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. 



Injective indecomposable modules over Laurent polynomial rings
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.




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**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.


