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This question already has an answer here:

Calculate the ring of integers of quadratic number field $\mathbb{Q}(\sqrt{d})$

Solution:

Let $F$ be an algebraic number field. Then an element $b\in F$ is integral iff it monic irreducible polynomial has integer coefficients.

For example, $\sqrt{d}$ for integer $d$ is integral.

If $d\equiv 1 \mod 4$, then the monic irreducible polynomial of $(1+\sqrt{d})/2$ over $\mathbb{Q}$ is $x^2 -x + (1-d)/4 \in \mathbb{Z}[x]$, so $(1+\sqrt{d})/2$ is integral.

Thus the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$ contains the subring $\mathbb{Z}[\sqrt{d}]$, and the subring $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1 \mod 4$. We show that there are no other integral elements.

An element $a+b\sqrt{d}$ with rational $a$ and $b\neq0$ is integral iff its monic irreducible polynomial $x^2 -2ax +(a^2 -db^2)$ belongs to $\mathbb{Z}[x]$. Therefore, $2a$,$2b$ are integers. If $a=(2k+1)/2$, for $k\in\mathbb{Z}$, then it is easy to see that $a^2 - db^2 \in \mathbb{Z}$ iff $b=(2l+1)/2$ for some $l\in\mathbb{Z}$, and $(2k+1)^2 - d(2l+1)^2$ is divisible by 4. The latter implies that d is quadratic residue modulo 4, i.e. $d\equiv 1 \mod 4$. In turn, if $d\equiv 1 \mod 4$ then every element $(2k+1)/2 +(2l+1)\sqrt{d}/2$ is integral.

Thus, integral elements of $\mathbb{Q}(\sqrt{d})$ are equal to $\mathbb{Z}[\sqrt{d}]$ if $d\not \equiv 1 \mod 4$, and $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1 \mod 4$.


Can someone explain why $\sqrt{d}$ for integer $d$ is integral, and why the monic irreducible polynomial of an integral element $a+b\sqrt{d}$ is of the form $x^2 -2ax +(a^2 -db^2)$?

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marked as duplicate by Adam Hughes, Chappers, Rolf Hoyer, HK Lee, user147263 Apr 18 '15 at 4:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $\sqrt{d}$ is integral because if we take the poly $x^2-d$, this is a monic irreducible polynomial (If $d$ is not a square number) with integer coefficients, with $d$ as a root. This happens to be THE monic irreducible polynomial for $\sqrt{d}$, which means $\sqrt{d}$ is integral. $\endgroup$ – Ramified_Minds Apr 16 '15 at 14:51
  • $\begingroup$ $ x = a+b\sqrt d\,\Rightarrow\, (x-a)^2 = (b\sqrt d)^2\ \ $ $\endgroup$ – Bill Dubuque Apr 16 '15 at 15:26
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For your first question, if $d$ is an integer then $\sqrt{d}$ satisfied $x^2 - d$ so is an algebraic integer.

For your second question, just plug in $a + b\sqrt{d}$ for $x$ and observe that you get $0$. This polynomial is monic, and it's irreducible so long as $\sqrt{d}$ is not a rational (i.e. so long as $d$ is not a square), because a quadratic polynomial with no roots in $\mathbb{Q}$ is irreducible over $\mathbb{Q}$.

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  • $\begingroup$ Ok now I understand. Thank you!! $\endgroup$ – P.D. Apr 16 '15 at 14:56

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