Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 2

The maximum isThe product of two fields $K\times L$ attains it: its only nonzero proper ideals are $K\times 0$ and $0\times L$, and both are maximal ideals.

For the upper bound, suppose ${\mathfrak{m}}_1$ and ${\mathfrak{m}}_2$ are distinct maximal ideals of $R$. Their intersection cannot be nonzero, since a nonzero proper ideal is maximal and cannot be properly contained in either of two distinct maximal ideals. Hence ${\mathfrak{m}}_1\cap {\mathfrak{m}}_2=0$. Also ${\mathfrak{m}}_1+{\mathfrak{m}}_2=R$, so the Chinese remainder theorem givesBoth factors are fields, and this product has exactly two maximal ideals. Thus a third maximal ideal is impossible.

For a commutative ring $R$, the Jacobson radical isLet $A\subseteq B$ be integral. If $\mathfrak{n}$ is maximal in $B$, then its contraction $\mathfrak{n}\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak{n}$, andConversely, the Lying-over theorem puts a maximal ideal $\mathfrak{n}$ of $B$ above every maximal ideal $\mathfrak{m}$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak{m}$, provingThis is the Jacobson radical under an integral extension formula.

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

Yes. Let $\mathfrak{m}$ be maximal and setThen $L$ is a field generated as a $\mathbb{C}$-algebra by countably many elements. Since the polynomial ring in countably many variables has a countable monomial basis, $L$ has at most countable dimension as a $\mathbb{C}$-vector space.

Suppose $t\in L$ were transcendental over $\mathbb{C}$. The familywould be linearly independent over $\mathbb{C}$. Indeed, after multiplying a finite relation by ${\prod }_j(t-x_j)$, evaluation at $t=x_i$ forces the $i$th coefficient to vanish. This would be an uncountable linearly independent subset of the countable-dimensional vector space $L$, a contradiction.

Thus $L/\mathbb{C}$ is algebraic. Since $\mathbb{C}$ is an algebraically closed field, $L=\mathbb{C}$. If $x_i$ is the image of $T_i$, the quotient map is evaluation at $(x_1,x_2,\dots )$ and