I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5, 7 $
Please help.
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ and $f'(x) = x^3+1$. Hence, $f^′(x)= 0$ for $x=2$. Therefore in an algebraically closed field of characteristic $3$, $f(x)$ has multiple roots.
Best Answer
The hint is to use the following result : Let $f(x) \in K[x]$ where $K$ is some field. $\alpha$ is a multiple root of $f(x)$ if and only if $\alpha$ is a root of $f'(x)$, the formal derivative of $f(x)$.
In the algebraic closure, $\bar{F}$ of a field of characteristic $2$, $f(x) = x^4 + x = x(x^3 + 1)$ and $f'(x) = 1$. $f(x)$ has four roots in $\bar{F}$, counting multiplicity. However, $f'(x) = 1$ which has no roots. Therefore in an algebraically closed field of characteric 2, $f(x)$ does not have multiple roots.
Do something similar for the others.