3

I was reading the following example from Dummit & Foote, Algebra:

enter image description here enter image description here

Okay, I get it, except for one point: I get that by the minimality of the splitting field, any subfield of the splitting field containing all the roots of the given polynomial must be the whole splitting field provided that the subfield in question contains the base field. How do we know that in this case, $\mathbb{F}$ contains $\mathbb{F}_p$? I.e., how do we know that every element of $\mathbb{F}_p$ is a root of the polynomial $x^{p^n}-x$?

Thanks!

user134070
  • 679
  • $x^{p^n}$ is $(((x^p)^p)^p\ldots)^p$ where $p$ appears $n$ times. But $x^p=x$ in $\mathbb{F}_p$. – Suzet Jul 12 '18 at 07:48

4 Answers4

3

Here are two purely abstract reasons why $\mathbb{F}_p \subseteq \mathbb{F}$.

Reason 1: If $K$ is any field and $f$ any polynomial over $K$, the splitting field of $f$ over $K$ is, in particular, defined to be a field extension of $K$, so $K$ is automatically contained in the splitting field.

Reason 2: Every field of characteristic $p$ contains $\mathbb{F}_p$ as a subfield.

2

This is a corollary of Fermat's little theorem, which says that all $x\in \Bbb F_p$ we have $x^p = x$ (or, in some phrasings, that for $x\neq 0$ we have $x^{p-1} = 1$, but that's the same thing). Now note that $x^{p^n} = (\cdots((x^p)^p)^p\cdots)^p$. One may also quote Lagrange's theorem on the multiplicative group of the field to get the same result.

Arthur
  • 199,419
1

It is the field generated by the roots. If it contains a root $x$, it contains its inverse $y$ and$xy=1$.

Tsemo Aristide
  • 87,475
1

$\mathbf F$ contains (a field isomorphic to) $\mathbf F_p$ because any field contains $1$, hence all multiples of $1$, which make a field isomorphic to $\mathbf F_p$.

Also every element of $\mathbf F_p$ satisfies the equation $x^p=x$ (that's what lil' Fermat asserts), so iteratively $$x^{p^n}=\bigl(x^p\bigr)^{p^{n-1}}=x\vphantom{h}^{p^{n-1}}=x.$$

Bernard
  • 175,478