The set $S=1+I$ is a <multiplicative subset>, since $(1+a)(1+b)=1+(a+b+ab)$. The <localization of a ring> $S^{-1}R$ consists of fractions $r/s$, where
$$
\frac rs=\frac{r'}{s'}
\quad\Longleftrightarrow\quad
u(s'r-sr')=0\text{ for some }u\in S.
$$
Similarly, the <localization of a module> $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps are
$$
R\longrightarrow S^{-1}R,\qquad r\longmapsto r/1,
\qquad
M\longrightarrow S^{-1}M,\qquad m\longmapsto m/1.
$$
The <Localization of a Noetherian ring> is Noetherian. Since a finitely generated module over a Noetherian ring is a <Noetherian module>, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The <prime ideal correspondence for localization> identifies $\operatorname{Spec}S^{-1}R$ with the primes $P\in\operatorname{Spec}R$ disjoint from $S$. Here
$$
P\cap(1+I)=\varnothing
\quad\Longleftrightarrow\quad
P+I\ne R,
$$
so the <spectrum of localization away from one plus an ideal> is
$$
\boxed{\operatorname{Spec}S^{-1}R\cong
\{P\in\operatorname{Spec}R:P+I\ne R\}.}
$$

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a)^jm\in I^jM$ for every $j$, proving one inclusion. Conversely, suppose $m\in\bigcap_{j\geq1}I^jM$. The <Artin-Rees lemma> applied to $Rm\subseteq M$ says that for some $c$,
$$
I^nM\cap Rm=I^{n-c}(I^cM\cap Rm),\qquad n\geq c.
$$
Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus
$$
\boxed{\ker(M\to S^{-1}M)=\bigcap_{j\geq1}I^jM.}
$$

For failure without Noetherianity, take
$$
R=k[t^q:q\in\mathbb Q_{\geq0}],
\qquad I=(t^q:q>0).
$$
The strict chain $(t)\subsetneq(t^{1/2})\subsetneq(t^{1/4})\subsetneq\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2})^2$, one has $I^2=I$, and hence $\bigcap_jI^j=I\ne0$. But $R$ is an <integral domain>, so its localization map into $(1+I)^{-1}R$ is injective. This is the <non-Noetherian failure of the intersection formula for localization>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 148 / 2 / Solution

Let
$$
R=k[t^q:q\in\mathbb Q_{\geq0}]
$$
be the semigroup algebra inside $k[t^{\mathbb Q}]$, and let $I$ be generated by the $t^q$ with $q>0$. This <integral domain> is not Noetherian, since
$$
(t)\subsetneq(t^{1/2})\subsetneq(t^{1/4})\subsetneq\cdots.
$$
Moreover $I^2=I$, because $t^q=(t^{q/2})^2$. Thus $\bigcap_{j\geq1}I^j=I\ne0$, whereas the map $R\to(1+I)^{-1}R$ is injective because $R$ is a domain.

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