# (No) difference between irreducible $\bar{\mathbb{Q}}$-valued characters and irreducible $\mathbb{C}$-valued characters of a finite group?

I would be most grateful if someone would check my logic below... Many thanks!

Let $G$ be a finite group and let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}.$

Standard results from Representation Theory tells us the following:

(1) the number of irreducible $\mathbb{C}$-valued characters of $G$ is equal to the number of conj classes of $G.$

(2) an irreducible representation over a field $F$ remains irreducible over any extension field of $F$ if and only if it remains irreducible over the algebraic closure of $F.$

But (1) still goes through if we replace $\mathbb{C}$ by $\bar{\mathbb{Q}}$ (right?).

Therefore surely we have that

$$\{\text{irreducible \mathbb{C}-valued characters of G}\} = \{\text{irreducible \bar{\mathbb{Q}}-valued characters of G}\}?$$

• Yes, that's right, because every character of a finite group has algebraic character values. In fact, you can do even better: a degree $n$ character value is a sum of (at most) $|G|$ $n$-th roots of unity. – P... Oct 25 '17 at 16:51
• Oops: that should have said "at most $n$ $|G|$-th roots of unity"! – P... Oct 25 '17 at 17:20
• @P... Did you mean since $g^{|G|} = 1$ the minimal polynomial $P(x)$ of $\rho(g)$ is a (monic) polynomial of $\mathbb{Z}[x]$ and it divides $x^{|G|}-1$. Factorizing it we see that $\mathbb{Q}[\rho(g)] \cong \mathbb{Q}[x] /(P(x))$ is a product of some cyclotomic fields $\mathbb{Q}(\zeta_{k_j}),k_j \,\mid \, |G|$, thus $tr(\rho(g))$ is given by $Tr_{\mathbb{Q}(\zeta_{k_j})/\mathbb{Q}}(\zeta_{k_j})$ and is an algebraic integer $\in \mathbb{Q}(\zeta_{|G|})$ ? – reuns Oct 25 '17 at 23:16