Injective resolution for an integral domain

Question

How could one write an injective resolution for an arbitrary commutative integral domain $R$? Thanks in advance!

Answer 1

If the ring is Noetherian, then one has a very nice description of the minimal injective resolution of any finite $R$-module $M$, which looks like $\DeclareMathOperator{\injdim}{injdim}$ $\DeclareMathOperator{\coker}{coker}$ $\DeclareMathOperator{\Spec}{Spec}$ $\DeclareMathOperator{\Ext}{Ext}$ $\DeclareMathOperator{\ht}{ht}$

$$0 \to M \xrightarrow{\delta^0} E(M) = E^0 \xrightarrow{\delta^1} E^1 \xrightarrow{\delta^2} E^2 \to \ldots$$

where $E^i = E(\coker \delta^i)$ (where $E$ denotes injective envelope) and $\delta^{i+1}$ is the composite $E^i \to \coker \delta^i \hookrightarrow E^{i+1}$. One can also explicitly describe $E_i$ as a direct sum of indecomposable injectives: these correspond $1-1$ with primes $p \in \Spec R$, and are of the form $E(R/p)$. The multiplicity with which a particular $E(R/p)$ occurs in $E^i$ is called the $i^\text{th}$ Bass number of $M$ with respect to $p$, denoted $\mu_i(p, M)$, and is given by the formula $\mu_i(p,M) = \dim_{k(p)}\Ext^i_{R_p}(k(p), M_p)$, where $k(p)$ is the residue field at $p$. Thus $\displaystyle E^i = \bigoplus_{p \in \Spec R} E(R/p)^{\mu_i(p,M)}$.

Now, in the local, finite injective dimension case, we can say even more (in the non-local case, it can happen that the localizations have finite injective resolutions, but the global ring does not, e.g. in Nagata's example of a Noetherian ring of infinite Krull dimension). If $\injdim_R(R) < \infty$, then $R$ is Gorenstein, and then the Bass numbers are given by the extremely simple formula $\mu_i(p, R) = \delta_{i, \ht p}$. Thus we see that $\displaystyle E^i = \bigoplus_{\ht p = i} E(R/p)$, and that $R$ is Cohen-Macaulay.

One last remark: if $R$ is any domain, then the minimal injective resolution of $R$ must start as

$$0 \to R \to K \to E(K/R) \to \ldots$$

where $K$ is the field of fractions of $R$. This is because $K$ is torsionfree and divisible over $R$, hence $R$-injective (since $R$ is a domain), and $R \subseteq K$ is an essential extension, so $K = E(R)$.

Answer 2

If $Q$ is a divisible abelian group, then the right $R$-module $\operatorname{Hom}_{\mathbb{Z}}(R,Q)$ is injective (Anderson-Fuller, 18.5). The right module action is defined by $$ fr\colon x\mapsto f(rx). $$ Since every right $R$ module $M$ can be embedded, as an abelian group, in a divisible abelian group $Q$, by considering an epimorphism $\varphi\colon\mathbb{Z}^{(X)}\to M$ so that $$ M\cong \frac{\mathbb{Z}^{(X)}}{\ker\varphi}\hookrightarrow \frac{\mathbb{Q}^{(X)}}{\ker\varphi}, $$ we can consider $$ M\cong\operatorname{Hom}_R(R,M)\hookrightarrow \operatorname{Hom}_{\mathbb{Z}}(R,M)\hookrightarrow \operatorname{Hom}_{\mathbb{Z}}(R,Q). $$ There is in general no “canonical” way to embed a module (or just the ring) into an injective module.

Once you know how to embed any module into an injective module, you're guaranteed to find an injective resolution.