, since locally Noetherian schemes are quasi-separated (Proposition 6), the difference between requiring a morphism to be of finite presentation and requiring it to be of finite type also disappears.

Rational Maps

In §Algebraic Structure of Schemes, ⁋Definition 12, we defined a rational function on a scheme \(X\) as the equivalence class of a pair consisting of a domain \(U\) and a function \(f\in \Gamma(U, \mathcal{O}_X)\) on it. Meanwhile, viewing \(U\) as a locally ringed space in its own right, by the adjunction of §Affine Scheme, ⁋Theorem 13 we obtain

\[\Hom_\LRS(U, \Spec \mathbb{Z}[\x])\cong \Hom_{\cRing}(\mathbb{Z}[\x], \Gamma(U,\mathcal{O}_X))\]

and since the ring homomorphism \(\mathbb{Z}[\x]\rightarrow \Gamma(U, \mathcal{O}_X)\) on the right-hand side is uniquely determined by the image of \(\x\), the right-hand side can be connected via an additional isomorphism

\[\Hom_{\cRing}(\mathbb{Z}[\x], \Gamma(U,\mathcal{O}_X))\cong \Gamma(U, \mathcal{O}_X)\]

to give that supplying a function \(f\) on \(U\) is the same data as supplying a scheme morphism \(U \rightarrow \Spec \mathbb{Z}[\x]=\mathbb{A}^1_\mathbb{Z}\).

On the other hand, having defined scheme morphisms, we may now choose a general \(\Spec A\) as the base scheme instead of \(\Spec \mathbb{Z}\), so repeating this argument allows us to think of a rational function as a morphism defined on \(U\) with target \(\Spec A[\x]\). To write this down precisely, we must specify which open sets are admissible as domains, and when two morphisms on different domains are to be considered equivalent. First, we give a name to morphisms whose image is not trapped inside a small closed subset of the target.

Definition 19 A scheme morphism \(\varphi: X \rightarrow Y\) is dominant if the image of \(\varphi\) is dense in \(Y\), that is, \(\cl(\varphi(X))=Y\).

A surjective morphism is always dominant, but the converse does not hold. For instance, for an integral domain \(A\) that is not a field, the image of \(\Spec \Frac A \rightarrow \Spec A\) consists only of the generic point \((0)\), but as we saw immediately after §The Spectrum, ⁋Definition 7, the only closed subset of \(\Spec A\) containing \((0)\) is \(\Spec A\) itself, so this morphism is dominant. Dominance between affine schemes can be interpreted purely algebraically as follows.

Proposition 20 For a ring homomorphism \(\phi: B \rightarrow A\) and the corresponding scheme morphism \(\varphi=\Spec \phi:\Spec A \rightarrow \Spec B\), the identity

\[\cl\left(\varphi(\Spec A)\right)=Z(\ker\phi)\]

holds. Therefore, \(\varphi\) is dominant if and only if \(\ker\phi\subseteq \mathfrak{N}(B)\).

Proof

By the second result of §The Spectrum, ⁋Proposition 14, for any \(T\subseteq \Spec B\) we have \(\cl(T)=Z(I(T))\), so for \(T=\varphi(\Spec A)\) it suffices to compute \(I(T)\). Since the elements of \(T\) are precisely of the form \(\phi^{-1}(\mathfrak{q})\), we obtain

\[I(T)=\bigcap_{\mathfrak{q}\in \Spec A}\phi^{-1}(\mathfrak{q})=\phi^{-1}\left(\bigcap_{\mathfrak{q}\in \Spec A}\mathfrak{q}\right)\]

. On the other hand, applying the first result of §The Spectrum, ⁋Proposition 14 to \(S=\{0\}\), we know that the intersection of all prime ideals of \(A\) is \(\sqrt{(0)}=\mathfrak{N}(A)\), and for any \(b\in B\), \(\phi(b)\) being nilpotent is equivalent to having \(b^n\in\ker\phi\) for some \(n\), so

\[I(T)=\phi^{-1}(\mathfrak{N}(A))=\sqrt{\ker\phi}\]

holds. Now, applying the first result of §The Spectrum, ⁋Proposition 14 again to the identity \(Z(I(Z(S)))=Z(S)\), we have \(Z(\sqrt{\ker\phi})=Z(\ker\phi)\), which yields the desired formula.

