6 Algebraic integer

Algebraic integers

Definition

A complex number $\lambda$ is an algebraic integer if and only if $\lambda$ is an eigenvalue of some matrix, all of whose entries are integers.

All algebraic intergers form a ring:

Theorem

If $\lambda$ and $\mu$ are algebraic integers, then $\lambda \mu$ and $\lambda +\mu$ are also algebraic integers.

\begin{proof}Consider the matrix of the tensor product. We can define an endomorphism of $V$ by $(x\otimes y)(A\otimes B)=(xA)\otimes (yB)$. Then $\lambda +\mu$ and $\lambda \mu$ are eigenvalues of $A\otimes B$ and $A\otimes I_n+I_m\otimes B$ respectively. \end{proof}

Remark that $\chi (g)$ is a sum of roots of unity, and each root of unity is an algebraic integer. So we have the following corollary.

Corollary

$\chi (g)$ is an algebraic integer.

Proposition

If $\lambda$ is both a rational number and an algebraic integer, then $\lambda$ is an integer.

The degree of every irreducible character divides $|G|$

It is the most important conclusion of this section.

Theorem

If $\chi$ is an irreducible character of $G$, then $\chi (1)$ divides $|G|$.

To describe an irreducible character $\chi$, consider the corresponding irreducible module $U$. Recall $\theta :v\mapsto vr$ is a $FG$-map if $r\in Z(\mathbb{C}G)$ and $\bar{C}={\sum }_{g\in C}g\in Z(\mathbb{C}G)$. Then by Schur lemma, since $U$ irreducible, we know

$$u\bar{C}=\lambda u$$

for some $\lambda$.

Computation of $\lambda$

Suppose that $\rho$ is a representation of $U$. Then $u\bar{C}=\lambda u$ yields

$$\sum _{g\in C}\rho (g)=\lambda I_d,$$

where $d=\dim U$. So $|C|\chi (g)=\mathrm{tr}(\text{RHS})=\mathrm{tr}(\text{LHS})=\lambda \chi (1)$. Therefore,

$$\lambda =\frac{|G|}{|C_G(g)|}\frac{\chi (g)}{\chi (1)}.$$

$\lambda$ is an algebraic integer

Since $u{\sum }_{g\in C}\rho (g)=\lambda u$, one can say $\lambda$ is an eigenvalue of ${\sum }_{g\in C}\rho (g)$ whose entries are algebraic integers and so $\lambda$ is an algebraic integer. Since $\overline{\chi (g)}$ is also an algebraic integer, we know

$$\sum _{i=1}^k\frac{|G|}{|C_G(g_i)|}\frac{\chi (g_i)\overline{\chi (g_i)}}{\chi (1)}\text{\hspace{1em}(*)}$$

is an algebraic integer. Since $(\ast )=|G|/\chi (1)$ and $(\ast )$ is a rational number and algebraic integer, $(\ast )$ is an integer and so $\chi (1)$ divides $|G|$. Now we finish the proof.

Corollaries

Corollary

No simple group has an irreducible character of degree $2$.

\begin{proof} Let $G$ be a simple group. If $G\simeq C_p$ for some prime $p$, then all irreducible characters of $G$ are of degree $1$. Now we assume that $G$ is a non-abelian simple group. Recall that $|G|$ is even, by Cauchy’s theorem there exists $g\in G$ such that $o(g)=2$.

Assume that $\chi$ is an irreducible character of degree 2 and its corresponding representation is $\rho$. Then

$$\rho (g{)}^2=\mathrm{id}\in \mathrm{SL}(2,\mathbb{C})⟹\rho (g)=\left(\begin{array}{cc}\pm 1& 0\\ 0& \pm 1\end{array}\right).$$

Since $G^{\prime }=G$, $G$ has no non-trivial linear character and so $\det \rho (g)=1$. For any non-trivial irreducible representation $\psi$, $\ker \psi ⊲G$ and so $\psi$ is faithful. Hence $\rho (g)=\mathrm{diag}(-1,-1)$ and so $g\in Z(G)$. However, as $Z(G)=\{1\}$, $\rho (g)\ne I$ is impossible. \end{proof}

Corollary

Suppose that $p$ is a prime and the degree of every irreducible character of $G$ is a power of $p$. Then $G$ has an abelian normal $p$-complement $N$, whose order is coprime to $p$ and $[G:N]$ is a power of $p$.

\begin{proof} See here. Use induction and ^a33a1v. \end{proof}

Properties of $\chi (g)$ for general element of group

A condition which ensures $\chi (g)$ is an integer

In number theory, we know for a $n$-th root of unity, ${\sum }_{1⩽i⩽n,(i,n)=1}{\omega }^i$ is an algebraic integer. By this conclusion, we get the following theorem:

Theorem

Let $g$ be an element of order $n$ in $G$. Suppose that $g$ is conjugate to $g^i$ for all $i$ with $1⩽i⩽n$ and $(i,n)=1$. Then $\chi (g)$ is an integer for all characters $\chi$ of $G$.

Corollary

All the character values of symmetric groups are integers.

The $p^{\prime }$-part of group elements

By Bézout Lemma, we can prove the following lemma:

Lemma

Let $p$ be a prime number and let $g\in G$. Then there exist $x,y\in G$ such that

• $g=xy=yx$
• the order of $x$ is a power of $p$
• the order of $y$ is coprime to $p$

Moreover, both $x$ and $y$ are unique. We call $y$ the $p^{\prime }$-part of $g$.

Consider $\zeta =e^{2\pi i/n}$ and $\mathbb{Z}[\zeta ]=\{{\sum }_{j=1}^n{a}_j{\zeta }^j,a_j\in \mathbb{Z}\}$. Let $P$ be the maximal ideal of $\mathbb{Z}[\zeta ]$ containing $p\mathbb{Z}[\zeta ]$, then $P$ is also a prime ideal of $\mathbb{Z}[\zeta ]$ and $P\cap \mathbb{Z}=p\mathbb{Z}$. Thus, we can get the following conclusions:

Theorem

Let $g\in G$ and let $y$ be the $p^{\prime }$-part of $g$. If $\chi$ is any character of $G$, then

$$\chi (g)-\chi (y)\in P.$$

\begin{proof} Suppose that order of $g$ is $m=u{p}^v$ and $y=g^{b{p}^v}$ for some $b$. Let $\omega =e^{2\pi i/m}$. By freshman’s dream we can prove $\omega -{\omega }^{b{p}^v}\in p\mathbb{Z}[\zeta ]$. Then

$$\chi (g)-\chi (y)=\sum _j(w^j-w^{jb{p}^v})\in p\mathbb{Z}[\zeta ]\subset P.$$

Now we finish the proof. \end{proof}

Remarks.

• If $\chi (g)$ and $\chi (y)$ are both integers, then $\chi (g)\equiv \chi (y)(\mathrm{mod}p)$.
• If $\mathrm{ord}(g)=p^u$ and $\chi (g)\in \mathbb{Z}$, then $\chi (g)\equiv \chi (1)(\mathrm{mod}p)$.