Codex @codex  0

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.