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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 ideal 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.

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Dear @Alex, I think that your first sentence should read "[...] meaning that they cannot be written as a finite intersection of ideals properly containing them". (A prime ideal can be an intersection of infinitely many maximal ideals properly containing it.) Regards, –  Bruno Joyal Aug 21 '12 at 1:11
    
@Bruno Yes, thank you. –  Alex Becker Aug 21 '12 at 1:51

3 Answers 3

up vote 7 down vote accepted

The answer to the 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 noetherian and domain we have a counterexample.

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Dear evgeniamerkulova, Thanks for pointing that out; I think I may need to assume that $\mathfrak{m}/\mathfrak{m}^2$ has dimension greater than one. –  Akhil Mathew Mar 23 '11 at 13:33
    
How is $I$ m-primary? My (possible wrong) calculations shows it is only (x)-primary. –  Fredrik Meyer Aug 19 '11 at 8:27
    
Your answer seems wrong to me: $xy \in I$, but neither $x \in I$ nor $y \in \sqrt{I}$, so $I$ cannot be primary... –  Najib Idrissi Apr 26 '12 at 14:40
    
nik and frederik: you are both right: thankyou. I corected mistake by giving other example . –  evgeniamerkulova Aug 20 '12 at 21:52

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 $(x,y^2) \cap (x^2, y) = (x^2, y^2, xy)$.

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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 ${ \mathfrak{m}}$).

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So the implication fails for Noetherian domains in general as well. Can you think of a sufficiently strong condition so that the implication will hold? –  Alex Becker Mar 23 '11 at 6:39

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