Clarification of proof in Artin's "Algebra"

Question

I'm trying to understand a proof in Artin's Algebra

"Theorem 15.7.3.b: the irreducible factors of the polynomial $x^{p^r}-x$ over the field $F_p$ are the irreducible polynomials in $F_p[x]$ whose degrees divide $r$.

Proof, converse direction: Let $g$ be an irreducible polynomial over $F_p$ over degree $k$. Suppose $k|r$. Let $\beta$ be a root of $g$ in an extension field of $F_p$ - then $[F_p(\beta):F_p]=k$, so by the subfield criterion $F_{p^r}$ contains a subfield isomorphic to $F_p(\beta)$. Therefore $g$ has a root in $F_{p^r}$, and so $g$ divides $x^{p^r}-x$."

The only step I don't understand is the last one. How does knowing that $g$ has one root in $F_{p^r}$ allow us to deduce that $g$ divides $x^{p^r}-x$?

Answer 1

Note that $g$ and $x^{p^r}-x$ have a common root in $F_{p^r}$. Therefore, they are not relatively prime. So, since $g$ is irreducible, $g\mid x^{p^r}-x$.

Answer 2

$F_{p^r}$ is the splitting field for $x^{p^r} - x$ over $F_p$, so $F_{p^r}$ is a normal extension. A characterization of normal extensions is that every irreducible polynomial with one root in the field splits completely in the field. So $g$ actually has all of its roots in $F_{p^r}$.

Answer 3

The proof is only 6.5 lines in Artin's book, which surely needs a detailed exposition. We need to prove the statement in two directions.

Forward direction

We have a polynomial $x^q-x$ over a prime field $F$, that is $x^q-x \in F[x]$. If there exist an irreducible polynomial that divides $x^q-x$, then its degree must divide $r$.

Backward direction

If we pick an irreducible polynomial in $F[x]$, whose degree divides $r$, then this polynomial will divide $x^q-x$ in $F[x]$.

Warm up

Firstly, Artin used confusing notation as $F$ in this particular case is a prime field, a finite field.

Secondly, in the backward direction he did not introduce field $K$, which was specified in parts (a) and (c) of this theorem.

Thirdly, the way he used isomorphism in the backward direction is not explained in detail.

Proof of the backward direction

We first start with the original prime field $F$ and extend it with a root for polynomial $g$.

$$F \subset F(\beta) \tag{A}$$

The degree of $F(\beta)$ over $F$ as a vector (=linear) space will be $k$, the degree of irreducible polynomial $g$. The number of elements in field $F(\beta)$ is $p^k$, and this number by 15.7.3(d) has unique structure.

Then, we take polynomial $x^q-x$ and split it completely in some other unrelated field $K$ of order $q=p^r$. We can do this by 15.7.3(d) and (a).

Field $K$ by 15.7.4(e) contains a field of order $p^k$, call it $\mathcal{F'}_{k}$. As by (e) the fields $F(\beta)$ and $\mathcal{F}_{k}$ have the same order and isomorphic, we can carefully carry the structure of $F(\beta)$ into $\mathcal{F'}_{k}$.

We carry over a set $F(\beta)$ together with its subset $F$ and get the chain of field extensions

$$\overbrace{\mathcal{F'} \subset \mathcal{F'}_{k}}^{\text{(A) carried over by isomorphism}} \subset K$$

We put apostrophes to differentite these subfields of $K$ from the original $F$.

Now we can use 15.6.4(f). We have both polynomials $g$ and $x^q-x$ over $\bf{\mathcal{F'}}$, $g$ is irreducible. Both polynomials have a root in a field extension $K$, so $g$ divides $x^q-x$ in $\mathcal{F'}[x]$.

And as $F$ and $\mathcal{F'}$ are isomorphic, $g$ divides $x^q-x$ in the original $F[x]$ as well.

Note

We did not use 15.6.3 page 456, because it says nothing about the degree of the resulting field extension.

We did not use polynomials and quotient constructions of 15.6 to start with the original field and get two extra field extensions for the roots of $g$ and $x^q-x$.

We used isomorphism twice. First to carry the structure of $F(\beta)$ into $K$, then the divisibility property in the prime subfield of order p of $K$ back to the original $F$.