Is simple module over commutative ring always a field? M is a simple module if and only if $M\cong R/I$ for some I maximal ideal in R. If $R$ is commutative, can I say M is a field? I'm confused about this fact because when proving it I use the fact that M is field... But then why call it module?
 A: The natural thing to say is that $M$ becomes a vector space of dimension$~1$ over the field $K=R/I$. Of course any such vector space is isomorphic after choosing a basis to $K^1$, which can be identified in an obvious way with $K$ itself (so you can equip $K^1$ with an internal multiplication and it becomes a field). However this does involve the choice of a single basis vector (which then becomes the neutral element for multiplication), and different choices give different field structures on$~M$. So it would be confusing to say that $M$ is a field, though it can be made into a field in various ways.
The main point here is that being isomorphic does not mean having a canoncial isomorphism at hand.
A: $M$ is module isomorphic to $R/I$, but that doesn't mean it has suddenly become a field.
Now, you can treat the module $M$ as a ring by transferring the structure of $R/I$ back to $M$ through the bijection, but until you do that, $M$ does not have any binary multiplication operation, and hence it does not make any sense to call it a field (or a ring.)
