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This was a problem in my math book and I was wondering what the proof looks like:

Let $R = \{a + b\sqrt{2} | a,b\in \mathbb{Z}\}$. Show that $I = \{a+b\sqrt{2}\in R | a$ is divisible by $2\}$ is an ideal of $R$, but $J = \{a + b\sqrt{2}\in R | a$ is divisible by $3\}$ is not. And $R$ is a ring.

Thanks

user1234
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    Alright, I think I have it. All I have to is show that $I$ is an ideal (i.e it is an additive subgroup and holds for 'product absorption') and than show that $J$ is not because product absorption does not hold. – user1234 Dec 15 '13 at 23:42

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Here's a simple ideal test for your modules $\rm\:\! M_2,M_3,\,$ for $\rm\, M_a = [a,\sqrt{2}] = a\:\!\Bbb Z + \!\sqrt{2}\,\Bbb Z,\ a\in\Bbb Z$
Lemma $ $ Module $\rm M = {[a,\,\overbrace{b\!+\!\sqrt{d}}^{\large \alpha}}] = a\,\Bbb Z + \alpha\,\Bbb Z\,$ is an ideal of $\rm \,\Bbb Z[\sqrt d]\!$ $\rm\iff\! a\mid \alpha\bar\alpha = b^2\!-\!d$

Proof $\, $ The module $\rm\,M = [a,b\!+\!\sqrt{d}]\:$ is an ideal of $\rm\,R = \Bbb Z[\sqrt{d}]\!\iff\! M\,$ is closed under multiplication by elements of $\rm\,R\!\iff\!$ $\rm\sqrt{d}\ M \subseteq M\!\iff a\sqrt{d},\, \sqrt{d}(b\!+\!\sqrt{d}) \in M.\,$ Clearly $\rm\:a\sqrt{d}\, =\, -ab+a(b\!+\!\sqrt{d})\in M,\,$ and $\rm \sqrt{d}\,(b\!+\!\sqrt{d}) = (\sqrt{d}+b)b-(b^2-d)\,\in M\!$ $\rm \iff\! b^2-d \in M,\,$ i.e. iff $\,\rm {\rm norm}(\alpha) = \alpha\bar\alpha\in M\cap\Bbb Z = a\,\Bbb Z$. $\ \ \bf\small QED$

Remark $ $ This generalizes to an ideal test for a module $\rm\,[a,b\!+\!c\,\omega]\,$ in the ring of integers of a quadratic number field, c.f. section 2.3 in Lemmermeyer's Ideals in Quadratic Number Fields.

This is a special case of module normal forms for higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory. Such (triangular) normal forms greatly simplify membership tests.

Bill Dubuque
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