Algebriac Number Theory:1

Algebraic Number, Algebraic IntegerGeneralization of Algebraic IntegerAlgebraic Number Field

 

Algebraic Number, Algebraic Integer

Definition. Algebraic Number

Let $\alpha \in \mathbb{C}$$\alpha\in\mathbb C$ be a root of a polynomial $f(X)\in \mathbb{Q}[X]$$f(X)\in\mathbb Q[X]$, i.e. $f(\alpha )=0$$f(\alpha)=0$, then we call $\alpha$$\alpha$ an algebraic number.

The $\alpha \in \mathbb{C}$$\alpha\in\mathbb C$ defines an evaluation map ${\mathrm{ev}}_{\alpha }:\mathbb{Q}[X]\to \mathbb{C},f(X)⟼f(\alpha )$$\mathrm{ev_{\alpha}}:\mathbb Q[X]\to\mathbb C,f(X)\longmapsto f(\alpha)$. Since $\mathbb{Q}[X]$$\mathbb Q[X]$ is a PID, then $\mathrm{Ker}({\mathrm{ev}}_{\alpha })=(m_{\alpha }(X))$$\mathrm{Ker}(\mathrm{ev_{\alpha}})=(m_{\alpha}(X))$, with $m_{\alpha }(X)$$m_{\alpha}(X)$ being monic, and we define $m_{\alpha }(X)$$m_{\alpha}(X)$ to be the minimal polynomial of $\alpha$$\alpha$, and $\mathbb{Q}[\alpha ]\cong \mathbb{Q}[X]/(m_{\alpha }(X))$$\mathbb Q[\alpha]\cong\mathbb Q[X]/(m_{\alpha}(X))$.

Definition. Algebraic Integer

If the minimal polynomial of an algebraic number $\alpha$$\alpha$ has coefficients in $\mathbb{Z}$$\mathbb Z$, then we say $\alpha$$\alpha$ is an algebraic integer.

$$\begin{array}{}\text{(1)}& m_{\alpha }(X)=X^n+a_{n-1}{X}^{n-1}+...+a_0\end{array}$$

Proposition. The relation between algebraic number and algebraic integer:

Let $\alpha$$\alpha$ be an algebraic number, then there exists an integer $b\in \mathbb{Z}$$b\in \mathbb Z$, such that $b\alpha$$b \alpha$ is an algebraic integer.

In other words, every algebraic number could be represented as ${\displaystyle \frac{\beta }{b}}$$\dfrac{\beta}{b}$ for some algebraic integer $\beta$$\beta$.

Proof. Let $f=a_n{\alpha }^n+a_{n-1}{\alpha }^{n-1}+...+a_0$$f=a_n\alpha^n+a_{n-1}\alpha^{n-1}+...+a_0$ with integer coefficients and consider $a_n^{n-1}f$$a_n^{n-1}f$.

$$\begin{array}{}\text{(2)}& a_n^n{\alpha }^n+a_n^{n-1}{a}_{n-1}{\alpha }^{n-1}+...+a_n^{n-1}{a}_0=(a_n\alpha {)}^n+a_{n-1}(a_n\alpha {)}^{n-1}+...+a_n^{n-1}{a}_0\end{array}$$

Hence $a_n\alpha$$a_n\alpha$ is an algebraic integer.

Generalization of Algebraic Integer

Definition. Let $A$$A$ be an $R$$R$-algebra where $R$$R$ is a commutative ring. An element $x\in A$$x\in A$ is called integer over $R$$R$ if

$$\begin{array}{}\text{(3)}& \mathrm{\exists }{a}_0,...,a_{n-1}\in R,x^n+a_{n-1}{x}^{n-1}+...+a_0=0\end{array}$$

Example. Let $R=\mathbb{Z},A\subseteq \mathbb{C}$$R=\mathbb Z,A\subseteq\mathbb C$. Then $x\in A$$x\in A$ is integer over $\mathbb{Z}$$\mathbb Z$ if $x$$x$ is an algebraic integer.

