So I am trying to prove that the product of two affine varieties is the categorical product in the correspondent category. If $X$ and $Y$ are two affine varieties on $K^n$ and $K^m$ resp., then $X\times Y$ is the topological subspace with the topology inherited from $K^{n+m}$ in its Zariski topology. It is closed as well, for if $X$ is given by the zeros of $p_1,\dots,p_k$ and $Y$ by the zeros of $q_1,\dots,q_l$, then $X\times Y$ is given by the zeros of
$$ r_i(t_1,\dots,t_{n+m}) = \begin{cases} p_i(t_1,\dots,t_n) & 1\leq i \leq k \\ q_i(t_{n+1},\dots,t_{n+m}) & k+1\leq i \leq k+l . \end{cases} $$
(If $X$ and $Y$ were quasiaffine, we would write them as the difference of two closed sets and the product would still the difference of two closed sets).
Let's assume I have managed to prove it is also reducible if so are $X$ and $Y$ (I have not tried yet).
Now, it should be easy to see that the projections $\pi_X$ and $\pi_Y$ are morphisms, so let's work on the universal property.
Let $Z$ be another (quasi)affine variety with morphisms $\varphi_X:Z\rightarrow X$ and $\varphi_Y:Z\rightarrow Y$. I have to show that there exists a unique morphism $\varphi:Z\rightarrow X\times Y$ satisfying $\varphi\pi_X=\varphi_X$ and $\varphi\pi_Y=\varphi_Y$. My candidate is $\varphi(z)=(\varphi_X(z),\varphi_Y(z))$, but I have problems to show it is actually a morphism. First, I should show it is continuous. Let $U\subset X\times Y$ be open. Then I can write $\varphi^{-1}(U)$ as the intersection
$$ \varphi_X^{-1}(\pi_X(U))\cap \varphi_Y^{-1}(\pi_Y(U)) .$$
But since $X\times Y$ does not have the product topology, I cannot ensure that $\pi_X$ and $\pi_Y$ are open.
- Question 1: How Can I show that $\varphi$ is continuous?
Next, let $z_0\in Z$ and $(x_0,y_0)=\varphi(z_0)$. Suppose that $f:X\times Y\rightarrow K$ is regular at $(x_0,y_0)$ and let's try to show that $\varphi f$ is regular at $z_0$. We know there is a neighbourhood $U$ of $(x_0,y_0)$ and polynomials $p,q\in K[t_1,\dots,t_{n+m}]$ with $q$ vanishing nowhere on $U$ such that
$$f(x,y)=p(x,y)/q(x,y) \qquad \forall (x,y)\in U . $$
Assuming Question 1 is solved (i.e. $\varphi$ is continuous), then $V=\varphi^{-1}(U)$ is a neighbourhood of $z_0$ such that $q(\varphi(z))\neq 0$ for all $z\in V$. Let's try to show that $\varphi q$ and also $\varphi p$ are polynomials in $K[t_1,\dots,t_N]$.
In my opinion, the key step here is to regard both $p$ and $q$ as sums of products of elements in $K[t_1,\dots,t_n]$ and $K[t_{n+1},\dots,t_{n+n}]$, (the tensor product actually), and hence
$$p(\varphi(z))/(q(\varphi(z)) = \frac{\sum_i m_i(\varphi_X(z))m'_i(\varphi_Y(z))}{\sum_j n_j(\varphi_X(z))n'_j(\varphi_Y(z))} $$
Now, each $varphi_Xm_i$, $\varphi_Ym'_i$ (and similarl) are regular by hypothesis, so they can be written as quotients of polynomials on some neighbourhood of $z_0$. In the end I find
$$ f(\varphi(z)) = \frac{\sum_i p_i(z)q_i(z)}{\sum _j p'_i(z)/q'_j(z)}, $$
with $q_i(z),q'_j(z)\neq 0$ for all $z\in V'$ and $i,j$ with $V'$ some small neighbourhood of $z_0$ made up with the intersections of all the previous ones. Rewriting the above I end up with an expression of the form $\tilde p(z)/\tilde q(z)$ for $f(\varphi(z))$, with $\tilde q$ vanishing nowhere on a neighbourhood of $z_0$.
- Question 2. Is this reasoning correct?