The last assertion holds because \(Z(\ker\phi)=\Spec B\) is equivalent to \(\ker\phi\) being contained in every prime ideal of \(B\), which, for the same reason as above, is equivalent to \(\ker\phi\subseteq \mathfrak{N}(B)\).

In particular, if \(B\) is a reduced ring, then \(\Spec\phi\) being dominant is equivalent to \(\phi\) being injective. For example, as seen in Example 16, \(\iota: \mathbb{K}[\x] \rightarrow \mathbb{K}[\x,\y]\) is injective, so \(\Spec\iota\) is dominant; and for an ideal \(\mathfrak{a}\subseteq B\), the morphism \(\Spec B/\mathfrak{a} \rightarrow \Spec B\) defined by the quotient map is dominant if and only if \(\mathfrak{a}\subseteq \mathfrak{N}(B)\), that is, the closed set defined by \(\mathfrak{a}\) is all of \(\Spec B\).

We now specify the open subsets that will serve as domains. Just as in the case of rational functions, the domain must be sufficiently large in \(X\), and the topological formulation of this is the density condition. As one can guess from the argument above, in a reduced scheme, this condition alone preserves the information of the function.

Lemma 21 For a reduced scheme \(X\) and a dense open subset \(U\), the restriction map \(\Gamma(X, \mathcal{O}_X) \rightarrow \Gamma(U, \mathcal{O}_X)\) is injective.

Proof

Suppose that \(s\in \Gamma(X, \mathcal{O}_X)\) satisfies \(s\vert_U=0\). Pick any affine open subset of \(X\), denoted \(\Spec A\), and for convenience, let us denote the restriction of \(s\) to \(\Spec A\) again by \(s\in A\). Then, since \(X\) is reduced, \(A\) is a reduced ring (§Algebraic Structure of Schemes, ⁋Definition 1), and since any nonempty open subset of \(\Spec A\) is also an open subset of \(X\) and thus meets \(U\), the intersection \(U\cap \Spec A\) is dense in \(\Spec A\).

Now, for any \(\mathfrak{p}\in U\cap \Spec A\), since the germ of \(s\) in the stalk \(\mathcal{O}_{X,\mathfrak{p}}=A_\mathfrak{p}\) is \(0\), we have \(ts=0\) for some \(t\in A\setminus \mathfrak{p}\), and since \(\mathfrak{p}\) is a prime ideal, this yields \(s\in \mathfrak{p}\). That is, \(U\cap \Spec A\subseteq Z(s)\); since \(Z(s)\) is closed and \(U\cap \Spec A\) is dense, we have \(Z(s)=\Spec A\), which means that \(s\) belongs to every prime ideal of \(A\). Therefore, by the first result of §The Spectrum, ⁋Proposition 14, we have \(s\in \mathfrak{N}(A)=0\). Since this holds for any affine open subset of \(X\), we obtain \(s=0\).

In this lemma, the assumption that the scheme is reduced cannot be omitted. For \(X=\Spec \mathbb{K}[\x_1,\x_2]/(\x_2^2,\x_1\x_2)\) in §Algebraic Structure of Schemes, ⁋Example 11, the nilradical is the prime ideal \((\x_2)\), so \(X\) is irreducible (§Algebraic Structure of Schemes, ⁋Lemma 3), and therefore the nonempty open subset \(D(\x_1)\) is dense. However, from \(\x_1\x_2=0\), making \(\x_1\) invertible yields \(\x_2\vert_{D(\x_1)}=0\), so a non-\(0\) function vanishes on a dense open subset. Requiring the domain of a rational function to contain all associated points in §Algebraic Structure of Schemes, ⁋Definition 12 was precisely intended to prevent this.

