Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 7 / 4

Let $S=A\setminus \mathfrak{p}$, let $S$ act on $B$ through $\phi$, and write $\mathfrak{p}B$ for the extension of an ideal generated by $\phi (\mathfrak{p})$. Since $\kappa (\mathfrak{p})=S^{-1}(A/\mathfrak{p})$, the fiber ring has the canonical presentationsThe first isomorphism sends $b\otimes (\bar{a}/s)$ to $\phi (a)b/s$ modulo $\mathfrak{p}B$. Its inverse sends the class of $b/s$ to $b\otimes (\bar{1}/s)$. The tensor balancing relation, reduction modulo $\mathfrak{p}$ and the fraction relations show that these maps are well-defined and inverse. Thus the description is canonical and does not depend on chosen representatives.

Suppose $\mathfrak{q}$ is a prime ideal of $B$ with ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$. Then $\mathfrak{p}B\subseteq \mathfrak{q}$ and $\mathfrak{q}$ is disjoint from $\phi (S)$. Its localization $S^{-1}\mathfrak{q}$ is a proper prime, since the quotient is the localization of the integral domain $B/\mathfrak{q}$. It contains $\mathfrak{p}{S}^{-1}B$, and therefore gives a primeConversely, for a prime $\mathfrak{Q}\subset C$, take its contraction $\mathfrak{q}$ under the ring homomorphism $\psi :B\to C$, $b\mapsto b\otimes 1$. Contraction preserves primality. Every $\phi (a)$ with $a\in \mathfrak{p}$ maps to zero, while every $\phi (s)$ with $s\in S$ maps to a unit and hence cannot belong to a proper prime. Thus ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$.

These constructions are inverse. For a prime disjoint from $\phi (S)$, the fraction $b/1$ belongs to its localized extension exactly when some $s\in S$ has $\phi (s)b\in \mathfrak{q}$; primality then implies $b\in \mathfrak{q}$. In the other direction, every element of $C$ has a representative $b/s$, and membership in $\mathfrak{Q}$ is equivalent to membership of $b/1$ because the denominator is a unit. Hence $\mathfrak{Q}$ is recovered by extending its contraction and quotienting. We have proved the canonical fiber of the map on spectra correspondence:If a denominator maps to zero, the fiber ring is the zero ring and both sets are empty; the same argument includes this case.

Let $\mathfrak{q}$ be a prime ideal of $C$. Its quotient ring $D=C/\mathfrak{q}$ is a nonzero integral domain and a finite-dimensional $K$-vector space. For $0\ne a\in D$, multiplication by $a$ is an injective $K$-linear map $D\to D$, because $D$ has no zero divisors. An injective endomorphism of a finite-dimensional vector space is surjective. In particular $ab=1$ for some $b\in D$. Every nonzero element is therefore invertible, so $D$ is a field and $\mathfrak{q}$ is a maximal ideal.

To prove finiteness, select any $r$ distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$. Distinct maximal ideals are comaximal: ${\mathfrak{m}}_i+{\mathfrak{m}}_j$ strictly contains ${\mathfrak{m}}_i$, so must equal $C$. The Chinese remainder theorem gives a surjectionHere is the needed surjectivity explicitly. For each $j\ne i$, choose $a_{ij}\in {\mathfrak{m}}_j$ with $a_{ij}\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$, using comaximality. Then $e_i={\prod }_{j\ne i}{a}_{ij}$ is one modulo ${\mathfrak{m}}_i$ and zero modulo the other selected ideals. Given any residue tuple represented by $b_i$, the element ${\sum }_i{b}_i{e}_i$ maps to that tuple. Thus the displayed map really is surjective.

Each nonzero field quotient has $K$-dimension at least one, soThere cannot be more than ${\dim }_{K}C$ maximal ideals, because otherwise one could select a finite list exceeding this bound. ConsequentlyThis proves the description of the spectrum of a finite-dimensional commutative algebra. If $C=0$, its spectrum is empty and the same conclusion holds.

Choose algebra generators $b_1,\dots ,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equationRepeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomialsThus $B$ is a finitely generated module over $A$, with at most $D={\prod }_i{e}_i$ generators. This proves the relevant subalgebra generated by finitely many integral elements assertion rather than merely invoking it.

For a fixed prime ideal $\mathfrak{p}\subset A$, form the fiber ring $C=B{\otimes }_A\kappa (\mathfrak{p})$. Tensoring the finite generating list shows ${\dim }_{\kappa (\mathfrak{p})}C\le D$: its pure tensors with one span the fiber over the residue field $\kappa (\mathfrak{p})$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak{p}$. HenceNo Noetherian hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.