Morphisms Local Rings and Rational Maps of Affine Varieties — Examples

Let $X={\mathbf{A}}_k^2$ with coordinates $s,t$, and let $Y=V(y-x^2)\subseteq {\mathbf{A}}_k^3\mathrm{.}$ Define $\phi :X\to Y,(s,t)⟼(s,s^2,t)\mathrm{.}$

The point lies in $Y$ because its coordinates satisfy $y-x^2=s^2-s^2=0$. Therefore $\phi$ is a morphism. Since $k[Y]=k[x,y,z]\mathrm{/}(y-x^2),$ the induced pullback sends $x\mapsto s,y\mapsto s^2,z\mapsto t\mathrm{.}$ The defining relation pulls back to $s^2-s^2=0$, so the homomorphism respects the quotient relation exactly as the affine-morphism theorem predicts.

Assume the Axiom of Choice and let $k$ be an algebraically closed field. Let $H:=V(xy-1)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ The substitutions $x\mapsto t$, $y\mapsto t^{-1}$ and $t\mapsto x$, $t^{-1}\mapsto y$ define inverse $k$-algebra homomorphisms $k[H]=k[x,y]\mathrm{/}(xy-1)\cong k[t,t^{-1}]\mathrm{.}$ The Laurent polynomial ring is a domain, so A classical affine variety has a domain as its coordinate ring, and conversely shows that $H$ is a classical affine variety.

Give $D_{{\mathbf{A}}^1}(t)$ its principal-open regular-function structure. Projection to the first coordinate gives a morphism $p:H\to D_{{\mathbf{A}}^1}(t),(x,y)\mapsto x,$ because the coordinate function $x$ is regular and can never be $0$ on $H$.

Conversely, $q:D_{{\mathbf{A}}^1}(t)\to H,u\mapsto (u,u^{-1})$ is regular on the principal open $D_{{\mathbf{A}}^1}(t)$, since $u^{-1}$ is regular there by Regular functions on a principal open are the principal localization of the coordinate ring, and its coordinates satisfy $u\cdot u^{-1}=1$, so it lands in $H$. Direct substitution gives $p(q(u))=u\text{and}q(p(x,y))=(x,x^{-1})=(x,y),$ where the last equality uses $xy=1$. Thus $p$ and $q$ are inverse regular maps. They exhibit $H$ as the affine model of the punctured line and realize the displayed coordinate-ring isomorphism.

Assume $\mathrm{char}(k)\mathrm{\ne }2$, and let $C:=V(x^2+y^2-1)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ The line through $(-1,0)$ of slope $t$ meets $C$ again at $x=\frac{1-t^2}{1+t^2},y=\frac{2t}{1+t^2}\mathrm{.}$ This gives a rational map ${\mathbf{A}}^1⇢C$.

On the dense open subset of $C$ where $x\mathrm{\ne }-1$, the inverse parameter is $t=\frac{y}{x+1}\mathrm{.}$ Substituting the displayed parametrization shows that these two formulas are mutually inverse on the common domain where $1+t^2\mathrm{\ne }0$ and $x+1\mathrm{\ne }0$. Thus the affine circle is birational to the affine line.

Not every rational map on affine space extends to a regular map everywhere.

On ${\mathbf{A}}_k^2$ with coordinates $(x,y)$, the quotient $y\mathrm{/}x$ is regular on the principal open $D(x)$ and therefore defines a rational map $\eta :{\mathbf{A}}_k^2⇢{\mathbf{A}}_k^1\mathrm{.}$ Its domain of definition contains every point with $x\mathrm{\ne }0$.

It does not extend regularly across any point $(0,b)$ with $b\mathrm{\ne }0$. Indeed, an extension near such a point would give an open neighbourhood $W$ of $(0,b)$ and a regular function $g$ on $W$ that agrees with $y\mathrm{/}x$ on $W\cap D(x)$. By Principal opens form a basis for the Zariski topology on an affine variety, choose a principal open $D(h)\subseteq W$ with $(0,b)\in D(h)$. Then Regular functions on a principal open are the principal localization of the coordinate ring identifies $g{\mathrm{\mid }}_{D(h)}$ with some fraction $a\mathrm{/}{h}^n\in k[x,y{]}_h$, and on the smaller principal open $D(h)\cap D(x)=D(hx)$ the same theorem identifies the equality $g=y\mathrm{/}x$ with the localization identity $\frac{a}{h^n}=\frac{y}{x}\text{in }k[x,y{]}_{hx}\mathrm{.}$ Thus some power of $hx$ annihilates $xa-y{h}^n$. Since $k[x,y]$ is a domain, it follows that $xa=y{h}^n$ in $k[x,y]$. Evaluating at $(0,b)$ gives $0=bh(0,b{)}^n,$ which is impossible because $b\mathrm{\ne }0$ and $(0,b)\in D(h)$ means $h(0,b)\mathrm{\ne }0$. This contradiction shows that no regular extension exists near $(0,b)$.

Birational affine varieties need not be isomorphic.

Assume the Axiom of Choice and let $k$ be an algebraically closed field. Take $X={\mathbf{A}}_k^1$ with coordinate ring $k[t]$ and $U=D_X(t)={\mathbf{A}}_k^1\setminus \{0\}\mathrm{.}$ The explicit isomorphism in The hyperbola xy = 1 is isomorphic to the punctured affine line identifies $U$ with the classical affine variety $V(xy-1)\subseteq {\mathbf{A}}_k^2$, so $U$ may be used as an affine-variety target. The identity on the common nonempty open $U$, in one direction, and the inclusion $U\hookrightarrow X$, in the other, exhibit $X$ and $U$ as birational.

They are not isomorphic. By Regular functions on a principal open are the principal localization of the coordinate ring, $k[U]\cong k[t,t^{-1}],$ whereas $k[X]=k[t]$. Every unit of $k[t]$ is constant: if $fg=1$ in $k[t]$, then $\mathrm{deg}(f)+\mathrm{deg}(g)=0$, so $\mathrm{deg}(f)=0$. By contrast, $t$ is a nonconstant unit of $k[t,t^{-1}]$ with inverse $t^{-1}$. Therefore the two rings have different unit groups and cannot be isomorphic. Thus Affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms rules out an isomorphism of varieties.

Assume the Axiom of Choice, let $k$ be an algebraically closed field, and take $X={\mathbf{A}}_k^1$ with coordinate $t$. The origin corresponds to the maximal ideal $(t)\subseteq k[t]$. By The local ring at a point of an affine variety is the localization at its maximal ideal, ${\mathcal{O}}_{X,0}\cong k[t{]}_{(t)}=\{r\in k(t):r=\frac{f(t)}{g(t)}\text{ for some }f,g\in k[t]\text{ with }g(0)\mathrm{\ne }0\}\mathrm{.}$

Thus a rational function is regular at the origin exactly when it admits a presentation $f\mathrm{/}g$ with $g(0)\mathrm{\ne }0$; another presentation may have a denominator that vanishes at $0$. The maximal ideal consists of the elements admitting such a presentation with $f(0)=0$, and The residue field at a classical point identifies the residue field with $k$.

A bijective morphism of affine varieties need not be an isomorphism.

Let $C:=V(v^2-u^3)\subseteq {\mathbf{A}}_k^2,$ and define $\nu :{\mathbf{A}}_k^1\to C,t\mapsto (t^2,t^3)\mathrm{.}$ Every point of $C$ has the form $(a^2,a^3)$ for a unique $a\in k$, so $\nu$ is bijection on points.

Its pullback on coordinate rings is $k[C]=k[u,v]\mathrm{/}(v^2-u^3)⟶k[t],u\mapsto t^2,v\mapsto t^3\mathrm{.}$ This homomorphism is injective but not surjective, because $t$ is not in its image. Hence Affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms shows that $\nu$ is not an isomorphism.