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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

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  • $\begingroup$ 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. $\endgroup$ – user1234 Dec 15 '13 at 23:42
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Here is s simple ideal test for such modules. If $\rm\:M = [a,b\!+\!\sqrt{d}] = a\,\Bbb Z + (b\!+\!\sqrt{d})\,\Bbb Z\:$ is an ideal of$\,\Bbb Z[\sqrt d]$ then it must contain the norm $\rm\: N(b\!+\!\sqrt{d}) = (b\!+\!\sqrt{d})(b\!-\!\sqrt{d}) = b^2\!-\!d,\: $ so $\rm\ a\mid b^2\!-\!d.\:$ This necessary condition is also sufficient, as we prove below.

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 $\rm\,R\iff$ $\rm\sqrt{d}\ M \subseteq M\iff a\sqrt{d},\, (b\!+\!\sqrt{d})\sqrt{d} \in M.\:$ The first membership is clear since $\rm\:a\sqrt{d}\, =\, a(b\!+\!\sqrt{d})-ab\in M.\:$ For the second membership

$$\begin{eqnarray}\rm -\sqrt{d}\,(b\!+\!\sqrt{d}) &=\,&\rm (b\!-\!\sqrt{d})(b\!+\!\sqrt{d})-b(b\!+\!\sqrt{d})\\ &= &\rm\ b^2-d\, -\,b(b\!+\!\sqrt{d})\end{eqnarray}$$

The prior is in $\rm\ M = [a,b\!+\!\sqrt{d}]\iff a\mid b^2\!-\!d = N(b\!+\!\sqrt{d}),\:$ where $\rm\:N = $ norm.

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

This is a special case of module normal forms that generalize to 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 normal forms greatly simplify membership tests.

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  • $\begingroup$ Note: apply the criteron to $\ R = [2,\sqrt{2}] = 2\,\Bbb Z + \!\sqrt{2}\,\Bbb Z\ $ and $\ J = [3,\sqrt{2}] = 3\,\Bbb Z +\! \sqrt{2}\,\Bbb Z\ \ $ $\endgroup$ – Bill Dubuque Mar 24 '15 at 18:58

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