Splitting Fields: Examples

The empty family has no roots, and a nonzero constant also has no roots. In either case the subfield generated over $F$ by the empty root set is $F$ itself.

The empty family is vacuously split over $F$, while $c=c{\prod }_{j=1}^0(x-{\alpha }_j)$ is the required empty linear factorisation. Thus both cases satisfy the splitting-field definition.

By [F2], $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ over $\mathbb{R}$. Its two roots are $\pm \sqrt{2}$, and both lie in $\mathbb{Q}(\sqrt{2})$.

Eisenstein's criterion [F3] with the prime $2$ makes $x^2-2$ irreducible over $\mathbb{Q}$. It is monic and vanishes at $\sqrt{2}$, so [F6] identifies it as the minimal polynomial; [F4] then gives the basis $1,\sqrt{2}$.

The field generated by both roots is $\mathbb{Q}(\sqrt{2},-\sqrt{2})=\mathbb{Q}(\sqrt{2})$, so [F5] identifies it as the splitting field.

By [F1], ${\alpha }^3=2$ and ${\sqrt{3}}^2=3$. Using [F2], direct calculation gives ${\omega }^2+\omega +1=0$, $\omega \mathrm{\ne }1$, and $x^3-1=(x-1)(x-\omega )(x-{\omega }^2)$. Hence the cube roots of unity are exactly $1,\omega ,{\omega }^2$.

By [F3], the roots of $x^3-2$ are exactly $\alpha ,\alpha \omega ,\alpha {\omega }^2$. They generate $\mathbb{Q}(\alpha ,\omega )$ because $\alpha$ is a root and $\omega =(\alpha \omega )\mathrm{/}\alpha$, with $\alpha \mathrm{\ne }0$. Hence this field is the splitting field.

Eisenstein at $2$ makes $x^3-2$ irreducible over $\mathbb{Q}$. Independently, [F4] gives a splitting field spanned by at most $3!=6$ root monomials; an isomorphism from it to $\mathbb{Q}(\alpha ,\omega )$ supplied by [F6] fixes $\mathbb{Q}$ and carries roots to roots by direct evaluation, so it transports that spanning family to one of the stated kind here.

Over $\mathbb{Q}$, direct multiplication gives $x^4+2x^2-8=(x^2-2)(x^2+4)$. By [F1] and [F2], the first factor has roots $\pm \sqrt{2}$ and the second has roots $\pm 2i$.

All four roots lie in $\mathbb{Q}(\sqrt{2},i)$. Conversely, the field generated by them contains $\sqrt{2}$ and $i=(2i)\mathrm{/}2$, so it is exactly $\mathbb{Q}(\sqrt{2},i)$. The claim follows from [F3].

By [F1], the roots of $x^2-2$ are $\pm \sqrt{2}$, so [F4] makes its splitting field $\mathbb{Q}(\sqrt{2})$. Similarly the splitting field of $x^2-3$ is $\mathbb{Q}(\sqrt{3})$.

Their composite is the smallest field containing both, namely $\mathbb{Q}(\sqrt{2},\sqrt{3})$. By [F2] it splits the product, and by [F3] it is the splitting field of the stated family.

In characteristic $2$, $(x^2+x+1{)}^2=x^4+x^2+1$. The polynomial $p=x^2+x+1$ takes the value $1$ at both $0$ and $1$, so [F2] makes it irreducible.

By [F3], adjoining $u=x+(p)$ gives a field with the four distinct elements $0,1,u,u+1$ and relation $u^2+u+1=0$. Substituting $u+1$ into $p$ in characteristic $2$ also gives zero, so $p=(x-u)(x-(u+1))$.

Squaring the factorisation yields $x^4+x^2+1=(x-u{)}^2(x-(u+1){)}^2$. The two roots are distinct by step 1.2, and any root of the displayed product equals one of them because a field has no zero divisors. Uniqueness of factorisation in [F6] makes both displayed exponents exactly two; in particular [F4] makes both roots repeated. Finally [F5] identifies ${\mathbb{F}}_2(u)$ as their splitting field.

The positive real number $\alpha =\sqrt[3]{2}$ is a root of $x^3-2$, and every element of $\mathbb{Q}(\alpha )$ is real because it is obtained from rational numbers and $\alpha$ by field operations inside $\mathbb{R}$.

In [F1], the imaginary part of $\omega$ is $\sqrt{3}\mathrm{/}2>0$, while direct multiplication gives the imaginary part of ${\omega }^2$ as $-\sqrt{3}\mathrm{/}2<0$. Since $\alpha >0$, the roots $\alpha \omega$ and $\alpha {\omega }^2$ are both nonreal and hence neither lies in $\mathbb{Q}(\alpha )$.

Therefore $\mathbb{Q}(\alpha )$ contains one root but not all roots of $x^3-2$, so the polynomial does not split there.

Take both splitting fields to be $E=E^{\mathrm{\prime }}=\mathbb{Q}(\sqrt{2})$ for $x^2-2$. The identity map is one $\mathbb{Q}$-isomorphism $E\to E$.

The polynomial $x^2-2$ satisfies [F4] with the prime $2$, so it is irreducible. Apply [F3] to the identity on $\mathbb{Q}$ and the corresponding roots $\sqrt{2}$ and $-\sqrt{2}$. It gives a $\mathbb{Q}$-automorphism $\tau :E\to E$ satisfying $\tau (\sqrt{2})=-\sqrt{2}$.

The roots are distinct: if $\sqrt{2}=-\sqrt{2}$, then $2\sqrt{2}=0$, and multiplication by $1\mathrm{/}2$ would give $\sqrt{2}=0$, contradicting $(\sqrt{2}{)}^2=2$. Hence $\tau$ is not the identity. Thus there are at least two base-fixing isomorphisms, refuting uniqueness while leaving the existence result [F1] intact.