Solvability by Radicals and Kummer Theory — Examples

The two $\mathbb{Q}$-embeddings of $K$ send $\sqrt{d}$ to $\pm \sqrt{d}$, so [L1] gives $N_{K\mathrm{/}\mathbb{Q}}(\alpha )=(a+b\sqrt{d})(a-b\sqrt{d})=a^2-d{b}^2,$ and ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(\alpha )=(a+b\sqrt{d})+(a-b\sqrt{d})=2a\mathrm{.}$

Multiplication by $\alpha$ sends $1\mapsto a+b\sqrt{d},\sqrt{d}\mapsto bd+a\sqrt{d},$ so its matrix in the basis $(1,\sqrt{d})$ is the displayed matrix. That matrix has determinant $a^2-d{b}^2$ and trace $2a$, agreeing with step 1.1 as [L2] predicts.

By [L1], the $n$ distinct ${\mathbb{F}}_q$-embeddings are $1,{\sigma }_q,\dots ,{\sigma }_q^{n-1}$.

Applying [L2] to the embeddings of step 1.1 gives $N_{{\mathbb{F}}_{q^n}\mathrm{/}{\mathbb{F}}_q}(\alpha )={\prod }_{i=0}^{n-1}{\sigma }_q^i(\alpha )={\prod }_{i=0}^{n-1}{\alpha }^{q^i}={\alpha }^{1+q+\cdots +q^{n-1}},$ and ${\mathrm{Tr}}_{{\mathbb{F}}_{q^n}\mathrm{/}{\mathbb{F}}_q}(\alpha )={\sum }_{i=0}^{n-1}{\sigma }_q^i(\alpha )=\alpha +{\alpha }^q+\cdots +{\alpha }^{q^{n-1}}\mathrm{.}$

If $t\in F$, then one could write $t=u(t^p)\mathrm{/}v(t^p)$ with coprime polynomials $u,v\in {\mathbb{F}}_p[s]$. Clearing denominators gives the polynomial identity $tv(t^p)-u(t^p)=0.$ But the first term has degree congruent to $1$ modulo $p$, while every term of $u(t^p)$ has degree divisible by $p$, impossible. Hence $t\mathrm{\notin }F$.

Since $t^p\in F$ and $t\mathrm{\notin }F$, the polynomial $x^p-t^p$ is irreducible over $F$ by [L2]. Therefore $K=F(t)$ has degree $p$ over $F$ by [L3], and it is purely inseparable because $t^p\in F$. The only $F$-embedding of $K$ into an algebraic closure is the inclusion.

Now [L4] gives $N_{K\mathrm{/}F}(t)=t^p,{\mathrm{Tr}}_{K\mathrm{/}F}(t)=pt=0,$ and the same embedding formula shows ${\mathrm{Tr}}_{K\mathrm{/}F}(x)=0$ for every $x\in K$.

If $(x,y)\in {\mathbb{Q}}^2$ satisfies $x^2+y^2=1$, then $\alpha =x+yi\in K^{\times }$ has norm one. By [L1], there is $\beta =u+vi\in K^{\times }$ with $x+yi=\frac{\beta }{\sigma (\beta )}=\frac{u+vi}{u-vi}=\frac{(u+vi{)}^2}{u^2+v^2}=\frac{u^2-v^2}{u^2+v^2}+\frac{2uv}{u^2+v^2}i\mathrm{.}$ If $u=0$ then $x=-1$, excluded in the statement. So with $t=v\mathrm{/}u\in \mathbb{Q}$ one gets $x=\frac{1-t^2}{1+t^2},y=\frac{2t}{1+t^2}\mathrm{.}$

Conversely, every rational $t$ gives ${\left(\frac{1-t^2}{1+t^2}\right)}^2+{\left(\frac{2t}{1+t^2}\right)}^2=1,$ so the displayed formula indeed parametrizes rational points on the unit circle. Clearing denominators gives the classical Pythagorean triple formulas.

The field $F\mathrm{/}\mathbb{Q}$ has degree $2$. If $2={\beta }^3$ for some $\beta \in F$, then taking norms from $F$ to $\mathbb{Q}$ gives $4=N_{F\mathrm{/}\mathbb{Q}}(2)=N_{F\mathrm{/}\mathbb{Q}}(\beta {)}^3,$ which is impossible because no rational cube equals $4$. So $2$ is not a cube in $F$. Hence $x^3-2$ has no root in $F$ and is irreducible there.

