Factor \(6\) in two different ways in \(R=\ZZ[\sqrt{-5}]\). Explain the discrepancy using ideal factorization, and factor the principal ideal \((6)\) into prime ideals.
Solution
There are two element factorizations \[ 6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}). \] These are genuinely different factorizations into irreducibles; \(R\) is not a UFD. Unique factorization is restored at the level of ideals.
Set \[ \mathfrak p_2=(2,1+\sqrt{-5}), \qquad \mathfrak p_3=(3,1+\sqrt{-5}), \qquad \overline{\mathfrak p}_3=(3,1-\sqrt{-5}). \]
1. These are prime ideals of norms \(2,3,3\), respectively.
The quotient maps \[ R\to\FF_2, \qquad a+b\sqrt{-5}\mapsto a-b\pmod2, \] and \[ R\to\FF_3, \qquad a+b\sqrt{-5}\mapsto a-b\pmod3 \] (or \(a+b\) for the conjugate ideal) have kernels \(\mathfrak p_2\), \(\mathfrak p_3\), and \(\overline{\mathfrak p}_3\). Hence the quotients are fields, so the ideals are maximal and prime. Their norms are the corresponding quotient sizes.
2. The rational primes factor as\[ (2)=\mathfrak p_2^2, \qquad (3)=\mathfrak p_3\overline{\mathfrak p}_3. \]
Modulo \(2\), the polynomial \(x^2+5\) becomes \[ x^2+1=(x+1)^2, \] so \(2\) ramifies. Modulo \(3\), \[ x^2+5=x^2-1=(x-1)(x+1), \] so \(3\) splits. The displayed prime ideals are exactly the ideals corresponding to these linear factors.
3. Therefore\[ (6)=\mathfrak p_2^2\mathfrak p_3\overline{\mathfrak p}_3. \]
Simply multiply the factorizations of \((2)\) and \((3)\).
4. The second element factorization produces the same prime-ideal factorization.
Since \(1+\sqrt{-5}\) lies in both \(\mathfrak p_2\) and \(\mathfrak p_3\), we have \[ (1+\sqrt{-5})\subseteq\mathfrak p_2\mathfrak p_3. \] Both ideals have norm \(6\): \[ |N(1+\sqrt{-5})|=1^2+5\cdot1^2=6, \qquad N(\mathfrak p_2\mathfrak p_3)=2\cdot3=6. \] Hence \[ (1+\sqrt{-5})=\mathfrak p_2\mathfrak p_3. \] By conjugation, \[ (1-\sqrt{-5})=\mathfrak p_2\overline{\mathfrak p}_3. \] Multiplying gives again \[ (6)=\mathfrak p_2^2\mathfrak p_3\overline{\mathfrak p}_3. \]
Thus nonunique factorization of elements is reconciled by unique factorization of nonzero ideals in this Dedekind domain.