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I'm doing a master's thesis about quadratic fields and I know that for squarefree m, the ring of integers for the quadratic field $\mathbb{Q}(\sqrt{m})$ is $\mathbb{Z}[\sqrt{m}]$ if $m \equiv 2,3 \pmod{4}$ and $\mathbb{Z}[\frac{1 + \sqrt{m}}{2}]$ if $m \equiv 1 \pmod{4}$. My question is this: Why is the ring of integers in $\mathbb{Q}(\sqrt{m}) = \mathbb{Z}[\frac{1+\sqrt{m}}{2}]$ when $m \equiv 1 \pmod{4}$ instead of just $\mathbb{Z}[\sqrt{m}]$?

EDIT: I have seen some of your posts and I know that it has to do with integral closure. My confusion comes when I see that the minimal polynomial of $3 + 6\sqrt{5}$ is $x^2 -6x -171$ even though $3 + 6\sqrt{5}$ does not seem to be an element of $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$. What I'm looking for is an element of $\mathbb{Z}[\sqrt{m}]$ which is not a quadratic integer. i.e. one whose minimal polynomial has a non-integer coefficient.

Arturo Magidin
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Sean Phillip Gould
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    By "why is the ring of integers = ..." , are you asking for a proof, a heuristic or something else? I mean, $\frac{1+\sqrt 5}{2}$ belongs to the ring of integers of $\mathbb Q(\sqrt 5)$, so that's a simple reason to expect the $\mathbb Z[\sqrt m]$ pattern to break. – Sarvesh Ravichandran Iyer Mar 11 '20 at 04:45
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    Welcome to Mathematics Stack Exchange. $\frac{1+\sqrt5}2$ is a root of a monic polynomial $x^2-x-1$ – J. W. Tanner Mar 11 '20 at 04:58
  • A possible answer: https://math.stackexchange.com/questions/1198188/why-is-quadratic-integer-ring-defined-in-that-way – justadzr Mar 11 '20 at 05:12
  • If you think $\sqrt{5}$ is not an element of $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$, that's wrong: $\sqrt{5}=-1+2(\frac{1+\sqrt{5}}{2})\in\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$. – C. Fujinomiya Mar 11 '20 at 06:09
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    You have this backwards! Note that $\mathbb Z[\sqrt{m}]\subset \mathbb Z[\frac{1+\sqrt m}2]$. What's important is that, when $m \equiv 2, 3\pmod 4$, $\frac{1+\sqrt m}2$ is not an algebraic integer: its minimal polynomial is $x^2-x+\frac{1-m}4$. – Mathmo123 Mar 11 '20 at 08:25
  • Mathmo123, can you prove that $\mathbb{Z}[\sqrt{m}] \subset \mathbb{Z}[\frac{1 + \sqrt{m}}{2}]$? – Sean Phillip Gould Mar 11 '20 at 20:56

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