Show that x -> x^p is an automorphism in K

[Hill]

Homework Statement

Let K be a finite field of characteristic p > 0. Show that x -> x^p is an automorphism in K.

Relevant Equations

s(x) = x^p

It is clear in case K is a prime field, because then s(x) is just an identity. If K is not prime, s(x) is identity on its prime subfield. But how to show the automorphism of s() for the rest of the field, when it's not prime?

 

[fresh_42]

Homework Statement: Let K be a finite field of characteristic p > 0. Show that x -> x^p is an automorphism in K.
Relevant Equations: s(x) = x^p

It is clear in case K is a prime field, because then s(x) is just an identity. If K is not prime, s(x) is identity on its prime subfield. But how to show the automorphism of s() for the rest of the field, when it's not prime?

$s(x\cdot y)=(x\cdot y)^p=x^p\cdot y^p=s(x)\cdot s(y)$ is obvious since $K$ is Abelian. To see that $s(x+y)=s(x)+s(y)$ use the binomial theorem. $s(x)$ is called Frobenius homomorphism.

 

[Hill]

$s(x\cdot y)=(x\cdot y)^p=x^p\cdot y^p=s(x)\cdot s(y)$ is obvious since $K$ is Abelian. To see that $s(x+y)=s(x)+s(y)$ use the binomial theorem.

Thank you. It shows that s() is homomorphism. But how to show that it is injective and surjective in K?

 

[PeroK]

Homework Statement: Let K be a finite field of characteristic p > 0. Show that x -> x^p is an automorphism in K.
Relevant Equations: s(x) = x^p

It is clear in case K is a prime field, because then s(x) is just an identity. If K is not prime, s(x) is identity on its prime subfield. But how to show the automorphism of s() for the rest of the field, when it's not prime?

What about:

Suppose $x^p = y^p$, then $(xy^{-1})^p = 1$. Therefore $xy^{-1}$ has order $1$ or $p$.

The order of the field is $p^n$, hence the order of the multiplicative group is $p^n -1$. As $p$ cannot divide $p^n - 1$, the order of $xy^{-1}$ is $1$. Hence $x = y$, and the mapping is injective.

 

[Hill]

What about:

Suppose $x^p = y^p$, then $(xy^{-1})^p = 1$. Therefore $xy^{-1}$ has order $1$ or $p$.

The order of the field is $p^n$, hence the order of the multiplicative group is $p^n -1$. As $p$ cannot divide $p^n - 1$, the order of $xy^{-1}$ is $1$. Hence $x = y$, and the mapping is injective.

Thank you. This is good. Now, for the surjective part ...?

 

[PeroK]

Thank you. This is good. Now, for the surjective part ...?

That follows automatically for a finite set.

 

[Hill]

That follows automatically for a finite set.

I am sorry. I guess it should be obvious, but I don't see it. Could you elaborate, how?

 

[PeroK]

I am sorry. I guess it should be obvious, but I don't see it. Could you elaborate, how?

A mapping from a finite set to itself is injective iff it is surjective. You only have to count the elements in the range.

 

[Hill]

A mapping from a finite set to itself is injective iff it is surjective. You only have to count the elements in the range.

Thanks a lot. Just wanted to tell you that I got it, but you have already answered.

 

[martinbn]

Just to point out that a homomorphism between fields is always injective. If not it will have a kernel, which is an ideal. Fields have no proper ideals.

 

[WWGD]

Notice too, the Binomial Theorem assumes commutativity of the product, as in $(a+b)^2= a^2+ab +ba + b^2$ assumes $ab=ba$, so this won't hold for general rings. Same for the argument $(ab)^p=a^pb^p$.

 

[mathwonk]

"Just to point out that a homomorphism between fields is always injective. If not it will have a kernel, which is an ideal. Fields have no proper ideals"
nice, and I guess you have to check the homomorphism is unitary, i.e. as here, that 1-->1. (so that the kernel is a proper ideal.) but of course you knew that.

 

[WWGD]

"Just to point out that a homomorphism between fields is always injective. If not it will have a kernel, which is an ideal. Fields have no proper ideals"
nice, and I guess you have to check the homomorphism is unitary, i.e. as here, that 1-->1. (so that the kernel is a proper ideal.) but of course you knew that.

I guess outside of the Complex Numbers, $1^x=1$ always holds.

 

[mathwonk]

you just checked it!

 

[nuuskur]

I am sorry. I guess it should be obvious, but I don't see it. Could you elaborate, how?

Pigeons and pigeon housing.