abstract algebra - Show that field extension of the rationals by distinct square roots of primes is Galois

Let $p_1, p_2, ..., p_n$ be distinct primes. Prove $\mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, ..., \sqrt{p_n})$ is Galois.

I think that we need to show it is normal and separable.

For normal, we can say that $\mathbb{Q}$ is a splitting field for the polynomial $f= (x^2-p_1)(x^2-p_2)...(x^2-p_n)$ which is obviously made up of factors that are irreducible over $\mathbb{Q}$

For separable, I don't know. Maybe something to do with Einstein's criterion

Answer

As noted in the comments, this follows because $\mathbb{Q}$ is a field of characteristic $0$. Because of this, any irreducible polynomial must have distinct roots. You may have recall seeing this in the form of a result that a polynomial has multiple roots (over a field of characteristic $0$) if and only if $f(x)$ and $f'(x)$ have a common zero, i.e. have a common factor.

Now each of the $x^2-p_i$ are irreducible over $\mathbb{Q}$, a field of characteristic $0$. Therefore, they have no repeated roots. You can also note that you know the roots of each polynomial, namely $\pm \sqrt{p_i}$, which are distinct. One could also use the fact that you know the degree of the field extension is $2^n$, where $n$ is the number of distinct primes used. Any automorphism is determined by how it acts on the $\sqrt{p_i}$'s. The map $\sqrt{p_i} \mapsto -\sqrt{p_i}$ and fixing all the other $\sqrt{p_j}$ for $j \neq i$ is an automorphism, this is routine to verify. Then you have $2^n$ possible automorphisms, the same as the degree of the extension.

Note: if you are confused why $\mathbb{Q}$ has characteristic $0$, note that $1+1+\cdots$ is never $0$ in $\mathbb{Q}$, so that $\mathbb{Q}$ has characteristic $0$ by definition.