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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?

aarbee
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  • Note the phrase "over a field of characteristic $p$..." http://en.wikipedia.org/wiki/Characteristic_(algebra) – Ryan Sep 05 '13 at 08:29
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    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 05 '13 at 08:29
  • 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 05 '13 at 11:49

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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 $3$.

egreg
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  • 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 05:52
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    @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 07:54
  • 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
  • @cmi I said that the only possible multiple root is $-8$; it may not be a root at all. – egreg Nov 21 '18 at 15:34
  • But you got it by solving that two coefficient equations . So the coefficientsshould always be zero. Why it is zero only when -23 and 47@egreg – cmi Nov 21 '18 at 15:37
  • @cmi Because $-8$ is a root in $F_p$ if and only if $23\cdot89=0$ in $F_p$, which means either $23=0$ or $89=0$, that is $p=23$ or $p=89$. – egreg Nov 21 '18 at 15:42
  • But you get the number $-8$ by solving $4a^3 +1$ and $6 - 3a^4$...then they should vanish when $a=-8$ irrespective of any field.@egreg – cmi Nov 21 '18 at 15:45
  • @cmi I can't understand what you're after: $a=-8$ is a *necessary* condition for a multiple root. In neither the $p=23$ or $p=89$ case it is actually a root. Problem solved. – egreg Nov 21 '18 at 15:51
  • I am really trying hard to understand the problem.. It has been 12 hours...I am a first reader of the field theory...How do you get this necessary condition can you please explain?@egreg – cmi Nov 21 '18 at 15:54
  • @cmi Please, read carefully the answer: multiply $4a^3+1=0$ by $3a$ and $6-3a^4=0$ by $4$; sum what you get and you get $3a+24=0$. – egreg Nov 21 '18 at 15:58
  • I got that...So I am telling elaborately where I am getting trouble.. If you are given two equation $A=0$ and $B=0$ , you solve them you get the root is $x$..so $x$ should vanish $A$ and $B$...That's how $-8$ should vanish $4a^3+1$ and $6-3a^4$ in any feild. Why they would vanish only those two kind of fields?@egreg – cmi Nov 21 '18 at 16:02
  • Hiii @egreg can you please respond? – cmi Nov 21 '18 at 16:13
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    @cmi Sorry, but it seems you’re not reading my comments. Please, take your time and rework your steps. – egreg Nov 21 '18 at 16:58