The Galois Correspondence — Examples

The field $\mathbb{Q}(\sqrt{2})$ has degree two, and $\sqrt{3}\mathrm{\notin }\mathbb{Q}(\sqrt{2})$: squaring an equation $\sqrt{3}=a+b\sqrt{2}$ forces $2ab=0$, and either case contradicts rationality. Thus $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis and the extension has degree four. Independent sign changes of the two square roots give four automorphisms. The field is the splitting field over $\mathbb{Q}$ of $(x^2-2)(x^2-3)$, whose four roots $\pm \sqrt{2},\pm \sqrt{3}$ are distinct, so [L2] makes the extension finite Galois with $\mathrm{\mid }G\mathrm{\mid }=4$; the four sign changes therefore exhaust $G$, which has exponent two and is thus $V_4$.

For $x=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$, invariance under ${\sigma }_2$, ${\sigma }_3$, or ${\sigma }_6$ respectively forces $(b,d)=(0,0)$, $(c,d)=(0,0)$, or $(b,c)=(0,0)$. Their fixed fields are therefore $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{2})$, and $\mathbb{Q}(\sqrt{6})$, which are distinct quadratic fields.

The degrees in step 2.1 equal the subgroup indices prescribed by [L1], and [L1] is a bijection, so the table includes every subgroup and every intermediate field, including both endpoints.

Eisenstein at $2$ makes $x^3-2$ irreducible over $\mathbb{Q}$, so by [L3] it is the minimal polynomial of $a$ and by [L4] $[\mathbb{Q}(a):\mathbb{Q}]=3$. Since $a$ is real, $\mathbb{Q}(a)\subseteq \mathbb{R}$, whereas the two roots $\omega ,{\omega }^2$ of $x^2+x+1$ are nonreal; that quadratic therefore has no root in $\mathbb{Q}(a)$, is irreducible over $\mathbb{Q}(a)$, and by [L3] and [L4] gives $[L:\mathbb{Q}(a)]=2$. By [L5], $[L:\mathbb{Q}]=[L:\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}]=6$. The three cube roots of $2$ are $a,\omega a,{\omega }^{2}a$, all in $L$, and they generate $L$ over $\mathbb{Q}$ because $\omega =(\omega a)a^{-1}$; hence $L$ is the splitting field of $x^3-2$ over $\mathbb{Q}$.

The displayed maps permute the three roots and preserve the defining relations, so they are automorphisms. Direct calculation gives $r^3=s^2=1$ and $srs=r^{-1}$, and the six maps $1,r,r^2,s,rs,r^{2}s$ are distinct. The three roots $a,\omega a,{\omega }^{2}a$ of $x^3-2$ are distinct, so that polynomial is separable and step 1.1 makes $L$ its splitting field; by [L2], $L\mathrm{/}\mathbb{Q}$ is finite Galois with $\mathrm{\mid }\mathrm{Gal}(L\mathrm{/}\mathbb{Q})\mathrm{\mid }=[L:\mathbb{Q}]=6$. The six maps therefore exhaust the automorphism group.

Each listed generator fixes its displayed field: $r$ fixes $\mathbb{Q}(\omega )$, $s$ fixes $\mathbb{Q}(a)$, while $rs$ sends $a$ to $\omega a$ and $\omega$ to ${\omega }^2$, so it fixes ${\omega }^{2}a$, and $r^{2}s$ sends $a$ to ${\omega }^{2}a$ and $\omega$ to ${\omega }^2$, so it fixes $\omega a$. Each of $a,\omega a,{\omega }^{2}a$ is a root of the irreducible $x^3-2$ and $\omega$ is a root of the irreducible $x^2+x+1$, so by [L3] and [L4] the four fields have degrees $3,3,3$ and $2$ over $\mathbb{Q}$, matching the indices of the corresponding subgroups. Each displayed field therefore sits inside the fixed field of its subgroup with the same finite degree over $\mathbb{Q}$, so the two coincide, and the fundamental theorem's bijection makes the table complete.

The subgroup $A_3$ is normal in $S_3$, while none of the three order-two subgroups is normal. By [L1], $\mathbb{Q}(\omega )\mathrm{/}\mathbb{Q}$ is Galois and the three cubic fields are not; the base and splitting fields give the two normal endpoints.

