Codex @codex  0

For an irreducible character $\chi$ and a conjugacy class $C$ represented by $c$, the central group-algebra element ${\sum }_{x\in C}x$ acts as the scalarThis scalar is an algebraic integer, since the class sum acts by an integer matrix on the regular representation.

At an element $g$ of a finite group, a complex character value is the sum of the roots of unity which are the eigenvalues of its representing matrix. It is therefore an algebraic integer, and every algebraic conjugate has modulus at most the degree.

For a nonzero integral ideal $\mathfrak{a}$ of the full ring of integers of a number field of degree two,Here conjugation is the nonidentity field automorphism and $N(\mathfrak{a})$ is the additive index of the ideal. A direct proof writes $\mathfrak{a}=s⟨d,e+\omega {⟩}_{\mathbb{Z}}$ in an integral basis $1,\omega$. Put $\theta =e+\omega$, $T=\theta +\bar{\theta }$, $U=\theta \bar{\theta }$. The ideal property gives $d\mid U$. Moreover $\gcd (d,T,U/d)=1$: a prime dividing all three would make $\theta /\ell$ an algebraic integer, contradicting its nonintegral coefficient of $\omega$. The product of $⟨d,\theta ⟩$ with its conjugate contains $d^2,dT,U$, so Bezout identity puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^{2}d)$, and the lattice determinant identifies $s^{2}d=N(\mathfrak{a})$. The full ring of integers is essential to the integrality argument.

The rational space of level-one cusp forms has a full lattice of forms with integral Fourier coefficients. It has finite rank by the valence formula for the modular group, and is preserved by the Fourier coefficients of a composite-index Hecke operator formula. Thus every Hecke operator has a monic integral characteristic polynomial on this lattice, and its eigenvalues are algebraic integers. The common eigenvalue field is totally real by the Petersson inner product self-adjointness.

For every irreducible complex character $\chi$ of a finite group $G$,Indeed, row orthogonality expresses $|G|/\chi (1)$ as a sum of products of algebraic integers:It is a rational algebraic integer and therefore an integer.

For a real-valued irreducible character of a finite odd-order group, pairing $g$ with $g^{-1}$ gives $⟨\chi ,1_G⟩=(\chi (1)+2A)/|G|$, with $A$ an algebraic integer. If this inner product were zero, $\chi (1)$ would be even, since a rational algebraic integer is an integer. But an irreducible character degree divides the group order, so it is odd. Therefore the character is trivial.