Primary ideals of Noetherian rings which are not irreducible

Question

It is known that all prime ideals are irreducible (meaning that they cannot be written as an finite intersection of ideals properly containing them). While for Noetherian rings an irreducible ring is always primary, the converse fails in general. In a recent problem set I was asked to provide an example of a primary ideal of a Noetherian ring which is not irreducible. The example I came up with is the ring $\mathbb{Z}_{p^2}[\eta]$ where $p$ is prime and $\eta$ is a nilpotent element of order $n > 2$, which has the $(p,\eta)$-primary ideal $(p)\cap (\eta) = (p\eta)$. But this got me thinking: how severe is the failure of primary ideals to be irreducible in Noetherian rings? In particular, are primary ideals of a Noetherian domain irreducible, or is a stronger condition on the ring required? I'd love to see suitably strong criteria for all primary ideals of a Noetherian ring to be irreducible, or examples of primary ideals of "well-behaved" rings which are not irreducible.

Answer 1

Answer to question "are primary ideals of a Noetherian domain irreducible ?" is "no". For example take for domain $R=K[x,y]$ polynomials over field $K$. Ideal $I=(x^2,xy,y^2)$ is $(x,y)$ primary but reducible because $I=(x,y^2)\cap (y,x^2)$. Since $R$ is notherian and domain we have conterexample.

Comment: I think Akhil Mathew proof and statement is not wright. For example take ring $S=K[x]_{(x)}$= local ring of affine line over $K$ with maximal ideal $\mathfrak{m}=(x)$. List of nontrivial ideals of $S$ are $\mathfrak{m}^n$. All them are $\mathfrak{m}-$ primary. Because they are totally ordered with inclusion they are all ireducible; in particular case $\mathfrak m^2$ is irreducible and claim that it is "finite intersection of ideals between $\mathfrak{m}$ and $\mathfrak{m}^2$ " seems not wright for me because there is no ideal between $\mathfrak{m}$ and $\mathfrak{m}^2$.

Answer 2

There is a beautiful characterization of prime, radical, irreducible and primary ideals among monomial ones in $k[x_1, \dots, x_n]$:

Theorem. Let $I$ be a monomial ideal of $k[x_1, \dots, x_n]$ and let $\mathcal{B}$ be its minimal basis. Then:

1. $I$ is maximal iff $\mathcal{B}=\{x_1, \dots, x_n \}$;
2. $I$ is prime iff $\mathcal{B} = \{ x_{i_1}, \dots, x_{i_r} \}$;
3. $I$ is radical iff $\mathcal{B}$ is made up of square-free monomials;
4. $I$ is irreducible iff $\mathcal{B} = \{ x_{i_1}^{a_1}, \dots, x_{i_r}^{a_r} \}$;
5. $I$ is primary iff $\mathcal{B} = \{ x_{i_1}^{a_1}, \dots, x_{i_r}^{a_r}, m_1, \dots, m_s \}$, where $m_1,\dots, m_s$ are monomials in the variables $x_{i_1}, \dots, x_{i_r}$.

So in this case it is very easy to produce a counter-example: $(x^2, y^2, xy)$. Its radical is maximal, so it is primary, but is reducible because $(x,y^2) \cap (x^2, y) = (x^2, y^2, xy)$.

Answer 3

If I am not blundering here, the implication primary implies irreducible is in general false. For instance, let $R$ be a noetherian local domain of embedding dimension $>1$, and $\mathfrak{m}$ a maximal ideal. Then $\mathfrak{m}/\mathfrak{m}^2$ is a finite-dimensional vector space. We can thus write $\mathfrak{m}^2$ as a finite intersection of ideals strictly between $\mathfrak{m}$ and $\mathfrak{m}^2$. However, $\mathfrak{m}^2$ is $\mathfrak{m}$-primary (the support of $R/\mathfrak{m}^2$ is just ${ \mathfrak{m}}$).