# Injective indecomposable modules over Laurent polynomial rings

What does the injective envelope of $$\mathbb C[x,x^{-1}]/(p(x,x^{-1}))$$ as a $$\mathbb C[x,x^{-1}]$$-module look like where $$p(x,x^{-1})$$ is an irreducible element? I’m sure this is well known, but when Googling I mostly found stuff for ordinary polynomial rings.

I am particularly interested in the possibile dimensions of the injective indecomposables over $$\mathbb C$$, i.e., can they be countable.

• k should be C. Sorry. I'll fix. – Benjamin Steinberg Oct 24 '18 at 14:29
• It does seem that the injective envelope of arbitrary cyclic modules over a PID should be well-known! – rschwieb Oct 24 '18 at 14:32
• @rschwieb, If you think about $\mathbb Z$ where you get $\mathbb Q$ and the Prufer p-groups, then it seems not so easy. That is why I want $\mathbb C$, so we have fewer irreducibles., which I realize I forgot to say. – Benjamin Steinberg Oct 24 '18 at 14:41
• I would have thought the Prüfer groups would have an analogue for any prime $p$ in a PID, but I don't understand them well :) Sorry I can only make comments like these instead of resolving the problem... – rschwieb Oct 24 '18 at 15:01
• Probably they do. Maybe you just invert the powers of the prime $p$. – Benjamin Steinberg Oct 24 '18 at 15:03

If $$R$$ is a PID and $$P$$ is a nonzero prime ideal, then $$E(R/P)=K/P_P$$, where $$K$$ is the fraction field of $$R$$ and $$R/P$$ is viewed as a submodule of $$K/P_P$$ via the evident map. Indeed, one readily checks that $$R/P\cong R_P/P_P$$ is an essential submodule of $$K/P_P$$ and that $$K/P_P$$ is divisible, hence injective (injective and divisible are the same since $$R$$ is a PID).

This fact can be generalized to arbitrary ideals in Dedekind domains:

Theorem. Let $$R$$ be a Dedekind domain with fraction field $$K$$ and let $$I$$ be an ideal of $$R$$ different from $$0$$ and $$R$$. Let $$P_1,\dots,P_t$$ denote the primes containing $$I$$ and let $$S$$ be the multiplicative set $$R\setminus (P_1\cup\dots\cup P_t)$$. Then the natural map $$R/I\to K/S^{-1}I$$ is an injective envelope of $$R/I$$.

Proof. Observe that the image of any $$s\in S$$ in $$R/I$$ is invertible. Indeed, it is not contained in any maximal ideal of $$R/I$$. Thus, $$R/I$$ is $$S$$-divisble, and it follows that the natural map $$R/I\to S^{-1}(R/I)=S^{-1}R/S^{-1}I$$ is an isomorphism of $$R$$-modules.

Next, I claim that $$K/S^{-1}I$$ is the injective envelope of $$S^{-1}R/S^{-1}I$$ viewed as an $$S^{-1}R$$-module. This is a generalization of the fact mentioned above: The ring $$S^{-1}R$$ is a PID with finitely many primes, namely, $$S^{-1}P_1,\dots,S^{-1}P_t$$ and $$S^{-1}I$$ is a nonzero ideal contained in their product. Using this, it is routine to check that $$S^{-1}R/S^{-1}I$$ is an essential $$S^{-1}R$$-submodule of $$K/S^{-1}I$$, and the latter is injective over $$S^{-1}R$$ because it is divisible.

To finish the proof it is enough to show that if $$M$$ is an $$R$$-module such that $$S^{-1}M$$ is injective as an $$S^{-1}R$$-module, then $$S^{-1}M$$ is injective as an $$R$$-module. This follows by writing down the diagram definition of injectivity, noting that the diagram maps into its localization relative to $$S$$, and applying the injectivity of $$S^{-1}M=S^{-1}(S^{-1}M)$$ over $$S^{-1}R$$ to the latter diagram.

