Unique factorization of ideals in a Dedekind domain

Remarks

One proof shows first that every nonzero ideal contains a product of nonzero primes, using the Noetherian condition. Then it shows that every nonzero prime $P$ is invertible: there is a fractional ideal $P^{-1}$ with $P P^{-1} = R$; here integral closedness and dimension $1$ are used. Dividing $I$ by a prime that contains it gives a strictly larger ideal, and the Noetherian condition ends the process. Uniqueness follows by cancelling invertible primes, since $P \supseteq P_1 \cdots P_r$ forces $P = P_i$ for some $i$.

Example: in $R = \Z[\sqrt{-5}]$, the element $6$ factors in two ways as $2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$, so $R$ is not a UFD. With $P = (2, 1 + \sqrt{-5})$, $Q = (3, 1 + \sqrt{-5})$ and $Q' = (3, 1 - \sqrt{-5})$, one has $P^2 = (2)$ and $Q Q' = (3)$, so $(6) = P^2 Q Q'$, and the two factorizations of $6$ come from regrouping these prime ideals.

Conversely, an integral domain in which every nonzero proper ideal is a product of prime ideals is a Dedekind domain. The dimension hypothesis is needed: $k[x, y]$ over a field $k$ is Noetherian and integrally closed, but it has dimension $2$. In it, the only prime ideal that contains $(x, y^2)$ is $(x, y)$, and no power of $(x, y)$ equals $(x, y^2)$, so $(x, y^2)$ is not a product of prime ideals.

Going deeper

For a student

The theorem shows that the nonzero fractional ideals of a Dedekind domain form a free abelian group with the nonzero prime ideals as a basis. The quotient by the principal fractional ideals is the ideal class group, which measures how far $R$ is from being a principal ideal domain. Locally, the exponent $e_i$ is the valuation of $I$ in the discrete valuation ring $R_{P_i}$. In number theory, the factorization of $pR$ for a prime number $p$ describes how $p$ splits in the number field.