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suppose $p$ a prime ideal. If $p=a \cap b$, $a,b$ ideals. Is it true that we have to have $p=a$ or $p=b$?

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Do you know about co-maximal ideals? –  Mathematician Feb 16 '13 at 5:30
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3 Answers

up vote 5 down vote accepted

True. Suppose that $\mathfrak{p}$ is neither $\mathfrak{a}$ nor $\mathfrak{b}$. Pick $x \in \mathfrak{a} \backslash \mathfrak{p}$, $y \in \mathfrak{b} \backslash \mathfrak{p}$. Consider $xy \in \mathfrak{a} \cap \mathfrak{b} = \mathfrak{p}$. This implies that $x$ or $y$ must lie in $\mathfrak{p}$, contradiction.

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Simple and clear, thanks. –  user62508 Feb 16 '13 at 5:37
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An ideal $I$ satisfying this condition -- namely that if for ideals $\mathfrak{a}$ and $\mathfrak{b}$ of $R$, $I = \mathfrak{a} \cap \mathfrak{b} \implies I = \mathfrak{a}$ or $I = \mathfrak{b}$ -- is called irreducible. Irreducible ideals occur in the theory of primary decomposition. For the bare rudiments of this theory, see e.g. Chapter 10 of my commutative algebra notes.

A few samples:

In a PID, an ideal is irreducible iff it is a prime power.

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Prime ideals are irreducible (the OP's question).

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In a Noetherian ring, irreducible ideals are primary.

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A proper ideal in a Noetherian ring is a finite intersection of irreducible ideals.

Combining the last two results one gets Noether's theorem, the basic result on primary decomposition.

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+1, nice answer. Is there any relation between irreducible ideals and irreducible elements of the ring? –  Eric Naslund Feb 16 '13 at 5:53
    
@Eric: I don't see one. An ideal in a GCD domain generated by an irreducible element is prime hence irreducible, but even in a PID there are nonprime irreducible ideals. (Needless to say, the terminology "irreducible ideal" was not my doing!) –  Pete L. Clark Feb 16 '13 at 8:00
    
Browsing through §10 of Pete's notes I am once again amazed by the quality of his text. –  Georges Elencwajg Feb 16 '13 at 8:20
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Since $ab\subset a\cap b$, we see that $ab\subset p$. Now use the definition of prime to see that $a\subset p$ or $b\subset p$.

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