Let $A$ be a noetherian integral domain and $\mathfrak{p}\subset A$ a prime ideal with residue field $k(\mathfrak{p}):= A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$.

I've seen in many places the symbol $E(k(\mathfrak{p}))$ denoting the injective hull of this field and i've seen a non-constructive proof for the existence of injective hulls in the general case. However i've seen very few few actual injective hulls (among them the prufer groups $\mathbb{Q}_p/\mathbb{Z}_{p}$ as hulls of the finite fields $\mathbb{F}_p$).

Does the following hold in the general (for a noetherian integral domain): $$E(k(\mathfrak{p})) \cong Frac(\widehat A_\mathfrak{p})/ \widehat A_\mathfrak{p}$$

Edit: This is apparently wrong. Thouh the main question still stands:

What is an explicit description of the injective hull of the residue field?

  • $\begingroup$ Do you even know if $Frac(k[[x,y]])/k[[x,y]]$ is injective as $k[x,y]$-module? ($k$ being a field) $\endgroup$
    – YCor
    Commented Jan 19, 2017 at 21:57
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    $\begingroup$ @YCor No. But I hope it is. $\endgroup$ Commented Jan 19, 2017 at 21:58
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    $\begingroup$ It is generally hopeless to describe $E$ beyond the case of 1-dimensional $A$. Firstly, $E[\mathfrak{m}^n]$ has the same finite length as $A/\mathfrak{m}^n$ and the union of these is $E$, so if $A$ is the local ring at a maximal ideal of a finite type $k$-algebra then $E$ has countable $k$-dimension. But $E$ is also the injective hull for $\widehat{A}$, and beyond dimension 1 your suggestion for such $\widehat{A}$ has uncountable $k$-dimension and is not $\mathfrak{m}$-power torsion. See Exercise 18.7 in Matsumura's Commutative Ring Theory for the local ring of affine space at the origin. $\endgroup$
    – nfdc23
    Commented Jan 19, 2017 at 22:35
  • $\begingroup$ @nfdc23 In the exercise you quote isn't it just the continuous dual of $\widehat A$? if so i'm inclined not to believe that "it is generally hopeless". I'm most likely wrong though... (I'll just add in case it wasn't clear - an explicit description could be some sort of local cohomology module) $\endgroup$ Commented Jan 19, 2017 at 22:53
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    $\begingroup$ I don't know what you mean by "continuous dual". Anyway, if by "explicit" you permit speaking in terms of local cohomology modules (personally I wouldn't call that "explicit" -- dualizing complexes are abstract) then there are very nice answers via Grothendieck's local duality (which is particular clean for Gorenstein local rings, in which case the ring is a dualizing complex over itself). See SGA2, Exp. IV, Thms 4.7 and 5.4 and Chapters IV--V in Hartshorne's Residues and Duality (especially section 5, Proposition 6.1, and Theorem 9.1(vi)). In a nutshell, teach yourself local duality. $\endgroup$
    – nfdc23
    Commented Jan 20, 2017 at 4:31

1 Answer 1


In the case Ycor mentioned the mentioned module is not even injective: Consider $A=K[x,y]$ for a field $K$ and let $p=(x,y)$. Then the map $$p\to Frac(\hat{A_p})/{\hat{A_p}}$$ given by sending $x$ to $0$ and $y$ to $1/x$ cannot be extended to $A$, and the module is not injective.


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