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](https://math.stackexchange.com/questions/1310709/is-an-irreducible-ideal-in-r-irreducible-in-rx) which can't be migrated because it is too old.



Is an irreducible ideal in $R$ also irreducible in $R[x]$?
I prove your question for (not necessarily Noetherian) commutative ring. Irreducible ideals in non-Noetherian ring are complicate (see [this question](https://mathoverflow.net/questions/87870/is-the-radical-of-an-irreducible-ideal-irreducible/88215#88215)). For Noetherian ring, see [our paper](http://arxiv.org/abs/1405.1136) for a study of the index of reducibility. The proof is elementary but quite long.


**Edit:** I realise that the proof also work for [another question](https://mathoverflow.net/questions/162739/graded-irreducible-ideals-are-irreducible/216106#216106). If$(0)$ is an irreducible ideal of $R$ then it is a graded irreducible ideal of the graded ring $R[X]$. 


We can assume that $I = 0$. Suppose $ 0 \in R[X]$ is reducible as the intersection of two proper ideals. We can assume these two ideals are principal, so $0 = (f) \cap (g)$ with
$$f = X^r (a\_0 + a\_1X + \cdots + a\_nX^n), a\_0 \neq 0,$$
$$g = X^s (b\_0 + b\_1X + \cdots + b\_mX^m), b\_0 \neq 0.$$
Choose $f$ and $g$ so that $m+n$ is **minimal**.


**Fact:** Let $f = X^rf'$ and $g = X^sg'$. Then $(f) \cap (g) = 0$ if and only if $(f') \cap (g') = 0$.


**Proof.** The "if part" is clear since $(f) \subseteq (f')$ and $(g) \subseteq (g')$.
For the "only if" part, suppose $(f') \cap (g') \neq 0$. We have $(X^{r+s} f') \cap (X^{r+s}g') = X^{r+s}((f') \cap (g')) \neq 0$. Thus $(f)\cap (g) \neq 0$ since $(X^{r+s}f') \subseteq (f)$ and $(X^{r+s}g') \subseteq (g)$. 


Using the above fact we can assume 
$$f = a\_0 + a\_1X + \cdots + a\_nX^n, a\_0 \neq 0$$
$$g = b\_0 + b\_1X + \cdots + b\_mX^m, b\_0 \neq 0.$$


If $m+n = 0$ then $f, g \in R$ and this contradicts the assumption that $(0)$ is irreducible there. 


So $m+n> 0$. Without loss of generality we assume that $m \ge n$ and thus $m>0$. Choose $0 \neq c \in (a\_0) \cap (b\_0)$ in $R$. Then $c = da\_0 = eb\_0$ for some $d, e \in R$. Replacing $f$ and $g$ by $df$ and $eg$, respectively, we can assume henceforth that $a\_0 = b\_0$.


By the minimality of $m+n$ we have the following.


**Claim 1:** Let $r$ be an element of $R$ such that $ra\_0 = 0$. Then $rf = 0$ and $rg = 0$.


**Proof.** Assume $ra\_0 = 0$ and $rf \neq 0$. Since $(f) \cap (g) = 0$ also $(rf) \cap (g) = 0$. However, since $ra\_0 = 0$, $rf = X^tf'$ with $\mathrm{deg}(f')<n$. By the Fact, $(f') \cap (g) = 0$ in contradiction to minimality of $m+n$. The same argument applies to $g$.


Applying Claim 1 inductively one can prove:


**Claim 2:** If a polynomial $h = c\_0 + c\_1X + \cdots + c\_k X^k$ satisfies $hf = 0$ (resp. $hg = 0$), then each coefficient $c\_i$ satisfies $c\_if = 0$ (resp. $c\_ig = 0$).


Combining Claims 1 and 2 we have


**Claim 3:** $hf = 0$ if and only if $hg = 0$.


We continue the proof. Let $g' = g-f$. Since $a\_0 = b\_0$, we get $g' = X^{m-m'}g' '$ for some $m' < m$ and polynomial $g' '$ of degree $m'$.
By minimality of $m+n$ and the Fact, we have $(f) \cap (g' ') \neq 0$. Thus there are polynomials $u,v,w$ such that $0 \neq w = uf = vg' '$, and thus $X^{m-m'}uf = vg' = v(g-f)$. Now, if $vg \neq 0$ then $vg = (u+v)f \in (f) \cap (g)$, a contradiction. Therefore $vg = 0$. By Claim 3 we have $vf = 0$ so $w = 0$. This is also a contradiction. The proof is complete.


