Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 11G (Groups, Rings and Modules)

Because $f$ is irreducible, the ideal $(f)$ is maximal in the polynomial ring $k[X]$. Henceis a field extension of $k$. Let $\alpha =X+(f)\in F$. Evaluation in the quotient gives $f(\alpha )=0$, and the factor theorem in $F[X]$ givesfor some $g\in F[X]$. This is the standard adjoining a root of an irreducible polynomial construction.

Proceed by induction on $d=\mathrm{deg}f$. A nonconstant polynomial over a field has an irreducible factor $h$. Part (a) gives an extension $k_1/k$ containing a root ${\alpha }_1$ of $h$, hence of $f$. The factor theorem giveswhere $f_1$ is monic of degree $d-1$. Apply the induction hypothesis over $k_1$ and compose the resulting field extensions. The final field $F$ contains roots ${\alpha }_1,\dots ,{\alpha }_d$ andEquivalently, every polynomial has a splitting field.

Put $q=p^n$. Every extension of $k={\mathbb{F}}_p$ has characteristic $p$, so the iterated Frobenius endomorphism satisfiesThe root set is $K=\{a\in F:a^q=a\}$. It contains $0$ and $1$. If $a,b\in K$, thenso it is closed under subtraction and multiplication. If $a\in K$ is nonzero, thenThus $K$ is closed under multiplicative inverses as well, and is therefore a field. This is the polynomial characterization of a finite field; in fact it is the finite field with $q$ elements, since $X^q-X$ has degree $q$ and no repeated roots.