Splitting field

The splitting field of a polynomial over $K$ is the smallest field extension of $K$ in which the polynomial is a product of linear factors.

For a polynomial $f\in K[x]$ of degree $n$, its splitting field $L$ has $[L:K]\mid n!$, including in positive characteristic when the polynomial is inseparable. Prove this by induction on the degree. If $f$ is irreducible, adjoining a root gives degree $n$ and leaves a polynomial of degree $n-1$ to split, so the tower law for field extensions gives the divisibility. If $f=gh$ with positive degrees $r,s<n$, split $g$ first and then $h$ over the resulting field. The degree divides $r!s!$, which divides $(r+s)!$. This separate reducible case avoids the invalid step of assuming that the degree of an arbitrarily chosen irreducible factor must divide $n$.

A root of an irreducible polynomial in an extension field defines a base-field embedding of the polynomial quotient by evaluation. Conversely the image of the residue class of the variable determines the embedding and is a root. This counts embeddings through separable minimal polynomials.

Every nonconstant polynomial $f\in K[X]$ has a splitting field of finite degree over $K$. It is constructed by adjoining a root of an irreducible factor and repeating over the enlarged field. Any two splitting fields of $f$ over $K$ are isomorphic by an isomorphism that fixes $K$. Inside one fixed algebraic closure, the field generated by all roots is unique as a subfield.

If $\alpha =\sqrt[3]{2}$ is the positive real root and ${\zeta }_3=e^{2\pi i/3}$, then the roots of $X^3-2$ areTheir splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\alpha ,{\zeta }_3)$.

A finite extension $L/K$ is normal exactly when it is the splitting field over $K$ of some polynomial. For the forward direction, multiply the minimal polynomials of finitely many generators of $L$; normality puts all their roots in $L$. The reverse direction follows because a $K$-embedding permutes the roots of a polynomial and hence preserves its splitting field.

Let $g\in K[T]$ in characteristic $p$, let $f(T)=g(T^p)$, let $L$ split $g$ over $K$, and let $M$ split $f$ over $L$. If $y_i$ are the roots of $g$ and $x_i^p=y_i$, then the Frobenius endomorphism makes each $x_i$ unique andbecause $y_i=x_i^p$. Thus $M$ is already the splitting field of $f$ over $K$, and every root of $f$ is purely inseparable over $L$.

Let $f\in {\mathbb{F}}_q[X]$ have irreducible factors of degrees $d_1,\dots ,d_r$. A root of a degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i$ divides $n$. Consequently the splitting field of $f$ is