Computation of Algebraic Sets and Coordinate Rings (15.1.24)

Let $k$ be an infinite field, and let $V=\mathcal{Z}(xy-z)$. Prove that $V$ is isomorphic to $\mathbb{A}^2$ and provide an explicit isomorphism $φ$ from $V$ to $\mathbb{A}^2$ and associated $k$-algebra isomorphism $Φ$ from $k[V]$ to $k[\mathbb{A}^2]$ and their inverses. Is $V=\mathcal{Z}(xy-z^2)$ isomorphic to $\mathbb{A}^2$?

Proof: We see $V$ is the set of all points of the form $(x,y,xy)$, so there is an isomorphism $φ~:~V→W$ given by $φ(x,y,xy)=(x,y)$ with inverse $ψ~:~W→V$ given by $ψ(x,y)=(x,y,xy)$. We see $(xy-z)⊆\mathcal{I}(V)$ naturally and by reducing modulo $(xy-z)$ any $f∈\mathcal{I}(V)$ to a polynomial in $k[x,y]$ we see the remainder must be $0$ as it would annihilate all of $\mathbb{A}^2$, hence $\mathcal{I}(V)⊆(xy-z)$. Thus this induces the isomorphism $Φ~:~k[W]→k[V]$ where $k[V]=k[x,y,z]/(xy-z)$ and $k[W]=k[x,y]$ given by $Φ(f(x,y))=\overline{f(x,y)}$ with inverse $Ψ~:~k[V]→k[W]$ given by $Ψ(\overline{f(x,y,z)})=f(x,y)$.

For the second part, we shall show $k[V]≇k[W]$ where $W=\mathbb{A}^2$ so that $V≇W$ as algebraic sets. First, we have $(xy-z^2)⊆\mathcal{I}(V)$. Assume $\mathcal{I}(V)⊈(xy-z^2)$ and let $f∈\mathcal{I}(V) \setminus (xy-z^2)$ be minimal with respect to the lexicographic ordering $z > y > x$, so $f=z·g_1(x,y)+g_2(x,y)$.

Let $k^2$ be the subset of nonzero squares in $k$, necessarily infinite since each square has only a finite number of square roots, and for each nonzero $b∈k$ let $k^2/b = \{v/b~|~v∈k^2\}$. Note $V$ is the set of points of the form $(x,y,\sqrt{xy})$ whenever $xy$ is square, so for each nonzero $b∈k$ we have an infinite number of $a∈k^2/b$ such that $$f(a,b,\sqrt{ab})=f(a,b,-\sqrt{ab})=$$ $$(\sqrt{ab})g_1(a,b)+g_2(a,b)=(-\sqrt{ab})g_1(a,b)+g_2(a,b)=0$$ Since $\sqrt{ab}$ is nonzero, for each such $b$ we have $g_1(x,b)∈k[x]$ has an infinite number of zeros $x=a$, so $g_1(x,b)=0$ for these $b$. Since there are an infinite number of such $b$, we have $g_1(x,y)∈k(x)[y]$ has an infinite number of roots $y=b$ in $k⊆k(x)$, hence $g_1(x,y)=0$ and by the same argument we find $g_2(x,y)=0$. Hence $f=0$ and we have shown $\mathcal{I}(V)=(xy-z^2)$.

We see $k[W]=k[x,y]$ is a UFD (cf. 9.3.8), and we shall prove $k[V]≇k[W]$ by showing $k[V]=k[x,y,z]/(xy-z^2)$ is not a UFD. Assume otherwise and observe the element $x∈k[V]$, which we shall prove is prime; we may write any element in $k[V]$ uniquely as a polynomial in $k[x,y][z]$ of degree $≤1$, so assume $x=(g_1+z·g_2)(g_3+z·g_4)$. Then $x=g_1g_3+xy·g_2g_4+z(g_1g_4+g_2g_3)$ is the unique reduced form, so $g_1g_3+xy·g_2g_4=x$ and $g_1g_4+g_2g_3=0$. Taking the first relation modulo $x$ we may assume without loss that $x~|~g_1$, and now taking the second relation modulo $x$ necessarily either $x~|~g_2$ or $x~|~g_3$; the former implies $x~|~(g_1+z·g_2)$, and the latter that the relation $g_1g_3+xy·g_2g_4$ involves adding monomial terms of degree strictly larger than $x$, hence the expression does not equal $x$. Thus $x$ must be prime. But $xy=z^2$ would imply $x~|~z$, a contradiction by observing unique representation. Hence $k[V]$ is not a UFD and not isomorphic to $k[W]$, and so neither is $V$ isomorphic to $\mathbb{A}^2$.$~\square$