I prove your question for (not necessarily Noetherian) commutative ring. Irreducible ideals in non-Noetherian ring are complicate (see [this question][1]). For Noetherian ring, see [our paper][2] for a study of the index of reducibility. The proof is elementary but quite long.


We can assume that $I = 0$. Suppose $ 0 \in R[X]$ is reducible then it is the intersection of two proper ideals $I_1, I_2$. We can assume these two ideals are principal, so $0 = (f) \cap (g)$. We write $f$ and $g$ in the following form
$$f = X^r (a_0 + a_1X + \cdots + a_nX^n), a_0 \neq 0$$
$$g = X^s (b_0 + b_1X + \cdots + b_mX^m), b_0 \neq 0.$$
Here we choose $f$ and $g$ such that $m+n$ is **minimal**. Replace $f$ and $g$ by $X^sf$ and $X^rg$, respectively. Since $X$ is indeterminate we can reduce to the case 
$$f = a_0 + a_1X + \cdots + a_nX^n, a_0 \neq 0$$
$$g = b_0 + b_1X + \cdots + b_mX^m, b_0 \neq 0.$$
Now, if $m+n = 0$ then $f, g \in R$ this contradicts with our assumption that $(0)$ is irreducible. So $m+n> 0 $ we assume that $m \ge n$ and so $m>0$. Choose $0 \neq c \in (a_0) \cap (b_0)$ in $R$. So $c = da_0 = eb_0$ for some $d, e \in R$. Replace $f$ 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 $ra_0 = 0$ (and hence $rb_0 = 0$). Then $rf = 0$ and $rg = 0$.

**proof.** If $ra_0 = 0$ and $rf \neq 0$, then we replace $f$ by $rf$.

Using Claim 1 and the induction one can prove the following.

**Claim 2:** Let $h = c_0 + c_1X + \cdots + c_k X^k$ be a polynomial satisfying that $hf = 0$ (resp. $hg = 0$). Then for all $i = 0, ..., k$ we have $c_if = 0$ (resp. $c_ig = 0$).

By Claims 1 and 2 we have.

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


We continue our proof. Let $g' = g-f$. Since $a_0 = b_0$, we have $g'$ can be written as
$$g' = X^{m -m'} (b_0' + b_1'X + \cdots b_{m'}X^{m'}), b_0' \neq 0, m' < m.$$
By the minimality of $m+n$ we have $(f) \cap (g') \neq 0$. Thus there are polynomials $u,v,w$ such that
$$0 \neq w = uf = v(g-f).$$
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.



  [1]: http://mathoverflow.net/questions/87870/is-the-radical-of-an-irreducible-ideal-irreducible/88215#88215
  [2]: http://arxiv.org/abs/1405.1136