Finite field isomorphic to $\mathbb F_{p^n}$.

Question

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 $n$.

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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

Answer 1

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.

Answer 2

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.