Edit. Concerning your question about the dimension of $$E(R/P)$$, if $$R$$ is a PID containing a field $$k$$ and $$P$$ is a nonzero prime ideal such that $$\dim_k (R/P)=1$$ (in your question $$k=\mathbb{C}$$), then $$\dim_k E(R/P)$$ is countable.

Proof. Suppose $$P=pR$$. As explained above, $$E(R/P)=K/P_P$$, where $$K$$ is the fraction field of $$R$$. I claim that the set $$\{1,p^{-1},p^{-2},\dots\}$$ spans $$K/P_P$$ as a $$k$$-vector space. (In fact, it is a $$k$$-basis.)

Given $$r\in K$$, there is some positive integer $$n$$ such that $$r\in p^{-n} R_P$$. Write $$r=p^{-n}a$$ with $$a\in R_P$$. Since the natural map $$k\to R/P\to R_P/P_P$$ is an isomorphism, there is $$\alpha_{-n}\in k$$ such that $$\alpha_n$$ and $$a$$ have the same image in $$R_P/P_P$$. Thus, $$a-\alpha_{-n}\in P_P=pR_P$$ and we can write $$r=\alpha_{-n} p^{-n}+r'$$ with $$r'=p^n(a-\alpha_{-n})\in p^{-n+1} R_P$$. Applying the same argument to $$r'$$, we see that there is $$\alpha_{-n+1}\in k$$ such that $$r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +r''$$ with $$r''\in p^{-n+2}R_P$$. Proceeding by induction, we eventually find that $$r=\alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0+r_1$$ with $$r_1\in P_P$$. Thus, $$r\equiv \alpha_{-n} p^{-n}+\alpha_{-n+1}p^{-n+1} +\dots+ \alpha_0p^0\bmod P_P$$, which is what we want.

Remark. One can elaborate this argument further to show that $$E(R/P)=K/P_P$$ can be described as the set of formal power series $$\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0$$ with $$\alpha_0,\dots,\alpha_{-\ell}\in k$$ and $$r\in\mathbb{N}$$. The action of $$R$$ is given as follows: Given $$r\in R$$, write it as $$r=\beta_0+\beta_1p+\dots+\beta_t p^t+r_{t+1}$$ with $$r_{t+1}\in p^{t+1}R$$ and $$\beta_0,\dots,\beta_t\in k$$ for $$t$$ sufficiently large (i.e. $$t>\ell$$). Compute the formal product $$(\alpha_{-\ell} p^{-\ell}+\dots+\alpha_0 p^0)(\beta_0+\beta_1p+\dots+\beta_t p^t)$$ and truncate the positive $$p$$-powers.

• I’m having trouble parsing this. Do I view K,P as R-modules, take the quotient $K/P$ as an R-module and then localize at P? Is it easy to see what this amounts to concretely? In my original example what I really want to know is whether the injective indecomposables can have countable dimension over $\mathbb C$ so I was hoping for something more explicit. – Benjamin Steinberg Oct 24 '18 at 17:06
• @Benjamin You are correct in your understanding, with the slight difference that only the $R$-module$P$ is localized at the prime ideal $P$. (Localizing $K$ at $P$ gives back $K$.) I agree that this is not as explicit as in the case $R=\mathbb{Z}$ mentioned in the comments. I will think more about it. – Uriya First Oct 24 '18 at 17:18
• My real interest is in the dimension over $\mathbb C$. Maybe some parentheses on the localization might help. I have plus one but I really would like to see a prufer group like description in this particular case preferably with a basis over $\mathbb C$. I will add that to the question to be clear. – Benjamin Steinberg Oct 24 '18 at 17:25
• by the way I think the slight difference is superfluous since localization is exact and so localizing $(K/P)$ at $P$ amounts to first localizing $K$ at $P$ and $P$ at $P$ and then moding out, but localizing $K$ at $P$ does nothing. – Benjamin Steinberg Oct 24 '18 at 18:00
• @BenjaminSteinberg The difference is indeed superfluous. I added an answer to your question about whether the dimension is countable (it is). – Uriya First Oct 24 '18 at 18:23