Recall that an abelian group $M$$M$ could be an $R$$R$-module iff there exists a ring homomorphism $f:R\to {\mathrm{End}}_{\mathrm{Ab}}(M)$$f:R\to\mathrm{End}_{\mathrm{Ab}}(M)$.

Then we say $M$$M$ is a faithful $R$$R$-module if $f$$f$ is injective. Notice that $\mathrm{Ker}(f)={\mathrm{ann}}_R(M)$$\mathrm{Ker}(f)=\mathrm{ann}_R(M)$.

Lemma.

View $A$$A$ as an $R[x]$$R[x]$ module via left multiplication, then $A$$A$ is faithful. Indeed, every submodule of $A$$A$ containing $1$$1$ is faithful.

Proof. Let $M$$M$ be a submodule of $A$$A$ containing $1$$1$, $\mathrm{\forall }s\in R[x],s\cdot 1=s\ne 0$$\forall s\in R[x],s\cdot1=s\ne 0$, hence ${\mathrm{ann}}_{R[x]}(M)=\{0\}$$\mathrm{ann}_{R[x]}(M)=\{0\}$. $◻$$\square$

Proposition. $\mathrm{\forall }x\in A$$\forall x\in A$, the following statements are equivalent.

$(i)$$(i)$ $x$$x$ is integer over $R$$R$

$(ii)$$(ii)$ $R[x]$$R[x]$ is a finitely generated $R$$R$-module.

$(iii)$$(iii)$ $x$$x$ is contained in a faithful sub $R[x]$$R[x]$-module of $A$$A$ and the submodule is finitely generated over $R$$R$.

Proof.

$(i)⟹(ii)$$(i)\implies (ii)$ Since $x^n+a_{n-1}{x}^{n-1}+...+a_0=0$$x^n+a_{n-1}x^{n-1}+...+a_0=0$, hence $x^{n+1}=-(a_0+...+a_{n-1}{x}^n)$$x^{n+1}=-(a_0+...+a_{n-1}x^n)$.

$(ii)⟹(iii)$$(ii)\implies (iii)$ is obvious.

$(iii)⟹(i)$$(iii)\implies (i)$ Since $x\in M$$x\in M$ and $M$$M$ is an $R[x]$$R[x]$-module, we have $xM\subseteq M$$xM\subseteq M$.

Assume $M$$M$ is generated by $b_1,...,b_m$$b_1,...,b_m$, then ${\displaystyle x=\sum _{i=1}^m{r}_i{b}_i}$$\displaystyle x=\sum_{i=1}^m r_i b_i$, and consider $x{b}_i$$xb_i$, which can be represented via a matrix $T$$T$.

$$\begin{array}{}\text{(4)}& x{b}_i=\sum _{j=1}^m{t}_{ij}{b}_j\end{array}$$

Hence we have

$$\begin{array}{}\text{(5)}& \begin{array}{c}(x\cdot 1_{m\times m}-T)\left(\begin{array}{c}{b}_1\\ .\\ .\\ .\\ b_m\end{array}\right)=\left(\begin{array}{c}0\\ .\\ .\\ .\\ 0\end{array}\right)\end{array}\end{array}$$

and consider multiplying by $(x\cdot 1_{m\times m}-T{)}^{\vee }$$(x\cdot 1_{m\times m}-T)^\vee$ on both sides, we get $f(x)=\det (x\cdot 1_{m\times m}-T)=x^n+...+a_0$$f(x)=\det(x\cdot 1_{m\times m}-T)=x^n+...+a_0$ which is monic.

$$\begin{array}{}\text{(6)}& \begin{array}{c}f(x)I\left(\begin{array}{c}{b}_1\\ .\\ .\\ .\\ b_m\end{array}\right)=\left(\begin{array}{c}0\\ .\\ .\\ .\\ 0\end{array}\right)⟹f(x)b_i=0\end{array}\end{array}$$

Since the module $M$$M$ is finitely generated by $b_i$$b_i$, hence $\mathrm{\forall }m\in M,f(x)m=0$$\forall m\in M,f(x)m=0$.

