Let $R$ be a commutative Noetherian ring and $I\subset R$ an ideal that is irreducible in the sense that if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$. Is (the ideal generated by) $I$ irreducible in the polynomial ring $R[x]$?

The answer to the question is "yes" since the number of irreducible components has a homological characterization which can be found in *Computational methods in commutative algebra and algebraic geometry* by W. V. Vasconcelos.

**Proposition 3.1.7**. Let $(R,\mathfrak{m})$ be a local commutative Noetherian ring. Let $\mathfrak{p}$ be an associated prime of an $R$-module $M$ and denote $\Delta_\mathfrak{p}(M)$ the submodule of $M$ whose elements are annihilated by $\mathfrak{p}$. The number of irreducible $\mathfrak{p}$-primary components in an irredundant irreducible decomposition of $0\subset M$ is $\dim_{k(\mathfrak{p})}(\Delta_\mathfrak{p}(M))_\mathfrak{p}$.

The minimal number of irreducible intersectands of $I$ equals the $R_{P}/P_P$-vector space dimension of the socle of $R[x]_{P}/I_{P}$. Now when adjoining indeterminates the field $R[x]_P/PR[x]_P$ grows, but the socle dimension is the same.

Now the real question is: Is there a simpler proof (without localization(?)) or do we have to use homological invariants of ideal decompositions?

Background: This comes from Exercise 3.6 in Eisenbud's book on commutative algebra which asks for a characterization of irreducible monomial ideals. I'm wondering if a reader who has only read Chapters 1,2,3 would be able to do that exercise. This is a repost of my math.se question which can't be migrated because it is too old.