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Let $\mathbb{Z}[\sqrt{2}]=\{a+b \sqrt{2}: a,b \in \mathbb{Z} \}$. Then

(a) Prove that $\mathbb{Z}[\sqrt{2}]$ is a commutative ring.

(b) Determine the units of $(\mathbb{Z}[\sqrt{2}])\neq 0 $.

(c) Find a subring $S$ of $\mathbb{Z}[\sqrt{2}]$ such that $S\neq \mathbb{Z} $ and $S \neq \mathbb{Z}[\sqrt{2}] $.

I do not know how to do parts (b) and (c).

Zev Chonoles
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Janeth Benavides
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1 Answers1

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For b: write $(a + b\sqrt{2})(c + d\sqrt{2}) = 1$. Distribute the left hand side. What can you say about the relationship between $a, b, c$ and $d$? For c: what's a subring of $\Bbb{Z}$ that is not equal to $\Bbb{Z}$? Expand this subring to one contained in $\Bbb{Z}[\sqrt{2}]$ but not contained in $\Bbb{Z}$ itself.

Vik78
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    I do not get what you mean with the "For c" part. There is none, at least not in the sense intended by the question. – quid Sep 02 '16 at 00:00
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    Are you defining a ring to have an identity? If not $\Bbb{Z}$ certainly has a proper nontrivial subring. – Vik78 Sep 02 '16 at 00:06
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    Yes, rings in this context (most contexts in fact, but certainly in algebraic number theory) must have an identity. And even if not the way it is written it is obviously meant that $Z \subsetneq S \subsetneq Z [\sqrt{2}]$ – quid Sep 02 '16 at 00:09
  • I was not saying that you should use a subring of $\Bbb{Z}$ as the answer. I was trying to nudge OP to find a subring $S$ not contained in $\Bbb{Z}$ by first finding a proper subring $R$ of $\Bbb{Z}$ and then letting $S$ be the set of $(a,b) \in \Bbb{Z}[\sqrt{2}]$ with $a, b \in R$. I don't define a ring as having an unity, personally. I don't see why you assume that the OP defined a ring that way, seeing as they never indicated one way or the other. – Vik78 Sep 02 '16 at 00:17
  • Because it is extremely wide spread in this context. Deviating from this without making it explicit lures the asker into making an error. Note that there is a tag [tag:rngs] for 'rings that do not have an identity' // Plus, in doubt, one should give an example that will certainly work. OP hardly could be required to exhibit a subring without identity. Thus one with identity will always be good; – quid Sep 02 '16 at 00:25
  • Even if she needed a unity, my suggestion would work with a tiny modification-- let $S$ be the set of $(a, b)$ with $a \in \Bbb{Z}$, $b \in R$. By the way, this is a basic intro to rings problem. I wouldn't really call it "algebraic number theory" and if the OP is at that level I doubt they have heard of an rng. – Vik78 Sep 02 '16 at 00:54
  • Yes, that would work. But I knew the complete description of all subrings of $\mathbb{Z}[\sqrt{2}]$ before reading your hint, and (still) had no idea how it would be hepful. // On the rest, well its a maximal order of an algebraic number field and what is effectively asked for is a non maximal order. It is (simple) alg NT. – quid Sep 02 '16 at 00:59
  • This is a very poor answer. The hint for (b) is useless, and also for (c). Playing with rings without unit within a homework exercise in basic algebra is a non-sense. (-1) – user26857 Sep 02 '16 at 07:13