5

  • Context: This is exercise $24$ from Dummit and Foote's section on Algebraic Geometry. Let $V=\mathcal{Z}(xy-z^2)\subset \mathbb{A}^3$. Where we work over the base field $k$.

  • If $k$ is a finite field, then $V\simeq W \Longleftrightarrow \operatorname{card}(V) = \operatorname{card}(W) $ (Establish a biyection and prove it is a morphism by Lagrange interpolation on each coordinate).

  • In the case $k$ is an algebraically closed field, $\mathcal{I(V)} = (xy-z^2)$ (here we use $k$ is algebraically closed and Nullstellensatz) so it's enough to show $k[x,y] \not \simeq k[x,y,z]/(xy-z^2)$, which is clear from noting one is an UFD and the other is not ($z^2=x\cdot y = z\cdot z$).

  • My question is how can we prove or disprove $k[x,y] \simeq k[x,y,z]/\mathcal{I}(\mathcal{Z}((xy-z^2)))$ when $k$ is infinite but not algebraically closed? (e.g. $k=\mathbb{R}$)

I tried so far:

Division as in The answer to this question about ideals in not algebraically closed fields ,I'm not familiarized with the dimension of a domain (also I couldn't find a source online for this topic). But I tried the division approach:

If $f\in I:= \mathcal{I}(\mathcal{Z}(xy-z^2))$ then we can perform division on $k(y,z)[x]$ and get \begin{align*} f=g(xy-z^2) + r' \text{ with } r'\in \frac{1}{y}k(z)[y] \end{align*} Since $A := A_1\times A_2$ with $A_1 = k\setminus\{0\}$, $A_2=k$ and since $A$ is the cartesian product of infinite sets (cf. this question) and $f|_A=r'|_A=0$ we have $r'=0$ as polynomial, so $f\in I$ and the result follow as in the algebraically closed field approach.

Is this argument correct?

PD: I was able to solve this problem for $V=\mathcal{Z}(xy-z)$, using the projection $\pi:V\to \mathbb{A}^2, (x,y,z)\mapsto (x,y)$.

RyeCatcher
  • 464
  • 3
  • 14

1 Answers1

2

Here's a sketch of a proof that given $I = (xy - z^2)$ and $V = \mathcal{Z}(I)$, then $I = \mathcal{I}(V)$, following a similar approach as part (3) of the Examples on p. 660.

For any polynomial $f(x, y, z) \in k[x, y, z]$, we can write

$$ f(x, y, z) = f_0(x, y) + f_1(x, y) z + (xy - z^2) g(x, y, z). $$ (This follows from the theory of Gröbner bases from Sec. 9.6, in particular Theorem 23.)

So now if $f \in \mathcal{I}(V)$, then it follows that

$$ f_0(x, y) + f_1(x, y) z = 0 $$ for all $(x, y, z)$ such that $xy = z^2$.

Now pick some one-dimensional subset of $V$ (i.e., some parameter $t \in k$ and functions $x(t), y(t), z(t)$ such that $x(t) y(t) = z(t)^2$ for all $t$, being careful to not assume too much about $k$) so that the polynomial above turns into a polynomial in $t$ with infinitely many roots, and so you can use $k$ being infinite to conclude that the polynomial in $t$ is $0$, i.e. all its coefficients are $0$.

Then if you pick your functions $x(t), y(t), z(t)$ carefully, you can then conclude that the coefficients of $f_0$ and $f_1$ must then also be $0$, and thus that $f \in I$.

Fred Akalin
  • 869