Morphisms Local Rings and Rational Maps of Affine Varieties

If $f\in {\mathcal{O}}_X(U)$ and $V\subseteq U$ is open, then every point of $V$ has an open neighbourhood $W\subseteq U$ on which [L1] gives a quotient formula $f=a\mathrm{/}b$. Then $W\cap V$ is an open neighbourhood of that point inside $V$, the denominator $b$ is still nowhere zero there, and $f{\mathrm{\mid }}_V=a\mathrm{/}b$ on $W\cap V$. Hence restriction maps are well defined.

If $f,g\in {\mathcal{O}}_X(U)$ and $f{\mathrm{\mid }}_{U_i}=g{\mathrm{\mid }}_{U_i}$ for every $i\in I$, then for each $x\in U$ some $U_i$ contains $x$, so $f(x)=g(x)$. Therefore $f=g$ on $U$. This is the uniqueness clause.

Suppose for each $i\in I$ we are given $f_i\in {\mathcal{O}}_X(U_i)$, and suppose $f_i=f_j$ on $U_i\cap U_j$ for all $i,j$. Define $f:U\to k$ by $f(x)=f_i(x)$ whenever $x\in U_i$. This is well defined by the overlap hypothesis.

Fix $x\in U$. Choose $i$ with $x\in U_i$. Since $f_i$ is regular on $U_i$, [L1] gives an open neighbourhood $W\subseteq U_i$ of $x$ and elements $a,b\in k[X]$ with $b$ nowhere zero on $W$ and $f_i=a\mathrm{/}b$ on $W$. On that same neighbourhood, the glued function $f$ equals $f_i$, so $f=a\mathrm{/}b$ on $W$. By [L1], $f$ is regular at $x$.

Step 1.1 gives restriction, step 1.2 gives uniqueness of gluing, and steps 1.3 and 2.1 give existence of gluing. Therefore $U\mapsto {\mathcal{O}}_X(U)$ is a sheaf of $k$-algebras on $X$.

Apply [L3] to the element $1\in k[X]$. By [L1], the target is ${\mathcal{O}}_X(X)$, and by [L2] the source is canonically $k[X]$. The resulting isomorphism is exactly the evaluation map from $k[X]$ to global regular functions on $X$.

Therefore evaluation induces a ring isomorphism $k[X]\stackrel{\sim }{\to }{\mathcal{O}}_X(X)$.

Let $\phi :X\to Y$ be a morphism. By [L1], pullback sends global regular functions on $Y$ to global regular functions on $X$, and [L2] identifies those two rings with $k[Y]$ and $k[X]$. Therefore $\phi$ determines a $k$-algebra homomorphism ${\phi }^{\ast }:k[Y]\to k[X]$, $h\mapsto h\circ \phi$. This is the desired pullback homomorphism attached to $\phi$.

Conversely, let $\alpha :k[Y]\to k[X]$ be a $k$-algebra homomorphism. Write ${\bar{y}}_1,\dots ,{\bar{y}}_m$ for the coordinate classes of $Y$ in [L3]. By [L2], each $\alpha ({\bar{y}}_i)$ is a global regular function on $X$, hence an honest function $X\to k$. Define ${\phi }_{\alpha }(x):=(\alpha ({\bar{y}}_1)(x),\dots ,\alpha ({\bar{y}}_m)(x))\in {\mathbf{A}}_k^m$. This produces a candidate point of affine $m$-space for each $x\in X$.

For every polynomial $g\in I(Y)$, its class is $0$ in $k[Y]$ by [L3]. Hence $g(\alpha ({\bar{y}}_1),\dots ,\alpha ({\bar{y}}_m))=\alpha (g)=0$ in $k[X]$. Evaluating at $x\in X$ gives $g({\phi }_{\alpha }(x))=0$. Since [L4] says that the common zero set of $I(Y)$ is exactly $Y$, the point ${\phi }_{\alpha }(x)$ lies in $Y$. Thus ${\phi }_{\alpha }:X\to Y$ is well defined.

