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?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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. |
|||
|
|
