# Show that the extension $F \vert \mathbb{Q}$ is Galois

Let $E \subset \mathbb{R}$ a field containing $\mathbb{Q}$ such that the extension $E \vert \mathbb{Q}$ is Galois, and let $F := E(\sqrt{-1})$. Show that the extension $F \vert \mathbb{Q}$ is Galois.

Try:

To prove that the extension $F \vert \mathbb{Q}$ is Galois, we need to find a polynomial $f \in \mathbb{Q}[t]$ such that $F$ is a splitting field of $f$ over $\mathbb{Q}$. Consider

$$f(t):=t^2+1 \in \mathbb{Q}[t]$$

A splitting field of $f$ over $\mathbb{Q}$ is $\mathbb{Q}(i)$, and so is $F:= E(\sqrt{-1})$ since $\mathbb{Q} \subset E$. But since splitting fields are unique up to isomorphism, then

$$\mathbb{Q}(i) \simeq F$$

Now, since the extension $\mathbb{Q}(i) \vert \mathbb{Q}$ has degree $deg(f)=2$, it is Galois, so $F \vert \mathbb{Q}$ is Galois since its degree is $2$ as well.

There must be something in the argument above that is wrong, since I haven't used the fact that $E \vert \mathbb{Q}$ is Galois at all. So I'd like to know what and where went wrong. Also, if it is not too much to ask, some clue or hint that would point me into the right path to solve the problem. Thanks in advance!

• I do not think $E/\mathbb{Q}$ being galois is necessary... Look at $E=\mathbb{Q}(\sqrt[4]{2})$.. $E/\mathbb{Q}$ is not galois but $E(\sqrt{-1})/\mathbb{Q}$ is galois..
– user311526
Apr 28, 2016 at 17:19
• You said in your proof that $F/\mathbb{Q}$ has degree $2$, but that's clearly false. Apr 28, 2016 at 17:21
• @topgeomj That's true! So is my answer fine then? Apr 28, 2016 at 17:21
• Your answer is not fine as said by @CaptainLama... I just said $E$ being galois is not necessary..
– user311526
Apr 28, 2016 at 17:22
• The fact that $E/\mathbb{Q}$ is Galois is not necessary in the strong logical sense, but some hypothesis is necessary. For instance $E=\mathbb{Q}(\sqrt[7]{2})$ gives a counter-example if $E/\mathbb{Q}$ is not Galois. Apr 28, 2016 at 17:23

In general if $F$ is a field and $K/F$ and $L/F$ are Galois extensions contained in some bigger extension $M/F$, then the composite extension $KL/F$ is Galois. To see this, note that if $K/F$ and $L/F$ are Galois, then they are splitting fields of separable polynomials $f(x),g(x) \in F[x]$. Therefore $KL/F$ is the splitting field of the separable polynomial $f(x)g(x) \in F[x]$, and is thus Galois.

In your case, you have $E/\mathbb{Q}$ is Galois and $F$ is the composite of $E/\mathbb{Q}$ and $\mathbb{Q}(i)/\mathbb{Q}$, which is Galois. Therefore $F/\mathbb{Q}$ is Galois. So the hypothesis $E \subset \mathbb{R}$ is not necessary.

Suppose that $E$ is a splitting field of $p$, then $F$ is the splitting field of $p(x)(x^2+1)$, thus is normal and henceforth Galois since extension of $Q$ are separable.

• Does this have any thing to do with $2$? Is it still true if we add any other root of unity in general?
– user311526
Apr 28, 2016 at 18:25

$E$ is the splitting field over $\mathbb{Q}$ of some polynomial $f(t)$. Then the polynomial $g(t)=(t^2+1)f(t)$ splits in $F[t]$. Thus we just need to prove that $F=\mathbb{Q}(a_1,\dots,a_n,i,-i)$, where $a_1,\dots,a_n$ are the roots of $f(t)$ (and which are real by assumption).

Let $K=\mathbb{Q}(a_1,\dots,a_n,i,-i)\subseteq F$. Suppose $b\in F$; its minimal polynomial $h(t)$ over $E$ has degree $1$ or $2$. In the first case $b\in E=\mathbb{Q}(a_1,\dots,a_n)\subseteq K$. If the degree is $2$ then $b$ is not real, because otherwise it would belong to $E$, the fixed field under conjugation, which is the only non identical automorphism of $F$ over $E$.

Therefore $b=c_1+ic_2$, with $c_1,c_2\in E=\mathbb{Q}(a_1,\dots,a_n)$. Hence $b\in K$, as we wanted to show.

• Does this have any thing to do with $2$? Is it still true if we add any other root of unity in general?
– user311526
Apr 28, 2016 at 18:25
• @topgeomj The important fact is that $b\in F\setminus E$ must have a minimal polynomial with only complex roots. Apr 28, 2016 at 20:04