Irreducible implies primary (Noetherian)

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Statement

In a Noetherian ring, any irreducible ideal is primary.

Relation with other results

Proof

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.