# field of characteristic $p$ and polynomial over it

Could any one tell me for which prime $p$ the polynomial $x^4 +x+6$ has a root of multiplicity $>1$ over a field of characteristic $p$?

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Your question in fact is: for which primes does the polynomial $\,x^4+x+6\pmod p\,$ have a zero derivative (why?)

Added a polynomial over a field has a multiple root iff this is also a root of its (formal) derivative.

Further added A polynomial $\,p(x)\,$ over a field has $\,\alpha\,$ as a root iff we can write $\,p(x)=(x-\alpha)^mg(x)\,,\,m\in\mathbb{N}\,,\,g(x)\,$ a polynomial, and $\,m\,$ is called the multiplicity of the root $\,\alpha\,$. When $\,m=1\,$ the root is called a simple one, and when $\,m>1\,$ it is called a multiple root. Derivating the above expression we get $$p'(x)=m(x-\alpha)^{m-1}g(x)+(x-\alpha)^mg'(x)$$and we see $\,\alpha\,$ is a multiple root (i.e., $\,m>1\,$) iff $\,p'(\alpha)=0\,$ , and this means both $\,p(x)\,\text{and}\,p'(x)\,$ have $\,\alpha\,$ a common root.

Now, if $\,p(x)\,$ is an irreducible polynomial with $\,\alpha\,$ a one of tis roots, this means $\,p(x)\,$ divides any polynomial that has $\,\alpha\,$ as one of its roots, and from the above it follows that in this case $\,p(x)\,$ divides its own derivative $\,p'(x)\,$, which is impossible unless $\,p'(x)=0\,$ (the zero polynomial), because $\,\deg p(x) >\deg p'(x)\,$ , and this means we actually have $p(x)=h(x^p)\,$ , for some polynomial $\,h(x)\,$ (we're in the ring $\,\mathbb{F}_q[x]\,\,,\,q=p^k\,,\,p\,$ a prime , as in characteristic zero it is impossible for a non-constant polynomial to have a zero derivative polynomial).

All the above stuff is pretty standard basic ring theory or even more limited: polynomial ring theory.

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I really don't understand,would you please explain. – La Belle Noiseuse Jun 16 '12 at 12:44
@Mex, can you prove what I wrote in Added? It is the gist of this. It's pretty simple, even if you never saw it before: $\,p(x)\,$ has a multiple root $\,\alpha\,$ (multiplicity 2 or more) iff we can write $\,p(x)=(x-\alpha)^mg(x)\,\,,\,m\geq 2\,$ , and now calculate the derivative (all the above over a field, of course, and though we can weaken this restriction it isn't necessary here) – DonAntonio Jun 16 '12 at 12:59
No I havent seen these before.plz explain.. – La Belle Noiseuse Jun 16 '12 at 13:08
It's true that if $q(x)$ is irreducible over $F$, then it has multiple roots if and only if its derivative is zero. But that would require us to check whether the polynomial is irreducible mod $p$ as well. Why not state it in terms of the gcd directly? $q(x)$ has multiple roots module $p$ if and only if $\gcd(q,q')=0$ or $\deg(\gcd(q,q'))\gt 0$, where the gcd is taken in $\mathbb{F}_p[x]$. – Arturo Magidin Jun 16 '12 at 20:09
The OP wrote he really doesn't understand and hasn't seen this stuff, so I didn't feel like getting into this. After all, all he needed to do is to derivate and check his polynomial's derivative is never zero, no matter what prime $\,p\,$ he chooses to work with... – DonAntonio Jun 16 '12 at 23:51

Root of multiplicity exceeding 1 is equivalent to discriminant zero. The discriminant is (essentially) the resultant of the polynomial and its derivative. So: if you don't know about resultants, look the term up; then calculate the resultant of your polynomial and its derivative; you get a root of multiplicity greater than 1 (possibly in some extension field) if and only the resultant is a multiple of $p$.

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