Finding roots of a polynomial within a splitting field.

Question

Let $\mathscr{K}$ be a field and $a, b\in \mathscr{K}[x]$ with $a = x^{3}-2$ and $b = x^{2} + x + 1$. Let $\mathscr{F}$ a splitting field for $ab$ over $\mathscr{K}$. For $c\in \mathscr{K}[x]$, let $\Delta_{c}$ be the set of roots of $c$ in $\mathscr{F}$.

Show that for $char\mathscr{K}\neq 2, 3$, $|\Delta_{a}| = 3$ and $|\Delta_{b}| = 2$.

So, for $a=x^{3}-2$, it's clear that this polynomial indeed has 3 roots and hence $|\Delta_{a}| = 3$. Likewise for $b=x^{2}+x+1$ and $|\Delta_{b}| = 2$.

However, I feel as though this may be incorrect and I'm not sure how to utilize the characteristic of $\mathscr{K}$ or the fact that $\Delta_{c}$ is defined to be the set of roots over the splitting field $\mathscr{F}$?

Answer 1

This problem is not as trivial as what you might have thought. A degree $n$ irreducible polynomial may not have $n$ distinct roots. You need to explain why $x^3-2$ has $3$ distinct roots, and why $x^2+x+1$ has $2$. This is where the characteristic comes into play.

Your claim is equivalent to proving that $a$ and $b$ are separable when $\text{char}(K)\not\in\{2,3\}$. We can make use of the formal derivative:

Let $f\in F[x]$. Then $f$ is separable if and only if $f$ and $f'$ are coprime.

We simply compute that $a'=3x^2$ and $b'=2x+1$. Now, if $\text{char}(K)=3$, then $3x^2=0$, so $\text{gcd}(x^3-2, 3x^2)=x^3-2\ne 1$. If $\text{char}(K)= 2$, then $2x+1=1$, so $\gcd(x^2+x+1,2x+1)=x^2+x+1\ne 1$. Otherwise, $\text{gcd}(a,a')=1$ and $\text{gcd}(b,b')=1$, so the claim is proved.

Answer 2

Your answer is correct. If the characteristic is 2 or 3, $a$ becomes $x^3$ or $(x+1)^3$ respectively, so the assertion becomes untrue.