I understand that variants of this question have been asked before on this site, but I can't seem to find a satisfactory answer for my particular question.
Notation: we work over an algebraically closed field $k$, and $Z(S)\subseteq \mathbb{A}^n$ denotes the zero locus of a subset $S\subseteq k[x_1,\dots,x_n]$.
Given Zariski closed subsets $X_i,Y_i\in\mathbb{A}^1$ for $1\leq i \leq n$, I'd like to show that $\bigcup_{k=1}^{n}X_i \times Y_i$ is Zariski closed in $\mathbb{A}^1\times \mathbb{A}^1=\mathbb{A}^2$ (where we identify these two sets in the obvious way).
Attempt at a solution: it suffices to show that $X_1\times Y_1$ is closed, as a finite union of closed sets is closed. Also, we know $X_1=Z(U)$ and $Y_1=Z(W)$ for some subsets $U,W\subseteq k[x]$. This is where I am stuck. Obviously, I'd like to write $X_1 \times Y_1=Z(Y)$ for some $Y\subseteq k[x_1,y_1],$ but I can't seem to find $Y$. All solutions I have found online are not quite satisfactory, mostly appealing to the fact that the product of lines and points is again a Zariski closed set.
Any help is appreciated, thank you.
Best Answer
What's going on here is greatly clarified if you think of the two coordinate spaces of the product as being separate, totally unrelated spaces. So instead of thinking of $U$ and $V$ as both being subsets of the same polynomial ring $k[x]$, they are now subsets of two separate polynomial rings: $U\subseteq k[x_1]$ and $V\subseteq k[y_1]$. Now it's easy to guess what $Y$ should be: just take the union of $U$ and $V$ inside $k[x_1,y_1]$.
In terms of your original notation, this means $$Y=\{f(x_1):f(x)\in U\}\cup\{g(y_1):g(x)\in V\}.$$ I'll leave it to you to verify that this $Y$ works.