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.

Its easy to proof that any non-zero field homomorphism is injective:

Proof Assume that $\exists a, b\in F: a\neq b~~and~~\psi(a)=\psi(b)$ then: $$\psi(1)=\psi((a-b)^{-1}(a-b))=\psi((a-b)^{-1})\cdot 0,$$ $$\forall x\in F:~~\psi(x)=\psi(x\cdot 1)=\psi(x)\cdot\psi(1)=\psi(x)\cdot0=0.$$ So, $\psi\equiv 0.$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$

But lets consider Frobenius homomorphism of algebraically closed field $F$: $$F\ni x\stackrel{\Phi}{\longmapsto} x^p\in F.$$ Equation $$x^p-a=0$$ have p different roots. So, $\Phi$ isn't injective. Where is mistake?

Thanks.

share|improve this question
2  
They're not all different; $a^\frac{1}{p}$ is a $p$-fold root of $x^p-a$ over an algebraically closed field of characteristic $p.$ –  jspecter Aug 12 '11 at 6:22
1  
One way to quickly recognize that the roots are not distinct is to note that the (formal) derivative is $px^{p - 1} = 0$. See the Wikipedia article on separable polynomials. –  Dylan Moreland Aug 12 '11 at 15:30
    
Once you've worked this out, you should try proving that Frobenius is an automorphism if and only if $F$ is perfect (so don't assume algebraically closed anymore). It is a neat little exercise and these sorts of ideas appear in it. –  Matt Aug 12 '11 at 15:37
add comment

1 Answer 1

up vote 8 down vote accepted

Over a field of characteristic $p$ such an equation has either a single root or no roots at all. If $\xi$ is one solution of the equation $x^p-a=0$, then $\xi^p=a$. Consequently $$ x^p-a=x^p-\xi^p=(x-\xi)^p, $$ and $\xi$ is a root of multiplicity $p$.

The mistake was assuming that the $p$ roots would be distinct.

share|improve this answer
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.