Let $h\in k[Y]$. By [L3], $h$ is a polynomial expression in the coordinate classes ${\bar{y}}_i$. Pulling that expression back along ${\phi }_{\alpha }$ replaces each ${\bar{y}}_i$ by $\alpha ({\bar{y}}_i)$, so ${\phi }_{\alpha }^{\ast }(h)=\alpha (h)$. In particular ${\phi }_{\alpha }^{\ast }(h)$ is regular on $X$ by [L2], and [L1] shows that ${\phi }_{\alpha }$ is a morphism.

Starting from a morphism $\phi$, the map constructed from ${\phi }^{\ast }$ has the same pullback on the coordinate classes ${\bar{y}}_i$, so it has the same value as $\phi$ at every point of $X$. Starting from a homomorphism $\alpha$, step 2.2 shows that the pullback of the constructed morphism is exactly $\alpha$. Therefore the two constructions are inverse to each other.

For the identity morphism on $X$, the pullback is the identity on $k[X]$ by the formula in step 1.1. If $\psi :Y\to Z$ and $\phi :X\to Y$ are morphisms, then for every $h\in k[Z]$ one has $(\psi \circ \phi {)}^{\ast }(h)=h\circ \psi \circ \phi ={\phi }^{\ast }({\psi }^{\ast }(h))$, so pullback reverses composition. Together with step 3.1 this proves the stated contravariant correspondence.

Fact [L1] supplies the object assignment $X\mapsto k[X]$ for affine algebraic sets.

Let $X$ and $Y$ be classical affine varieties. Fact [L2] gives a bijection $\mathrm{Mor}(X,Y)\cong {\mathrm{Hom}}_{k\text{-alg}}(k[Y],k[X])$ that respects identities and composition in the contravariant order. Therefore, on the full subcategory of classical affine varieties, the assignment of step 1.1 is contravariantly full and faithful.

This is exactly the claimed morphism-level completion of the affine algebraic-set dictionary on classical affine varieties.

Let ${\bar{y}}_1,\dots ,{\bar{y}}_m\in k[Y]$ be the coordinate classes. By [L1] and [L3], the pullbacks ${\phi }^{\ast }({\bar{y}}_i)$ and ${\psi }^{\ast }({\bar{y}}_i)$ are elements of $k[X]$. Since $\phi =\psi$ on $U$, these two functions agree on $U$ for every $i$.

Fix $i$, and put $h_i={\phi }^{\ast }({\bar{y}}_i)-{\psi }^{\ast }({\bar{y}}_i)\in k[X]$. The function $h_i$ vanishes on $U$. If $D_X(h_i)$ were nonempty, then [L2] would make both $U$ and $D_X(h_i)$ dense open subsets of $X$, so they would meet. That contradicts [L4], because $h_i$ is zero on $U$ and nonzero on $D_X(h_i)$. Hence $D_X(h_i)=\mathrm{\varnothing }$, so $h_i=0$ on all of $X$.

Step 2.1 shows that every coordinate function of $Y$ has the same pullback under $\phi$ and $\psi$. Therefore the two maps have the same $m$ coordinate functions on $X$, so $\phi (x)=\psi (x)$ for every $x\in X$. Thus $\phi =\psi$.

Let $f_x\in {\mathcal{O}}_{X,x}$ be a germ, represented by a regular function $f$ on an open neighbourhood $U$ of $x$. By [L4], choose a principal open neighbourhood $D_X(s)$ with $x\in D_X(s)\subseteq U$. Then [L5] identifies $f{\mathrm{\mid }}_{D_X(s)}$ with some fraction $a\mathrm{/}{s}^n\in k[X{]}_s$. Because $x\in D_X(s)$, the denominator $s$ does not vanish at $x$, so [L2] and [L3] give $s\mathrm{\notin }{\mathfrak{m}}_x$. Define $\mathrm{\Phi }(f_x):=a\mathrm{/}{s}^n\in k[X{]}_{{\mathfrak{m}}_x}$. This gives a candidate image for the germ $f_x$.