On the other hand, there are largely two conditions that the domain \(U\) of a rational map can satisfy. One is to take \(U\) to be a dense open subset, just as in classical algebraic geometry ([Algebraic Varieties] §Rational Maps, ⁋Definition 5), and the other is to view a rational map as a generalization of §Algebraic Structure of Schemes, ⁋Definition 12 and impose the condition that it contains all associated points. As seen above, for a non-reduced scheme, requiring it to be a dense open subset is generally a weaker condition than requiring it to contain all associated points; however, for a locally Noetherian reduced scheme, these two conditions coincide. To verify this, we first show that in \(X\), every irreducible component \(C\) has a generic point (since \(C\) meets an affine open subset \(\Spec A\), it suffices to take the closure in \(X\) of the point obtained by applying §The Spectrum, ⁋Proposition 16), and from this deduce that an open subset of \(X\) is dense if and only if it contains the generic point of every irreducible component of \(X\). Then, for a reduced ring \(A\), the associated points of \(\Spec A\) are always minimal prime ideals, that is, the generic points of the irreducible components; conversely, since we have already verified in §Algebraic Structure of Schemes, §§Associated Primes that the minimal prime ideals of a Noetherian ring are always associated prime ideals, these two conditions coincide.

Definition 22 A rational map from a scheme \(X\) to a scheme \(Y\) is an equivalence class of pairs \((U,\alpha)\) consisting of a dense open subset \(U\) of \(X\) and a morphism of schemes \(\alpha: U \rightarrow Y\). Here, two pairs \((U,\alpha)\) and \((V,\beta)\) are equivalent if there exists an open subset \(W\) contained in \(U\cap V\) and dense in \(X\) such that \(\alpha\vert_W=\beta\vert_W\).

A rational map is written with a dashed arrow as \(\varphi: X \dashrightarrow Y\), where the dashed arrow indicates that \(\varphi\) may not be defined at all points of \(X\). That the above definition is indeed an equivalence relation is because the intersection of two dense open subsets is again a dense open subset; while this is a weaker condition than the classical requirement that two representatives agree on the entirety of \(U\cap V\) ([Algebraic Varieties] §Rational Maps, ⁋Definition 5), the two coincide if \(X\) is reduced and \(Y=\Spec B\) is affine. This is because two morphisms \(\alpha,\beta: U\cap V \rightarrow Y\) correspond by §Affine Scheme, ⁋Theorem 13 to ring homomorphisms \(B \rightarrow \Gamma(U\cap V, \mathcal{O}_X)\), and since \(U\cap V\) is a reduced scheme and \(W\) is dense in it, \(\Gamma(U\cap V, \mathcal{O}_X) \rightarrow \Gamma(W, \mathcal{O}_X)\) is injective by Lemma 21.

A rational map \(\varphi: X \dashrightarrow Y\) is dominant if its representative \((U,\alpha)\) is a dominant morphism in the sense of Definition 19. This does not depend on the choice of representative, because if \(W\subseteq U\) is an open subset dense in \(X\), then \(W\) is dense in \(U\) as well, and since \(\alpha\) is continuous, \(\cl(\alpha(U))=\cl(\alpha(W))\) holds. The reason dominant rational maps are important becomes apparent when composing two rational maps \(\varphi: X\dashrightarrow Y\) and \(\psi: Y \dashrightarrow Z\). Let us choose representatives \((U,\alpha)\) and \((V,\beta)\) of the two rational maps. Then the actual domain of their composition \(\beta\circ \alpha\) is the part where \(\alpha\) enters the domain of the latter function, namely \(\alpha^{-1}(V)\), so for this map to define a rational map, \(\alpha^{-1}(V)\) must be dense in \(X\). A condition that can naturally be required then is the dominance of \(\alpha\); requiring this ensures that \(\alpha(U)\) meets \(V\), so \(\alpha^{-1}(V)\) does not become empty. However, in the world of schemes, this condition is still somewhat insufficient. For example, if we let \(X\) be the disjoint union of \(\mathbb{A}^1_\mathbb{K}\) and \(\Spec\mathbb{K}\), let \(Y=\mathbb{A}^1_\mathbb{K}\), and let \(\alpha\) be the morphism that is the identity map on the first component and sends the second component to a closed point \(p\), then \(\alpha\) is dominant, but for \(V=Y\setminus \{p\}\), \(\alpha^{-1}(V)\) is only \(\mathbb{A}^1_\mathbb{K}\setminus\{p\}\) in the first component, which fails to meet the second component, a nonempty open subset of \(X\). This problem arises because \(X\) is split into multiple pieces; if \(X\) is irreducible, every nonempty open subset is dense, so this problem disappears. Moreover, since we have already seen in the counterexample immediately following Lemma 21 that \(X\) needs to be reduced, for convenience we consider the case where \(X,Y\) are both integral schemes. (§Algebraic Structure of Schemes, ⁋Proposition 4) Then, in this case, for a dominant rational map \(\varphi: X\dashrightarrow Y\) and an arbitrary rational map \(\psi: Y \dashrightarrow Z\), their composition \(\psi\circ\varphi\) is well-defined by the representative \((\alpha^{-1}(V), \beta\circ \alpha)\), and it is not difficult to show that this does not depend on the choice of representatives.