Since $M$$M$ is a faithful $R[x]$$R[x]$ module, $f(x)=0$$f(x)=0$. Hence $x$$x$ is integer over $R$$R$.

Corollary.

Let $\alpha ,\beta$$\alpha,\beta$ be two integers over $R$$R$, then $\alpha +\beta ,\alpha \beta$$\alpha+\beta,\alpha\beta$ are integers over $R$$R$. Hence the integers over $R$$R$ in $A$$A$ form a ring.

Proof.

By the previous proposition, that means $R[\alpha ],R[\beta ]$$R[\alpha],R[\beta]$ are finitely generated by $({\alpha }_i),({\beta }_i)$$(\alpha_i),(\beta_i)$, then $R[\alpha ,\beta ]$$R[\alpha,\beta]$ is generated by $({\alpha }_i{\beta }_j)$$(\alpha_i\beta_j)$.

Easy to see that as an $R[\alpha +\beta ]$$R[\alpha+\beta]$-module and $R[\alpha \beta ]$$R[\alpha\beta]$ module, $R[\alpha ,\beta ]$$R[\alpha,\beta]$ is faithful since it contains $1$$1$. Hence $\alpha +\beta ,\alpha \beta$$\alpha+\beta,\alpha\beta$ are integers over $R$$R$. $◻$$\square$

Definition. Let $A$$A$ be an integral domain, then we say $A$$A$ is integrally closed if the integer ring of $\mathrm{Frac}(A)$$\mathrm{Frac}(A)$ over $A$$A$ is $A$$A$.

Example. In section 7 of https://www.researchgate.net/publication/378858835_ODE_An_Algebraic_Approach, you will see how to use differential rings to prove the holomorphic function ring is integrally closed.

Algebraic Number Field

Definition. A field $K$$K$ satisfying $[K:\mathbb{Q}]<\mathrm{\infty }$$[K:\mathbb Q]<\infty$ is called an algebraic number field. Hence we can always treat $K$$K$ as a subfield of $\mathbb{C}$$\mathbb C$. We denote the algebraic integer ring in $K$$K$ over $\mathbb{Z}$$\mathbb Z$ as ${\mathcal{O}}_K$$\mathcal O_K$.

Lemma. ${\mathcal{O}}_{\mathbb{Q}}=\mathbb{Z}$$\mathcal O_\mathbb Q=\mathbb Z$.

Proof. Consider $f(\alpha )={\alpha }^n+...+a_0=0$$f(\alpha)=\alpha^n+...+a_0=0$ with $a_0...a_{n-1}\in \mathbb{Z}$$a_0...a_{n-1}\in\mathbb Z$. Assume $\alpha \notin \mathbb{Z}$$\alpha\notin\mathbb Z$ then $\alpha ={\displaystyle \frac{p}{q}},gcd(p,q)=1$$\alpha=\dfrac{p}{q},\gcd(p,q)=1$. Then

$$\begin{array}{}\text{(7)}& {\displaystyle \frac{p^n}{q^n}}+a_{n-1}{\displaystyle \frac{p^{n-1}}{q^{n-1}}}...+a_0=0⟹{\displaystyle \frac{p^n}{q}}=-(a_{n-1}{p}^{n-1}+...+a_0{q}^{n-1})\end{array}$$

