Primary ideals of Noetherian rings which are not irreducible

Solution 1:

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.

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