# Multiple roots of a polynomial over a field of characteristic $p$

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 $p$.

$p=2, 3, 5, 7$

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.

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You should also mention that $f(2) = 0$ in characteristic $3$. –  Robert Israel Jun 5 '12 at 19:58
Just compute the gcd of $f$ and $f'$ using Euclid's algorithm. If the outcome is not a constant, then they have a common root, and hence $f$ has a multiple root. The $p=3$ manipulations are, of course, correct. –  Jyrki Lahtonen Jun 5 '12 at 20:04

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.

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Thanks for the help. –  preeti Jun 5 '12 at 9:57
Please see if the solution is correct. –  preeti Jun 5 '12 at 19:39
$\alpha$ must be assumed to be a root of $f$. –  Robert Israel Jun 5 '12 at 19:49

The discriminant of $f(x)$ is $55269 = (3^3)(23)(89)$. So the characteristics in which $f(x)$ has a multiple root are $3$, $23$ and $89$. In fact, $f(x) = x(x+1)^3$ in characteristic $3$, $(x^2+7x+8)(x+8)^2$ in characteristic $23$ and $(x+8)^2(x+28)(x+45)$ in characteristic $89$.

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a multiple root of $x^4 + x + 6$ in $\mathbf F_p$ is the exact same thing as a root common to $x^4 + x + 6$ and its derivative $4x^3 + 1$. But, on $\mathbf Z$, we have the following formula: $$4(x^4 + x + 6) = (4x^3 + 1) x + (3x + 24).$$

So, for $p \neq 2$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $3\alpha + 24 = 0$.

So, for $p \neq 2, 3$, a root $\alpha$ of $x^4 + x + 6$ is multiple (in $\mathbf F_p$) iff $\alpha + 8 = 0$.

So the $p\neq 2, 3$ answering the questions are exactly the prime factors ($\neq 2, 3$) of $f(-8) = 4094 = 2\cdot 23\cdot 89$ (and, for those $p$, the multiple root is $-8$).

You have to check the two remaining cases by hand. The factorisations into prime factors of $f$ on $\mathbf F_2$ and $\mathbf F_3$ are $x(x+1)(x^2+x+1)$ and $x(x+1)^3$, respectively.

So the prime numbers answering the question are 3, 23 and 89 (and the multiple roots are $-1$ (triple), $-8$ and $-8$, respectively).

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