That is impossible. $◻$$\square$

Proposition. Let $K=\mathbb{Q}[\sqrt{m}]$$K=\mathbb Q[\sqrt m]$ for a square-free (that is, $(m)$$(m)$ is a radical ideal) number $m$$m$, then

$$\begin{array}{}\text{(8)}& \begin{array}{c}{\mathcal{O}}_K=\{\begin{array}{l}\{{\displaystyle \frac{u+v\sqrt{m}}{2}}:u\equiv v\mathrm{mod}2,u,v\in \mathbb{Z}\}=\{a+{\displaystyle \frac{1+b\sqrt{m}}{2}}:a,b\in \mathbb{Z}\},\text{ if }m\equiv 1\mathrm{mod}4\\ \{a+b\sqrt{m}:a,b\in \mathbb{Z}\},\text{ if }m\equiv 2\text{ or }3\mathrm{mod}4\end{array}\end{array}\end{array}$$

Proof.

Let $\alpha =a+b\sqrt{m}\in K$$\alpha=a+b\sqrt m\in K$ be an algebraic integer, and $\sigma :a+b\sqrt{m}⟼a-b\sqrt{m}$$\sigma:a+b\sqrt m\longmapsto a-b\sqrt m$ , hence

$$\begin{array}{}\text{(9)}& {\alpha }^n+...+a_0=0=\sigma (\alpha {)}^n+...+\sigma (a_0)=\sigma ({\alpha }^n)+...+\sigma (a_0)=0\end{array}$$

i.e. $\sigma (\alpha )\in {\mathcal{O}}_K$$\sigma(\alpha)\in \mathcal O_K$ as well, hence $\alpha +\sigma (\alpha )=2a,\alpha \sigma (\alpha )=a^2-m{b}^2\in {\mathcal{O}}_K$$\alpha+\sigma(\alpha)=2a,\alpha\sigma(\alpha)=a^2-mb^2\in \mathcal O_K$.

Since $2a,a^2-m{b}^2\in \mathbb{Q}\cap {\mathcal{O}}_K$$2a,a^2-mb^2\in\mathbb Q\cap\mathcal O_K$, hence $2a,a^2-m{b}^2\in \mathbb{Z}$$2a,a^2-mb^2\in\mathbb Z$

Hence

$$\begin{array}{}\text{(10)}& (2a{)}^2-m(2b{)}^2=4(a^2-m{b}^2)\in \mathbb{Z}⟹m(2b{)}^2\in \mathbb{Z}\end{array}$$

Hence $2b\in \mathbb{Z}$$2b\in\mathbb Z$, since if $2b={\displaystyle \frac{p}{q}},gcd(p,q)=1⟹{\displaystyle \frac{m{p}^2}{q^2}}\in \mathbb{Z}⟹q^2|m$$2b=\dfrac{p}{q},\gcd(p,q)=1\implies \dfrac{mp^2}{q^2}\in\mathbb Z\implies q^2|m$ but $m$$m$ is a square-free number.

Now we know that $2a,2b\in \mathbb{Z}$$2a,2b\in\mathbb Z$, so we could assume $a={\displaystyle \frac{u}{2}},b={\displaystyle \frac{v}{2}},u,v\in \mathbb{Z}$$a=\dfrac{u}{2},b=\dfrac{v}{2},u,v\in\mathbb Z$.

Then

$$\begin{array}{}\text{(11)}& a^2-m{b}^2\in \mathbb{Z}⟹u^2-m{v}^2\in 4\mathbb{Z}⟺u^2\equiv m{v}^2\mathrm{mod}4\end{array}$$

If $m\equiv 1\mathrm{mod}4$$m\equiv 1\mod 4$, then $u^2\equiv v^2\mathrm{mod}4⟹u^2\equiv v^2\mathrm{mod}2⟺u\equiv v\mathrm{mod}2$$u^2\equiv v^2\mod 4\implies u^2\equiv v^2\mod 2\iff u\equiv v\mod 2$ by $\pi :\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$\pi:\mathbb Z/4\mathbb Z\to\mathbb Z/2\mathbb Z$.

