Past exam of the commutative algebra course of the University of Oxford / 2023 / 1

An ideal $\mathfrak{q}$ is $\mathfrak{p}$-primary if it is a proper primary ideal whose radical of an ideal is $\mathfrak{p}$. Explicitly,Equivalently, $ab\in \mathfrak{q}$ and $b\notin \mathfrak{p}$ imply $a\in \mathfrak{q}$: multiplication by every element outside $\mathfrak{p}$ is injective on the quotient ring $R/\mathfrak{q}$.

A minimal primary decomposition is a finite expressionin proper primary ideals such that the prime ideals ${\mathfrak{p}}_j=\sqrt{{\mathfrak{q}}_j}$ are pairwise distinct and no primary ideal can be omitted. More explicitly, for every $j$,Thus minimality means both irredundancy and distinct radicals of ideals; it does not assert that the primary ideals themselves are uniquely determined. For $I=R$, one may use the empty intersection convention.

In the terminology of a minimal primary decomposition, an associated prime of a primary decomposition of $I$ is a prime ideal occurring as $\sqrt{{\mathfrak{q}}_j}$ in such a decomposition. The first uniqueness theorem for primary decomposition says that this collection is independent of the chosen minimal primary decomposition. Among these prime ideals, an isolated prime of a primary decomposition is minimal under inclusion, equivalently a minimal prime over an ideal $I$; an embedded prime ideal strictly contains another member of the collection. ThusWhen $R$ is a Noetherian ring, these are precisely the associated primes of a module $R/I$. In that case the equivalent definition iswhere $(I:x)$ is a colon ideal. The decomposition terminology applies when a minimal primary decomposition exists; a general ring need not provide one for every ideal.

Suppose first that $\mathfrak{a}$ is $\mathfrak{m}$-primary. Since $R$ is a Noetherian ring, choose a finite generating set of an ideal $\mathfrak{m}=(x_1,\dots ,x_d)$. The equality of radicals of ideals $\sqrt{\mathfrak{a}}=\mathfrak{m}$ supplies positive integers $e_i$ with $x_i^{e_i}\in \mathfrak{a}$. SetEvery product of $n$ generators has some $x_i$ appearing at least $e_i$ times. Such products generate the product of ideals ${\mathfrak{m}}^n$, so ${\mathfrak{m}}^n\subseteq \mathfrak{a}$. Also $\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}=\mathfrak{m}$. If $\mathfrak{m}=0$, take $n=1$ directly.

Conversely, suppose ${\mathfrak{m}}^n\subseteq \mathfrak{a}\subseteq \mathfrak{m}$. Taking radicals of ideals givesHence $\sqrt{\mathfrak{a}}=\mathfrak{m}$. The only maximal ideal containing $\mathfrak{a}$ is $\mathfrak{m}$: any such maximal ideal contains $\sqrt{\mathfrak{a}}$ and therefore equals $\mathfrak{m}$. By the correspondence theorem for ideals, the quotient ring $R/\mathfrak{a}$ is a local ring with maximal ideal $\mathfrak{m}/\mathfrak{a}$. Every element outside this maximal ideal is a unit.

If $ab\in \mathfrak{a}$ and $a\notin \mathfrak{a}$, then $b$ cannot be outside $\mathfrak{m}$, since multiplying by its inverse in the quotient ring would give $a\in \mathfrak{a}$. Thus $b\in \mathfrak{m}$ and $b^n\in \mathfrak{a}$, proving the required primary ideal condition. We have proved the maximal-primary ideal criterion

Put $A=\mathbb{C}[X,Y]$, $P=(X)$ and $Q=(X^2,Y)$. The quotient ring $A/P\cong \mathbb{C}[Y]$ is an integral domain, so $P$ is a prime ideal and hence a primary ideal. Put $\mathfrak{m}=(X,Y)$. ThenThe polynomial ring $A$ is a Noetherian ring by the Hilbert basis theorem, and $\mathfrak{m}$ is a maximal ideal since $A/\mathfrak{m}\cong \mathbb{C}$. Therefore the maximal-primary ideal criterion from c makes $Q$ $\mathfrak{m}$-primary. In particular its radical of an ideal is $\mathfrak{m}$, distinct from $P$.

Neither primary ideal can be omitted: $X\in P\setminus Q$, while $Y\in Q\setminus P$. Thus the given intersection is already a minimal primary decomposition. To express the ideal by generators, write an element of $P\cap Q$ as $Xh=X^{2}u+Yv$. Reducing modulo the prime ideal $(X)$ gives $Y\bar{v}=0$ in the integral domain $\mathbb{C}[Y]$, so $v=Xw$. Cancelling $X$ in the integral domain $A$ gives $h=Xu+Yw$. HenceBy the first uniqueness theorem for primary decomposition, the associated primes of a module $A/J$ are exactly the two radicals of ideals. Their strict inclusion $P⊊\mathfrak{m}$ givesOne can also see both associated primes of a module directly through annihilators:For the first colon ideal, reducing $hY\in J$ modulo $(X)$ forces $h\in (X)$, and the reverse inclusion follows from $XY\in J$. For the second colon ideal, cancel $X$ in $hX\in X(X,Y)$. Both represented elements of the quotient module $A/J$ are nonzero.

Because $(r)$ is a proper prime ideal, each $(r^k)$ is proper and its radical of an ideal isIndeed, $(r^k)\subseteq (r)$ gives one inclusion of radicals of ideals, while $(rc{)}^k\in (r^k)$ gives the other.

We prove the saturation condition for a primary ideal: if $ab\in (r^k)$ and $a\notin \mathfrak{p}$, then $b\in (r^k)$. For $k=1$ this is the defining property of a prime ideal. For $k>1$, write $ab=r^{k}c$. Since $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$. The element $r$ is a non-zero-divisor, so cancellation givesInduction gives $b_1\in (r^{k-1})$, hence $b\in (r^k)$. Together with the computed radical of an ideal, this proves

Since the radical of an ideal $\sqrt{\mathfrak{q}}$ is $(r)$, some positive power of $r$ belongs to $\mathfrak{q}$. Choose the least such exponent $k$. Then $(r^k)\subseteq \mathfrak{q}$ and $r^j\notin \mathfrak{q}$ for $0\le j<k$, with $r^0=1$ excluded because the primary ideal $\mathfrak{q}$ is proper.

Take $x\in \mathfrak{q}$. If $x\notin (r^k)$, choose the largest $j\in \{0,\dots ,k-1\}$ for which $x\in (r^j)$, and write $x=r^{j}u$. Maximality of $j$ implies $u\notin (r)=\sqrt{\mathfrak{q}}$: otherwise $x\in (r^{j+1})$, a contradiction, including when $j=k-1$. The primary ideal condition applied to $r^{j}u\in \mathfrak{q}$ therefore forces $r^j\in \mathfrak{q}$, contrary to the choice of $k$.

Consequently every $x\in \mathfrak{q}$ lies in $(r^k)$, andTogether with i, this classifies the primary ideals for a regular principal prime. The exponent is unique: if $(r^k)=(r^{k+1})$, then $r^k=r^{k+1}c$, and cancellation of the non-zero-divisor $r^k$ would give $1=rc$, contradicting that $(r)$ is proper. No Noetherian hypothesis is needed.