Problem 5.2.7
As hinted, we consider the representations of \(G\) over the field \(\overline{\mathbb{Q}}\). Since this field is algebraically closed, its representations are characterized by Proposition 3.5.8 (together with Maschke's theorem). In particular, \(|G| = \sum_i (\dim V_i)^2\) where \(V_i\) ranges over the irreducible representations. Every irreducible representation over \(\overline{\mathbb{Q}}\) corresponds to a representation over \(\mathbb{C}\) in an obvious way. The proof of Theorem 4.5.1 shows that it applies to any algebraically closed field whose characteristic doesn't divide the order of the group, so in particular \(\langle V, V \rangle = 1\) for the irreps of \(G\) over \(\overline{\mathbb{Q}}\); this obviously holds in \(\mathbb{C}\) as well, so the \(V_i\) are irreducible there also. Since the sum-of-squares relation also holds in \(\mathbb{C}\), these representations are all the irreducible representations of \(G\) over \(\mathbb{C}\); that is, every irrep of \(G\) over \(\mathbb{C}\) can be presented so that its matrices have only algebraic entries. Choose arbitrary bases \(B_i = \{b_{ij}\}_{j=1}^{\dim V_i}\) for the \(V_i\)'s considered as representations over \(\overline{\mathbb{Q}}\), and let \(K\) be the smallest Galois extension of \(\mathbb{Q}\) that contains the entries of all matrices of all the group elements in the representations \(V_i\) with respect to the bases \(B_i\). Again by Maschke's theorem and Proposition 3.5.8, all finite-dimensional complex representations of \(G\) decompose as a direct sum of the \(V_i\)'s. By choosing a basis for such a representation in the obvious way from the decomposition into irreducibles and the \(B_i\)'s previously chosen, we see that all matrix entries belong to the field \(K\), as desired.
Again, we follow the hint. Assume \(\chi_V(g) \ne 0\) for all \(g \in G\). Then \(|\chi_V(g)|^2 > 0\) for all \(g \in G\). By orthonormality of irreducible characters, \(\sum_{g \in G} |\chi_V(g)|^2 = |G|\). Now \(|\chi_V(1)|^2 = (\dim V)^2 > 1\), so \(\sum_{g \in G \setminus \{1\}} |\chi_V(g)|^2 < |G| - 1\). Thus, the arithmetic mean of the numbers \(|\chi_V(g)|^2\) for \(g \ne 1\) is strictly less than 1. By the inequality of means, \(\left(\prod_{g \in G \setminus \{1\}} |\chi_V(g)|^2\right)^{1/(|G| - 1)} < 1\). Since all \(|\chi_V(g)|^2 > 0\), we must in fact have \(0 < \prod_{g \in G \setminus \{1\}} |\chi_V(g)|^2 < 1\).
Now observe that: \begin{align*} \prod_{g \in G \setminus \{1\}} |\chi_V(g)|^2 &= \prod_{g \in G \setminus \{1\}} \chi_V(g) \overline{\chi_V(g)} \\ &= \prod_{g \in G \setminus \{1\}} \chi_V(g) \chi_V(g^{-1}) \\ &= \prod_{g \in G \setminus \{1\}} \chi_V(g) \prod_{g \in G \setminus \{1\}} \chi_V(g^{-1}) \\ &= \left[\prod_{g \in G \setminus \{1\}} \chi_V(g)\right]^2 \end{align*} Write \(\beta = \prod_{g \in G \setminus \{1\}} \chi_V(g)\); we see that \(\beta\) is real and \(|\beta| \in (0, 1)\).
From part (a), we may choose a basis of \(V\) so that all entries of the matrices for elements of \(G\) are algebraic. This implies that the numbers \(|\chi_V(g)|\) are all algebraic and belong to a common field \(K \subseteq \mathbb{C}\) that is a Galois extension of \(\mathbb{Q}\), as in part (a). Let \(\sigma \in \operatorname{Gal}(K/\mathbb{Q})\); then applying \(\sigma\) to all the matrix entries produces another irreducible representation \(\sigma(V)\) of \(G\), in which \( \prod_{g\in G\setminus\{1\}} \chi_{\sigma(V)}(g)\) is a real number whose absolute value lies in \((0, 1)\); but \begin{align*} \prod_{g \in G\setminus\{1\}} \chi_{\sigma(V)}(g) &= \prod_{g \in G\setminus\{1\}} \tr \sigma(\rho(g)) \\ &= \prod_{g \in G\setminus\{1\}} \sigma(\tr \rho(g)) \\ &= \prod_{g \in G\setminus\{1\}} \sigma(\chi_V(g)) \\ &= \sigma\left(\prod_{g \in G\setminus\{1\}} \chi_V(g) \right) \\ &= \sigma(\beta) \end{align*} thus all conjugates of \(\beta\) have absolute value strictly between 0 and 1. Now the results of section 4.4 imply that \(\chi(g)\) is the sum of roots of unity for all \(g \in G\); each root of unity is an algebraic integer, so each \(\chi(g)\) is also an algebraic integer. The number \(\beta\), being a product of such values, is also an algebraic integer. This implies that its minimal polynomial has integer constant term. But this constant term is the product of the algebraic conjugates of \(\beta\), which is strictly between 0 and 1 in absolute value; thus, a contradiction has been reached.
Remark 5.2.8 The proof along the lines of the Remark is similar to 5.2.7(b), so it is left as an exercise for the reader.