most recent 30 from http://mathoverflow.net 2013-05-19T05:07:53Z http://mathoverflow.net/feeds/question/88956 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88956/tensor-products-sub-algebras-sub-modules-and-inclusions Ago Szekeres 2012-02-19T19:59:30Z 2012-02-19T20:22:00Z

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

http://mathoverflow.net/questions/88956/tensor-products-sub-algebras-sub-modules-and-inclusions/88959#88959 Florian Eisele 2012-02-19T20:13:01Z 2012-02-19T20:13:01Z

<p>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. </p>