For $f = x^7+x+1$, find all primes $p$ for which $f_p$ ($f$ considered modulo $p$) is not separable

Question

Question from David Cox Galois Theory.

For $f = x^7+x+1$, find all primes $p$ for which $f_p$ ($f$ considered modulo $p$) is not separable. How to do this problem?

I have computed $gcd(f, f')=1$. But I don't know how to proceed.

Answer 1

I am not sure what you computed, but you indeed can solve this by computing $\gcd(f, f')$.

First, $f_7$ is separable. This is true since $f_7' = 1$. Now we suppose $p \neq 7$, and $f_p' = 7x^6 + 1$.

Then $\gcd(f,f') = \gcd(x^7+x+1, x^6+7^{-1}) = \gcd(x^6+7^{-1}, x^7+x+1 - (x^7 + 7^{-1}x))$, which is $\gcd(x^6+7^{-1}, (1-7^{-1})x + 1) = \gcd(x^6+7^{-1}, 7^{-1}\cdot6x+1)$

If $7^{-1}\cdot6x + 1 = 0$ has no solution, which happens when $6$ is not invertible, i.e., $p = 2, 3$, then $f$ is separable (why?).

Now we look at the case when $p \neq 2,3,7$. In this case $7^{-1}\cdot6x + 1 = 0$ has solution $x = -7\cdot6^{-1}$. Therefore the gcd is not $1$ if this is also a solution of $x^6 + 7^{-1} = 0$. Solve this, and you get the answer of this question.

The answer will be three primes, one of them has two decimal digits, and the other two have three digits.