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Let $A$ be a not neccessarily commutative algebra, and let $B \subset A$ be a subalgebra of $A$. Moreover, let $M$ be an $A$-bimodule, and let $N \subset M$ be a $B$-sub-bimodule. The tensor product $N \otimes_{B} N$ has a natural inclusion in $M \otimes_{A} M$, and it seems to me that this inclusion should be injective, but I can't prove it.

Am I right here, or does one need to make extra assumptions? Is there a clean/non-messy way to prove all this?

The question boils down to showing that $$ (N \otimes_B N) \cap \lbrace m_1a \otimes m_2 - m_1 \otimes am_2 | m_i \in M\, a \in A \rbrace. $$ is equal to $$ \lbrace n_1b \otimes n_2 - m_1 \otimes bm_2 | n_i \in N,b \in B \rbrace. $$ But I can't see how to do this.

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  • $\begingroup$ By "isomorphism" you mean "injective"? $\endgroup$ – darij grinberg Feb 19 '12 at 20:06
  • $\begingroup$ As to why it should be an isomorphism, I don't know, it just seemed natural that it should be. Maybe more restrictions are needed? $\endgroup$ – Ago Szekeres Feb 19 '12 at 20:07
  • $\begingroup$ Well, the question you ask in the first paragraph definitely does not boil down to the last paragraph. You should fix more typos. $\endgroup$ – darij grinberg Feb 19 '12 at 20:08
  • $\begingroup$ Yes, I mean injective (what I meant by isomorphism was isomorphic to the image of $N \otimes_B N$ in $M \otimes_A M$ - a bad choice of terminology I admit). $\endgroup$ – Ago Szekeres Feb 19 '12 at 20:26
  • $\begingroup$ I think (but I am not an algebraist) that the moral is: tensoring, even with $A=B$, often does not preserve monomorphisms -- see Florian's comment below -- and this is of course where Tor enters the fray. $\endgroup$ – Yemon Choi Feb 20 '12 at 1:29
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Here is a counterexample: $A=k[x]/x^2$, $B=k$, $M=N=A$ as an $(A,A)$-bimodule. Then $\dim_k M\otimes_A M = \dim_k M = 2$, but $\dim_k N\otimes_B N = 4$, so $N\otimes_B N$ cannot possibly embed into $M\otimes_A M$. And of course there are also examples where the natural map $N\otimes_B N\longrightarrow M\otimes_A M$ is not surjective.

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  • $\begingroup$ And there are also examples with $B=A$ (which make life hard for Hopf algebraists). $\endgroup$ – darij grinberg Feb 19 '12 at 20:25
  • $\begingroup$ Then do there exist sufficent conditions for the inclusion to be injective? $\endgroup$ – Ago Szekeres Feb 19 '12 at 20:28
  • $\begingroup$ @Ago: In some special cases this might be true (e.g. take $A=B=k$). But in general you would always expect $M\otimes_B M$ to be bigger than $M\otimes_A M$. $\endgroup$ – Florian Eisele Feb 19 '12 at 20:32
  • $\begingroup$ But I asking that $N$ be a $B$-sub-module, and so, elements of the form $na$, for $n \in N$, and $a \in A$, may not be in $N$, so $n_1 a \otimes n_2 − n_1 \otimes an_2$ may not be in $N \otimes N$. So if I require that $na$ is in $N$ iff $a \in B$, would I get injectivity? $\endgroup$ – Ago Szekeres Feb 19 '12 at 20:41
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    $\begingroup$ @Ago: If $A=B$, then your condition would be satisfied trivially. Still there are examples where the natural map is not an embedding (think of $A=B=k[x]/x^2$, $M=k[x]/x^2$, $N=\langle x \rangle$; Here $N\otimes_B N$ is one-dimensional, $M\otimes_A M$ is two-dimensional, but the natural map between the two is the zero map). $\endgroup$ – Florian Eisele Feb 19 '12 at 21:24

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