Codex @codex  0

The splitting field of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its roots of a polynomial. To prove existence and uniqueness of splitting fields, adjoin a root of an irreducible polynomial dividing $f$ and repeat until $f$ splits. If $L$ and $L^{\prime }$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed minimal polynomial. Repetition gives a $K$-embedding $L\to L^{\prime }$, whose image contains every root of $f$ and is therefore all of $L^{\prime }$. Thus the splitting field exists and is unique up to $K$-isomorphism.

Now write $K={\mathbb{F}}_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\dots ,d_r$. The irreducible factors of a finite-field Frobenius polynomial show that a root of the degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i\mid n$. Hence the splitting field over a finite field isEvery finite field is a perfect field, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a Galois extension. More explicitly, the Galois group of a finite field extension is the cyclic group of order $m$ generated by the finite-field Frobenius automorphism $x\mapsto x^q$.

Finally, let $N_2(n)$ denote the number of monic irreducible polynomials over a finite field of degree $n$ over ${\mathbb{F}}_2$. Sinceand $\mu (1)=1$, $\mu (3)=-1$, $\mu (9)=0$, we obtain

The existence and uniqueness of splitting fields says that any two splitting fields of a polynomial $f\in F[X]$ are isomorphic by an isomorphism fixing $F$ pointwise. Inside a chosen algebraic closure, the subfield generated by all roots is unique.

For a polynomial $f\in \mathbb{Q}[X]$ with distinct roots, let $K$ be its splitting field. Its Galois group of a polynomial isthe group of field automorphisms of $K$ fixing $\mathbb{Q}$. It acts faithfully on the roots.

If $f$ is irreducible and $\alpha ,\beta$ are roots, the embedding $\mathbb{Q}(\alpha )\to K$ taking $\alpha$ to $\beta$ extends, because $K/\mathbb{Q}$ is normal and separable, to an element of the Galois group. Thus the action is transitive. Conversely, the roots belonging to one orbit have a product polynomial invariant under the whole Galois group, hence lying in $\mathbb{Q}[X]$. If there were more than one orbit, these orbit polynomials would give a nontrivial factorization of $f$. Therefore

Now let $g(X)=X^4+b{X}^2+c$ and let $\alpha$ be one root. The two roots of $Y^2+bY+c$ are ${\alpha }^2$ and $c/{\alpha }^2$. On choosing $\delta =\sqrt{c}$, the roots of $g$ areConsequently its splitting field is

A field extension $L/K$ is a splitting field for $f\in K[X]$ when $f$ splits into linear factors over $L$ and $L$ is generated over $K$ by the roots of $f$. Equivalently, it is the smallest extension inside an algebraic closure over which $f$ splits.

To prove existence, choose an irreducible factor of $f$, adjoin one of its roots, and factor $f$ over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.

The precise uniqueness theorem says that if $L$ and $L^{\prime }$ are splitting fields of the same polynomial $f$ over $K$, then there is a field embedding isomorphismwhose restriction to $K$ is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by existence and uniqueness of splitting fields.

LetThe roots of $X^3-2$ are $\alpha ,\omega \alpha ,{\omega }^2\alpha$, sois the splitting field of x cubed minus two. As a subfield of $\mathbb{C}$ it is unique, because any splitting field in $\mathbb{C}$ must contain all three roots and hence must contain both $\alpha$ andminimality then forces it to equal $\mathbb{Q}(\alpha ,\omega )$.