# Subgroups and corresponding subfields of galois group of $x^6 + x^3 + 1$

I think I have found the correct Galois group of $f = x^6 + x^3 + 1$ over $\mathbb{Q}$ to be $C_6$, the splitting field being $\mathbb{Q}(w)$ where $w$ is a complex 9th root of unity. Now I am trying to find the corresponding fixed subfields for the subgroups of $C_6$ but I am having difficulty.

I know they should be fields of order $3$ and $2$ corresponding to $C_3$ and $C_2$. I want match up automorphisms in the Galois group with elements in the subgroup and then see what the fixed field of these look like. However I'm not quite sure what the automorphisms of $\mathbb{Q}(w)$ look like, and therefore I am stuck.

Thank you for any help

Edit: With help from the comments I've managed to find the fixed subfield corresponding to $C_2$ by looking at the field $\mathbb{Q}(\omega + \omega^{-1})$ and showing that is a degree 3 extension and hence must correspond to $C_2$. But now I am having difficulty finding the subfields corresponding to $C_3$, in particular if we take the subgroup generated by $\sigma(\omega) = \omega^4$ I'm not sure what elements that are fixed by this look like?

• An automorphism is fully determined once you know where it sends $w$. The alternatives are $w,w^2,w^4,w^5,w^7$ and $w^8$. Can you identify the usual complex conjugation among them? Another hint: if an automorphism $\sigma$ has the proerty that $\sigma(w)/w$ is a third root of unity, then the order of $\sigma$ is... Nov 15, 2014 at 13:33
• It is important to say that $\omega$ is a primitive $9$th root of unity, because if you just say that it is a $9$th root of unity, you can get different fields depending on which precise root you take (maybe this is what you meant by a "complex" root, but you should avoid this terminology: $1$ is just as well an element of $\mathbf{C}$ as $e^{2i\pi/9}$). Nov 15, 2014 at 13:40
• @JyrkiLahtonen Thanks for your answer. So there is an automorphism corresponding to complex conjugation which is an order 2 subgroup of the Galois group (sending $w \to w^8), but then how do we determine what the fixed subfield of this is? Nov 15, 2014 at 14:40 • Wooster, what numbers are equal to their complex conjugates? Ok, you do want to describe the field. Hint: check that$w+w^8$is a real number. Observe that$w^8=w^{-1}$Can you show that$[\Bbb{Q}(w+w^{-1}):\Bbb{Q}]=3$? A hint for that would be to show that$K=\Bbb{Q}(w+w^{-1})$is not all of$L=\Bbb{Q}(w)$but$w$is a zero of a quadratic polynomial with coefficients in$K$, so$[L:K]=2$. Nov 15, 2014 at 22:07 ## 1 Answer As you already know, the elements of the Galois group are exactly:$\newcommand{\Q}{\mathbb{Q}} w \mapsto w, w \mapsto w^2, w \mapsto w^4, w \mapsto w^5, w \mapsto w^7, w \mapsto w^8$.$w \mapsto w^2$is a generator, and we have $$w \mapsto w^2 \mapsto w^4 \mapsto w^8 \mapsto w^7 \mapsto w^5 \mapsto w.$$ Thus we are looking for the fixed fields of$w \mapsto w^4$and of$w \mapsto w^8 = w^{-1}$. The fixed field of$\boldsymbol{\sigma: w \mapsto w^{-1}}$As you have observed,$\alpha = w + w^{-1}$is fixed by$\sigma$and satisfies $$\alpha^3 = w^3 + w^{-3} + w^1 + w^{-1} = (w^6 + 1)/(w^3) + \alpha = \alpha - 1$$ Since$\alpha^3 - \alpha + 1$is irreducible over$\Q$, the desired fixed field is$\Q(\alpha) = \Q(w + w^{-1})$. The fixed field of$\boldsymbol{\tau: w \mapsto w^4}\beta = w^3$satisfies$\beta^2 + \beta + 1$, which is irreducible over$\Q$of degree$2$. So the desired fixed field is$\Q(\beta) = \Q(w^3)$. A Remark The generators for a subfield can end up being really simple: in our case$w + w^{-1}$and$w^3$. But how do you find them? The first thing to do is try looking for a single term or two terms added together that is fixed under the automorphism in question. In our case,$w^3$is a single term so you would have found it quickly;$w + w^{-1}$would not take much longer. Alternatively, expecially if the first method fails, you can write an arbitrary element$z \in \Q(w) / \Q$as$z = a w^5 + b w^4 + c w^3 + d w^2 + e w + f$, manually set$\sigma(z) = z$and then solve for equations in$a, b, c, d, e, f$. You will use the minimal polynomial for$w$(or in general, the minimal polynomial for a generator of whatever extension you are considering) to simplify powers of$w$above$w^5$, and then equate each coefficient$w^0, w^1, \ldots, w^5$. This will give you a set of things fixed by the autorphism, and picking one such list of$a, b, c, d, e, f$chances are good you will get a generator of the desired field. A Second Remark This method should always work, though it could potentially be more tedious. Say we have an extension$F(\alpha) / F$of degree$n$with Galois group$G$. First, enumerate all the automorphisms of the extension by casework on where they send$\alpha$, and use them to classify the Galois group$G$. Then enumerate the subgroups, and for each subgroup$H$, look for elements$z \in F(\alpha)$that are fixed by$H$. Once you have an element$z$with$[F(z) : F] = [G : H] = n / |H|$, you know that$z$generates the corresponding extension. This works for any$H$;$H$need not be normal in$G$. For a Galois extension$K / F$with group$G$the size of a subgroup$H$always equals the degree$[K : E]$, where$E$is the fixed field of$H$. In other words, the number of automorphisms of$K / F$fixing$E$equals the degree of$K / E$. Equivalently,$\boldsymbol{K/E}$is Galois for any$H$. What normality of$H$corresponds to is that$\boldsymbol{E / F}$is Galois, i.e. that the number of automorphisms of$E / F$fixing$F$equals the degree$[E : F]$. In this case the Galois group of$E / F$is the quotient group$G / H$. A good example is the extension$K / \Q$, where$K$is the splitting field of$x^3 - 2$. If$\omega$is a cube root of unity, the subextensions are generated by$\omega, \sqrt[3]{2}$,$\omega \sqrt[3]{2}$, and$\omega^2 \sqrt[3]{2}$, and have degree 2, 3, 3, and 3, respectively over$\Q$. The Galois group is$S_3$and the corresponding subgroups are of index 2 (normal), index 3 (not normal), index 3 (not normal), and index 3 (not normal), respectively. So the index of the subgroup always equals the degree of the extension. Normality doesn't factor in unless we consider the automorphism group of a subextension rather than the subgroup fixing it. • Thank you for the detailed answer. Just one quick question, this would fail if the subgroups of$C_6$happened not to be normal right? Because then we wouldn't be able to use the fact that the degrees of the intermediate extensions should be$6$divided by the order of their Galois group? Dec 30, 2014 at 14:33 • @Wooster No, the degrees of the intermediate extensions are always equal to$6\$ divided by the order of their corresponding subgroup. I have edited a second remark into my answer explaining where I think you are confused. Let me know if that helps. Dec 30, 2014 at 20:35
• Yes that is very helpful. Thank you very much! I understand now Dec 30, 2014 at 20:38