4
$\begingroup$

1) Let $p$ prime and $n\geq 1$ an integer. Show that there is a finite field of order $p^n$ in an algebraic closure $\mathbb F_p^{alg}$ and that all finite field is isomorphic to exactly one field $\mathbb F_{p^n}$

2) Let $\mathbb F_q$ a finite field and $n\geq 1$ an integer. Let $\mathbb F_q^{alg}$ an algebraic closure. Show that there is a unique extension field of $\mathbb F_q$ of degree $n$.


1) By Fermat theorem, for all $\alpha \in \mathbb F_p$, $$\alpha ^p\equiv \alpha \pmod p.$$ In particular, since $\mathbb F_p^{alg}$ is a closure algebraic, if $\beta \in\mathbb F_p^{alg}$, the minimal polynomial $$p(X)=a_0+a_1X+...+a_nX^n\in\mathbb F_p[X]$$ split over $\mathbb F_p^{alg}$. In particular, $$p(X)=a_0+...+a_nX^n=a_0^p+a_1^pX+...+a_n^p X^n,$$ and thus, $P'(X)=0\in\mathbb F_p(X)$. Therefore $f\in\mathbb F_p[X^p]$, in particular, $$P(X)=P(X^p)=P(X)^p$$ and thus \begin{align*}Frob:\mathbb F_p&\longrightarrow \mathbb F_p\\ x&\longmapsto x^p\end{align*} is surjective and thus bijective.

Q1) I'm really not sure about my implication of $Frob$ surjective.

Q2) How can I continue ?

2) By $1$) $\mathbb F_q$ is isomorphic to an $\mathbb F_{p^n}$. We have that $X^{q^n}-X$ split over $\mathbb F_{q^n}$.

Q3) It's in written in my course, but I can't prove it. Any idea ?

Q4) I don't know how to continue

$\endgroup$
2
$\begingroup$

Q1) As you noticed in your point 1) you have $x^p=x$ for any $x \in \mathbb F_p$. Hence the Frobenius isomorphism is equal to the identity map on $\mathbb F_p$ which is obviously injective and surjective.

Q2) Usually you prove the existence of the field $\mathbb F_{p^n}$ by saying that $\mathbb F_{p^n}$ is the splitting field of the polynomial $P(X)=X^{^n}-X$ over $\mathbb F_p$. $P$ has indeed $p^n$ roots as $P^\prime(X)=-1$ and do not vanish. How is defined in your course the field $\mathbb F_{p^n}$?

Q3) The unicity is a consequence that for a field the multiplicative group is cyclic. If $\alpha$ is a generator, you have therefore $\alpha^{p^n-1}-1$. And one can prove that the splitting field of a polynomial is unique up to isomorphism.

You'll find all this in most of field courses.

$\endgroup$
2
$\begingroup$

Consider $F = \{ a \in \mathbb F_p^{alg} : a^{p^n} = a \}$, that is, the zeros of $f(X)=X^{p^n}-X$. Since $f'(X)=-1$, all zeros of $f$ are distinct and so $F$ has exactly $p^n$ elements.

$F$ is a field because $x \mapsto x^p$ is a ring homomorphism in characteristic $p$.

This proves that there exists a field having $p^n$ elements.

Now, by Lagrange's theorem, we have $x^{p^n-1}=1$ for all non-zero elements in a finite field having $p^n$ elements. This means that such a field is the splitting field of $X^{p^n}-X$ and so is unique up to isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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