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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.

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  • $\begingroup$ How does the zero locus of $k[x_1]$ and $k[y_1]$ relate to the zero locus of $k[x_1,y_1]$. If you can answer that, you may be able to answer your question. $\endgroup$ – Justin Benfield Sep 1 '16 at 18:32
  • $\begingroup$ Would this prove the claim in your title? By this I mean: is it obvious that all closed sets in the product topology look like this? [I think you mean to write $\subseteq$ instead of $\in$ in the third paragraph but this is minor.] $\endgroup$ – Hoot Sep 1 '16 at 21:13
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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.

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The Zariski closed subsets in $A^1$ are the finite subsets and $A^1$.

If all the subsets $X_i$ and $Y_i$ are finite, you can suppose without really restricting the generality that $X_i=\{x_i\}, Y_i=\{y_i\}$, $\bigcup_i X_i\times Y_i=Z(I)$ where $I$ is the ideal generated by $(X-x_i,Y-y_i)$.

If one $X_i$ and one $Y_j$ is $A^1$, then $A\times A=\bigcup_i X_i\times X_j$.

If you have some $X_i=A^1$ and no $Y_i$ is $A^1$, then $\bigcup_iX_i\times Y_i$ is the union of a finite subset $F$ and lines defined $Y=y_i$ if $X_i=A^1$. You conclude by saying that a finite union of Zariski closed subsets is closed.

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  • $\begingroup$ It looks like your ideal $I = (P)$ in the first case cuts out a union of coordinate lines. The ideal defining $X_i \times Y_i$ should be the non-principal ideal $(X - x_i, Y - y_i)$, and the ideal defining $\cup_i X_i \times Y_i$ should be the product of all such ideals. $\endgroup$ – Ravi Fernando Sep 1 '16 at 19:58
  • $\begingroup$ thanks for the correction. $\endgroup$ – Tsemo Aristide Sep 1 '16 at 20:01

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