Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 1 / 2

If $M=0$, all its localizations are zero. Conversely, take $0\ne m\in M$. Its annihilator $I={\mathrm{Ann}}_R(m)$ is a proper ideal, so it is contained in a maximal ideal $P$, which is prime. The zero criterion for module localization says that $m/1=0$ in $M_P$ only if some $s\notin P$ satisfies $sm=0$. This would put $s\in I\subseteq P$, a contradiction. Therefore $M_P\ne 0$, provingIn fact maximal ideals suffice, and neither finite generation nor Noetherianity is needed for this implication.

Set $C=N/\phi (M)$. The natural map $N_P\to C_P$ is onto. Its kernel is the image of ${\phi }_P$: a fraction $n/s$ maps to zero precisely when $tn\in \phi (M)$ for some $t\notin P$. If $tn=\phi (m)$, then $n/s={\phi }_P(m/(ts))$. This proves the localized cokernel identity directly:Thus ${\phi }_P$ is a surjection for every $P$ exactly when $C_P=0$ for every $P$. Part (i) makes this equivalent to $C=0$, or to $\phi$ being a surjection. Hence surjectivity is detected at all prime localizations, and again maximal ideals alone would suffice.

Let $S=R\setminus P$. If $Q\subseteq P$ is a prime ideal, then $Q\cap S=\varnothing$. Its extension isIt is proper: $1=q/s$ would give $u(s-q)=0$ for some $u\in S$, and hence $us=uq\in Q$, impossible because $u,s\notin Q$. If a product $(a/s)(b/t)$ lies in $Q{R}_P$, clearing a denominator gives $uab\in Q$ for some $u\in S$. Primeness and $u\notin Q$ imply $a\in Q$ or $b\in Q$. Thus $Q{R}_P$ is prime. Its contraction is $Q$, since $r/1\in Q{R}_P$ implies $ur\in Q$ for some $u\in S$, forcing $r\in Q$.

Conversely, contract a prime ideal $\mathfrak{q}$ of $R_P$ to $Q=\{r:r/1\in \mathfrak{q}\}$. This is a prime ideal of $R$. It contains no element of $S$, since such an element becomes a unit and cannot belong to the proper ideal $\mathfrak{q}$. Hence $Q\subseteq P$. Also $r/s\in \mathfrak{q}$ exactly when $r/1\in \mathfrak{q}$, because $s/1$ is a unit, so $\mathfrak{q}=Q{R}_P$. These extension and contraction operations are inverse, giving the prime ideal correspondence for localization:

An associated prime of a module is an annihilator of a nonzero element. Suppose $Q={\mathrm{Ann}}_R(m)$ is prime and $Q\subseteq P$. Then $m/1\ne 0$ in $M_P$: a denominator $s\notin P$ killing $m$ would belong to $Q\subseteq P$. MoreoverIndeed, if $(a/s)(m/1)=0$, some $t\notin P$ has $tam=0$, so $ta\in Q$; primeness and $t\notin Q$ imply $a\in Q$. The reverse containment is immediate. This proves one inclusion in the localization of associated primes formula.

For the other, let $\mathfrak{q}={\mathrm{Ann}}_{R_P}(m/s)$ be an associated prime of $M_P$ and contract it to $Q$. By part (iii), $Q$ is prime, $Q\subseteq P$ and $Q{R}_P=\mathfrak{q}$. Since $s$ is invertible, $\mathfrak{q}$ is also the annihilator of $m/1$. The Noetherian ring hypothesis makes $Q=(a_1,\dots ,a_r)$ finitely generated. For each $a_i$, the equality $(a_{i}m)/1=0$ supplies $t_i\notin P$ with $t_i{a}_{i}m=0$. Put $t={\prod }_i{t}_i$, taking $t=1$ for the zero ideal. Then $Q$ annihilates $tm$, while $tm/1\ne 0$ because $t/1$ is a unit and $m/1\ne 0$.

If $b(tm)=0$, then $b/1$ annihilates $m/1$, so $b\in Q$. Therefore ${\mathrm{Ann}}_R(tm)=Q$, and $Q\in {\mathrm{Ass}}_R(M)$. We concludeThe single common denominator $t$ is the key point: one must turn a localized annihilator into the annihilator of one actual nonzero element of $M$, not merely assert that annihilators commute with localization.