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Question as in title. I only really need the special case where one of the open embeddings is the identity, but the more general case would be useful to know.

Edit - By product I mean: given $U\to X$ and $V\to Y$ open embeddings, is the canonical map $U\times V \to X\times Y$ an open embedding?

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There could be more than one interpretation to your question. If $U\hookrightarrow X$ and $V\hookrightarrow Y$ are open immersions, is $U\times V\hookrightarrow X\times Y$ an open immersion? Or the question could be, given $U, V\hookrightarrow X$ open immersions, is $U\times V\to U\hookrightarrow X$ an open immersion? –  Matt Mar 1 '11 at 0:27
    
@Matt I mean the product in the category of schemes of the two maps $U \to X$ and $V \to Y$. I'll edit to make it clear. –  David Roberts Mar 1 '11 at 0:36
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up vote 5 down vote accepted

Yes. In fact, this can be seen as follows. Open immersions are preserved under base-change; thus $U \times V \to X \times V$ is an open immersion; similarly so is $X \times V \to X \times Y$; thus the composition is.

Most of the nice properties of schemes (immersions, separatedness, properness, finite type, quasicompactness, quasiseparatedness, smoothness, flatness, quasifiniteness, surjectivity, radicialness) are preserved the basic operations of base-change and composition. This implies that they are preserved under products as you ask. (Much of made of this in EGA.)

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Thanks. I had guessed it would be, but I thought my topological intuition would let me down. –  David Roberts Mar 1 '11 at 0:46
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