= Solution
Set $S=R\setminus P$, a <multiplicative subset>. The <localization at a prime ideal> is $R_P=S^{-1}R$, whose elements are fractions $r/s$. Equality of two fractions means that $u(rs'-r's)=0$ for some $u\in S$. Likewise the <localization of a module> is $M_P=S^{-1}M$, with
$$
\frac m s=\frac{m'}{s'}\quad\Longleftrightarrow\quad u(s'm-sm')=0\text{ for some }u\in S.
$$
Addition uses a common denominator, and the <module> action is
$$
\boxed{\frac r s\,\frac m t=\frac{rm}{st}.}
$$
The equivalence relations make these operations well-defined, so $M_P$ is an $R_P$-module.
If $M\neq0$, choose $0\neq m\in M$. Its <annihilator> is proper, so it lies in a <maximal ideal> $P$. The element $m/1$ cannot vanish in $M_P$: vanishing would mean $sm=0$ for some $s\notin P$, contrary to $\operatorname{Ann}_R(m)\subseteq P$. The converse is immediate by localizing the zero <module>. Thus
$$
\boxed{M=0\quad\Longleftrightarrow\quad M_P=0\text{ for every prime }P.}
$$
This proof of <localization detects zero elements> actually applies to arbitrary <modules>, without finite-generation or <Noetherian> assumptions. We will use that extra generality for an <Ext functor> <module> below.
An <injective module> $E$ has the extension property: for each inclusion $A\hookrightarrow B$, every map $A\to E$ extends to a map $B\to E$. Equivalently, $\operatorname{Hom}_R(-,E)$ is exact. A <projective module> $Q$ has the lifting property: for each surjection $B\twoheadrightarrow C$, every map $Q\to C$ lifts to $Q\to B$. Equivalently, $\operatorname{Hom}_R(Q,-)$ is exact, or $Q$ is a direct summand of a <free module>.
For the <local criterion for injectivity over a Noetherian ring>, recall the <Baer criterion>: $E$ is injective precisely when every map from an <ideal> $J\subseteq R$ into $E$ extends to $R$. Through the <short exact sequence> $0\to J\to R\to R/J\to0$, this is equivalent to
$$
\operatorname{Ext}^1_R(R/J,E)=0\quad\text{for every ideal }J.
$$
We also need <localization of Ext over a Noetherian ring>. Because $R$ is <Noetherian>, the <module> $R/J$ has a free resolution with every term finitely generated: all successive kernels are finitely generated, so this can be built recursively. For a finite free term $F$, the natural map
$$
S^{-1}\operatorname{Hom}_R(F,M)\cong\operatorname{Hom}_{S^{-1}R}(S^{-1}F,S^{-1}M)
$$
is an isomorphism, as is clear from a finite basis. <Exactness of localization> lets us pass to cohomology of the Hom complex. Therefore
$$
\boxed{\bigl(\operatorname{Ext}^1_R(R/J,M)\bigr)_P\cong\operatorname{Ext}^1_{R_P}(R_P/JR_P,M_P).}
$$
This explains the finiteness hypothesis needed for the localization argument, rather than assuming that localization preserves injectivity automatically.
If $M$ is injective, the left-hand side vanishes for every $J$ and $P$. Every <ideal> $L$ of $R_P$ is $JR_P$, where $J$ is its contraction to $R$: if $r/s\in L$, then $r/1=s(r/s)\in L$, and conversely localization of a member of the contraction stays in $L$. Hence all <ideal> tests for $M_P$ vanish, and the <Baer criterion> over $R_P$ makes $M_P$ injective.
Conversely, suppose every $M_P$ is injective. For each <ideal> $J$, the displayed <Ext functor> localization is zero at every prime. The zero-detection argument above gives $\operatorname{Ext}^1_R(R/J,M)=0$, without needing this Ext <module> to be finitely generated. Applying the <Baer criterion> over $R$ proves
$$
\boxed{M\text{ injective}\quad\Longleftrightarrow\quad M_P\text{ injective over }R_P\text{ for every prime }P.}
$$
In fact, under the <Noetherian> <ring> hypothesis this equivalence holds for arbitrary $M$; the printed finite-generation assumption is more than is needed.
The <global dimension> is
$$
\boxed{\operatorname{gldim}R=\sup\{\operatorname{pd}_R N:N\text{ an }R\text{-module}\},}
$$
where <projective dimension> is the smallest length of a <projective resolution>, or infinity if there is no finite one. If $\operatorname{gldim}R=0$, every <module> is projective. Given an inclusion $A\hookrightarrow B$, the quotient $B/A$ is then projective, so the exact sequence $0\to A\to B\to B/A\to0$ splits. A retraction $\rho:B\to A$ exists. For any <module> $E$ and any map $f:A\to E$, the composition $f\rho:B\to E$ extends $f$. Thus every <module> is also injective. \b[Global dimension zero makes all <modules> both projective and injective], as recorded by <global dimension zero and split exact sequences>.
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