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?

  • $\begingroup$ Note the phrase "over a field of characteristic $p$..." en.wikipedia.org/wiki/Characteristic_(algebra) $\endgroup$ – Ryan Sep 5 '13 at 8:29
  • 2
    $\begingroup$ 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. $\endgroup$ – Arthur Sep 5 '13 at 8:29
  • $\begingroup$ 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. $\endgroup$ – 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 $3$.

  • $\begingroup$ 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 $\endgroup$ – cmi Nov 21 '18 at 5:52
  • 1
    $\begingroup$ @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 $-8$. $\endgroup$ – egreg Nov 21 '18 at 7:54
  • $\begingroup$ How can you get the root $-8$? can you please explain @egreg $\endgroup$ – cmi Nov 21 '18 at 14:43
  • $\begingroup$ If $a = -8 $ is a root of that two equation , then $-8$ should vanish the remainder in any field @egreg $\endgroup$ – cmi Nov 21 '18 at 14:49
  • 1
    $\begingroup$ @cmi Sorry, but it seems you’re not reading my comments. Please, take your time and rework your steps. $\endgroup$ – egreg Nov 21 '18 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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