What is the concrete meaning of fixing an extension field through a subgroup of automorphisms in $x^3-2$?

Question

The automorphisms corresponding to the extension fields of the splitting polynomials of $x^3-2$ are enumerated in this answer:

we know that the automorphism of $\mathbb{Q}(\sqrt[3]{2}, \omega_3)$, which fix $\mathbb{Q}$ are determined by the action on $\sqrt[3]{2}$ and $\omega_3$, where $\omega_3$ is a third root of unity. It's trivial to conclude that such an automorphism sends $\sqrt[3]{2}$ to a root of $x^3 - 2$ and $\omega_3$ to a root of $x^2 + x + 1$. Making all possible combinations we get:

$$ e : \begin{array}{lr} \sqrt[3]{2} \mapsto \sqrt[3]{2}\\ \omega_3 \mapsto \omega_3 \end{array} \quad \quad r: \begin{array}{lr} \sqrt[3]{2} \mapsto \omega_3\sqrt[3]{2}\\ \omega_3 \mapsto \omega_3 \end{array} \quad \quad r^2: \begin{array}{lr} \sqrt[3]{2} \mapsto \omega_3^2\sqrt[3]{2}\\ \omega_3 \mapsto \omega_3 \end{array} $$ $$ f : \begin{array}{lr} \sqrt[3]{2} \mapsto \sqrt[3]{2}\\ \omega_3 \mapsto \omega_3^2 \end{array} \quad \quad fr: \begin{array}{lr} \sqrt[3]{2} \mapsto \omega_3^2\sqrt[3]{2}\\ \omega_3 \mapsto \omega_3^2 \end{array} \quad \quad fr^2: \begin{array}{lr} \sqrt[3]{2} \mapsto \omega_3\sqrt[3]{2}\\ \omega_3 \mapsto \omega_3^2 \end{array} $$

These are shown as symmetries in the dihedral group $D3$ ($\zeta,$ in place of $\omega,$ and $zeta^2$ on either side of the face of the triangle; $2^{1/3}$ elements at the vertices):

Now by checking which elements of the basis remain fixed by a subgroup of $ G,$ you can determine the corresponding fixed fields.

I understand that (composition from R to L):

• $r$ fixes $\omega$ and $\omega^2.$
• $f$ fixes $2^{1/3}.$
• $fr$ fixes $2^{1/3}(1+\omega).$
• $fr^2$ fixes $2^{2/3}(1+\omega).$

But I am still not sure what "fixing" means - How can you check that these automorphisms "fix" these subfields?

It would be possibly enlightening to see how $fr$ fixes $2^{1/3}(1+\omega),$ and $fr^2$ fixes $2^{2/3}(1+\omega).$ I don't quite see without a concrete example what is meant by "positions" for example, although I understand it has to do with permutations.

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Notes in relation to the accepted answer:

The expression of the fixed elements above, which was extracted from here, was reformulated using the minimal polynomial $x^2 + x + 1,$ which for a root of it, $\omega,$ obeys $\omega^2 + \omega + 1 =0;$ and hence, $\omega + 1 = -\omega^2.$ Therefore, $2^{1/3}(1+\omega)=-2^{1/3}\omega^2.$

the automorphisms detailed above can be summarized into:

$$\begin{align} &r: 2^{1/3} \mapsto \omega 2^{1/3}\\ &r: \omega \mapsto \omega \end{align} $$

$$\begin{align} f: &2^{1/3} \mapsto 2^{1/3}\\ f: & \omega \mapsto \omega^2 \end{align} $$

As shown in the answer, there is a mistake in the list of fixed points above. The correct correspondence is:

$$\begin{align} &\omega 2^{1/3} \text{ is a fixed point of } fr.\\ &\omega^2 2^{1/3} \text{ is a fixed point of } fr^2. \end{align}$$

Finally a reminder of of the rules to apply automorphisms to the elements of the basis following the rules:

$$\begin{align} \phi(a+b)&=\phi(a)+\phi(b)\\ \phi(ab)&=\phi(a)\phi(b)\\ \phi(z)&=z, \forall z\in F\text{ base field} \end{align}$$

So if we look into the rotations of $\omega_3,$

$$\begin{align} r(\omega_3)&=\omega_3\\ r^2(\omega_3)&=\omega_3 \end{align}$$

by definition of the automorphism. But also,

$$\begin{align} r(\omega_3^2)&=r(\omega_3)r(\omega_3)=\omega_3\omega_3=\omega_3^2\\ r^2(\omega_3^2)&=r(r(\omega_3^2))=r(\omega_3^2)=\omega_3^2 \end{align}$$

Hence, $r$ fixes $\omega_3$ and $\omega_3^2.$ This corresponds to $\mathbb Q(\omega_3).$

Answer 1

An element $z$ is said to be a fixed point of an automorphism $\sigma$, if $\sigma(z)=z$, and that's what you are looking for here.

