# Proving that a field extension is Galois

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?

• 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. – Lord Shark the Unknown Apr 26 at 3:34
• Hm, I see. Thanks. – Eevee Trainer Apr 26 at 3:42
• 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 $\mathbb Q(\sqrt[3]2)$ – Don Thousand Apr 26 at 3:58
• 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 $\Bbb Q(\sqrt 3,\sqrt{11}) = \Bbb Q(\sqrt 3 + \sqrt{11})$. – Andreas Caranti Apr 26 at 12:31