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Let $k$ be any field and suppose $M$ is a $k \times k$ bimodule. Can we say $M$ is also a $k$-module (i.e a vector space over $k$)? (by considering the inclusion $k \hookrightarrow k \times k$ or is there a "natural" way in which $M$ is a $k$-module?

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These seem like quite separate questions. –  Dylan Moreland Jun 27 '12 at 15:47
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(1) Does $(\alpha+\beta)\cdot v=\alpha \cdot v+\beta\cdot v$ hold? (2) Yes. –  anon Jun 27 '12 at 16:02
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@user10 Sorry, when you say $k \times k$ bimodule, what do you mean? –  Dylan Moreland Jun 27 '12 at 16:10
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I should say that I thought you either meant $(k, k)$-bimodule [a left and right $k$-module structure that were compatible] or $k \times k$-bimodule as a shorthand for $(k \times k, k \times k)$-bimodule. –  Dylan Moreland Jun 27 '12 at 16:26
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@user10 Actually, it seems to me that you still have a left $k \times k$-module structure, and then you can do what I said above. Of course, you have to decide between the left and right action, so maybe this isn't so natural. –  Dylan Moreland Jun 27 '12 at 16:39

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If "$k\times k$ bimodule $M$" refers to a bimodule like this $_kM_k$, then yes. For any bimodule $_AM_B$, it is both a left $A$ and a right $B$ module.

If you mean a module over a product ring like this $_{k\times k}M$, then consider what happens if you define $k\cdot m:=(k,1)m$. (I'm saying look for something that goes wrong :) )

If you want to try $k\cdot m:=(k,0)$ then you will get a module, but it doesn't have to be unital.

Combining both of these, if you mean $_{k\times k}M_{k\times k}$, you can pass to a single side and restrict to the action of one factor, if you wish.

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