I am trying to wrap my head around these new concepts. Suppose I have a scenario where $V$ is a free module and also $V$ is a finite dimensional module, if V is simple, must $V$ be dimension $1$?
Def:
A module is said to be simple if any submodule of such module is either $\{0\}$ or itself.
My thought is that if we have a $A$-module $V$ which happens to be also finite dimensional then say $V=\langle v_1,\cdots,v_n\rangle$, with the scalars come from $A$, then surely if we take any proper subset of such basis vectors, their span would be a proper submodule of $V$? Hence $V$ is simple only if $n=1$? (of course the converse is not true, if $n=1$ then it wouln't necessarily be simple)
I feel like there is something wrong with my argument, I must have confused something from modules with Linear Algebra.
Best Answer
Nope. Simplicity of a module of an algebra over a field does not say much about its dimension.
Consider the matrix ring $R=M_2(F)$ for a field $F$.
Then $F\times F$, viewed as a right $R$ module using matrix multiplication, is a simple module that is $2$ dimensional over $F$.
Another example with the same idea: let $V$ be an infinite dimensional vector space, and $E$ be its ring of linear transformations. Then $V$ is in fact a simple $E$ module, and yet $V$ has infinite $F$ dimension.
This all assumes you meant "dimension" in the way it is most commonly used: as an invariant of a vector space. If in fact you meant some other type of dimension, you'll have to specify. There are several you could be talking about. For example, if by dimension you meant "composition length", then yes, a simple module has composition length $1$.