Irreducible implies primary (Noetherian)

Statement
In a Noetherian ring, any irreducible ideal is primary.

Relation with other results

 * Irreducible implies prime (PID) uses a very similar approach
 * Irreducible implies prime (Dedekind)

Hands-on proof
Let $$R$$ be a Noetherian ring and $$P$$ an irreducible ideal in $$R$$. Suppose $$ab \in P$$ but $$a \notin P$$. We need to show that there exists a $$n$$ such that $$b^n \in P$$.

Define $$P_i$$ as the ideal of all elements $$x$$ such that $$xb^i \in P$$. Clearly we get an ascending chain of ideals in $$R$$:

$$P = P_0 \le P_1 \le P_2 \le \ldots$$

Since $$R$$ is Noetherian, there exists a natural number $$n$$ such that $$P_n = P_m$$ for all $$m \ge n$$. In other words, there exists a $$n$$ such that if $$b^{n+1}x \in P$$, then $$b^nx \in P$$ for any $$x \in R$$.

Now let $$Q = P + (a)$$ and $$R = P + (b^n)$$. Clearly, $$Q \cap R$$ contains $$P$$. We want to show that $$Q \cap R = P$$.

For this, suppose $$y \in Q \cap R$$. Then we can write:

$$y = p_1 + ua = p_2 + vb^n$$

where $$p_i \in P$$. This tells us that:

$$ua - vb^n \in P$$.

multiplying both sides by $$b$$ we get:

$$uab - vb^{n+1} \in P$$

By our assumption, $$ab \in P$$ and hence $$uab \in P$$.

This gives us:

$$vb^{n+1} \in P$$

but from the property of $$n$$, $$vb^n \in P$$ and hence $$y = p_2 + vb^n \in P$$.

Thus any element of $$Q \cap R$$ is in $$P$$, so $$Q \cap R = P$$.

Comparison with the proof for principal ideal rings
In the case of principal ideal rings, we employ a very similar proof, except that the $$n$$ is replaced by 1.