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.

  • $\begingroup$ k should be C. Sorry. I'll fix. $\endgroup$ Oct 24, 2018 at 14:29
  • 1
    $\begingroup$ It does seem that the injective envelope of arbitrary cyclic modules over a PID should be well-known! $\endgroup$
    – rschwieb
    Oct 24, 2018 at 14:32
  • $\begingroup$ @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. $\endgroup$ Oct 24, 2018 at 14:41
  • $\begingroup$ 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... $\endgroup$
    – rschwieb
    Oct 24, 2018 at 15:01
  • $\begingroup$ Probably they do. Maybe you just invert the powers of the prime $p$. $\endgroup$ Oct 24, 2018 at 15:03

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Oct 24, 2018 at 17:06
  • $\begingroup$ @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. $\endgroup$ Oct 24, 2018 at 17:18
  • $\begingroup$ 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. $\endgroup$ Oct 24, 2018 at 17:25
  • $\begingroup$ 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. $\endgroup$ Oct 24, 2018 at 18:00
  • $\begingroup$ @BenjaminSteinberg The difference is indeed superfluous. I added an answer to your question about whether the dimension is countable (it is). $\endgroup$ Oct 24, 2018 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.