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Given two algebras $A$ and $B$, and two ideals $I, J \subseteq B$ with non-empty intersection, is it true that $$ (A \otimes I) \cap (A \otimes J) = A \otimes (I \cap J)? $$ (Where both sides of the equality are conatined in $A \otimes B$.) If so, then why?

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  • $\begingroup$ I don't understand this. Inside what set are $A\otimes I$ and $B\otimes J$ subsets? $\endgroup$ Dec 13, 2010 at 19:30
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    $\begingroup$ If "algebras" means "$k$-algebras" for a field $k$, then yes. More generally, if $A$ and $C$ are two $k$-vector subspaces of a $k$-vector space $U$, and if $B$ and $D$ are two $k$-vector subspaces of a $k$-vector space $V$, then $\left(A\otimes B\right)\cap \left(C\otimes D\right) = \left(A\cap C\right)\otimes \left(B\cap D\right)$. For a proof, consider bases of $A\cap C$, $A\diagup C$ and $C\diagup A$ and similarly for $B$ and $D$. On the other hand, if you are NOT over a field $k$, then your question doesn't even make sense: $A\otimes I$ is not (in general) a subset of $A\otimes B$. $\endgroup$ Dec 13, 2010 at 19:52
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    $\begingroup$ And if $A$ <i>is</i> flat over the base ring, then it will preserve monos and pullbacks of pairs of monos = intersections. The algebra structure of $B$ doesn't really enter into this at all. $\endgroup$
    – Todd Trimble
    Dec 13, 2010 at 19:55
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    $\begingroup$ Any kind of basis. I guess your algebras are finitely-dimensionally filtered or something like that; there is no reason to use AC in most applications. $\endgroup$ Dec 13, 2010 at 20:44
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    $\begingroup$ Besides, you can always restrict yourself to finite-dimensional subspaces of $A$, $B$, $C$, $D$ instead of considering the whole $A$, $B$, $C$, $D$ (remember, every tensor is a sum of finitely many pure tensors, and every pure tensor comes from one-dimensional subspaces), and again there is no need for Hamel bases. $\endgroup$ Dec 13, 2010 at 20:46

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