Since $\omega \in F$, all roots of $x^3-2$ are $\sqrt[3]{2},\omega \sqrt[3]{2},{\omega }^2\sqrt[3]{2},$ so they all lie in $F(\sqrt[3]{2})$. The irreducibility from step 1.1 and [L2] therefore make $F(\sqrt[3]{2})\mathrm{/}F$ cyclic of degree $3$.

The displayed cube roots satisfy $u^3+v^3=1,u^3{v}^3=1,$ so our choice $uv=1$ is compatible. Now $(u+v{)}^3=u^3+v^3+3uv(u+v)=1+3(u+v)\mathrm{.}$ Therefore $x^3-3x-1=(u+v{)}^3-3(u+v)-1=0.$

Put $x_0=u+v$, $x_1=\omega u+{\omega }^{2}v$, and $x_2={\omega }^{2}u+\omega v$, and let $\sigma$ cycle $x_0,x_1,x_2$. The definition [L1] gives $R_{\sigma ,\omega }(x_0)=x_0+{\omega }^2{x}_1+\omega x_2=3u,$ and $R_{\sigma ,{\omega }^2}(x_0)=x_0+\omega x_1+{\omega }^2{x}_2=3v\mathrm{.}$ Thus $u$ and $v$ are the two nontrivial resolvents divided by $3$, while step 1.1 is the load-bearing check that their symmetric combination is a genuine root of the cubic.

Here $a=0$, $b=-10$, $c=0$, and $d=1$, so [L1] gives $R_f(y)=y^3+10y^2-4y-40.$ Direct substitution shows that $y=2$ is a root, and polynomial division yields the displayed factorization.

Use the resolvent root $y=2$ to choose the pairing in [L2]. Seek a factorisation $f(x)=(x^2+ux+v)(x^2-ux+w)\mathrm{.}$ The paired products give $v+w=y=2$, while the constant term gives $vw=1$, so $v=w=1$. Comparing the $x^2$ coefficient gives $v+w-u^2=-10$, hence $u^2=12$. Taking $u=2\sqrt{3}$ yields $f(x)=(x^2+2\sqrt{3}x+1)(x^2-2\sqrt{3}x+1)\mathrm{.}$

Solving the two quadratic factors in step 2.1 gives $-\sqrt{3}\pm \sqrt{2}$ and $\sqrt{3}\pm \sqrt{2}$, which are exactly $\pm (\sqrt{2}+\sqrt{3}),\pm (\sqrt{2}-\sqrt{3})\mathrm{.}$ Thus the resolvent root leads to the four displayed quartic roots.

The five roots of $x^5-2$ are $\sqrt[5]{2},{\zeta }_5\sqrt[5]{2},{\zeta }_5^2\sqrt[5]{2},{\zeta }_5^3\sqrt[5]{2},{\zeta }_5^4\sqrt[5]{2}\mathrm{.}$ Therefore [L1] makes the splitting field a subfield of $\mathbb{Q}({\zeta }_5,\sqrt[5]{2})$.

The field $\mathbb{Q}({\zeta }_5,\sqrt[5]{2})$ is radical over $\mathbb{Q}$: first adjoin ${\zeta }_5$, a root of $x^5-1$, and then adjoin $\sqrt[5]{2}$, a root of $x^5-2$. Hence [F1] says that $x^5-2$ is solvable by radicals.

Suppose $t=b^p-b$ for some $b\in {\mathbb{F}}_p(t)$. Write $b=u\mathrm{/}v$ in lowest terms. At the pole $t=\mathrm{\infty }$, if $b$ has pole order $m>0$ then $b^p-b$ has pole order $pm$, a multiple of $p$; if $b$ has no pole there, neither does $b^p-b$. Both alternatives contradict the simple pole of $t$ at infinity. Therefore no such $b$ exists.

Let $\alpha$ be a root in an algebraic closure and let $m_{\alpha }$ have degree $d$ over $F$. Every conjugate of $\alpha$ is a root of $x^p-x-t$, hence has the form $\alpha +i$ with $i\in {\mathbb{F}}_p$ and therefore lies in $F(\alpha )$. The polynomial $m_{\alpha }$ thus splits in $F(\alpha )$, and it is separable because it divides a polynomial with derivative $-1$. By [L3], $F(\alpha )\mathrm{/}F$ is Galois. Its automorphisms inject into the additive group of ${\mathbb{F}}_p$ by $\sigma (\alpha )=\alpha +i$, so $d$ is $1$ or $p$. Step 1.1 excludes $d=1$, hence $d=p$ and $m_{\alpha }=x^p-x-t$. Thus this polynomial is irreducible.

