Ring k[[x]] when viewed as a module over k[x] I had a question that I hope makes sense.  If I set $R = k[x]$ and then have $M=k[[x]]$ be an R-Module, then I can look at the ideal $(x)$ in $R$ and embed it into the formal power series ring as an R-submodule.  What I am curious about what $M/(x)$ looks like, since all of the elements of $(x)$ consist of finite sums.  Does anyone have any idea?  Thanks!
 A: It's a little easier to think about $k[[x]]/k[x]$, which is not so far from
$k[[x]]/x k[x]$.
The most interesting thing I know to say about the quotient $k[[x]]/k[x]$ is that its $k[x]$-module structure naturally extends to a vector space structure over $k(x)$.  
Indeed, if $p(x) \in k[x]$ is irreducible but prime to $x$ (i.e. has non-zero
constant term), then multiplication
by $p(x)$ is invertible on $k[[x]]$, and hence also on the quotient 
$k[[x]]/k[x]$.  
If $f(x) \in k[[x]]$ is such that $x f(x) \in k[x]$, then $f(x)$ itself is a polynomial.  Hence multiplication by $x$ is injective on $k[[x]]/k[x]$.  On the other hand,
if $f(x) \in k[[x]]$, with constant term $a$, then $f(x) -a \in x k[[x]]$,
while $f(x)$ and $f(x) -a $ have the same image in $k[[x]]/k[x]$.
Thus multiplication by $x$ is also surjective on $k[[x]]/k[x]$.
We have proved that multiplication by every irreducible element of $k[x]$ is invertible on $k[[x]]/k[x],$ and this implies that $k[[x]]/k[x]$ is naturally a $k(x)$-vector space.

Turning now to your quotient $k[[x]]/xk[x],$ it has $k[[x]]/k[x]$ as a quotient,
with kernel equal to $k[x]/xk[x]$.
In other words, thought of as a $k[x]$-module, it is an extension of a $k(x)$-vector space by the torsion module $k[x]/xk[x]$.

Added: There is a blunder in the above argument: if $p(x)$ is irreducible but prime to $x$, then multiplication by $p(x)$ is certainly invertible on $k[[x]]$,
but this doesn't imply that multiplication by $p(x)$ is invertible on $k[[x]/k[x]$, just that multiplication by $p(x)$ is surjective on this quotient.
[Compare with the fact that e.g. multiplication by $3$ on $\mathbb Q$ is
invertible, but multiplication by $3$ on $\mathbb Q/\mathbb Z$ is surjective, but not injective. The problem is that mult. by $3$ is not invertible on $\mathbb Z$.]
What is true is that $k[[x]]/k[x]_{(x)}$ --- where $k[x]_{(x)}$ is the localization of $k[x]$ at the prime ideal $(x)$ --- is a vector space over $k(x)$.  The point is now that if $p(x)$ is irreducible and prime to $x$, then
multiplication by $p(x)$ is invertible on $k[x]_{(x)}$, and hence multiplication by $p(x)$ is invertible on $k[[x]]/k[x]_{(x)}$.  (And the proof that mult. by $x$ is a bijection carries over.)
So what we deduce is that $k[[x]]/k[x]$ is an extension of the $k(x)$-vector
space $k[[x]]/k[x]_{(x)}$ by the quotient $k[x]_{(x)}/k[x]$, which is a divisible module over $k[x]$, every element of which is annihilated by some power of an irreducible polynomial $p(x)$ that is prime to $x$.
And $k[[x]]/xk[x]$ is an extension of this module by the torsion module
$k[x]/xk[x]$.
