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 goes roughly as follows: Localize at the associated prime $P$ of $I$. 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 only the field $R[x]_P/PR[x]_P$ should grow, but the dimension is the same. This is explained in Section 3 of Computational methods in commutative algebra and algebraic geometry by W. V. Vasconcelos. I could post more details once I find my copy of the book.
Now the real question is: Is there a simpler proof (say without localization) or do we have to use homological invariants of ideal decompositions?