Eisenstein makes $x^4-2$ irreducible, so $[\mathbb{Q}(a):\mathbb{Q}]=4$. Since $\mathbb{Q}(a)\subseteq \mathbb{R}$ and $i\mathrm{\notin }\mathbb{R}$, adjoining $i$ doubles the degree. Thus $1,a,a^2,a^3,i,ia,i{a}^2,i{a}^3$ is a basis of the degree-eight splitting field $L$.

The displayed maps preserve $a^4=2$ and $i^2=-1$, permute the roots $a,ia,-a,-ia$, and satisfy $r^4=s^2=1$ and $srs=r^{-1}$; their eight composites are distinct. Those four roots are distinct, so $x^4-2$ is separable and step 1.1 makes $L$ its splitting field; by [L2], $L\mathrm{/}\mathbb{Q}$ is finite Galois with $\mathrm{\mid }\mathrm{Gal}(L\mathrm{/}\mathbb{Q})\mathrm{\mid }=[L:\mathbb{Q}]=8$. The eight composites therefore exhaust the automorphism group and give $D_4$.

Applying the generators to the eight basis coefficients verifies that every field in the table is fixed by its displayed subgroup. Their degrees over $\mathbb{Q}$ are respectively $8,4,4,4,4,4,2,2,2,1$.

Those degrees equal the subgroup indices required by [L1], so each containment in step 3.1 is equality and the table is complete. The four reflection subgroups are nonnormal and give the nonnormal quartic fields; the remaining subgroups are normal and give the normal strict fields or endpoints.

Eisenstein at $2$ makes $x^3-2$ irreducible, so $[\mathbb{Q}(a):\mathbb{Q}]=3$. Characteristic zero makes the polynomial separable, and $a\mathrm{\ne }0$ because $a^3=2$.

The other roots are $\omega a$ and ${\omega }^{2}a$ for nonreal cube roots of unity $\omega ,{\omega }^2$, whereas $\mathbb{Q}(a)\subseteq \mathbb{R}$. Thus the minimal polynomial does not split in $\mathbb{Q}(a)$, so the extension is not normal.

By [L1], a $\mathbb{Q}$-automorphism must send $a$ to a root of $x^3-2$ that lies in $\mathbb{Q}(a)$. Step 2.1 leaves only $a$, and fixing the generator fixes all of $\mathbb{Q}(a)$. Hence the automorphism group has exactly its identity element.

Put $u=t^p$. The element $u$ is not a $p$-th power in the rational-function field ${\mathbb{F}}_p(u)$, as the valuation at the prime $u$ of a $p$-th power is divisible by $p$. Hence $x^p-u$ is irreducible over ${\mathbb{F}}_p(u)={\mathbb{F}}_p(t^p)$, while in ${\mathbb{F}}_p(t)[x]$ it equals $(x-t{)}^p$.

Step 1.1 gives degree $p$ and shows that every element of the extension has a power in the base field, so the extension is purely inseparable and not separable. By [L1] it is normal. This includes the smallest prime $p=2$.

A base-field automorphism must send $t$ to another root of its minimal polynomial, but step 1.1 shows that $t$ is the unique root. Thus every such automorphism fixes $t$ and is the identity on ${\mathbb{F}}_p(t)$.

The only rational-root candidates are $1$ and $-1$, and the polynomial takes the values $-1$ and $3$ there. It has no rational root, so the cubic is irreducible.

For a depressed cubic $x^3+px+q$, the discriminant is $-4p^3-27q^2$; here it is $-4(-3{)}^3-27=81=9^2$, which is nonzero.

Steps 1.1 and 1.2 give an irreducible separable cubic with square discriminant, so [L1] gives Galois group $A_3$. Its order is three, equal to the splitting-field degree.

Eisenstein at $2$ proves that $x^3-2$ is irreducible over $\mathbb{Q}$.

Its depressed-cubic discriminant is $-27(-2{)}^2=-108$, which is nonzero and negative, hence is not a square in $\mathbb{Q}$.

By [L1], steps 1.1 and 1.2 give Galois group $S_3$, agreeing with the explicit six automorphisms of the splitting-field lattice example.

Modulo $2$ the polynomial is $x^4+x+1$. It has no root in ${\mathbb{F}}_2$, and the only irreducible quadratic $x^2+x+1$ does not divide it, so the reduction is irreducible. The reduction test makes $x^4-x-1$ irreducible over $\mathbb{Q}$.

The resolvent formula gives $R(y)=y^3+4y-1$. Its only possible rational roots are $1$ and $-1$, neither of which is a root, so the resolvent cubic is irreducible.

