Lemma 47.5.3 (08Y4)—The Stacks project

Table of contents
Part 3: Topics in Scheme Theory
Chapter 47: Dualizing Complexes
Section 47.5: Injective hulls
Lemma 47.5.3 (cite)

Lemma 47.5.3. Let $R$ be a ring. Let $M$, $N$ be $R$-modules and let $M \to E$ and $N \to E'$ be injective hulls. Then

for any $R$-module map $\varphi : M \to N$ there exists an $R$-module map $\psi : E \to E'$ such that

\[ \xymatrix{ M \ar[r] \ar[d]_\varphi & E \ar[d]^\psi \\ N \ar[r] & E' } \]

commutes,
if $\varphi $ is injective, then $\psi $ is injective,
if $\varphi $ is an essential injection, then $\psi $ is an isomorphism,
if $\varphi $ is an isomorphism, then $\psi $ is an isomorphism,
if $M \to I$ is an embedding of $M$ into an injective $R$-module, then there is an isomorphism $I \cong E \oplus I'$ compatible with the embeddings of $M$,

In particular, the injective hull $E$ of $M$ is unique up to isomorphism.

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). $\square$

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