Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

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 $\,\Bbb F_{64}\,$ .

share|improve this answer
    
@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 ?? –  Theorem Oct 16 '12 at 22:58
    
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... –  DonAntonio Oct 17 '12 at 2:58
    
Sir sorry for less information , here the field is $F_2$ , $f\in F_2[x]$ –  Theorem Oct 17 '12 at 5:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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