The Stacks project

Proof. Let $M$ be an $R$-module. By More on Algebra, Section 15.55 the category of $R$-modules has enough injectives. Choose an injection $M \to I$ with $I$ an injective $R$-module. Consider the set $\mathcal{S}$ of submodules $M \subset E \subset I$ such that $E$ is an essential extension of $M$. We order $\mathcal{S}$ by inclusion. If $\{ E_\alpha \} $ is a totally ordered subset of $\mathcal{S}$, then $\bigcup E_\alpha $ is an essential extension of $M$ too (Lemma 47.2.3). Thus we can apply Zorn's lemma and find a maximal element $E \in \mathcal{S}$. We claim $M \subset E$ is an injective hull, i.e., $E$ is an injective $R$-module. This follows from Lemma 47.3.5. $\square$

Proof. Part (1) follows from the fact that $E'$ is an injective $R$-module. Part (2) follows as $\mathop{\mathrm{Ker}}(\psi ) \cap M = 0$ and $E$ is an essential extension of $M$. Assume $\varphi $ is an essential injection. Then $E \cong \psi (E) \subset E'$ by (2) which implies $E' = \psi (E) \oplus E''$ because $E$ is injective. Since $E'$ is an essential extension of $M$ (Lemma 47.2.2) we get $E'' = 0$. Part (4) is a special case of (3). Assume $M \to I$ as in (5). Choose a map $\alpha : E \to I$ extending the map $M \to I$. Arguing as before we see that $\alpha $ is injective. Thus as before $\alpha (E)$ splits off from $I$. This proves (5). $\square$

Proof. Part (1) follows from Lemma 47.5.3. Part (2) follows from part (1) and the definition of injective hulls.

Proof of (3). Set $A = \text{End}_ R(E, E)$ and $I = \{ \varphi \in A \mid \mathop{\mathrm{Ker}}(\varphi ) \not= 0\} $. The statement means that $I$ is a two sided ideal and that any $\varphi \in A$, $\varphi \not\in I$ is invertible. Suppose $\varphi $ and $\psi $ are not injective. Then $\mathop{\mathrm{Ker}}(\varphi ) \cap \mathop{\mathrm{Ker}}(\psi )$ is nonzero by (2). Hence $\varphi + \psi \in I$. It follows that $I$ is a two sided ideal. If $\varphi \in A$, $\varphi \not\in I$, then $E \cong \varphi (E) \subset E$ is an injective submodule, hence $E = \varphi (E)$ because $E$ is indecomposable.

Proof of (4). Consider the ring map $R \to A$ and let $\mathfrak p \subset R$ be the inverse image of the maximal ideal $I$. Then it is clear that $\mathfrak p$ is a prime ideal and that $R \to A$ extends to $R_\mathfrak p \to A$. Thus $E$ is an $R_\mathfrak p$-module. It follows from Lemma 47.3.3 that $E$ is injective as an $R_\mathfrak p$-module. $\square$

Proof. By Lemma 47.2.4 the inclusion $R/\mathfrak p \subset \kappa (\mathfrak p)$ is an essential extension. Then Lemma 47.5.3 shows (2) holds. For $f \in R$, $f \not\in \mathfrak p$ the map $f : \kappa (\mathfrak p) \to \kappa (\mathfrak p)$ is an isomorphism hence the map $f : E \to E$ is an isomorphism, see Lemma 47.5.3. Thus $E$ is an $R_\mathfrak p$-module. It is injective as an $R_\mathfrak p$-module by Lemma 47.3.3. Finally, let $E' \subset E$ be a nonzero injective $R$-submodule. Then $J = (R/\mathfrak p) \cap E'$ is nonzero. After shrinking $E'$ we may assume that $E'$ is the injective hull of $J$ (see Lemma 47.5.3 for example). Observe that $R/\mathfrak p$ is an essential extension of $J$ for example by Lemma 47.2.4. Hence $E' \to E$ is an isomorphism by Lemma 47.5.3 part (3). Hence $E$ is indecomposable. $\square$

Proof. Let $\mathfrak p$ be the prime ideal found in Lemma 47.5.6. Say $\mathfrak p = (f_1, \ldots , f_ r)$. Pick a nonzero element $x \in \bigcap \mathop{\mathrm{Ker}}(f_ i : E \to E)$, see Lemma 47.5.6. Then $(R_\mathfrak p)x$ is a module isomorphic to $\kappa (\mathfrak p)$ inside $E$. We conclude by Lemma 47.5.6. $\square$

Proof. The second statement is Lemma 47.5.8. For the first statement, let $I$ be an injective $R$-module. We will use transfinite recursion to construct $I_\alpha \subset I$ for ordinals $\alpha $ which are direct sums of indecomposable injective $R$-modules $E_{\beta + 1}$ for $\beta < \alpha $. For $\alpha = 0$ we let $I_0 = 0$. Suppose given an ordinal $\alpha $ such that $I_\alpha $ has been constructed. Then $I_\alpha $ is an injective $R$-module by Lemma 47.3.7. Hence $I \cong I_\alpha \oplus I'$. If $I' = 0$ we are done. If not, then $I'$ has an associated prime by Algebra, Lemma 10.63.7. Thus $I'$ contains a copy of $R/\mathfrak p$ for some prime $\mathfrak p$. Hence $I'$ contains an indecomposable submodule $E$ by Lemmas 47.5.3 and 47.5.7. Set $I_{\alpha + 1} = I_\alpha \oplus E_\alpha $. If $\alpha $ is a limit ordinal and $I_\beta $ has been constructed for $\beta < \alpha $, then we set $I_\alpha = \bigcup _{\beta < \alpha } I_\beta $. Observe that $I_\alpha = \bigoplus _{\beta < \alpha } E_{\beta + 1}$. This concludes the proof. $\square$