# The Galois group is $\mathbb{Z_{4}}$ if and only if $\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\in\mathbb{Q}$

This question is from Lang's Algebra Chapter VI Exercise Q8

Let $f(x)=x^4+ax^2+b$ be an irreducible polynomial over $\mathbb{Q}$, with roots $\pm\alpha$, $\pm\beta$ and splitting field $K$.

I have shown that the Galois Group is either $\mathbb{Z_{4}}$ or $\mathbb{Z_{2}}\times\mathbb{Z_{2}}$ or $D_{8}$

The second part of this question asks me to show the Galois group is $\mathbb{Z_{4}}$ if and only if $\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\in\mathbb{Q}$

I have also shown that if the Galois group is $\mathbb{Z_{4}}$, then $\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\in\mathbb{Q}$. However, I don't know how to show the opposite.

I have tried that since there are only four possibilities of $\sigma(\alpha)$ and four possiblities for $\sigma(\beta)$, by $\sigma(\frac{\alpha}{\beta}-\frac{\beta}{\alpha})=\frac{\alpha}{\beta}-\frac{\beta}{\alpha}$ we can only have following three possibilities:

1) $\sigma(\alpha)=-\alpha$, $\sigma(\beta)=-\beta$

2) $\sigma(\alpha)=\beta$, $\sigma(\beta)=-\alpha$

3) $\sigma(\alpha)=-\beta$, $\sigma(\beta)=\alpha$

Clearly, the second case is what we want, but I don't know how to get rid of the case 1) and case 3).

There is a similar post here Galois group of a biquadratic quartic, but I don't know how to connect this post to my question.

Any hints or explanations are really appreciated!!

EDIT 1:

By the hints from Jyrki, I have shown that the case 1) and case 3) are actually the power of case 2), case 1) is the second power, and the case 3) is the third power.

Moroever, in case 2), the order of $\sigma$ is $4$ so that we have the cyclic group of order $4$

Then, I found I skipped one step here. I have not shown that the extension degree is four. I think, to show this, I have to show that $K(\alpha, \beta)=K(\alpha)$ so that since $f(x)$ is irreducible with $\alpha$ being a root, the degree is $4$. However, I have no idea how to show this right now.

Any ideas?

Thank you!

EDIT 2:

The answer from Jyrki is definitely right, the reason of this edition here is to add one more little point in the whole proof.

After all the arguments, we could only say that the Galois group has a cyclic subgroup, in our case, say $<\sigma_{2}>$ of order four, but it cannot conclude that the Galois group is $<\sigma_{2}>\cong \mathbb{Z_{4}}$. The reason behind is that the automorphism $\sigma_{2}$ was picked from the Galois group, and without knowing the degree of the splitting field over $\mathbb{Q}$, we cannot conclude the size of the Galois Group. In other words, if the splitting field is of degree $8$ over $\mathbb{Q}$, we could still get $<\sigma_{2}>$, but it is the subgroup of the Galois Group $\cong D_{8}$

Thus, the degree here is important. The splitting field is $\mathbb{Q}(\alpha,\beta)$, so the degree is $4$ if and only if $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)$, since if you think about the intermediate field $\mathbb{Q}(\alpha)$ between $\mathbb{Q}(\alpha,\beta)$ and $\mathbb{Q}$, the degree between $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha,\beta)$ can only be $1$ or $2$ (cannot be $3$ as the polynomial cannot only have one rational root, cannot be $4$ as otherwise the Galois group is of order $16$, which not divides $24=|S_{4}|$, but the Galois group should be a subgroup of $S_{4}$, as $f(x)$ irreducible).

Then, if the degree between $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha,\beta)$ is $2$, we get the Galois group to be $D_{8}$, if it is $1$, we get it to be $\mathbb{Z_{4}}$ or $\mathbb{Z_{2}\times Z_{2}}$. In current case it is $\mathbb{Z_{4}}$, but the point is that if the degree is $1$, $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)$ (The "only if" part is really similar).

Therefore, to prove the degree is $4$, we need to show $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)$. To show this, we need to show that $\alpha$ generates $\beta$.

First note that, since $\pm\alpha$, $\pm\beta$ are four roots of $f(x)=x^4+ax^2+b$. We could factor $f(x)$ into $f(x)=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)$, but it will finally give us $f(x)=x^{4}-(\alpha^{2}+\beta^{2})x^{2}+\alpha^{2} \beta^{2}$. Since $f(x)\in\mathbb{Q}[x]$, $-(\alpha^{2}+\beta^{2})=-\alpha^{2}-\beta^{2}\in\mathbb{Q}$. Thus, $-\alpha^{2}-\beta^{2}=d$ for some $d\in\mathbb{Q}$. Thus, $\beta^{2}=-d-\alpha^{2}$. Then, $\frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}=\frac{2\alpha^{2}+d}{\alpha \beta}$. Since $\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\in\mathbb{Q}$, then $\frac{2\alpha^{2}+d}{\alpha \beta}\in\mathbb{Q}$, and thus $\frac{2\alpha^{2}+d}{\alpha \beta}=c$ for some $c\in\mathbb{Q}$. Thus, $\frac{2\alpha^{2}+d}{c\alpha}=\beta$.

Thus, $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)$, since $f(x)$ is irreducible of order $4$ with $\alpha$ as a root over $\mathbb{Q}$, then $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)$ is of degree $4$ over $\mathbb{Q}$.

Now, we finish the whole proof.

• The Galois group acts on the set of zeros transitively, so we know that there exists an automorphism $\sigma$ such that $\sigma(\alpha)=\beta$. You can then deduce what $\sigma(\beta)$ must be. Then you need to show that no automorphisms other than powers of that particular $\sigma$ preserve that difference of ratios. – Jyrki Lahtonen Apr 11 '18 at 12:54
• @JyrkiLahtonen so you mean that for case 1) and 3), the Galois group does not act on the set of roots transitively? – JacobsonRadical Apr 11 '18 at 12:57
• No, I don't mean that. I want you to work with a particular automorphism with $\sigma(\alpha)=\beta$, and see where that leads. – Jyrki Lahtonen Apr 11 '18 at 12:58
• Then, how to show there is no other automorphism? @JyrkiLahtonen – JacobsonRadical Apr 11 '18 at 12:59
• @daruma my proof only shows that the Galois group contains a cyclic subgroup of order 4, so it still can be D8, so I cannot conclude the Galois group is C4 without knowing the degree – JacobsonRadical Apr 11 '18 at 14:54

## 1 Answer

Let $G$ be the Galois group $G=Gal(K/\Bbb{Q})$. We know that elements of $G$ permute the roots $\{\pm\alpha,\pm\beta\}$. Furthermore, 1) because $K=\Bbb{Q}(\alpha,\beta)$ knowing the images of $\alpha$ and $\beta$ determines an automorphism uniquely, 2) because $f(x)$ is known to be irreducible the permutation action is transitive.

• Transitivity implies that there exists an automorphism $\sigma\in G$ such that $\sigma(\alpha)=\beta$. Obviously then $\sigma(-\alpha)=-\beta$, so $\sigma(\beta)=\pm\alpha$.
• If $\sigma(\beta)=\alpha$, then, denoting $z=\alpha/\beta-\beta/\alpha$, $$\sigma(z)=\frac{\sigma(\alpha)}{\sigma(\beta)}-\frac{\sigma(\beta)}{\sigma(\alpha)}=\frac\beta\alpha-\frac\alpha\beta=-z.$$ Because $z\in\Bbb{Q}$ it must be fixed under $\sigma$. So $z=\sigma(z)=-z$ implying that $z=0$. But if $z=0$, then $\alpha^2=\beta^2$ and hence $\alpha=\pm\beta$ which is absurd.
• Therefore $\sigma(\beta)$ must be $-\alpha$ and consequently also $\sigma(-\beta)=\alpha$. Thus $\sigma$ permutes the roots of $f$ in a 4-cycle: $$\alpha\mapsto\beta\mapsto-\alpha\mapsto-\beta\mapsto\alpha.$$ Furthermore, we easily verify that this 4-cycle fixes $z$, so this fits the bill. We also conclude that $\sigma$ is of order four in $G$.
• If $\tau\in G$ is an automorphism that maps $\alpha$ to itself, we see that the composition $\sigma\tau$ also maps $\alpha$ to $\beta$. Repeating the earlier argument we see that $\sigma\tau$ must also act on the four zeros as the same 4-cycle. But, an automorphism in $G$ is uniquely determined by how it permutes the roots, so we can conclude that $\sigma\tau=\sigma$. Hence $\tau$ must be the identity.
• If $\tau\in G$ is arbitrary, then $\tau(\alpha)=\sigma^k(\alpha)$ for some choice of $k\in\{0,1,2,3\}$. This implies that $\sigma^{-k}\tau$ maps $\alpha$ to itself. By the previous bullet we can conclude that $\sigma^{-k}\tau=1_G$. Hence $\tau=\sigma^k$.
• We have shown that $G$ is generated by $\sigma$. Therefore $G\simeq C_4$. This, of course, also implies that $[K:\Bbb{Q}]=4$ and $K=\Bbb{Q}(\alpha)$.
• Thank you! I made some new edition, with respect to your last step. All others are definitely right! – JacobsonRadical Apr 12 '18 at 17:34
• @JacobsonRadical I solved that part by proving that the Galois group must have order four. That's what my three last bullets were all about! There can be only four automorphisms. Therefore the field extension has degree four also. You had it from the beginning yourself actually! – Jyrki Lahtonen Apr 12 '18 at 18:16
• Oh! I am sorry that i misunderstood your proof for the last three bullets. You are right. Thank you for all the help!!! – JacobsonRadical Apr 12 '18 at 20:06
• Thanks for answer! In point 4 you are writing that "$\sigma \tau$ must also act on the four zeros as the same $4$-cycle". But in order to conclude this you have to know how $\tau$ acts on $\beta$, but we only know that $\tau: \alpha \mapsto \alpha$. Could you explain this moment, please? – ZFR Apr 9 at 23:49
• Sorry, but I guess that this point is quite useless. If $\tau \in G$ such that $\tau(\alpha)=\alpha$ and since $z\in \mathbb{Q}$ then $\tau(z)=z$ and you can conclude that $\tau(\beta)=\beta$. General point: Any element of this Galois group which maps $\alpha$ to itself should be identity. – ZFR Apr 9 at 23:59