# False proof of $R$ Noetherian, $I$ irreducible hence $I$ prime

Can you tell me what's wrong with my proof? Thanks.

Claim: If $R$ is a Noetherian ring and $I$ is an irreducible ideal in $R$ then $I$ is prime

Proof: Let $xy \in I$. We want to show that either $x\in I$ or $y \in I$. By contradiction assume neither $x\in I$ nor $y \in I$. Then $I + \langle x \rangle$ and $I + \langle y \rangle$ are two ideals properly containing $I$ and $I + \langle x \rangle \cap I + \langle y \rangle = I + \langle xy \rangle = I$, i.e. $I$ is reducible which is a contradiction to our assumption.

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Wish I could delete this question : / – Rudy the Reindeer Jul 23 '12 at 14:00

How do you know that: $$I + \langle x \rangle \cap I + \langle y \rangle = I + \langle xy \rangle$$? That seems like a big leap, especially since it is not true in general that $\langle x\rangle\cap\langle y\rangle = \langle xy \rangle$ (which is the case $I=\{0\}$.)
Doh. It's not even true in $\mathbb Z$: $\langle 10 \rangle \cap \langle 5 \rangle = \langle \mathrm{lcm}(5, 10)\rangle = \langle 10 \rangle$. Thanks. – Rudy the Reindeer Jul 23 '12 at 13:59