Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 / 3

A module $E$ is injective if every map $A\to E$ extends across every inclusion $A\hookrightarrow B$. Baer criterion says it suffices to test inclusions of left ideals $I\hookrightarrow R$.

Necessity is immediate. Conversely, order all extensions of a given map $A\to E$ to intermediate submodules of $B$. A maximal one exists by Zorn's lemma. If its domain $C$ is not $B$, choose $b\notin C$ and let $I=\{r:rb\in C\}$. The map $I\to E$, $r\mapsto f(rb)$, extends to $R$ by the hypothesis; its value at $1$ extends $f$ to $C+Rb$, contradicting maximality. Thus $C=B$.

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If $R$ is left Noetherian and $I\to {⨁}_{\alpha }{E}_{\alpha }$ is a map from a left ideal, finitely many generators of $I$ have support in one finite set of summands. The map therefore lands in a finite direct sum of injectives and extends to $R$. Baer's criterion proves that the full direct sum is injective.

Conversely, let $I_1\subseteq I_2\subseteq \cdots$ and put $I=\bigcup I_n$. Embed each $R/I_n$ in an injective module $E_n$. The maphas finite support. If the direct sum is injective, it extends to $R$; the extension's value at $1$ has finite support, forcing $a\in I_n$ for every sufficiently large $n$ and every $a\in I$. Hence the chain stabilizes. This is the Bass-Papp theorem.

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Over a commutative PID, Baer's criterion reduces to maps $(r)\to E$. Such a map extends to $R$ exactly when every equation $re=x$ with $r\ne 0$ is solvable. Thus injective modules are exactly the divisible modules.

Let $Q$ be the fraction field. The indecomposable injectives arefor one representative $p$ of each associate class of irreducibles. The latter is the $p$-primary Prüfer module, the union of the cyclic modules generated by $p^{-n}+R$. The structure theorem for divisible modules decomposes every divisible module into copies of $Q$ and these Prüfer modules, proving that the list is complete.

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Because $N$ is essential in $M$, every associated prime of $M$ occurs in $N$. Since $JN=0$, all these primes contain $J$. For a finitely generated module over a commutative Noetherian ring,Hence $J\subseteq \sqrt{\mathrm{ann}M}$; finite generation of the ideal $J$ gives $J^{n}M=0$ for some $n$.

Now $R/J$ is essential in its injective hull. For $x\in E(R/J)$, the finitely generated module $Rx+R/J$ has essential submodule $R/J$, so the result just proved gives $J^{n}x=0$ for some $n$. The reverse inclusion is tautological, and therefore

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