The discriminant is the resolvent discriminant $-4(4{)}^3-27(-1{)}^2=-283$, a negative nonsquare in $\mathbb{Q}$ and in particular nonzero.

The derivative of $x^4-x-1$ is $4x^3-1$, a nonzero element of $\mathbb{Q}[x]$, and $\mathbb{Q}$ has characteristic zero, hence not two.

Steps 1.1 and 1.4 with [L2] make $x^4-x-1$ separable. The hypotheses of [L1] are then supplied by steps 1.1, 1.2, 1.3, and 1.4, so the Galois group is $S_4$.

No integer divisor of $12$ is a root, so there is no rational linear factor. Modulo $5$, one has $x^4+8x+12=(x-4)(x^3+4x^2+x+2)$; the cubic has no root in ${\mathbb{F}}_5$ and is irreducible. A monic factorization into two rational quadratics would reduce to a quadratic-by-quadratic factorization modulo $5$, contradicting the displayed irreducible factorization. Gauss's lemma therefore makes the quartic irreducible over $\mathbb{Q}$.

The resolvent is $y^3-48y-64$. Modulo $5$ it is $y^3+2y+1$, whose values at all elements of ${\mathbb{F}}_5$ are nonzero; hence the cubic resolvent is irreducible over $\mathbb{Q}$.

Its discriminant, and hence the quartic discriminant, is $-4(-48{)}^3-27(-64{)}^2=331776={576}^2$, which is nonzero.

Steps 1.1, 1.2, and 1.3 satisfy [L1], so the quartic has Galois group $A_4$.

After substituting $x+1$, the polynomial becomes $x^4+5x^3+10x^2+10x+5$, which is Eisenstein at $5$. Thus the original polynomial is irreducible.

The resolvent formula gives $R(y)=y^3-y^2-3y+2=(y-2)(y^2+y-1)$, so it has exactly one rational root.

If $\zeta$ is a root, then ${\zeta }^5=1$ and $\zeta \mathrm{\ne }1$. The roots are the distinct elements $\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4$, all in $\mathbb{Q}(\zeta )$, so this degree-four simple extension is the splitting field. The embedding $\zeta \mapsto {\zeta }^2$ is an automorphism and has order four on the exponents modulo $5$; it therefore generates the full Galois group, which is $C_4$.

Step 2.1 proves the group directly, while step 1.2 places it in the unique-root branch described by [L1]; the quadratic resolvent splitting field makes the quartic reducible there, as the $C_4$ row requires.

Put $\alpha =\sqrt{2}+\sqrt{3}$. Then $({\alpha }^2-5{)}^2=24$, so ${\alpha }^4-10{\alpha }^2+1=0$. Since ${\alpha }^{-1}=\sqrt{3}-\sqrt{2}$, one recovers $\sqrt{2}=(\alpha -{\alpha }^{-1})\mathrm{/}2$ and $\sqrt{3}=(\alpha +{\alpha }^{-1})\mathrm{/}2$, so $\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and the polynomial is the degree-four minimal polynomial of $\alpha$.

Its four distinct conjugates are $\sqrt{2}+\sqrt{3}$, $\sqrt{2}-\sqrt{3}$, $-\sqrt{2}+\sqrt{3}$, and $-\sqrt{2}-\sqrt{3}$, all in the biquadratic field. Thus this is the splitting field and [L1] gives Galois group $V_4$.

The resolvent is $y^3+10y^2-4y-40=(y+10)(y-2)(y+2)$, so it splits completely over $\mathbb{Q}$, agreeing with the $V_4$ row.

The extension in [L1] has degree $3$ but automorphism-group order $1$, so the two numbers are unequal.

The same witness satisfies the correct statements $1\le 3$ and $1\mid 3$, showing that equality, not the bound or divisibility, is the failed assertion.

Choose any order-two subgroup from [L1]. It is a subgroup in the finite Galois correspondence, but it is not normal in $S_3$, and its fixed cubic field is not normal over $\mathbb{Q}$.

The strict cubic fixed field in step 1.1 is therefore a counterexample to the universal normality claim.

The polynomials in [L1] and [L2] both have degree three and are separable and irreducible, but their Galois groups have orders $3$ and $6$.

Groups of different finite orders are not isomorphic, so the common polynomial degree does not determine the Galois group.

In [L1], the trivial subgroup is strictly contained in an order-two subgroup, but its fixed field is the entire biquadratic field and strictly contains the quadratic fixed field of the larger subgroup.

The subgroup containment in step 1.1 produces the reverse strict field containment, so it refutes inclusion preservation.