But I think something is wrong with the data. Let's check the action of $fr$ on the number $z=2^{1/3}(1+\omega)=-2^{1/3}\omega^2$: $$ r(z)=-r(2^{1/3})r(\omega)^2=-\omega2^{1/3}\omega^2=-2^{1/3}, $$ and therefore $$ fr(z)=f(r(z))=f(-2^{1/3})=-2^{1/3}\neq z. $$ Meaning that $z$ is not a fixed point of $f\circ r$.

On the other hand $$ r^2(z)=-r(2^{1/3})=-\omega 2^{1/3}, $$ and therefore $$ fr^2(z)=f(r^2(z))=-f(\omega2^{1/3})=-f(\omega)f(2^{1/3})=-\omega^22^{1/3}=z. $$ So $z$ is a fixed point of the automorphism $fr^2$.

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The reason Mark Bennet (in a comment) and I asked about the order of composition is that in this Galois group we have the relation $rf=fr^2$. This Galois group is not abelian, so the order of composition matters. If automorphisms were applied left-to-right, then we would have $(z)rf=z$.

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There are many ways to find fixed points for $fr$.

1. In terms of Galois theory the fixed points are related to intermediate fields. By definition $f$ fixes all the elements of $\Bbb{Q}(\root3\of2)$. By the above calculation $fr^2$ fixes all the elements of $\Bbb{Q}(z)=\Bbb{Q}(\omega^2\root3\of2)$. It stands to reason that the we should see the intermediate field generated by the third root of $x^3-2$, namely $\Bbb{Q}(\omega\root3\of2)$ also. Indeed, $$r(\omega\root3\of2)=r(\omega)r(\root3\of2)=\omega^2\root3\of2$$ and hence $$f(r\omega\root3\of2)=f(\omega)^2f(\root3\of2)=\omega^4\root3\of2=\omega\root3\of2.$$ So $\omega\root3\of2$ is a fixed point of $fr$.
2. A general fact about group actions is that if $z$ is a fixed point of $g$ then $h(z)$ is a fixed point of $hgh^{-1}$: $$(hgh^{-1})(h(z))=h(g(h^{-1}(h(z))))=h(g(z))=h(z).$$ In this Galois group we can use the relation $rf=fr^2$ I mentioned above. It implies that $$rfr^{-1}=(rf)r^{-1}=(fr^2)r^{-1}=fr.$$ Applying the general observation to $g=f$, its fixed point $z=\root3\of2$, and $h=r$, it follows that $r(z)=\omega\root3\of2$ must be a fixed point of $rfr^{-1}=fr$.
3. Then there is the lower technology, boring way, but one that is guaranteed to work. You know the effect of the automorphism $fr$ to the elements of the basis $\mathcal{B}=\{1,\root3\of2,\root3\of4,\omega,\omega\root3\of2,\omega\root3\of4\}$. You can then write the matrix $M$ of $fr$ with respect to $\mathcal{B}$. The fixed points are exactly the eigenspace of the eigenvalue $\lambda=1$. Leaving the calculations to you. That eigenspace is 3-dimensional. Given that $(fr)^2=e$, the eigenvalues satisfy $\lambda^2=1$, so $\lambda=-1$ is the other eigenvalue. The corresponding eigenspace is also 3-dimensional. That is hardly a surprise given that if $w$ belongs to eigenvalue $-1$, then so does $zw$ for all the fixed points $z$.

Answer 2

For any set $X$, any function $F : X \to X$, and any $x \in X$, to say that $F$ fixes $x$ means that $F(x)=x$. The intuition behind this terminology is that $F$ moves the "positions" of the elements of $X$; elements whose position does not change are said to be "fixed".

In this case, $X = \mathbb Q(\sqrt[3]{2},\omega_3)$, and $F : X \to X$ is one of the various automorphisms discussed in your post. The element $x$ is any of the various roots, e.g. $F = r$ fixes $x = \omega$. Also the element $x$ is any rational number: every automorphism of $\mathbb Q(\sqrt[3]{2},\omega_3)$ fixes every element of the subfield $\mathbb Q$.