Lemma. Let $Y=(y_{ij})$ be a $n\times n$ matrix. Then $\det Y$ is an irreducible polynomial of $y_{ij}.$
Proof. Consider a special case $Y=Y_1=tE_n+\sum\limits_{j=1}^nx_je_{j,j+1}$ where $e_{j,j+1}$ are matrix units and $j$ is defined modulo $n.$ Then $\det Y_1=t^n+(-1)^{n-1}x_1\dots x_n$ is an irreducible polynomial of $t,x_1,\dots,x_n.$
Suppose that $\det Y$ factors into a product of a few irreducible polynomials. Then they are homogenous, too. Hence, after a substitution $Y=Y_1$ they could not become non-zero constants, so $\det Y$ is irreducible.
The question: Why couldn't homogenous divisors become non-zero constants after the substitution?
I'm stuck a bit here. Could you please help me?
Best Answer
The substutition $Y = Y_1$ is taking some variables and setting them equal to $0$, and taking some other variables, and setting them all equal to each other. So the claim is that if $f(x_1, \ldots, x_n)$ is homogeneous non-constant polynomial, then any polynomial obtained by setting variables equal to each other or by substituting 0 for a variable is identically zero or non-constant. This seems clear if you just write $f$ out as a sum of a bunch of terms.