Definition 23 For integral schemes \(X, Y\), a dominant rational map \(\varphi: X \dashrightarrow Y\) is a birational map if there exists a dominant rational map \(\psi: Y \dashrightarrow X\) such that \(\psi\circ\varphi\) and \(\varphi\circ\psi\) are equal to the rational maps defined by \(\id_X\) and \(\id_Y\), respectively. Two integral schemes \(X, Y\) are said to be birationally equivalent if there exists such a birational map \(\varphi: X\dashrightarrow Y\).

If an isomorphism means that two schemes have completely identical structures, birational equivalence means that two schemes have the same structure on dense open subsets. The following proposition makes this precise.

Proposition 24 For integral schemes \(X, Y\), given a dominant rational map \(\varphi: X \dashrightarrow Y\), the following two conditions are equivalent.

1. \(\varphi\) is a birational map.
2. There exist a non-empty open subset of \(X\), \(\widetilde{U}\), and a non-empty open subset of \(Y\), \(\widetilde{V}\), such that an isomorphism \(\widetilde{U} \rightarrow \widetilde{V}\) is a representative of \(\varphi\).

Proof

First, suppose that \(\varphi\) is a birational map, and choose a representative of \(\varphi\), \((U,\alpha)\), and a representative of \(\psi\) playing that role, \((V,\beta)\). Then, from \(\psi\circ\varphi=\id_X\), there exists a non-empty open subset of \(X\), \(W_1\subseteq \alpha^{-1}(V)\), such that \((\beta\circ \alpha)\vert_{W_1}=\id_{W_1}\), and from \(\varphi\circ\psi=\id_Y\), there exists a non-empty open subset of \(Y\), \(W_2\subseteq \beta^{-1}(U)\), such that \((\alpha\circ \beta)\vert_{W_2}=\id_{W_2}\). Now, we set

\[\widetilde{U}=W_1\cap \alpha^{-1}(W_2),\qquad \widetilde{V}=W_2\cap \beta^{-1}(W_1)\]

Since \(W_1\) is dense in \(U\) and \(\varphi\) is dominant, \(\cl(\alpha(W_1))=\cl(\alpha(U))=Y\), and therefore the non-empty open subset \(W_2\) meets \(\alpha(W_1)\), so that \(\widetilde{U}\neq\emptyset\).

