Exhibit a pair of algebraic integers in QQ(sqrt(-5)) which are relatively prime but for which Bezout’s identity does not hold

Show that 3 and 1 + \sqrt{-5} are relatively prime as algebraic integers in \mathbb{Q}(\sqrt{-5}), but that there do not exist algebraic integers \lambda,\mu \in \mathbb{Q}(\sqrt{-5}) such that 3\lambda + (1+\sqrt{-5})\mu = 1.

Let a+b\sqrt{-5} be an arbitrary integer in \mathbb{Q}(\sqrt{-5}). Since -5 \equiv 3 \not\equiv 1 mod 4, a and b are integers. Note that N(a+b\sqrt{-5}) = a^2 + 5b^2. If this element has norm 3, then taking this equation mod 5 we have a^2 \equiv 3. Note, however, that the squares mod 5 are 0, 1, and 4. In particular, no algebraic integer in \mathbb{Q}(\sqrt{-5}) has norm 3.

Now N(3) = 9 and N(1+\sqrt{-5}) = 6; if these elements have a factorization, then some factor must be a unit. Thus both 3 and 1 + \sqrt{-5} are irreducible as integers in \mathbb{Q}(\sqrt{-5}). By Theorem 7.7 in TAN, the units in this ring are precisely \pm 1, so that 3 and 1 + \sqrt{-5} are not associates.

If \delta|3 and \delta|(1+\sqrt{-5}), then since N(\delta) \neq 3, N(\delta) = \pm 1, and thus by Lemma 7.3 in TAN \delta is a unit. That is, every common divisor of 3 and 1+\sqrt{-5} is a unit, so that these elements are relatively prime.

Suppose now that there are integers a,b,c,d such that, with \lambda = a+b\sqrt{-5} and \mu = c+d\sqrt{-5}, we have 3\lambda + (1+\sqrt{5})\mu = 1.Comparing coefficients, we have 3a+c-5d = 1 and 3b+c+d = 0. Mod 3, these equations reduce to c+d \equiv 1 and c+d \equiv 0, a contradiction. So no such \lambda and \mu exist.

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