Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 101 (Commutative Algebra)

An ideal of definition of a Noetherian local ring $(A,\mathfrak{m})$ is an $\mathfrak{m}$-primary ideal $I$; equivalently, $\sqrt{I}=\mathfrak{m}$, or some power ${\mathfrak{m}}^r$ is contained in $I$.

If $M$ is a finitely generated module of dimension $d$, the Hilbert–Samuel functionagrees for all sufficiently large $n$ with a polynomial of degree $d$. Its leading term iswhere $e_I(M)$ is the Hilbert–Samuel multiplicity.

Assume first that $f\ne 0$ and letbe the least total degree of a nonzero homogeneous part of $f$. The associated graded ring iswhere $f_d$ is the initial homogeneous form. Multiplication by the nonzero polynomial $f_d$ is injective in $k[x,y]$, so the degree-$j$ component has dimensionSumming the components of degrees below $n$ givesThus the Hilbert polynomial isIts leading coefficient is the order of vanishing, or multiplicity, of the plane curve $f=0$ at the origin. If $f=0$, no relation is imposed and $\chi =n(n+1)/2$ for every $n$.

A monomial of weighted degree $n$ is $x^{n-2j}{y}^j$ with $0\le j\le \lfloor n/2\rfloor$. ThereforeThis is a quasipolynomial of period two, not eventually one polynomial: it equals $n/2+1$ for even $n$ and $(n+1)/2$ for odd $n$. The usual eventual-polynomial theorem for a graded algebra assumes a standard graded algebra, generated in degree one. Here the generator $y$ has degree two, so there is no contradiction.

The ring of p-adic integers is the inverse limitEquivalently, each element has a unique convergent expansion ${\sum }_{j\ge 0}{b}_j{p}^j$ with digits $0\le b_j<p$. It is a complete discrete valuation ring with maximal ideal $(p)$ and residue field ${\mathbb{F}}_p$.

Addition preserves the condition becauseFor multiplication, write $c_n={\sum }_{i+j=n}{a}_i{b}_j$. Given $q$, choose $N$ so that $a_i,b_i\in p^q{\mathbb{Z}}_p$ for $i\ge N$. If $n\ge 2N$, every pair $i+j=n$ has $i\ge N$ or $j\ge N$, so every summand lies in $p^q{\mathbb{Z}}_p$. Hence $\mathrm{ord}(c_n)\to \infty$, proving that ${\mathbb{Z}}_p⟨T⟩$ is a subring of the formal power series ring ${\mathbb{Z}}_p[[T]]$.

The $(p)$-adic completion isA compatible system of polynomials determines coefficients $a_n\in {\mathbb{Z}}_p$. For each $q$, its reduction has finite degree, so all but finitely many $a_n$ lie in $p^q{\mathbb{Z}}_p$. This is exactly $\mathrm{ord}(a_n)\to \infty$. Conversely, every such restricted series reduces modulo $p^q$ to a polynomial and hence defines a compatible system. Therefore

An element lies in ${\mathfrak{p}}^{\ast }$ exactly when all its homogeneous components lie in $\mathfrak{p}$. Suppose $ab\in {\mathfrak{p}}^{\ast }$ but $a,b\notin {\mathfrak{p}}^{\ast }$. Choose the least-degree components $a_i\notin \mathfrak{p}$ and $b_j\notin \mathfrak{p}$. In the degree $i+j$ component of $ab$, every term other than $a_i{b}_j$ contains a lower component of $a$ or $b$ and hence lies in $\mathfrak{p}$. Since the whole component lies in $\mathfrak{p}$, it follows that $a_i{b}_j\in \mathfrak{p}$, contradicting primality. Thus

First, $\sqrt{{\mathfrak{q}}^{\ast }}=\mathfrak{p}$. Indeed, each homogeneous element of $\mathfrak{p}=\sqrt{\mathfrak{q}}$ has a power in $\mathfrak{q}$, and that power is homogeneous and hence lies in ${\mathfrak{q}}^{\ast }$; finite homogeneous generators of the Noetherian ideal $\mathfrak{p}$ give the assertion for every element.

