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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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    As @George said (but deleted(?) the comment before I had managed to mark it as a good one): Two power series of $M$ represent the same element of $M/(x)$, iff their coefficients differ only at finitely many powers $x^i$ not including the constant. This object is not a ring, because $(x)$ is not an ideal of $M$.2011-09-08
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    @Jyrki Lahtonen: Sorry for deleting it. I had deleted it out of failure to come up with a nicer description.2011-09-08
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    @George: No harm done. I was just about to suggest that you write it as an answer. I don't think there's anything more to be said.2011-09-08
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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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    Thanks for the information, however I'm confused on a few parts. When say it naturally extends to a vector space structure over $k(x)$ are you saying that it has a vector space structure over the rational functions? and if so, don't we already know this because $k(x)$ is a field? Also, in your second paragraph you say that multiplication of a polynomial with non-zero constant term is invertible on $k[[x]]$. Isn't it only invertible by multiplying with an element from $k[[x]]$ and not with a polynomial? Because don't we not have this multiplication available with a $k[x]$-module structure?2012-03-01
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    @Luke: Dear Luke, The quotient $k[[x]]/k[x]$ is *a priori* just a $k[x]$-module, but I am claiming that this structure is actually a $k(x)$-module structure. (This is like having something given to you a an abelian group, but then discovering that is uniquely divisible, and hence is secretly a $\mathbb Q$-vector space.) I don't understand what you mean when you write "don't we already know this"? Certainly *if* we had a $k(x)$-module structure, it is a vector space --- by definition, $k(x)$-modules are the same thing as $k(x)$-vector spaces (since $k(x)$ is a field). ...2012-03-02
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    The point is that we have to show that the $k[x]$-module structure actually does extend to a $k(x)$-module structure. Regarding your second question: I am not saying that the inverse lies in $k[x]$; I'm just saying that multiplication by $p(x)$ induces a bijection on $k[[x]]$. E.g. $3$ is not invertible in $\mathbb Z$, but mutliplication by $3$ is invertible as an endomorphism of $\mathbb Z/5\mathbb Z$, or (and this is closer to the situation we have) as an endomorphism of the $5$-adic integers $\mathbb Z_5$. Regards,2012-03-02
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    cool, that makes sense! Thanks!2012-03-02
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    @Luke: Dear Luke, There was a mistake in the above argument (based on an unconscious substitution of $k[x]_{(x)}$ for $k[x]$ on my part), which is now corrected. Regards,2012-03-06
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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