2
$\begingroup$

Does there exist two non isomorphic minimal field extension ( root field) of $f= \frac {x^{64}-x}{x(x-1)} \in F_2[x]$ .

I may be using wrong word here saying minimal field extension but in german its said as "minimal würzelkörper" .

Any help would be appreciated .

Attempt : I tried factorizing using maple into irreducible factors . And using the fact that if we have a field of degree $p^n$ then the subfield must have degree $p^m$ where $m$ divides $n$ , and if this subfield contains the root of the irreducible polynomial then its the minimal root field ( field extension ) . Where am i going wrong ?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

I think you may be referring to "splitting field" = (the/a) minimal extension field containing all the roots of a given polynomial over some certain field, which you haven't provided. Anyway:

$$\frac{x^{64}-x}{x(x-1)}=\frac{x(x-1)(x^{62}+x^{61}+...+x+1)}{x(x-1)}=x^{62}+x^{61}+...+x+1$$

Now, $\,62=2^6\Longrightarrow\,$ the splitting field of $\, x^{64}-x\,$ over the prime field $\,\Bbb F_2:=\Bbb Z/2\Bbb Z\,$ is the (unique, up to isomorphism of fields) field $\,\Bbb F_{2^6=64}\,$ of degree $\,6\,$ over $\,\Bbb F_2\,$.

Since in the above simplification we've cancelled the linear factors for $\,x=0,1\,$ and we must have these two elements in any field, we can see that the minimal extension field of $\,\Bbb F_2\,$ that contains ll the roots of the original rational function is the forementioned field $\,\Bbb F_{64}\,$ .

$\endgroup$
3
  • $\begingroup$ @DanAntonio : Sir, I am just talking about the extension field not the splitting field. How can i find two extensions $F(\alpha)$ and $F(\beta)$ which are not isomorphic ?? $\endgroup$
    – Theorem
    Commented Oct 16, 2012 at 22:58
  • 1
    $\begingroup$ Extension of what field? What does the expression $\,\frac{x^{64}-x}{x(x-1)} \,$ have to do with all this? In general, for a field $\,\Bbb F\,$ and two elements $\,\alpha, \beta\,$ in some algebraic extension, the fields $\,\Bbb F(\alpha)\cong\Bbb F(\beta)\,$ iff $\,\alpha,\beta\,$ are roots of the same irreducible polynomial in $\,\Bbb F[x]\,$ . If one of $\,\alpha,\beta\,$ is not algebraic over $\,\Bbb F\,$ then things can be even harder... $\endgroup$
    – DonAntonio
    Commented Oct 17, 2012 at 2:58
  • $\begingroup$ Sir sorry for less information , here the field is $F_2$ , $f\in F_2[x]$ $\endgroup$
    – Theorem
    Commented Oct 17, 2012 at 5:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .