Affine Algebraic Sets and Coordinate Rings — Examples

Let $k$ be an algebraically closed field and let $n\ge 1$. Since every algebraically closed field is infinite, the lemma A polynomial vanishing at every tuple from an infinite subdomain is the zero polynomial shows that no nonzero polynomial in $k[x_1,\dots ,x_n]$ vanishes on all of $k^n={\mathbf{A}}_k^n$. Therefore $I({\mathbf{A}}_k^n)=0,$ and the coordinate-ring definition gives $k[{\mathbf{A}}_k^n]=k[x_1,\dots ,x_n]\mathrm{.}$

At the boundary case $n=0$, the affine space ${\mathbf{A}}_k^0$ is the one-point set. Its ambient polynomial ring is just $k$, its vanishing ideal is again $0$, and its coordinate ring is therefore $k$.

Let $k$ be an algebraically closed field and let $X=V(y-x^2)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ Division by the monic polynomial $y-x^2$ in the variable $y$ writes every $f\in k[x,y]$ uniquely as $f=q(y-x^2)+r(x)\mathrm{.}$ If $f$ vanishes on $X$, then $r(t)=f(t,t^2)=0$ for every $t\in k$. The field $k$ is infinite, so $r=0$ and $f\in (y-x^2)$. Hence $I(X)=(y-x^2)$ and $k[X]=k[x,y]\mathrm{/}(y-x^2)\mathrm{.}$ Define $\varphi :k[t]\to k[X]$ by $\varphi (t)=\bar{x}$. By the universal property of the polynomial ring, this is a $k$-algebra homomorphism, and every element of $k[X]$ can be written using only $\bar{x}$ because $\bar{y}={\bar{x}}^2$.

Conversely, define $\psi :k[X]\to k[t]$ on polynomial classes by substituting $x\mapsto t$ and $y\mapsto t^2$. The relation $y-x^2$ maps to $0$, so $\psi$ is well defined. The composites satisfy $\psi (\varphi (t))=t,\varphi (\psi (\bar{x}))=\bar{x},\varphi (\psi (\bar{y}))=\varphi (t^2)={\bar{x}}^2=\bar{y},$ so $\varphi$ and $\psi$ are inverse isomorphisms. Hence $k[X]\cong k[t]=k[{\mathbf{A}}_k^1]\mathrm{.}$

Set-theoretically, the maps $t⟼(t,t^2),(x,y)⟼x$ identify $X$ with the affine line, showing that the parabola is cut out by a nontrivial equation in the plane but has the same intrinsic affine geometry as ${\mathbf{A}}_k^1$.

Let $k$ be an algebraically closed field and let $X=V(xy)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ By Zero loci in affine space are the closed sets of the classical Zariski topology, $V(xy)=V(x)\cup V(y),$ so $X$ is the union of the two coordinate axes. Each axis is a proper closed subset of $X$, hence $X$ is reducible.

If a polynomial $f\in k[x,y]$ vanishes on $X$, then $f(x,0)$ vanishes at every element of the infinite field $k$, so it is the zero polynomial and $y$ divides $f$. Likewise, $f(0,y)=0$ identically, so $x$ divides $f$. Since $x$ and $y$ are coprime, $xy$ divides $f$. Thus $I(X)=(xy)$.

Its coordinate ring is $k[X]=k[x,y]\mathrm{/}(xy)\mathrm{.}$ The classes $\bar{x}$ and $\bar{y}$ are both nonzero: for instance, $x$ does not vanish at $(1,0)$ and $y$ does not vanish at $(0,1)$, so neither lies in the ideal $(xy)$. But $\bar{x}\bar{y}=\overline{xy}=0.$ Thus $k[X]$ has zero divisors, exactly as the irreducible-domain criterion predicts.

Assume the Axiom of Choice, and let $k$ be an algebraically closed field. Inside ${\mathbf{A}}_k^1$ with coordinate $t$, the punctured affine line is ${\mathbf{A}}_k^1\setminus \{0\}=D_{{\mathbf{A}}_k^1}(t)\mathrm{.}$ Therefore Regular functions on a principal open are the principal localization of the coordinate ring identifies its regular-function ring with the principal localization $k[t{]}_t,$ the Laurent polynomial ring obtained by adjoining $t^{-1}$.

By On the affine line, the classical Zariski topology is cofinite, the closed subsets of ${\mathbf{A}}_k^1$ are exactly the finite sets and the whole line. Hence $\{0\}$ is closed, so its complement ${\mathbf{A}}_k^1\setminus \{0\}$ is open; but that complement is not all of ${\mathbf{A}}_k^1$ and is infinite, so it is not closed. This is the basic quasi-affine example: it is affine as a principal open, but not as a closed subset of the ambient affine line.

Over an infinite field, the Zariski topology on ${\mathbf{A}}_k^1$ is Hausdorff.

Let $a\mathrm{\ne }b$ be two points of ${\mathbf{A}}_k^1$. By On the affine line, the classical Zariski topology is cofinite, every nonempty open subset of ${\mathbf{A}}_k^1$ is the complement of a finite set. If $U$ and $V$ are nonempty open neighborhoods of $a$ and $b$, then $U\cap V={\mathbf{A}}_k^1\setminus (F\cup G)$ for finite sets $F$ and $G$. Because $k$ is infinite, $F\cup G$ is still a proper subset of $k$, so $U\cap V\mathrm{\ne }\mathrm{\varnothing }$.

Thus no two distinct points admit disjoint neighborhoods. The refuted statement is therefore false: the classical Zariski topology on ${\mathbf{A}}_k^1$ is not Hausdorff.

The underlying zero locus of an ideal determines the same quotient ring as the ideal itself.

In $k[x]$, the ideals $(x)$ and $(x^2)$ cut out the same point set because A zero locus depends only on the generated ideal and its radical gives $V(x)=V(x^2)\mathrm{.}$

The quotient rings are nevertheless different: $k[x]\mathrm{/}(x)\cong k$ is reduced, while in $k[x]\mathrm{/}(x^2)$ the class $\bar{x}$ is nonzero and satisfies ${\bar{x}}^2=0$. So the second quotient remembers a nilpotent thickening that the common zero locus does not see. This refutes the statement.

By definition, $V(1)=\mathrm{\varnothing }\mathrm{.}$ Conversely, every polynomial vanishes on the empty set, so $I(\mathrm{\varnothing })=k[x_1,\dots ,x_n]\mathrm{.}$ Therefore the coordinate ring of the empty affine algebraic set is $k[\mathrm{\varnothing }]=k[x_1,\dots ,x_n]\mathrm{/}k[x_1,\dots ,x_n]=0.$

This is the exact boundary case promised by the affine-set convention: the empty algebraic set matches the unit ideal and the zero ring.