# Finite field as a splitting field of some irreducible polynomial

In many texts that I've read regarding finite fields, it always appears to be simply stated that a finite field is a splitting field of some irreducible polynomial, without proof. What are some good sources that actually provide an explicit proof? Now on p. $3$ of the lecture notes http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/finitefields.pdf of Keith Conrad, Lemma $2.1$ states that a field of power $p^n$ (cardinality of a finite field) is a splitting field of the polynomial $x^{p^n} - x$, but I'm not sure if this is irreducible. Might this be of any help?

• The given polynomial has $x = 1$ as a root. – Hoot Jan 2 '15 at 0:38
• Right, so it is not irreducible, but certainly a factor of it could be irreducible, correct? – Libertron Jan 2 '15 at 0:39
• $x^{p^n}-x$ isn't irreducible, but you can verify that the set of roots over $\mathbb{F}_p$ forms a field. – yoyo Jan 2 '15 at 0:40
• Actually, $x^{p^n}-x$ is the product of all irreducible polynomials over $\mathbf F_p$ whose degree divides $n$. – Bernard Jan 2 '15 at 0:48
• Certainly. There's an irreducible dividing every polynomial! You may be looking for the claim that there's irreducible $f$ dividing $x^{p^n}-x$ such that the splitting field of $x^{p^n}-x$ is $\mathbf{F}_p(\alpha)$ for $\alpha$ some root of $f$. This is true, and you can give an existence proof via the fact that finite fields have cyclic multiplicative group or via the primitive element theorem. But none of this is necessary to show that the roots of $x^{p^n}-x$ form a field of order $p^n$. – Kevin Carlson Jan 2 '15 at 0:49

No, $x^{p^n} - x$ is not in general irreducible. What it is, though, is the product of all irreducible polynomials over $\mathbb{Z}/p$ whose degrees divide $n$. The field with $p^n$ elements will be the splitting field of any factor of $x^{p^n}-x$ that is a multiple of any of those irreducible polynomials of degree $n$.
There will be many such irreducible polynomials of degree $n$, always at least one and usually nearly $p^n/n$ of them.
In particular, even when $x^{p^n}-x$ is not itself irreducible, it is a factor of itself and does have an irreducible factor of degree n. It's the splitting field both of itself and of that factor.