I think the explicit definition is known, so I omit it. Now my question is, that in a book I read that $d \in \mathbb{Z} \setminus \{1\}$ has to be square-free, i.e. if $x^2\mid d$ for $x \in \mathbb{N}$ we have $x = 1$. First of all I was not sure why we impose this restriction. The answer was, that then the mapping $\overline{x + y \sqrt{d}} := x- y\sqrt{d}$ is well defined, since $1$ and $\sqrt{d}$ are linearly independent over $\mathbb{Q}$. But if we take for example $d = 12$, then $\sqrt{d} = 2\sqrt{3}$ which is still linearly independent over $\mathbb{Q}$. Is there any other problematic which I should know of? I think that the square-free restriction is a bit too tight, since it also excludes cases like $d = 12$.
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2AFAIK, the square free restriction is just to make things notationally simpler. Note that $\mathbb Q(\sqrt{m^2d}) = \mathbb Q(\sqrt{d})$, and $\mathbb Z[\sqrt{m^2}d]=\mathbb Z[m\sqrt d]$ – Mathmo123 Jan 15 '17 at 13:00
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@Mathmo123 Ah I see...nice. – TheGeekGreek Jan 15 '17 at 13:06
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One reason to prefer $\,\Bbb Z[\sqrt{d}]$ over $\,\Bbb Z[n\sqrt{d}]$ is that it may have much simpler factorization theory. For example the former domain may be a UFD (has unique prime factorizations), but the latter is not a UFD if $\,|n|> 1\,$ since the same proof as in $\,\Bbb Z\,$ shows RRT = Rational Root Test is true in a UFD, therefore fractional roots of monic polynomials are integral, i.e. they must lie in the UFD. But that fails for the latter domain: if $\,w= n\sqrt d\,$ then over $\,\Bbb Z[w]\,$ the monic $\,x^2-d\,$ has fractional root $\, w/n \not\in \Bbb Z[w]\,$ when $\,|n|>1.\,$ Said more technically: UFDs are integrally closed.
Bill Dubuque
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