Now [L2] makes $F(\alpha )\mathrm{/}F$ cyclic of degree $p$. Its full root set is $\alpha +i$ for $i\in {\mathbb{F}}_p$, since $(\alpha +i{)}^p-(\alpha +i)={\alpha }^p-\alpha =t$.

The field $F$ contains ${\mu }_3$, so adjoining cube roots is exactly the Kummer situation of [L1]. To show that the classes of $2$ and $3$ are independent modulo cubes, suppose $2^a{3}^b=c^3\text{in }{F}^{\times }\mathrm{.}$ Taking norms from $F$ to $\mathbb{Q}$ gives $2^{2a}{3}^{2b}=N_{F\mathrm{/}\mathbb{Q}}(c{)}^3\mathrm{.}$ The left side is a rational cube only when $3\mid 2a$ and $3\mid 2b$, hence only when $3\mid a$ and $3\mid b$. Therefore the classes of $2$ and $3$ each have order $3$ and generate a subgroup isomorphic to $(\mathbb{Z}\mathrm{/}3{)}^2$.

By [L1], the field generated by the corresponding cube roots is a Kummer extension with Galois group $\mathrm{Hom}((\mathbb{Z}\mathrm{/}3{)}^2,{\mu }_3)\cong (\mathbb{Z}\mathrm{/}3{)}^2\mathrm{.}$ Since $K$ is exactly that field, the stated conclusion follows.

Let $\alpha =\sqrt[5]{2}$, $E=\mathbb{Q}(\alpha )$, and $F=\mathbb{Q}({\zeta }_5)$. By [L2] with the prime $2$, $x^5-2$ is irreducible and $[E:\mathbb{Q}]=5$. By [L3], $F\mathrm{/}\mathbb{Q}$ is Galois of degree $4$. The degree of $E\cap F$ divides both $5$ and $4$ by [L5], so $E\cap F=\mathbb{Q}$. Hence [L4] gives $[K:\mathbb{Q}]=20$, $[K:F]=5$, and $[K:E]=4$.

The minimal polynomial of $\alpha$ over $F$ is therefore $x^5-2$. Since ${\zeta }_5\alpha$ is another root in $K$, [L6] gives an $F$-automorphism $\sigma$ of $K$ with $\sigma (\alpha )={\zeta }_5\alpha$ and $\sigma ({\zeta }_5)={\zeta }_5$; it has order $5$. Similarly, $[K:E]=4$ shows that the minimal polynomial of ${\zeta }_5$ over $E$ is the degree-four cyclotomic polynomial. Its root ${\zeta }_5^2$ therefore gives by [L6] an $E$-automorphism $\tau$ with $\tau (\alpha )=\alpha$ and $\tau ({\zeta }_5)={\zeta }_5^2$; it has order $4$.

These automorphisms do not commute: $\tau \sigma {\tau }^{-1}(\alpha )=\tau ({\zeta }_5\alpha )={\zeta }_5^2\alpha ={\sigma }^2(\alpha )\mathrm{\ne }\sigma (\alpha )\mathrm{.}$ So the Galois group of $x^5-2$ is nonabelian, even though [L1] makes the polynomial solvable by radicals. This disproves the claim.

The polynomial $x^5-2$ has degree five, so it is a quintic.

By [L1] it is solvable by radicals, so the statement that every quintic is insoluble by radicals is false.

By [L2], the product over the distinct embeddings of $t$ is just $t$ itself.

But [L1] and [L2] give $N_{{\mathbb{F}}_p(t)\mathrm{/}{\mathbb{F}}_p(t^p)}(t)=t^p,$ and $t^p\mathrm{\ne }t$ in the rational function field. Therefore the displayed claim is false.

By [L1], the trace map ${\mathrm{Tr}}_{{\mathbb{F}}_p(t)\mathrm{/}{\mathbb{F}}_p(t^p)}\text{\hspace{0.17em}⁣}:{\mathbb{F}}_p(t)\to {\mathbb{F}}_p(t^p)$ has image $\{0\}$.

The codomain is not the zero field, so this map is not surjective. Therefore the displayed claim is false.