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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!

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    Except that Matt E proved me wrong by studying the $k[x]$-module structure in his nice answer!2011-09-09

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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 $k[x]/xk[x]$.

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    That's a great point. Not a huge issue though. The main thrust was that since elements are identified if they differ by only finitely many terms, then we can extend to$k(x)$by simply throwing out which ever terms of a representative element would go to negative exponents, and then shifting the rest downward. Thanks!2012-03-08