A question about prime ideals: If $p=a \cap b$ then $p=a$ or $p=b$? Suppose $p$ a prime ideal. If $p=a \cap b$, $a,b$ ideals. Is it true that we have to have $p=a$ or $p=b$?
 A: True. Suppose that $\mathfrak{p}$ is neither $\mathfrak{a}$ nor $\mathfrak{b}$. Pick $x \in \mathfrak{a} \setminus \mathfrak{p}$, $y \in \mathfrak{b} \setminus \mathfrak{p}$. Consider $xy \in \mathfrak{a} \cap \mathfrak{b} = \mathfrak{p}$. This implies that $x$ or $y$ must lie in $\mathfrak{p}$, contradiction.
A: 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.
A: 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 $b\subset p$.