In \(\widetilde{U}\), for any point \(x\), we have \(\alpha(x)\in W_2\), and since \(x\in W_1\), it follows that \(\beta(\alpha(x))=x\in W_1\), that is, \(\alpha(x)\in \beta^{-1}(W_1)\). Thus \(\alpha(\widetilde{U})\subseteq \widetilde{V}\), and in the same way, in \(\widetilde{V}\), for any point \(y\), we have \(\beta(y)\in W_1\) and \(\alpha(\beta(y))=y\in W_2\), so \(\beta(\widetilde{V})\subseteq \widetilde{U}\). For the two morphisms \(\alpha\vert_{\widetilde{U}}: \widetilde{U} \rightarrow \widetilde{V}\) and \(\beta\vert_{\widetilde{V}}: \widetilde{V} \rightarrow \widetilde{U}\) obtained this way, their compositions are the identity maps by \(\widetilde{U}\subseteq W_1\) and \(\widetilde{V}\subseteq W_2\), respectively, so \(\alpha\vert_{\widetilde{U}}\) is an isomorphism, and since \(\widetilde{U}\) is dense in \(X\), \((\widetilde{U}, \alpha\vert_{\widetilde{U}})\) is a representative of \(\varphi\).

Conversely, suppose that an isomorphism \(\alpha: \widetilde{U} \rightarrow \widetilde{V}\) is a representative of \(\varphi\), and let its inverse be \(\beta: \widetilde{V} \rightarrow \widetilde{U}\). Then, since \(\widetilde{V}\) is dense in \(Y\), the composition of \(\beta\) and the inclusion map of the open subset \(\widetilde{U} \hookrightarrow X\) defines a rational map \(\psi: Y \dashrightarrow X\), and since \(\beta\) is surjective, \(\psi\) is dominant. Moreover, \(\psi\circ\varphi\) restricts on \(\widetilde{U}\) to \(\beta\circ \alpha=\id_{\widetilde{U}}\), and \(\varphi\circ\psi\) restricts on \(\widetilde{V}\) to \(\alpha\circ \beta=\id_{\widetilde{V}}\), so these two are equal to the rational maps defined by \(\id_X\) and \(\id_Y\), respectively.

If an integral scheme \(X\) is locally Noetherian, the domain of definition of a rational function in §Algebraic Structure of Schemes, ⁋Definition 12 is precisely a non-empty open subset of \(X\). Indeed, for a non-empty affine open subset of \(X\), \(\Spec A\), since \(A\) is an integral domain (§Algebraic Structure of Schemes, ⁋Definition 1), the annihilator of an element other than \(0\) is always \((0)\), and therefore the only associated point of \(\Spec A\) is \((0)\). In other words, the only associated point of \(X\) is the generic point \(\eta\), so the condition that the domain of definition must contain all associated points becomes equivalent to the condition that the domain of definition is non-empty. We have already seen that the collection of rational functions \(K(X)\) obtained in this way coincides with \(\mathcal{O}_{X,\eta}\cong\Frac A\) and forms a field (§Algebraic Structure of Schemes, §§Rational Functions), and we call this the function field of \(X\).

Corollary 25 For two birationally equivalent integral locally Noetherian schemes \(X, Y\), we have \(K(X)\cong K(Y)\).

Proof

By Proposition 24, there exists a birational map having an isomorphism \(\alpha: \widetilde{U} \rightarrow \widetilde{V}\) as a representative. Since the generic point of \(X\), \(\eta_X\), belongs to the non-empty open subset \(\widetilde{U}\) and stalks do not change under restriction to an open subscheme, \(K(X)=\mathcal{O}_{X,\eta_X}=\mathcal{O}_{\widetilde{U}, \eta_X}\), and for the same reason, \(K(Y)=\mathcal{O}_{\widetilde{V}, \eta_Y}\). Meanwhile, since \(\widetilde{V}\) is a non-empty open subset of \(Y\), its generic point is \(\eta_Y\), and since \(\alpha\) is an isomorphism, \(\alpha(\eta_X)=\eta_Y\). Therefore, the isomorphism between stalks induced by \(\alpha\) gives \(K(X)\cong K(Y)\).

That is, birational equivalence preserves function fields. In the case of varieties, we verified in [Algebraic Varieties] §Rational Maps, ⁋Proposition 10 that the converse also holds.

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References

[Har] R. Hartshorne, Algebraic geometry. Graduate texts in mathematics. Springer, 1977.
[Vak] R. Vakil, The rising sea: Foundation of algebraic geometry. Available online.