# If Brauer characters are $\bar{\mathbb{Q}}$-linearly independent, why are they $\mathbb{C}$-linearly independent?

If Brauer characters are $\bar{\mathbb{Q}}$-linearly independent, why are they $\mathbb{C}$-linearly independent?

I think this is a linear algebra fact showing up when proving the irreducible Brauer characters on a finite group are linearly independent over $\mathbb{C}$. The proof I've seen observes that the characters take values in the ring of algebraic integers, and then proves linear independence over $\bar{\mathbb{Q}}$.

Why is it sufficient to only check linear independence over $\bar{\mathbb{Q}}$? It seems like something could go wrong when extending the field all the way up to $\mathbb{C}$.

The proof I'm reading is Theorem 15.5 in Isaacs' Character Theory of Finite Groups.

• This does not hold in the generality you presume (e. g. the constant functions $1,\sqrt 2$ are linearly independent over $\mathbb Q$ but not over $\mathbb Q(\sqrt 2)$). One needs that $\mathbb C$ is transcendental over $\overline {\mathbb Q}$ (and that the values of your functions lie in $\overline {\mathbb Q}$). – Claudius Jul 29 '16 at 11:26
• I might be easier if everyone had the same information as you: Where (give a source) did you stumble over this problem? – Claudius Jul 29 '16 at 11:34
• @user218931 It is Theorem 15.5 of Isaacs' Character Theory. – Kally Jul 29 '16 at 17:07

If $E/F$ is a field extension, we have $F^n\subset E^n$, and if a subset of $F^n$ is $F$-linearly independent, then it is also $E$-linearly independent. A nice, super easy way to see it: extend the subset to a basis for $F^n$. Form the matrix whose columns are elements of this basis. Its determinant is nonzero. But this shows that the columns form a basis for $E^n$ since the determinant has the same formula regardless of the field you work over.
The space of class functions of a finite group can be identified with $F^n$ in an obvious way ($n$= number of conjugacy classes).
Fact. Extension of scalars preserves linear independence. That is, if $X$ is a $k$-linearly independent subset of a $k$-vector space $V$, and $K/k$ is any field extension, then it will also be a $K$-linearly independent subset of the $K$-vector space $K\otimes_k V$ (where we identify $x$ with $1\otimes x$).
Proof. Write $K=\bigoplus_{t\in T}kt$ as vector spaces for some $k$-basis $T$ of $K$. Then if $\sum c_i x_i=0$ for some $x_1,x_2,\cdots,x_n\in X$ and scalars $c_1,c_2,\cdots,c_n\in K$, we may write $c_i=\sum_{t\in T} c_{i,t}t$ for some restricted scalars $c_{i,t}\in k$, in which case the relation becomes $\sum_i c_{i,t}x_i=0$ for all $t$, but this implies each $c_{i,t}=0$ and hence every $c_i=0$.
It applies here with $V$ the vector space of functions spanned by characters and $\mathbb{C}/\overline{\mathbb{Q}}$.