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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 its monic irreducible polynomial has integer coefficients.

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

If $d\equiv 1 \bmod 4$, 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 $d\equiv 1 \bmod 4$. 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 a quadratic residue modulo $4$, i.e. $d\equiv 1 \bmod 4$. In turn, if $d\equiv 1 \bmod 4$ 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 $d\not \equiv 1 \bmod 4$, and $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1 \bmod 4$.

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 $x^2 -2ax +(a^2 -db^2)$?

user26857
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LLL
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  • $\sqrt{d}$ 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. – Ramified_Minds Apr 16 '15 at 14:51
  • $ x = a+b\sqrt d,\Rightarrow, (x-a)^2 = (b\sqrt d)^2\ \ $ – Bill Dubuque Apr 16 '15 at 15:26
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    how did you conclude that the minimal polynomial is in $\mathbb{Z}[x]$, then $2a,2b$ are integers?. I understood why $2a$ is an integer but couldn't understand why $2b$ is an integer. – permutation_matrix May 22 '21 at 19:13
  • @permutation_matrix Let $b=\frac{s}{t}$ with $s$ and $t$ coprime and $t=\prod p_i^{\alpha_i}$ be a prime factorization of $t$. Then $4db^2=\frac{4ds^2}{t^2}=\frac{4ds^2}{\prod p_i^{2\alpha_i}}$. Since $4db^2$ is an integer we see $4$ divides $\prod p_i^{2\alpha_i}$. Our only other hope at cancellation is to use $d$. But $d$ is squarefree! So we could only cancel off one factor of $p_i$ using $d$. Hence, $\prod p_i^{2\alpha_i}=4$. That is, $b=\frac{s}{2}$. – user2154420 May 08 '23 at 23:31

1 Answers1

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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 $\mathbb{Q}$.

hunter
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