Conversely, let $a\mathrm{/}s\in k[X{]}_{{\mathfrak{m}}_x}$. By [L2] and [L3], $s(x)\mathrm{\ne }0$, so $x\in D_X(s)$. On that principal open, the function $y\mapsto a(y)\mathrm{/}s(y)$ is regular by [L5], hence defines a germ at $x$. Put $\mathrm{\Psi }(a\mathrm{/}s):=(a\mathrm{/}s{)}_x$. This gives the inverse candidate on localisation fractions.

The value in step 1.1 is independent of the representative. Indeed, if the same germ is also represented by $g$ on a neighbourhood $V$, choose by [L4] a principal open $D_X(t)$ with $x\in D_X(t)\subseteq U\cap V$ on which $f=g$. By [L5], the restrictions of $f$ and $g$ give equal fractions in $k[X{]}_t$. Since $t\mathrm{\notin }{\mathfrak{m}}_x$ by [L2] and [L3], the equality criterion [L6] makes their images equal in the further localisation $k[X{]}_{{\mathfrak{m}}_x}$. Thus $\mathrm{\Phi }$ is well defined.

If $a\mathrm{/}s=b\mathrm{/}u$ in $k[X{]}_{{\mathfrak{m}}_x}$, then [L6] gives $t\mathrm{\notin }{\mathfrak{m}}_x$ with $t(au-bs)=0$ in $k[X]$. On the principal open $D_X(stut)$ every factor $s$, $u$, and $t$ is nonzero. Evaluating the relation there gives $t(y)(a(y)u(y)-b(y)s(y))=0,$ and division by $t(y)\mathrm{\ne }0$ in the field $k$ yields $a(y)u(y)=b(y)s(y)$. Therefore $a\mathrm{/}s=b\mathrm{/}u$ as ordinary functions on $D_X(stut)$. Since $x\in D_X(stut)$ by [L2] and [L3], the two regular functions have the same germ at $x$. Thus $\mathrm{\Psi }$ is well defined.

Start with a germ $f_x$. Choosing a principal-open representative as in step 1.1, the construction of $\mathrm{\Psi }(\mathrm{\Phi }(f_x))$ recovers exactly the germ of that same representative, so $\mathrm{\Psi }(\mathrm{\Phi }(f_x))=f_x$.

Start with a fraction $a\mathrm{/}s\in k[X{]}_{{\mathfrak{m}}_x}$. The germ constructed in step 1.2 is represented on $D_X(s)$ by the regular function corresponding under [L5] to the same fraction $a\mathrm{/}s$. Applying $\mathrm{\Phi }$ therefore returns $a\mathrm{/}s$. Hence $\mathrm{\Phi }(\mathrm{\Psi }(a\mathrm{/}s))=a\mathrm{/}s$.

Steps 2.3 and 2.4 show that $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse ring homomorphisms. By [L7], the target is local, so ${\mathcal{O}}_{X,x}$ is canonically identified with the localisation $k[X{]}_{{\mathfrak{m}}_x}$ as a local ring.

By [L1], write $U=D_X(\bar{f})$. Since $U$ is nonempty, the element $\bar{f}$ is not zero in the domain $k[X]$ of [L3]. By [L2], the coordinate ring of $U$ is $k[X{]}_{\bar{f}}$.

Every fraction in $k[X{]}_{\bar{f}}$ is represented by $a\mathrm{/}{\bar{f}}^n$ with $a\in k[X]$. Sending that fraction to the same formal quotient in $\mathrm{Frac}(k[X])$ defines an injective ring map $k[X{]}_{\bar{f}}\hookrightarrow \mathrm{Frac}(k[X])$. Conversely, if $a\mathrm{/}b\in \mathrm{Frac}(k[X])$ with $b\mathrm{\ne }0$, then $b\mathrm{/}1$ is nonzero in the domain $k[X{]}_{\bar{f}}$, so the same quotient may be read as an element of $\mathrm{Frac}(k[X{]}_{\bar{f}})$. Hence each field of fractions contains the other in the obvious way.

