Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 148 (Algebra)

An element $t\in T$ is an integral element over $R$ when it satisfies a monic equationThe extension $R\subseteq T$ is an integral extension when every $t\in T$ is integral over $R$.

Let $P\in \mathrm{Spec}R$ and localize both rings at $S=R\setminus P$. The extension $R_P\subseteq S^{-1}T$ remains integral. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}T$. The contraction of a maximal ideal under an integral extension is maximal, and the local ring $R_P$ has unique maximal ideal $P{R}_P$, so $\mathfrak{n}\cap R_P=P{R}_P$. Contracting $\mathfrak{n}$ back to $T$ produces $Q\in \mathrm{Spec}T$ with $Q\cap R=P$. This proves the Lying-over theorem and hence the surjectivity of

Suppose $Q_1\subseteq Q_2$ and both contract to $P$. Quotient by $Q_1$ and localize the resulting integral domain at the nonzero elements of $R/P$. The localized ring is an integral domain integral over the field $\mathrm{Frac}(R/P)$, and is therefore itself a field. The localization of $Q_2/Q_1$ must consequently be zero; since the localized ring is a domain, $Q_2/Q_1=0$. Thus the incomparability theorem for integral extensions gives

The Krull dimension of a ring is the supremum of the lengths of its strict chains of prime ideals. Incomparability makes the contraction of every strict prime chain in $T$ strict, so $\dim T\le \dim R$. Conversely, start over the bottom prime of any chain in $R$ using lying over and lift each subsequent inclusion using the Going-up theorem. Hence

For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and putThenand characteristic zero allows division by three. ThereforeDivision by the monic quadratic shows that $T$ is free over $k[u,v]$ with basis $1,z$. In particular, $u,v$ are algebraically independent and the requested Noether normalization of the quadratic surface xy plus yz plus zx is

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 $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe 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 $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,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

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. 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.

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.

For the $R$-$R$-bimodule $M$, the Hochschild chain complex iswith boundaryThenThe Hochschild cochain complex is $C^n(R,M)={\mathrm{Hom}}_k(R^{{\otimes }_{k}n},M)$ withand $H{H}^n(R,M)=H^n(C^{\bullet }(R,M),\delta )$.

A derivation into a bimodule is a $k$-linear map $d:R\to M$ satisfyingIt is an inner derivation when $d=d_m$ for some $m\in M$, where $d_m(r)=rm-mr$. The degree-one cocycle equation is precisely the Leibniz rule, and the degree-one coboundaries are the inner derivations. Therefore the First Hochschild cohomology as outer derivations is

Let $\mu :R{\otimes }_{k}R\to R$ be multiplication and $\Omega =\ker \mu$, equipped with the outer bimodule structure. The universal bimodule derivationsatisfies $D(rs)=rD(s)+D(r)s$. Composition with $D$ definesFor a derivation $d$, its inverse image under this map isIf ${\sum }_i{r}_i{s}_i=0$, the Leibniz rule shows that this formula is both left and right $R$-linear. Moreover every element of $\Omega$ is ${\sum }_{i}D(r_i)s_i$, proving existence and uniqueness.

For $d_m(r)=rm-mr$, the corresponding map isThese are exactly the maps $\Omega \to M$ which extend to bimodule maps $R{\otimes }_{k}R\to M$.

Choose homogeneous algebra generators $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$ for $S$; they exist because $S$ is Noetherian. For a finitely generated graded module $V={⨁}_j{V}_j$, its Poincare series of a graded module isThe Hilbert-Serre theorem states thatfor a Laurent polynomial $Q(t)\in \mathbb{Z}[t,t^{-1}]$.

We prove this by induction on $r$. For $r=0$, $S=k$ and $V$ is finite-dimensional, so its series is a Laurent polynomial. For $x=x_r$ of degree $d=d_r$, multiplication gives the exact sequence of graded modulesAdditivity of the Hilbert series yieldsBoth modules on the right are finitely generated over $k[x_1,\dots ,x_{r-1}]$, so the induction hypothesis proves the formula.

For the finitely generated commutative algebra $R$, let $R_j$ be the stated degree filtration. Its associated graded ringis a standard graded algebra generated by the initial forms of $x_1,\dots ,x_n$. SinceHilbert–Serre shows that the quantity ${\dim }_k{R}_j$ agrees with a polynomial for all sufficiently large $j$. The growth of a finitely generated commutative algebra has polynomial degree $\dim R$, independently of the chosen finite generating set.

Forwith the four displayed generators, a basis of $R_j$ consists of the Laurent monomials $Y_1^a{Y}_2^b$ satisfying $|a|+|b|\le j$. There are $4r$ points with $|a|+|b|=r>0$, and one point at the origin. Hence the growth of the two-variable Laurent polynomial algebra is