Prove that $p(x)=x^5-x^2+1$ is separable over all fields.

When the field is finite or of characteristic zero it is automatically true, since any polynomial is separable. The definition of separability requires to look at the irreducible components, which seems pretty hard.

The only other tool I know of is showing that the derivative is relatively prime to the polynomial, i.e. that $\gcd(x^5-x^2+1,5x^4-2x)=1$ over all fields, and it suffices to show $\gcd(x^5-x^2+1,5x^3-2)=1$, since $0$ is never a root of $p(x)$.

  • $\begingroup$ A standard way to check modulo which primes $f$ and $f'$ have a common factor is to compute their resultant, if you know how. Or equivalently, to compute the discriminant of $f$. In this case, $$ \operatorname{Disc}(f) = \operatorname{Res}(f, f') = 3017 = 7 \cdot 431 $$ $\endgroup$ – Hurkyl Apr 5 '14 at 22:39

That idea leads you to the solution.

If the characteristic is $5$ you get $\gcd$ $=1$.

Now assume that characteristic is not $5$. Then you can divide by $5$ (cleaner to multiply by $5$) and compute the gcd.

We get, $\gcd(x^5-x^2+1,5x^3-2)=\gcd(5x^5-5x^2+5,5x^3-2)=\gcd(3x^2-5,5x^3-2)$

If characteristic is $3$ we get the $\gcd=1$. Assume characteristic is not $3$ either.

The $\gcd(3x^2-5,5x^3-2)=\gcd(3x^2-5,15x^3-6)=\gcd(25x-6,3x^2-5)=\gcd(25x-6,75x^2-125)=\gcd(18x-125,25x-6)=\gcd(25\cdot18x-25\cdot125,25x-6)=\gcd(3017,25x-6)$.

We have that $3017=7\cdot 431$. For these primes we get $\gcd\neq1$.

Actually, we get $\gcd=25x-6$. This tells us what the possible multiple root would be.

Let us do the characteristic $7$ case. What this $\gcd$ is telling us is that the solution of $25x-6=0$, i.e. $x=5$, is a root of our polynomial and its derivative.

Divide $x^5-x^2+1$ by $x-5$. We get $x^5-x^2+1=(x-5)^2(x^3+3x^2+5x+2)$.

Using the old definition of separable:

In this case we ($char =7$) we have $x-5$ is an irreducible factor with only one root. We will need to do with the factor $x^3+3x^2+5x+2$, the same study we did for the original polynomial.

Computing the $\gcd$ with its derivative (we proceed as before, multiplying by leading coefficients to ease the division, and taking remainder mod $7$)

$\gcd(x^3+3x^2+5x+2,3x^2+6x+5)=\gcd(3x^3+2x^2x+6,3x^2+6x+5)=\gcd(3x+1,3x^2+6x+5)=\gcd(x+4,0)=x+4$. So $x+4$ is a multiply factor of $x^3+3x^2+5x+2$. To not have to compute any further, notice that because this polynomial is of degree $3$ it should split completely. So, I guess it would be called separable according to the old definition (if I am reading it correctly).

The case of $char=431$ is all yours. But the work is just the same.

  • $\begingroup$ What if char=$7$? then $\gcd(p,p')=0$, and then one of $p,p'$ is $0$, which seems wrong. What am I missing? $\endgroup$ – Emolga Feb 9 '14 at 16:51
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    $\begingroup$ Notice the $\gcd(0,A(x))=A(x)$, not zero. $\endgroup$ – user127249 Feb 9 '14 at 16:55
  • $\begingroup$ and still, char=$7$ makes the proposition ultimately false? Or $p(x)$ isn't irreducible in this case? $\endgroup$ – Emolga Feb 9 '14 at 19:31
  • $\begingroup$ @Leullame Yes. Remember the problem is not being irreducible. Itis about the multiplicity of the roots in an splitting field. In characteristic $7$ we have that $x=5$ is a root with multiplicity $2$. $\endgroup$ – user127249 Feb 9 '14 at 19:58
  • $\begingroup$ Got it. For the record, the proposition is using a different definition of separability, in which we look at each irreducible factor separately: wiki says it is older. With this it is true even in characteristic 7.(en.wikipedia.org/wiki/Separable_polynomial#Older_definition) $\endgroup$ – Emolga Feb 9 '14 at 20:18

But (for example using Maple) mod 7 the gcd is not 1.

  • $\begingroup$ Do you argue that the proposition is false? $\endgroup$ – Emolga Feb 9 '14 at 15:59
  • $\begingroup$ Yes, the gcd is x+2 and x=-2 is a root mod 7. $\endgroup$ – i. m. soloveichik Feb 9 '14 at 19:50

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