Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
You should also mention that $f(2) = 0$ in characteristic $3$. –  Robert Israel Jun 5 '12 at 19:58
2  
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

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
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
2  
$\alpha$ 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 $89$.

share|improve this answer

Here's an elementary answer:

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.