I am attempting to prove the following result.
Let $K$ be a field, $n \ge 3$, and $e_1,\ldots, e_n$ be positive integers which are not multiples of the characteristic. Then the polynomial $$ p(x) = \sum_{i=1}^n x_i^{e_i} $$ is irreducible in $K[x_1,\ldots, x_n]$.
I am attempting to expand the outline given in this answer into a proof. My attempt to do so is below.
Proof by induction.
For the base case, $n = 3$, we claim that irreducibility of $x^l + y^m + z^n$ in $k[x,y,z]$ is equivalent to irreducibility in $k(z)[x,y]$. Since the coefficients of our polynomial are all $1$, this follows from Gauss' Lemma; in particular, both are equivalent to irreducibility in $k(y,z)[x]$.
So we view our polynomial as $x^l + (y^m + z^n)$ in $k(z)[y][x]$ and want to satisfy Eisenstein's criterion. In particular, we can check by the derivative test that $a_0 \equiv (y^m + z^n)$ in $k(z)[y]$ is separable, since it is of the form $y^m + c$ for $c\neq 0$ and the characteristic does not divide $m$. Since $a_0$ is separable, its prime factors in $k(z)[y]$ are all distinct. Let $\mathfrak{p}$ be the prime ideal generated by one of these prime factors. The conditions of Eisenstein are satisfied, so our polynomial is irreducible in $k(z)[y][x]$ and in $k[x,y,z]$.
Now for the inductive step, also by Eisenstein. We write $$ p(x) = x_n^{e_n} + (x_1^{e_1} + \cdots + x_{n-1}^{e_{n-1}}) \in k[x_1, \ldots, x_{n-1}][x_n]. $$ Since by the inductive hypothesis the constant term is irreducible, it generates a prime ideal $(x_1^{e_1} + \cdots + x_{n-1}^{e_{n-1}})$ in the polynomial ring. We need to check that $$ x_1^{e_1} + \cdots + x_{n-1}^{e_{n-1}} \not\in \langle x_1^{e_1} + \cdots + x_{n-1}^{e_{n-1}}\rangle^2 = \langle(x_1^{e_1} + \cdots + x_{n-1}^{e_{n-1}})^2\rangle. $$ (Hagen von Eitzen) If it were, then setting $x_2,\ldots, x_{n-1}$ all to zero gives $$ x_1^{e_1} = x_1^{2e_1} f(x_1) $$ for some $f \in k[x_1]$, which is absurd. Hence the inductive step holds by Eisenstein.
- (Edit: Answered by Hagen von Eitzen) This last fact seems like it should be true, at least in characteristic not 2, but I don't see how to prove it. Or do I need another technique to do the inductive step?
- This proof does not use the full strength of the assumptions, as far as I can see. I have so far only needed one of the $e_i$ to not be multiples of the characteristic. Am I making a mistake in the base case, or is the statement true that generally?
Ad 1: If $x_1^{e_1}+\cdots x_n^{e_n}=(x_1^{e_1}+\cdots +x_n^{e_n})^2f(x_1,\ldots, x_n)$, then $x_1^{e_1}=x_1^{2e_1}f(x_1,0,\ldots,0)$ in $k[x_1]$, which is absurd for $e_1>0$ in all characteristics.