A simple fact on representation of finite groups, without Galois theory Let $G$ be a finite group and $\chi$ a complex character of $G$. We know that $|\chi(g)/\chi(1)| \leq 1$. I would like to show that if $0 < |\chi(g)/\chi(1)| < 1$, then $\chi(g) / \chi(1)$ is not an algebraic integer.
I would like to suppose it is an integer and of absolute value less than $1$, and show it is zero. Introduce the minimal polynomial $P$ of $\chi(g)/\chi(1)$ in $\mathbb{Z}[X]$. The constant term is an integer, and product of the roots.
Is there an easy way to see that the other roots $z$ of $P$ are $|z| \leq 1$ (typically I would like to say they are also sums of $\chi(1)$ roots of unity, divided by $\chi(1)$, but without using the Galois group). The division by $\chi(1)$ seems particularly cumbersome (otherwise than using the fact that we can take them out from a Galois automorphism)
 A: $\newcommand\GL{\mathrm{GL}}\newcommand\Tr{\mathrm{Tr}}\newcommand\Z{\mathbb Z}\newcommand\Q{\mathbb Q}$
Just to publicly set terminology, I assume by "character" you mean the character of some representation $\rho:G\to\GL(V)$, so $\chi(g)=\Tr\rho(g)$. In particular, $\chi(1)=\dim V=:n$.
Let $\alpha:=\chi(g)/\chi(1)$, so $\alpha=\lambda_1+\dots+\lambda_n$ with $\lambda_i$ some $N$th root of unity, where $N:=\#G$. Suppose that $\alpha$ is an algebraic integer, and let $P(x)\in\Z[x]$ be $\alpha$'s minimal polynomial. We want to show $|\beta|\le1$ for all roots $\beta$ of $P$. We'll basically just expand out the Galois theoretic proof enough that it will hopefully appear understandable to your students.
Let $L=\Q[x]/(P(x))$, a $d$-dimensional $\Q$-vector space, where $d:=\deg P$. Let $m_\alpha:L\to L$ be the $\Q$-linear map $m_\alpha(x)=\alpha x$ given by multiplication by $\alpha$. Similarly let $m_i(x)=\lambda_i x$ be multiplication by $\lambda_i$. We make two observations.
(1) The characteristic polynomial
$$Q(x):=\det(xI - m_\alpha)\in\Z[x]$$
is a degree $d$ monic polynomial vanishing at $\alpha$, so $Q(x)=P(x)$. Hence, the roots of $P(x)$ are the eigenvalues of $m_\alpha$.
(2) $$m_\alpha=\frac{m_1+\dots+m_n}n:L\to L.$$
Furthermore, the $m_i$'s are visibly simultaneously diagonalizable over $L$, so the eigenvalues of $m_\alpha$ are all of the form $\frac1n(\mu_1+\dots+\mu_n)$ with $\mu_i$ an eigenvalue of $m_i$. Finally, $\lambda_i^N=1\implies m_i^N=1\implies\mu_i^N=1$, so each $\mu_i$ is an $N$th root of unity which means the eigenvalues of $m_\alpha$ all have absolute value at most $1$.
(1) + (2) should do the trick.
A: The following is probably more convoluted than you would like (and too hard to motivate to your students) but I think it is a cute way to explicitly make the needed symmetry arguments without using any Galois theory that is not done by hand.  Suppose $\chi(g)$ is a sum of $n$th roots of unity; say $$\chi(g)=\sum_{i=1}^{\chi(1)}\zeta^{d_i}$$ where $\zeta$ is a primitive $n$th root of unity.  The idea is then to take the other sums of $n$th roots of unity divided by $\chi(1)$ that ought to be the Galois conjugates of $\chi(g)/\chi(1)$, and calculate that the polynomial with these conjugates as roots has integer coefficients.  That is, for each $m\in(\mathbb{Z}/(n))^\times$ let $$a_m=\sum_{i=1}^{\chi(1)}\zeta^{md_i}.$$  The claim is then that the polynomial $$Q(x)=\prod_m(\chi(1)x-a_m)$$ has integer coefficients.  Given this claim, $Q(x)$ is divisible by $P(x)$ since it has $a_1/\chi(1)=\chi(g)/\chi(1)$ as a root.  But the roots of $Q(x)$ are the numbers $a_m/\chi(1)$, which all have absolute value at most $1$ since $a_m$ is a sum of $\chi(1)$ roots of unity.
It remains to check that $Q(x)$ has integer coefficients.  Let us work in the ring $A=\mathbb{Z}[t]/(t^n-1)$, which we can also identify as the group ring $\mathbb{Z}[C_n]$ of a cyclic group of order $n$.  Define $$b_m=\sum_{i=1}^{\chi(1)}t^{md_i},$$ the analogue of $a_m$ in this ring, and $$R(x)=\prod_m(\chi(1)x-b_m)\in A[x].$$  For each $m\in(\mathbb{Z}/(n))^\times$, the $m$th power map is an automorphism of $C_n$ and thus induces an automorphism $\varphi_m:A\to A$ which satisfies $\varphi_m(t)=t^m$.  These automorphisms permute the factors of $R(x)$ so its coefficients are fixed by each $\varphi_m$.  But now an element of $R$ that is fixed by each $\varphi_m$ must be an integer linear combination of elements of the form $c_r=\sum_{\gcd(i,n)=r}t^i$ corresponding to the orbits of the action of $(\mathbb{Z}/(n))^\times$ on $C_n$.
Now let $F:A\to\mathbb{C}$ be the homomorphism sending $t$ to $\zeta$.  This homomorphism turns the coefficients of $R$ into the coefficients of $Q$, so we now know each coefficient of $Q$ is an integer linear combination of numbers of the form $F(c_r)$, which is just the sum of all the primitive $(n/r)$th roots of unity.  You can now show by induction that for any $d$, the sum of all primitive $d$th roots of unity is an integer.  This is because the sum of all $d$th roots of unity is $0$, and by the induction hypothesis the non-primitive roots in this sum also add up to an integer.
(The point of the ring $A$ here is to avoid needing any theory of cyclotomic fields; instead of needing to know about the automorphisms of $\mathbb{Q}(\zeta)$, you can just use automorphisms of $A$ which are easy and permute your nice monomial basis of $A$.  As for how you might motivate this, the idea is that you want to expand out the coefficients of $Q(x)$ as sums of powers of $\zeta$, pretending you know nothing about $\zeta$ other than that $\zeta^d=1$.  The point is then that the powers of $\zeta$ that appear in each coefficient will have to be symmetric with respect to the maps $\zeta\mapsto\zeta^m$, and thus be built out of sums of all primitive $d$th roots of unity for various values of $d$.  The ring $A$ is just how you make this "pretending" formal: instead of expanding out $Q$ in terms of powers of $\zeta$, you expand out $R$ in terms of powers of $t$.)