Step 2.1 gives $\mathrm{Frac}(k[U])=\mathrm{Frac}(k[X{]}_{\bar{f}})\cong \mathrm{Frac}(k[X])=k(X)$ by [L2] and [L4]. Therefore every nonempty affine open of $X$ has the same function field as $X$, and any two such opens have canonically isomorphic function fields.

The sets $W_{12}$ and $W_{23}$ are nonempty open subsets of $X$ by [L1]. By [L2], both are dense in $X$, so their intersection is nonempty.

The set $W_{12}\cap W_{23}$ is open in $X$. Since both $W_{12}$ and $W_{23}$ are affine opens, [L3] identifies their intersection as another affine open subset. On that common open we have ${\phi }_1={\phi }_2={\phi }_3$.

Step 2.1 gives a nonempty affine open subset of $U_1\cap U_3$ on which ${\phi }_1$ and ${\phi }_3$ agree. Therefore ${\phi }_1\sim {\phi }_3$.

Choose a representative morphism $\phi :U\to Y$ of $\eta$ on a nonempty affine open $U\subseteq X$, guaranteed by [L2]. By [L4], $\phi$ induces a $k$-algebra homomorphism ${\phi }^{\ast }:k[Y]\to k[U]$. Using [L6] and [L1], regard $k[U]$ as a subring of $k(X)$.

The map ${\phi }^{\ast }$ is injective. Let $0\mathrm{\ne }h\in k[Y]$. By [L5], the ring $k[Y]$ is a domain, so [L7] makes $D_Y(h)$ a nonempty open subset of $Y$. Let $\psi :\mathrm{Dom}(\eta )\to Y$ be the representative from [L3] whose image is dense in $Y$. By the definition of rational map, $\psi$ and $\phi$ agree on some nonempty open subset $W\subseteq U$. Since $W$ is a nonempty open subset of the classical affine variety $X$, [L8] makes $W$ dense in $X$, hence dense in the open subspace $\mathrm{Dom}(\eta )$. Therefore $\psi (W)$ is dense in $\psi (\mathrm{Dom}(\eta ))$, which is dense in $Y$. Because $\phi (W)=\psi (W)\subseteq \phi (U)$, the image $\phi (U)$ is dense in $Y$. It follows that $\phi (U)$ meets $D_Y(h)$, so for some $u\in U$ one has ${\phi }^{\ast }(h)(u)=h(\phi (u))\mathrm{\ne }0$. Hence ${\phi }^{\ast }(h)\mathrm{\ne }0$.

For $a\mathrm{/}b\in k(Y)=\mathrm{Frac}(k[Y])$, define ${\eta }^{\ast }(a\mathrm{/}b):={\phi }^{\ast }(a)\mathrm{/}{\phi }^{\ast }(b)\in \mathrm{Frac}(k[U])=k(X)$. This is well defined because step 2.1 makes ${\phi }^{\ast }(b)$ nonzero whenever $b\mathrm{\ne }0$.

If $\psi :V\to Y$ is another representative of $\eta$, then by the definition of rational map the two representatives agree on some nonempty affine open $W\subseteq U\cap V$. Their coordinate-ring pullbacks therefore agree on $k[Y]$, and after passing to fraction fields and using [L6] for $W$, the two constructions of step 3.1 agree in $k(X)$. Thus ${\eta }^{\ast }$ is independent of the chosen representative.

Let $\theta :Y⇢Z$ be another dominant rational map, represented on some nonempty affine open of $Y$. On a common domain where both pullbacks are defined, pullback of regular functions satisfies $(\theta \circ \eta {)}^{\ast }(h)={\eta }^{\ast }({\theta }^{\ast }(h))$ for every $h\in k[Z]$. By fraction-field arithmetic the same identity holds for every element of $k(Z)$. Therefore pullback on function fields is functorial.

