Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 2

The localization of a ring $S^{-1}R$ consists of pairs $(r,s)\in R\times S$ modulofor some $u\in S$. Addition and multiplication are

The localization is the zero ring exactly when $1/1=0/1$, which means that some $u\in S$ is zero. Thus $0\in S$. This holds exactly when $S$ contains a nilpotent element: if $s^n=0$ and $s\in S$, multiplicative closure gives $0\in S$, while zero itself is nilpotent.

Let $e=(1,0)$. Since $e$ becomes invertible in $S^{-1}(A\times B)$ and $e(0,b)=0$, every $(0,b)$ becomes zero. The mapis therefore a well-defined ring isomorphism, with inverse $a\mapsto (a,0)/1$.

For a subring $\mathbb{Z}\subseteq R\subseteq \mathbb{Q}$, letIf $a/b\in R$ is in lowest terms, Bézout gives integers $u,v$ with $ua+vb=1$, soEvery prime divisor $p$ of $b$ then satisfies $1/p=(b/p)/b\in R$. Conversely, every rational whose denominator uses only primes in $\mathcal{P}$ lies in $R$. Hence all intermediate rings are exactlywhere $S$ is generated by an arbitrary set $\mathcal{P}$ of primes. For example, $1/3=1-2/3$, so $\mathbb{Z}[2/3]=\mathbb{Z}[1/3]=\{3^n:n\ge 0{\}}^{-1}\mathbb{Z}$.

For $x\in L$, choose a monic equation over $K$,Choose nonzero $a\in R$ with $a{c}_i\in R$ for every $i$. Multiplying by $a^n$ shows that $b=ax$ satisfieswhose coefficients lie in $R$. Thus $b\in \overline{R}$ and

Extension and contraction give the prime ideal correspondence for localizationExplicitly, $\mathfrak{p}$ maps to $S^{-1}\mathfrak{p}$, while $\mathfrak{q}\subseteq S^{-1}R$ maps to $\{r:r/1\in \mathfrak{q}\}$. Primality follows by clearing denominators, and the two operations are inverse because an ideal in a localization contains $r/s$ exactly when it contains $r/1$.

The weak Hilbert Nullstellensatz says that if a field is a finitely generated algebra over $k$, then it is a finite algebraic extension of $k$. Since every maximal ideal is prime, the nilradical $\mathcal{N}(R)$ is contained in the Jacobson radical $J(R)$.

Conversely, let $f$ be nonnilpotent. Then $R_f\ne 0$, and it is a finitely generated $k$-algebra, so it has a maximal ideal $\mathfrak{n}$. Its contraction $\mathfrak{p}$ to $R$ avoids $f$. Moreover,is a finitely generated $k$-domain inside a finite algebraic extension of $k$, hence is itself a field. Thus $\mathfrak{p}$ is maximal and does not contain $f$. Therefore $f\notin J(R)$, proving