TITLE: Representation theory: An origin story.
QUESTION [6 upvotes]: I read that:

Dedekind made the observation: taking the multiplication table of a finite group $G$ and turning it into a matrix $X_G$ by replacing each entry $g$ of the table by a variable $x_g$. The determinant of $X_G$ factors into a product of irreducible polynomials, each of which occurs with multiplicity equal to its degree.

For a second I couldn't think of what 'multiplicity equal to its degree meant' and then I realised that they must mean I have $f_1^{a_1}f_2^{a_2}f_3^{a_3}$ where $f_i$ are irreducible, and each have degree $a_i$.
So I went to check this in an easy case, to make sure I understood, and I see that for $\Bbb Z_3$:
$$\begin{matrix}&0&1&2\\0&0&1&2\\1&1&2&0\\2&2&0&1\end{matrix} \mapsto \begin{bmatrix}x_0&x_1&x_2\\x_1&x_2&x_0\\x_2&x_0&x_1\end{bmatrix}$$
Where the determinant of this is $$-x_0^3-x_1^3-x_2^3+3x_0x_1x_2=-(x_0+x_1+x_2)(x_0^2+x_1^2+x_2^2-x_0x_1-x_0x_2-x_1x_2)$$
which I believe are irreducible. But the latter term has degree $2$ and multiplicity $1$.
What has gone wrong?

REPLY [2 votes]: The quadratic factor is in fact reducible, but you have to work over the complex numbers:
$$
\begin{aligned}
&x_0^2 + x_1^2 + x_2^2 - x_0 x_1 - x_0 x_2 - x_1 x_2
\\&=
(x_0 + \tfrac12 x_1 + \tfrac12 x_2)^2 + \tfrac34 (x_1 - x_2)^2
\\&=
A^2 + B^2
\\&=
(A+iB)(A-iB)
.
\end{aligned}
$$