Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 2

A multiplicative subset $S\subseteq R$ contains $1$ and satisfies $s,t\in S\Rightarrow st\in S$. The localization of a ring consists of fractions $r/s$ modulo the relationIts structure map $\iota :R\to S^{-1}R$ makes every $s\in S$ invertible. The universal property of localization says that if $f:R\to T$ is any ring homomorphism for which every $f(s)$ is a unit, there is one unique homomorphism $\tilde{f}:S^{-1}R\to T$ satisfying $\tilde{f}\iota =f$, namely

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The local criterion states that an $R$-module $M$ is flat if and only if $M_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$. It is enough equivalently to test maximal ideals.

If $M$ is flat, localization of an exact sequence and the natural isomorphismshow immediately that every $M_{\mathfrak{p}}$ is flat.

Conversely, let $N^{\prime }\to N$ be injective and let $K$ be the kernel ofAfter localization at any prime $\mathfrak{p}$, flatness of $M_{\mathfrak{p}}$ gives $K_{\mathfrak{p}}=0$. A module whose localization at every maximal ideal is zero must itself be zero: if $0\ne x\in K$, its annihilator is contained in a maximal ideal $\mathfrak{m}$, and then $x/1\ne 0$ in $K_{\mathfrak{m}}$. Hence $K=0$, tensoring by $M$ preserves every injection, and $M$ is flat. This proves that flatness is local.

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Put $S=A\setminus \mathfrak{p}$ and definewhere $A$ acts on $B$ through the given inclusion. By the prime ideal correspondence for localization, primes of $B_{\mathfrak{p}}$ correspond to primes $\mathfrak{q}$ of $B$ satisfying $\mathfrak{q}\cap S=\varnothing$, equivalently $\mathfrak{q}\cap A\subseteq \mathfrak{p}$.

The localized extension $A_{\mathfrak{p}}\subseteq B_{\mathfrak{p}}$ remains integral. If $\mathfrak{q}\cap A=\mathfrak{p}$, thenis an integral domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$. An integral domain integral over a field is a field, so $S^{-1}\mathfrak{q}$ is maximal.

Conversely, if $\mathfrak{n}$ is maximal in $B_{\mathfrak{p}}$, the contraction of a maximal ideal under an integral extension is maximal in the local ring $A_{\mathfrak{p}}$, hence equals $\mathfrak{p}{A}_{\mathfrak{p}}$. Contracting once more to $A$ gives $\mathfrak{p}$. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.

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TakeThis extension is not integral. Since every nonzero integer is already invertible in $\mathbb{Q}$,has the maximal ideal $(0)$. On the other hand, the only prime ideal of $B$ is $(0)$, whose contraction to $\mathbb{Z}$ is $(0)$ rather than $(2)$. The set of primes of $B$ lying over $\mathfrak{p}$ is therefore empty while $\mathrm{MaxSpec}{B}_{\mathfrak{p}}$ is not.

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Let $\mathfrak{m}$ be maximal in $R/I$ and let $\mathfrak{n}\subseteq R$ be its contraction. The ideal $\mathfrak{n}$ is maximal and contains $I$. It is disjoint from $S=1+I$: if $1+i\in \mathfrak{n}$ with $i\in I\subseteq \mathfrak{n}$, then $1\in \mathfrak{n}$. The prime ideal correspondence for localization therefore defines the proper ideal $S^{-1}\mathfrak{n}$, andis a field. Thus $f(\mathfrak{m})=S^{-1}\mathfrak{n}$ is maximal.

Conversely, let $\mathfrak{q}$ be the contraction to $R$ of a maximal ideal of $S^{-1}R$. Then $\mathfrak{q}$ is maximal among ideals disjoint from $S$. Since $\mathfrak{q}+I$ is disjoint from $S$—otherwise an equation $q+i=1+j$ would put $1+(j-i)\in \mathfrak{q}\cap S$—maximality gives $I\subseteq \mathfrak{q}$. If a proper ideal strictly contained $\mathfrak{q}$ in a maximal ideal $\mathfrak{n}$ of $R$, then $\mathfrak{n}$ would also contain $I$ and hence remain disjoint from $S$, a contradiction. Thus $\mathfrak{q}$ is maximal and contains $I$.

The two standard extension-contraction bijections, first for $R\to R/I$ and then for $R\to S^{-1}R$, now give inverse maps

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