By [L1], every dominant rational map $\eta :X⇢Y$ determines an injective $k$-homomorphism ${\eta }^{\ast }:k(Y)\hookrightarrow k(X)$.

Conversely, let $\sigma :k(Y)\hookrightarrow k(X)$ be an injective $k$-homomorphism. For the coordinate classes ${\bar{y}}_1,\dots ,{\bar{y}}_m\in k[Y]$ from [L3], write $\sigma ({\bar{y}}_i)=\frac{a_i}{d_i}$ with $a_i,d_i\in k[X]$ and $d_i\mathrm{\ne }0$ in the domain $k[X]$. Put $d=d_1\cdots d_m$ and $U=D_X(d)$. By [L5], each $a_i\mathrm{/}{d}_i$ is regular on $U$, so each $\sigma ({\bar{y}}_i)$ lies in $k[U]$.

Define a $k$-algebra homomorphism $\alpha :k[Y]\to k[U]$ by sending ${\bar{y}}_i$ to $\sigma ({\bar{y}}_i){\mathrm{\mid }}_U$. This is well defined because [L4] identifies $I(Y)$ as the defining ideal of $Y$, and if $g\in I(Y)$, then its class is $0$ in $k[Y]$ by [L3], hence $g(\sigma ({\bar{y}}_1),\dots ,\sigma ({\bar{y}}_m))=\sigma (g)=0$ in $k(X)$. Therefore every relation of $I(Y)$ is respected.

By [L6], the homomorphism $\alpha$ corresponds to a morphism $\phi :U\to Y$. Its rational-map class gives a rational map ${\eta }_{\sigma }:X⇢Y$.

The rational map ${\eta }_{\sigma }$ is dominant. If its image were not dense in $Y$, then some nonzero $h\in k[Y]$ would vanish on $\phi (U)$. Then $\alpha (h)=0$ in $k[U]$, so $\sigma (h)=0$ in $k(X)$ by step 2.1, contradicting injectivity of $\sigma$.

The pullback $({\eta }_{\sigma }{)}^{\ast }$ agrees with $\sigma$ on the coordinate classes ${\bar{y}}_i$, hence on the whole coordinate ring $k[Y]$, and therefore on its fraction field $k(Y)$ by [L2]. Thus the construction of steps 1.2-4.1 is inverse to the construction in step 1.1.

Steps 1.1 and 5.1 give the stated bijection between dominant rational maps $X⇢Y$ and injective $k$-homomorphisms $k(Y)\hookrightarrow k(X)$.

Suppose $\eta :X⇢Y$ is birational, with inverse $\theta :Y⇢X$. By [L2], both maps induce injective pullbacks ${\eta }^{\ast }:k(Y)\hookrightarrow k(X)$ and ${\theta }^{\ast }:k(X)\hookrightarrow k(Y)$. Because pullback is functorial, $(\theta \circ \eta {)}^{\ast }={\eta }^{\ast }\circ {\theta }^{\ast }$ and $(\eta \circ \theta {)}^{\ast }={\theta }^{\ast }\circ {\eta }^{\ast }$. Since the two compositions are identity rational maps by [L1], the two field homomorphisms are inverse isomorphisms.

Conversely, let $\sigma :k(Y)\stackrel{\sim }{\to }k(X)$ be a field isomorphism. By [L2], the map $\sigma$ comes from a dominant rational map $\eta :X⇢Y$, and ${\sigma }^{-1}$ comes from a dominant rational map $\theta :Y⇢X$. Applying [L2] again to the compositions shows that $\theta \circ \eta$ and $\eta \circ \theta$ induce the identity on function fields, so they are the identity rational maps. Hence $\eta$ is birational.

Steps 1.1 and 1.2 prove that $X$ and $Y$ are birationally equivalent exactly when $k(X)\cong k(Y)$ as $k$-extensions.