Okay, for an assignment I'm seeking to show that a field extension is Galois. However we never really went into detail on proving such things, at least with concrete examples, and I'm having trouble following along with what few examples I can find online. I've tried though, and here's what I've got.

Prove or disprove that $\Bbb Q(\sqrt 3,\sqrt{11})$ is Galois over $\Bbb Q$.

To show an extension is Galois, we need to show it is normal and separable. In this case, we're working in fields of characteristic $0$, so we have separability immediately. For normality, the extension must be algebraic (which it visibly is). Then it is sufficient to show that each irreducible polynomial in $\Bbb Q$ that has at least one root in $\Bbb Q(\sqrt 3,\sqrt{11})$ has all of its roots in there, by one of several definitions of normality.

But then the question became "how to do this"? Wikipedia did give a helpful idea: if we know the extension is separable (it is) and finite (it is), then it is normal if there exists a polynomial in the lower field, such that with its roots and the lower field as a whole we can generate the upper field.

In that light, we consider the (monic) polynomial

$$f(x) = \underbrace{(x^2 - 3)}_{min. \; poly. \\ for \; \sqrt 3}\underbrace{(x^2 - 11)}_{min. \; poly. \\ for \; \sqrt {11}}$$

Would $\Bbb Q(\sqrt 3,\sqrt{11})$ be the splitting field for this polynomial? And if so, is this sufficient to show the field in question is Galois over the rationals?

My main qualm with this - of course if there are other problems, point them out! - that comes to mind is that Wikipedia states it has to be irreducible over the lower field:

If $L$ is a finite extension of $K$ that is separable (for example, this is automatically satisfied if $K$ is finite or has characteristic zero) then the following property is also equivalent: There exists an irreducible polynomial whose roots, together with the elements of $K$, generate $L$. (One says that $L$ is the splitting field for the polynomial.)

My $f$ is obviously not irreducible - it can be factored into two nontrivial polynomials as shown. Yet a similar example for whether $\Bbb Q(\sqrt 2, \sqrt 3)$ is Galois over the rationals (link) itself uses a polynomial that is not irreducible. So what exactly gives there?

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    $\begingroup$ You are overthinking this. Normal extensions are splitting fields of polynomials. Your field is the splitting field of $f$, so is normal. As Wikipedia says, each normal and separable extension is a splitting field of an irreducible, but so what? You don't have to find such an irreducible polynomial to see your field is normal; reducible is fine. $\endgroup$ – Lord Shark the Unknown Apr 26 at 3:34
  • $\begingroup$ Hm, I see. Thanks. $\endgroup$ – Eevee Trainer Apr 26 at 3:42
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    $\begingroup$ That's correct. As far as I'm aware, the simplest examples of non-Galois extensions over $\mathbb Q$ are roots of degree 3 polynomials, such as $\mathbb Q(\sqrt[3]2)$ $\endgroup$ – Don Thousand Apr 26 at 3:58
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    $\begingroup$ I subscribe to the point of view of @LordSharktheUnknown, of course. Still, should you want to see what an irreducible polynomial looks like here, consider the minimal polynomial of $\sqrt{3} + \sqrt{11}$ over the rationals - in fact, it is not difficult to see that $\Bbb Q(\sqrt 3,\sqrt{11}) = \Bbb Q(\sqrt 3 + \sqrt{11})$. $\endgroup$ – Andreas Caranti Apr 26 at 12:31

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