Let $X \subset k^m , Y \subset k^n$ be algebraic sets ($k$ an algebraically closed field). Then $X\times Y \subset k^{m+n}$ is an algebraic set whose Zariski topology is finer than the product topology of the Zariski topologies of the factors. Could someone give me an example that demonstrates that it can be strictly finer?

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    $\begingroup$ I'm having trouble being totally rigorous here, but consider affine 2-space as product of affine 1-space with itself. The basic idea is that not all polynomials in x and y are products of a polynomial in x with a polynomial in y. To be a little more specific, consider the circle $x^2-y^2-1$ in 2-space. This should not be closed in the product space. There is a characterization of closed sets in the product space as intersections of finite unions of products of closed sets, if that helps. Note here that affine 1-space has the cofinite topology. $\endgroup$
    – Seth
    Commented Jan 25, 2015 at 18:45
  • $\begingroup$ @seth thanks for your help $\endgroup$
    – Mithras
    Commented Jan 25, 2015 at 19:20

1 Answer 1


(I will take $n=m=1$, you will adapt in higher dimensions.) Let $P = X-Y$ and $Z = V(P)$, so that $Z$ is closed. Now $Z = \{(x,x)\;|\; x\in \mathbf{A}_k^1\}$. Now (easy) $Z$ is closed in the product topology if and only if $\mathbf{A}_k^1$ is Hausdorff, and this is not the case as $k$ is infinite, being algebraically closed, and as (independently of $k$ finiteness) the closed sets of the affine line are the whole affine line and finite sets.

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    $\begingroup$ Remark : the following is purely general : a topological space $X$ is Hausdorff if and only if $\Delta_X = \{(x,x)\;|\;x\in X\}$ is closed in $X\times X$, the latter being endowed with the product topology. $\endgroup$
    – Olórin
    Commented Jan 25, 2015 at 18:57

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