Quadratic integer rings

Integral extensions of rings

This is a generalization to rings of algebraic extensions of fields.

Definition 332.1(Definition).

Suppose $S$ is a subring of a commutative ring $R$.

1. $r\in R$ is integral over $S$ if $r$ is the root of a monic polynomial in $S[x]$.
2. The ring $R$ is an integral extension of $S$ or just integral over $S$ if every $r\in R$ is integral over $S$.
3. The integral closure of $S$ in $R$ is the set of elements of $R$ that are integral over $S$.
4. The ring $S$ is said to be integrally closed in $R$ if $S$ is equal to its integral closure in $R$. The integral closure of an integral domain $S$ in its field of fractions is called the normalization of $S$. An integral domain is called integrally closed or normal if it is integrally closed in its field of fractions.
5. For any $r\in R$, $S[r]$ is the smallest subring of $R$ containing $S$ and $r$.

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Here’s the analog of Rmk 119.14:

Proposition 332.2(Proposition).

Suppose $S\subseteq R$. TFAE:

1. $r\in R$ is integral over $S$.

2. The subring $S[r]\subseteq R$ is a finitely generated $S$-module.

Proof.

If $r$ is the root of a monic, degree $n$ polynomial over $S$, then $S[r]$ is spanned as an $S$-module by $\{1,r,\dots ,r^{n-1}\}$. Conversely, if $S[r]$ is spanned as an $S$-module by finitely many elements, then at most finitely many powers of $r$, say $\{1,r,r^2,\dots ,r^{n-1}\}$, appear in the formulas for these elements. It follows that these powers of $r$ span $S[r]$ as an $S$-module, hence $r^n$ is an $S$-linear combination of lower powers of $r$.□

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Lemma 332.3(Lemma).

Let $R$ be an integral domain and $S\subseteq R$ be a UFD. Let $r\in R$ be integral over $S$; let $f(x)\in S[x]$ be a monic irreducible polynomial such that $f(r)=0$. Then, $S[r]\cong S[x]/(f(x))$.

Proof.

Let $\phi :S[x]\to S[r]$ be defined by $x\mapsto r$. We will first show that $\ker \phi$ is principally generated. We know that $\ker \phi$ is nonempty since $f(x)\in \ker \phi$. Let $g(x)$ be the smallest degree polynomial in $\ker \phi$. Since we can write $g(x)=c(g)g^{\prime }(x)$ and $g(r)=0$ implies $g^{\prime }(r)=0$, we can WLOG assume $c(g)=1$. Let $p(x)\in \ker \phi$. Let $\overline{S}$ be the field of fractions of $S$. We have

$$p(x)=\overline{q}(x)g(x)+\overline{m}(x),$$

where $\overline{q}(x),\overline{m}(x)\in \overline{S}[x]$ and $\text{deg}\overline{m}(x)<\text{deg}g(x)$. Write $\overline{m}(x)=\alpha m(x)$, where $\alpha \in \overline{S}$ and $m(x)\in S[x]$. Since $m(r)=0$, $m(x)=0$ to maintain the minimality of $g(x)$. It follows that $g(x)$ divides $p(x)$ in $\overline{S}[x]$. Since $S$ is a UFD, it follows (Thm 115.10.4) that $g(x)$ divides $p(x)$ in $S[x]$. Thus, $\ker \phi =(g(x))$.

Now, $g(x)$ divides $f(x)$. Since $f(x)$ is irreducible, it follows that $g(x)$ and $f(x)$ are associates, so $\ker \phi =(g(x))=(f(x))$.□

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It follows that if $r$ is algebraic over $S$, then $S[r]$ is a finitely generated $S$ module.

For example, If $\omega \in \mathbb{C}$ is integral over $\mathbb{Z}$, and $f(x)\in \mathbb{Z}[x]$ is a monic irreducible polynomial such that $f(\omega )=0$, then $\mathbb{Z}[\omega ]\cong \mathbb{Z}[x]/(f(x))$.

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Quadratic integer rings

Definition 332.4(Definition).

Let $D$ be a squarefree integer. ${\mathcal{O}}_{\mathbb{Q}(\sqrt{D})}$ is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}\sqrt{D}$, and is called a quadratic integer ring.

Proposition 332.5(Proposition).

${\mathcal{O}}_{\mathbb{Q}(\sqrt{D})}=\mathbb{Z}[\omega ]$, where

$$\omega =\{\begin{array}{ll}\sqrt{D}& D\equiv 2,3\mathrm{mod}4\\ \frac{1+\sqrt{D}}{2}& D\equiv 1\mathrm{mod}4.\end{array}$$

See Hughes (2015) for a proof.

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References

Hughes, A. (2015). Answer to “Why Is Quadratic Integer Ring Defined in That Way?” In Mathematics Stack Exchange. https://math.stackexchange.com/a/1198964/1677240