# Understanding 'root' in its context

For which of the following primes p, does the polynomial $x^4+x+6$ have a root of multiplicity$> 1$ over a field of characteristic $p$? $p=2/3/5/7$.

My book solves it using the concepts of modern algebra, which I am not very comfortable with.

I wonder if there is an intuition based method to solve this question.

Like, $x^2-2x+1$ would have $1$ as root with multiplicity$=2$. But in the given equation, everything is positive, so what is meant by root here? Is it not the value of $x$ when the graph crosses the $x$-axis?

• Note the phrase "over a field of characteristic $p$..." en.wikipedia.org/wiki/Characteristic_(algebra) – Ryan Sep 5 '13 at 8:29
• This is where "Characteristic $p$" comes in. For instance, in characteristic $5$, the following polynomials all evaluate to the same: $$x^4 + x + 6 \\\\ x^4 + x + 1 \\\\ x^4 -4x + 1 \\\\ x^4 + x - 4$$ because $5$ evaluates to $0$. This also means they all have the same roots with the same multiplicities. Hope that this helps you partway towards an understanding, at least. – Arthur Sep 5 '13 at 8:29
• Note that the test for a multiple root - that the polynomial shared a root with its formal derivative - still holds over characteristic $p$ and the division algorithm for polynomials can be used to find the highest common factor. – Mark Bennet Sep 5 '13 at 11:49

The question is slightly ambiguous, because a polynomial may only have its roots in an extension of the field where it's defined, for instance $x^2+1$ has no roots in $\mathbb{R}$, but it has roots in $\mathbb{C}$.

However, having a multiple root is equivalent to be divisible by $(x-a)^2$, where $a$ is in the field where the polynomial has its coefficients (or maybe, depending on conventions) in an extension field.

Polynomial division is carried out the same in every field: $$x^4+x+6 = (x-a)^2 (x^2+2ax+3a^2) + ((4a^3+1)x+(6-3a^4))$$ where $(4a^3+1)x+(6-3a^4)$ is the remainder. For divisibility we need the remainder is zero, so $$\begin{cases} 4a^3 + 1 = 0 \\ 6 - 3a^4 = 0 \end{cases}$$ We can immediately exclude the case the characteristic is $2$, because in this case the remainder is $x+c$ ($c$ some constant term). If the characteristic is $3$, then the constant term in the remainder is zero and the first equation becomes $$a^3+1=0$$ So, when $a=-1$, there is divisibility.

Note also that $0$ can never be a multiple root of the polynomial, so we can say $a\ne0$.

Assume the characteristic is neither $2$ nor $3$. We can multiply the first equation by $3a$ and the second equation by $4$; summing them up we get $$3a+24=0$$ which can be simplified in $a=-8$. Plugging it in the first equation, we get $$4(-8)^3+1=-2047=-23\cdot89$$ which is zero if and only if the characteristic is either $23$ or $89$.

Thus the only prime in your list that gives multiple roots is $p=3$: indeed $$x^4+x+6=x(x^3+1)=x(x+1)^3$$ when the characteristic is $3$.

• A filed of characteristics $P$ means a field of order $F_{P^K}$ ..Why you did not consider the other elements of $F_{p^k}$?@egreg – cmi Nov 21 '18 at 5:52
• @cmi It's not necessary; in the characteristic $2$ case, the polynomial is $x(x+1)(x^2+x+1)$, which hasn't multiple roots (in any extension). In the characteristic $3$ case it does have multiple roots. In characteristic not $2$ or $3$, the only possible multiple root is $-8$. – egreg Nov 21 '18 at 7:54
• How can you get the root $-8$? can you please explain @egreg – cmi Nov 21 '18 at 14:43
• If $a = -8$ is a root of that two equation , then $-8$ should vanish the remainder in any field @egreg – cmi Nov 21 '18 at 14:49