Now suppose $ab\in {\mathfrak{q}}^{\ast }$ and $a\notin \mathfrak{p}$. Choose the least homogeneous component $a_i\notin \mathfrak{p}$. If $b\notin {\mathfrak{q}}^{\ast }$, choose the least component $b_j\notin \mathfrak{q}$. The degree $i+j$ component of $ab$ differs from $a_i{b}_j$ by terms in $\mathfrak{q}$. Since it lies in $\mathfrak{q}$, we get $a_i{b}_j\in \mathfrak{q}$. The $\mathfrak{p}$-primary property and $a_i\notin \mathfrak{p}$ imply $b_j\in \mathfrak{q}$, a contradiction. Hence $b\in {\mathfrak{q}}^{\ast }$, proving that

Pass to the graded domain $R=S/{\mathfrak{p}}^{\ast }$ and let $\overline{\mathfrak{p}}=\mathfrak{p}/{\mathfrak{p}}^{\ast }$. This is a nonzero prime containing no nonzero homogeneous element. Localize at the multiplicative set $U$ of all nonzero homogeneous elements. Every nonzero homogeneous element of $U^{-1}R$ is a unit; its nonzero graded pieces are one-dimensional over the degree-zero field, so after reindexing degrees this localization is a Laurent polynomial ring $K[t,t^{-1}]$. The extended prime $U^{-1}\overline{\mathfrak{p}}$ is therefore a nonzero prime of height one.

Any prime strictly between $0$ and $\overline{\mathfrak{p}}$ would remain a nonzero prime strictly below it after localization, impossible in $K[t,t^{-1}]$. Contracting back proves that no prime lies strictly between ${\mathfrak{p}}^{\ast }$ and $\mathfrak{p}$. The graded height theorem, equivalently the same localization argument applied to saturated chains, then gives

Choose a finite presentation $F_1\to F_0\to M\to 0$ with $F_0,F_1$ finite free. Applying ${\mathrm{Hom}}_A(-,A)$ givesLet $N=F_0^{\ast }$ and let $P$ be the image in $F_1^{\ast }$. Thenis exact, $N$ is finite free, and $P$ is torsion-free because it is a submodule of the free module $F_1^{\ast }$ over the domain $A$.

The reflexive-module second-syzygy criterion says that the kernel of a map from a finite free module to a torsion-free module over a Noetherian domain is reflexive. Applying it to this sequence shows that $M^{\ast }$ is reflexive. Concretely, after localizing at the fraction field, every functional on $(M^{\ast }{)}^{\ast }$ represented generically by an element of $N$ has no denominator: torsion-freeness of $P$ forces its image to vanish already over $A$. Thus the natural evaluation map $M^{\ast }\to M^{\ast \ast \ast }$ is an isomorphism.

Put $g=y^2-zw$. Every generator of $I$ is divisible by $x^2$ and belongs to $(y,z{)}^2$, while the displayed decomposition in the next part will show thatIts two minimal primes, and hence its isolated associated primes, areThe corresponding isolated primary components are

ConsiderIt is $(x,y^2-zw)$-primary. A direct ideal-intersection calculation givesThis decomposition is irredundant and its radicals areThereforeIn particular, ${\mathfrak{p}}_3=(x,y^2-zw)$ is the embedded associated prime and the decomposition also confirms $\sqrt{I}=(x)\cap (y,z)$.

Restriction of scalars makes every $S^{-1}A$-module an $A$-module, and multiplication by $s\in S$ is invertible with inverse multiplication by $1/s$.

Conversely, suppose each multiplication map $s_M:m\mapsto sm$ is an automorphism of the $A$-module $M$. DefineThe universal property of localization shows that this is well-defined and gives the unique $S^{-1}A$-module structure extending the $A$-action. These constructions are inverse.

Under this structure the natural map $M\to S^{-1}M$ has inverseso it is an isomorphism.

Localization of the inclusion $N\hookrightarrow M$ gives an injection $S^{-1}N\hookrightarrow S^{-1}M$, whose image lies in $N^{\prime }$. Conversely, take $m/s\in N^{\prime }$. Since $s/1$ is a unit and $N^{\prime }$ is an $S^{-1}A$-submodule,By the definition of the inverse image, $m\in N$, and hence $m/s$ lies in the image of $S^{-1}N$. Therefore