2

Let $K$ be a field with $\text{char}K=n>0$. Define for $a \in K$ $p:=x^n-x-a \in K[x]$

  1. Show that $p$ is separable
  2. Show that if $p$ has a root in $K$, then $p$ splits into linear factors in $K[x]$

My attempt: A polynomial is separable if it has no repeated roots. This is equivalent with the condition that $p' \neq 0$

$p'(x)=\underbrace{nx^{n-1}}_{=0,\text{since } char =n } -1=-1$

Now $p$ can't be the zero polynomial, thus $p$ is separable.

Now I don't really know how to argue in part 2)

wanymose
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  • It's not true that a polynomial being inseparable is equivalent to the condition $p' \neq 0$. This holds only when p is irreducible. In general the equivalent condition for $p$ to be separable is that $gcd(p,p')=1$. In this case since $char(K)=n$ you in fact get that $p'=-1$, which has no roots in an algebraic closure and hence is coprime to $p$. – jl00 May 09 '23 at 23:48
  • @jl00 oh right, thank you. – wanymose May 10 '23 at 05:53
  • See also here and the links in there for more about Artin-Schreier polynomials. An obvious +1 to Kenanski Bowspleefi, but we have also covered this theme rather well on the site :-) – Jyrki Lahtonen May 10 '23 at 06:05

1 Answers1

2

Here's a hint. Recall that in characteristic $n$, we have $(a+b)^n = a^n + b^n$. Supposing that $r$ is a root of $p$ in $K$, can we use this to write down more roots?

Try $p(r+1)$. Can we get all the roots this way?

We have $p(r+1) = (r+1)^n - (r+1) - a = r^n + 1 - r - 1 - a = r^n - r - a = p(r) = 0$, so in fact $r+1$ is also a root of $p$. Replacing $r$ with $r+1$ and proceeding inductively, it follows that $r,r+1,r+2,\dots,r+n-1$ are all roots of $p$. These are distinct (since $K$ has characteristic $n$), so in fact $p$ has $n$ roots in $K$, that is, $p$ splits into linear factors in $K$. (Polynomials of this form are Artin-Schreier polynomials.)

Kenanski Bowspleefi
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