I am working on the problem from "Ideas, Varieties and Algorithms" by David Cox, John Little and Donal O'Shea. Here is the homework problem for my course.
Let $V \subset k^n$ and $W \subset k^m$ be two affine varieties, and let
$$\begin{aligned}
V \times W &= \{(x_1, \dots, x_n, y_1, \dots, y_m) \in k^{n + m}|\\&\quad(x_1, \dots x_n) \in k^n, (y_1, \dots y_m) \in k^m\}
\end{aligned}$$be their cartesian product. Prove that $V \times W$ is an affine variety in $k^{n + m}$. Hint: If $V$ is defined by $f_1, \dots, f_s \in k[x_1, \dots, x_n]$, then we can regard $f_1, \dots, f_s$ as polynomials in $k[x_1, \dots, x_n, y_1, \dots, y_m]$, and similarly for $W$. Show that this gives defining equations for the cartesian product.
My Attempt
Assume that $V \subset k^n$ and $W \subset k^m$ are affine varieties. That is:
$$\begin{aligned}
V &= \{(x_1, \dots, x_n) \in k^n |f_1=f_2=\cdots=f_s = 0 \}\\
W &= \{(y_1, \dots, y_m) \in k^m |g_1=g_2=\cdots=g_t = 0 \}
\end{aligned}$$
As already given
$$\begin{aligned}
V \times W &= \{(x_1, \dots, x_n, y_1, \dots, y_m) \in k^{n + m}|\\&\quad(x_1, \dots x_n) \in k^n, (y_1, \dots y_m) \in k^m\}
\end{aligned}$$
Using the hint for this problem, we have the following:
- If $V$ is defined by $f_1, \dots, f_s \in k[x_1, \dots, x_n]$, then we can regard $f_1, \dots, f_s$ as polynomials in $k[x_1, \dots, x_n, y_1, \dots, y_m]$
- Likewise, if $W$ is defined by $g_1, \dots, g_t \in k[y_1, \dots, y_m]$, then we can regard $g_1, \dots, g_t$ as polynomials in $k[x_1, \dots, x_n, y_1, \dots, y_m]$
I am not sure what to claim for this problem (e.g. for $V \cup W$, we claim $V \cup W = \mathbf{V}(f_ig_i : 1 \leq i \leq s, 1 \leq i \leq t)$), so I go on and write what I understand about this problem…
Set $(x_1, \dots, x_n, y_1, \dots, y_m) \in (V \times W)$, where $(x_1, \dots, x_n) \in V$ and $(y_1, \dots, y_m) \in W$. Then, at the same time $f_1, \dots, f_s$ vanish at $(x_1, \dots, x_n)$ and $g_1, \dots, g_t$ vanish at $(y_1, \dots, y_m)$.
For the next step, I thought for around hours or so and then gave the following try:
Since $f_1, \dots, f_s \in k[x_1, \dots, x_n, y_1, \dots, y_m]$, and $f_1, \dots, f_s$ vanish at $(x_1, \dots, x_n)$, the set of those polynomials also vanish at $(x_1, \dots, x_n, y_1, \dots, y_m)$. Likewise, since $g_1, \dots, g_t \in k[x_1, \dots, x_n, y_1, \dots, y_m]$ and $g_1, \dots, g_t$ vanish at $(y_1, \dots, y_m)$, the set of those polynomials also vanish at $(x_1, \dots, x_n, y_1, \dots, y_m)$. So $f_ig_j$ for $1 \leq i \leq s$ and $1 \leq j \leq t$ vanish at $(x_1, \dots, x_n, y_1, \dots, y_m)$.
I think my proof needs a bit of adjustment or extra reasoning, but I'm not sure about this.
Any advices or comments about my approach?
Best Answer
Looks like the correct approach. Note that what you're showing is that $$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \}.$$
If you feel uncomfortable because the defining equations should be polynomials in $x, y$ (instead of just in $x$ for the $f$'s and just in $y$ for the $g$'s), you might want to make things explicit.
Say $\hat {f_i}(x_1,\dots,x_n,y_1,\dots,y_m) = f_i(x_1,\dots,x_n)$ (i.e., $f_i$ explicitly considered as a polynomial in $x, y$) and similarly $\hat{g_j}(x_1,\dots,x_n,y_1,\dots,y_m) = g_j(y_1,\dots,y_m)$. Phrased this way, you're showing that $$V \times W = \{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid \hat{f_1}(x,y) = \dots = \hat{f_s}(x,y) = 0, \hat{g_1}(x,y) = \dots = \hat{g_t}(x, y) = 0 \},$$ but the right-hand side of this is of course still equal to $$\{(x_1, \dots,x_n,y_1, \dots, y_m) \in k^{n+m} \mid f_1(x) = \dots = f_s(x) = 0, g_1(y) = \dots = g_t(y) = 0 \},$$ so that is really just a notational issue.