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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.