So when $m\equiv 1\mathrm{mod}4$$m\equiv 1\mod 4$, we have $\alpha =a+b\sqrt{m}={\displaystyle \frac{u+v\sqrt{m}}{2}}$$\alpha=a+b\sqrt m=\dfrac{u+v\sqrt m}{2}$. When $u\equiv v\equiv 0\mathrm{mod}2$$u\equiv v\equiv 0\mod 2$,

$$\begin{array}{}\text{(12)}& \alpha =u^{\prime }+v^{\prime }\sqrt{m}\in \mathbb{Z}\left[{\displaystyle \frac{1+\sqrt{m}}{2}}\right]\end{array}$$

When $u\equiv v\equiv 1\mathrm{mod}2$$u\equiv v\equiv 1\mod 2$, $u=2k+1$$u=2k+1$, $v=2s+1$$v=2s+1$, $\alpha ={\displaystyle \frac{2k+1+(2s+1)\sqrt{m}}{2}}=k+s\sqrt{m}+{\displaystyle \frac{1+\sqrt{m}}{2}}=k-s+2s{\displaystyle \frac{1+\sqrt{m}}{2}}\in \mathbb{Z}\left[{\displaystyle \frac{1+\sqrt{m}}{2}}\right]$$\alpha=\dfrac{2k+1+(2s+1)\sqrt m}{2}=k+s\sqrt m+\dfrac{1+\sqrt m}{2}=k-s+2s\dfrac{1+\sqrt m}{2}\in \mathbb Z\left[\dfrac{1+\sqrt m}{2}\right]$ Since ${\mathcal{O}}_K$$\mathcal O_K$ is a free module and $\dim {\mathcal{O}}_K\ne 1$$\dim\mathcal O_K\ne 1$ and ${\mathcal{O}}_K\subseteq \mathbb{Z}\left[{\displaystyle \frac{1+\sqrt{m}}{2}}\right]$$\mathcal O_K\subseteq \mathbb Z\left[\dfrac{1+\sqrt m}{2}\right]$ and $\dim \mathbb{Z}\left[{\displaystyle \frac{1+\sqrt{m}}{2}}\right]=2$$\dim\mathbb Z\left[\dfrac{1+\sqrt m}{2}\right]=2$

We claim that ${\mathcal{O}}_K=\mathbb{Z}\left[{\displaystyle \frac{1+\sqrt{m}}{2}}\right]$$\mathcal O_K=\mathbb Z\left[\dfrac{1+\sqrt m}{2}\right]$ when $m\equiv 1\mathrm{mod}4$$m\equiv 1\mod 4$.

If $m\equiv 3\equiv -1\mathrm{mod}4$$m\equiv 3\equiv -1\mod 4$, then

$$\begin{array}{}\text{(13)}& u^2+v^2\equiv 0\mathrm{mod}4⟹u^2+v^2\equiv 0\mathrm{mod}2⟹u\equiv v\equiv 0\mathrm{mod}2\end{array}$$

Hence $a,b\in \mathbb{Z}$$a,b\in\mathbb Z$.

If $m\equiv 2\mathrm{mod}4$$m\equiv 2\mod 4$, then

$$\begin{array}{}\text{(14)}& u^2\equiv 2v^2\mathrm{mod}4⟹u^2\equiv 0\mathrm{mod}2⟹u\equiv 0\mathrm{mod}2\end{array}$$

Hence $u^2\equiv 0\equiv 2v^2\mathrm{mod}4$$u^2\equiv 0\equiv 2v^2\mod 4$, $v^2\equiv 0\mathrm{mod}2⟹v\equiv 0\mathrm{mod}2$$v^2\equiv 0\mod 2\implies v\equiv 0\mod 2$. Hence $a,b\in \mathbb{Z}$$a,b\in\mathbb Z$.

Hence ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{m}]$$\mathcal O_K=\mathbb Z[\sqrt m]$, if $m\equiv 2,3\mathrm{mod}4$$m\equiv 2,3\mod 4$.