$L= \mathbb{Q}( \sqrt 2, \sqrt 3)$ is normal extension of $\mathbb{Q}$ or not?

Question

$L= \mathbb{Q}( \sqrt 2, \sqrt 3)$ is normal extension of $\mathbb{Q}$ or not?

I noted that $L= \mathbb{Q}(\sqrt 2+\sqrt 3)$ using tower law. $[L:\mathbb{Q}]=4\ne 2$ so can't conclude from 'quadratic extensions are normal'. Suppose that the extension is normal. Then Irr($\sqrt 2+\sqrt 3)=p(x)$ must split in $L$. $p(x)=(x^2-2)^2-24$. So L must contain splitting field of $p$ over $\mathbb{Q}$, which is just $\mathbb{Q}(\sqrt 2+\sqrt 3)$. But this does not give me a contradiction.

So it seems that the extension is normal. To prove this: Take $a\in L$. Consider $q(x)= Irr (a, \mathbb{Q})$. It must be shown that $q$ splits in $L$. But I am not sure how to show this.

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Edit: (added by Kenny, see comments).

The OP is using this definition for a normal extension.

An algebraic extension $K/F$ is normal iff for every $k$ in $K$, $irr (k, F)$ splits in $K$.

This is equivalent to the following:

For every irreducible polynomial $p(x)$ in $F[x]$ with a root in $K$, $p(x)$ splits in $K$.

At the present moment, this is the only definition the OP is aware of. The OP is specifically looking for an argument that starts from this definition.

Answer 1

As discussed in the comments, there are several equivalent definitions of what it means for a finite field extension $L : F$ to be normal.

Definition 1. The field extension $L : F$ is normal iff $L$ is the splitting field for some polynomial $f(x) \in F[x]$

Definition 2. The field extension $L : F$ is normal iff every irreducible polynomial $p(x) \in F[x]$ with at least one root in $L$ splits completely in $L$.

As I understand it, you are familiar only with Definition 2.

However, the problem is much easier to solve if we work with Definition 1. Indeed, as @TheSilverDoe points out, your $L$ is the splitting field of the polynomial $(x^2 - 2)(x^2 - 3) \in \mathbb Q[x]$.

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So how do we prove that Definition 1 implies Definition 2?

Let's suppose that $L$ is the splitting field of some $f(x) \in F[x]$. Let $p(x) \in F[x]$ be an irreducible polynomial with a root $a \in L$. We want to prove that $p(x)$ splits completely over $L$.

I think the cleanest approach is to start by constructing a field $K$ such that (i) $L \subset K$ and (ii) $p(x)$ splits completely in $K$. (For example, we could take $K$ to be the splitting field of $p(x)$ over $L$.)

Let $b$ be some root of $p(x)$ in $K$. Our goal is to prove that $b$ is in fact in $L$.

At this point, we should draw a diagram, showing various important subfields of $K$ and the relationships between them.

Now we can make a few observations.

Observation 1: $[F(a) : F] = [F(b) : F]$.

To see this, note that $a$ and $b$ are both roots of the irreducible polynomial $p(x) \in F[x]$.

Therefore, there exists an isomorphism $g : F(a) \to F(b)$ that sends $a \mapsto b$ and fixes elements of $F$.

The claim that $[F(a) : F] = [F(b) : F]$ follows.

Observation 2: $[L(a) : F(a)] = [L(b) : F(b)]$.

To see this, observe that $L(a)$ is the splitting field of $f(x)$ over $F(a)$, and $L(b)$ is the splitting field of $f(x)$ over $F(b)$.

The isomorphism $g : F(a) \to F(b)$ induces an isomorphism $\hat g : F(a)[x] \to F(b)[x]$. Since $g$ fixes elements of $F$, and the polynomial $f$ has all of its coefficients in $F$, we have that $\hat g(f) = f$. Thus:

• $g: F(a) \to F(b)$ is an isomorphism.
• $L(a)$ is the splitting field of $f$ over $F(a)$.
• $L(b)$ is the splitting field of $\hat g(f)$ over $F(b)$.

Therefore, by the uniqueness theorem for splitting fields, $g$ extends to an isomorphism $\widetilde g : L(a) \to L(b)$. The claim that $[L(a) : F(a)] = [L(b) : F(b)]$ follows immediately.

Observation 3: $[L(a) : L] = 1$.

This is trivial, since $a$ is in $L$ by assumption.

Anyway, using our three observations together with the tower law, we can see that \begin{multline} [L(b) : L] = \frac{[L(b) : F]}{[L:F]} = \frac{[L(b) : F(b)] [F(b) : F]}{[L:F]} \\ = \frac{[L(a) : F(a)] [F(a) : F]}{[L:F]} = \frac{[L(a) : F]}{[L:F]} = [L(a) : L] = 1.\end{multline}

This proves that $b$ is in $L$.

Answer 2

Here is a direct way, using your definition only.

Let $\alpha\in L$, and let $f=Irr(\alpha,\mathbb{Q})$.

Write $\alpha=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$, where $a,b,c,d\in\mathbb{Q}$.

Set $g=(X-(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}))(X-(a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}))(X-(a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}))(X-(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}))$.

Expanding everything, one may see that $g\in\mathbb{Q}[X]$.

Since $g(\alpha)=0$, $f\mid g$. But $g$ splits in $L$, and thus so does $f$. QED.