Let $K$ be the splitting field of $f(x)=x^3+2x-4$ over $\mathbb Q$. List all fields $L$ with $\mathbb{Q} \subset L \subset K$ for which $L$ is a Galois extension of $\mathbb{Q}$.

Things I know:

Since this is about intermediate fields, we can try to apply the Fundamental Theorem of Galois theory.

EDIT: Note that we can compute that the Galois group $Gal(K/ \mathbb{Q})$ is $S_3$

So the corresponding Galois groups for $L$ are the normal subgroups of $S_3$

From here I'm not sure how to continue to find $L$.

Any help will be appreciated!


2 Answers 2


The polynomial is primitive ($\gcd$ of coefficients is $1$ as a polynomial over $\Bbb{Z}[x]$) and has no integer roots (it is easy to prove by elementary calculus or by drawing a graph). So it is irreducible in $\Bbb{Z}[x]$ and hence also in $\Bbb{Q}[x]$ due to Gauss Lemma.

Also from elementary calculus, it is obvious that it has a unique real root (function is monotonically increasing). Thus it has a unique real root and two complex roots which occur as conjugates.

Let $\alpha$ be it's unique real root.

So we have $[\Bbb{Q}(\alpha):\Bbb{Q}]=3$ and over $\Bbb{Q}(\alpha)$ , the polynomial factors into $(x-\alpha)(x-z)(x-\bar{z})$, where $z$ and $\bar{z}$ are the complex roots.

It is obvious that over $\Bbb{Q}(\alpha)$ the polynomial $(x-z)(x-\bar{z})$ is irreducible and hence $[\Bbb{Q}(\alpha,z):\Bbb{Q}(\alpha)]=2$. And since quadratic extensions of $\Bbb{Q}$ are always Galois (alternatively you can prove that any quadratic extension of $\Bbb{Q}$ is a splitting field of a separable polynomial), you have that over $\Bbb{Q}(\alpha,z)$ our polynomial splits and it is the splitting field as it is the smallest field containing $\alpha,z$.

Thus we have that $|\text{Gal}( \Bbb{Q}(\alpha,z)/\Bbb{Q})|=[\Bbb{Q}(\alpha,z):\Bbb{Q}]=3\cdot 2 = 6$ due to the fact that the compositum of two extensions of degrees $m,n$ such that $\gcd(m,n)=1$ has degree $mn$.

Now it remains to prove that this group is indeed non-abelian. Take the elements $\sigma,\tau\in\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})$ such that $\sigma=\begin{cases}\alpha\mapsto z\\ z\mapsto \bar{z}\\\bar{z}\mapsto \alpha\end{cases}$ and $\tau=\begin{cases} z\mapsto \bar{z}\\ \bar{z}\mapsto z\\ \alpha\mapsto \alpha\end{cases}$ . Then we can say that $\sigma\cdot\tau\neq \tau\cdot\sigma $ which proves that the Galois group is non-abelian and as it is of order $6$ it must be $S_{3}$.

Now the only proper normal subgroup of $S_{3}$ is the unique subgroup of order $3$. Here it is seen that $\sigma\in\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})$ does have order $3$. and hence the only such Galois extension $L$ of $\Bbb{Q}$ is the fixed field of $\langle \sigma\rangle $ and it is such that $[L:\Bbb{Q}]=\bigg|\frac{\text{Gal}(\Bbb{Q}(\alpha,z)/\Bbb{Q})}{\langle \sigma\rangle}\bigg|=\frac{6}{3}=2$.

So $L=K^{\langle \sigma\rangle}$. That is the fixed field of $\sigma$.

More explicitly $K=\Bbb{Q}(\alpha,\sqrt{D})$ where $D$ is the discriminant of the polynomial. Look at Dummit and Foote page $613$ for reference. And hence $L=\Bbb{Q}(\sqrt{D})$ which is the unique degree $2$ extension of $\Bbb{Q}$ contained in $K$.

A short explanation regarding the above. We have $\displaystyle \prod_{i<j} (\alpha_{i}-\alpha_{j})^{2}=D $ and define $\sqrt{D}= \prod_{i<j} (\alpha_{i}-\alpha_{j})$ where $\alpha_{i}$'s are roots of the polynomial . Note that $\sqrt{D}$ is just a symbol for the expression. For $\sigma\in S_{n}$ you have $\sigma(\sqrt{D})=\text{sgn}(\sigma)\sqrt{D}$ where $\text{sgn}$ is the signum of the permutation.

Hence it is clear from above that $\sigma\in A_{n}\subset S_{n}$ if and only if $\sigma(\sqrt{D})=\sqrt{D}$. Thus in our above case $A_{3}$ will fix $\Bbb{Q}(\sqrt{D})$ and it will be the unique quadratic extension of $\Bbb{Q}$ properly contained in $K$ .

Now as to how to compute the discriminant, you should take a look at page $613$ of Dummit and Foote.

  • $\begingroup$ Thank you very much! Just a minor question: does it have to be $\sigma$ here? Can we choose for example $\tau$, and do the same thing and say that $L=K^{<\tau>}$? $\endgroup$
    – Korn
    Commented Jun 24, 2022 at 18:25
  • $\begingroup$ $K^{\langle\tau\rangle}$ will be an extension of degree $3$ and it will not be Galois over $\Bbb{Q}$ as $\langle\tau\rangle$ of order $2$ is not a normal subgroup of $S_{3}$. But yeah sure. It will be an "extension" of $\Bbb{Q}$ contained inside $K$ . Just that it is not Galois. $\endgroup$ Commented Jun 24, 2022 at 18:33

The Galois group of your polynomial is not $\mathbb{Z}_{2}$, it's the symmetric group of order 6, $\mathbb{S}_3$. This is because the polynomial is irreducible in $\mathbb{Q}$ since it has no roots in the field and has degree $3$. Also the discriminant is $-464$, whose square root is not an element of the field.

Now if you want to know the list of subfields $L$ such $L|\mathbb{Q}$ is Galois, under the fundamental Theorem of Galois theory you just need to find the normal subgroups of the symmetric group of order 6

  • $\begingroup$ Oh right! Thank you for your help! Just one more thing, in this case if we find the normal subgroups of $S_3$, we will find the corresponding Galois groups for $L$. But how do we find what $L$ is? $\endgroup$
    – Korn
    Commented Jun 23, 2022 at 18:31
  • $\begingroup$ That's more difficult, since you dont know who generates the extension. If you can find the generators ( maybe in the undeterminate roots, knowing that one is real and the others are complex conjugated), then the extension is generated by the elements that get fixed by the automorphisms that generate the normal subgroup $\endgroup$ Commented Jun 23, 2022 at 18:34
  • 1
    $\begingroup$ The only normal subgroup of $S_3$ is $A_3$, and famously the fixed field of the alternating group is generated by the square root of the discriminant. So all you have to do is calculate the discriminant of this polynomial. $\endgroup$ Commented Jun 23, 2022 at 18:45
  • $\begingroup$ Didn't know that fact, where can i find a proof? $\endgroup$ Commented Jun 23, 2022 at 18:46
  • $\begingroup$ @Guillerminho77 It is in Dummit and Foote I believe . But I have to check it again. $\endgroup$ Commented Jun 23, 2022 at 18:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .