if you have an extension that is not Galois, can you compose it with another extension and make it Galois?

Question

I have $\mathbb Q (\sqrt[3]{5},\sqrt{2})$. I know that $\mathbb Q(\sqrt[3]{5})$ is not a Galois extension because although the splitting field of the polynomial $x^3 -5$ over $\mathbb Q$ is $\mathbb Q(\sqrt[3]{5})$, not all of the roots of this polynomial are in $\mathbb Q(\sqrt[3]{5})$. So if I have an extension that is not Galois, if I adjoin $\sqrt{2}$, can the result be Galois even though the first extension is not? Thanks for your time and consideration.

Answer 1

The splitting field of the polynomial $x^3 -5$ over $\mathbb Q$ is $\mathbb Q(\sqrt[3]{5})$.

It's true that $\mathbb Q(\sqrt[3]{5})$ is an extension of $\mathbb Q$ that contains one root of the polynomial $x^3 -5$. But $\mathbb Q(\sqrt[3]{5})$ is not a splitting field for $x^3 -5$ over $\mathbb Q$. A splitting field for $x^3 -5$ over $\mathbb Q$ is by definition the smallest extension of $\mathbb Q$ that contains all the roots of $x^3 -5$. The splitting field for $x^3 -5$ over $\mathbb Q$ is $\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3})$.

If I have an extension that is not Galois and I adjoin an additional element, can the resulting extension be Galois even though the original extension is not?

In general, yes. For example, $\mathbb Q(\sqrt[3]{5})$ is not a Galois extension over $\mathbb Q$, but if we adjoin the element $e^{2\pi i / 3}$, then we get $\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3})$, which is a Galois extension over $\mathbb Q$. $\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3})$ is Galois over $\mathbb Q$ because it is the splitting field of the polynomial $x^3 - 5$.

Is $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ a Galois extension over $\mathbb Q$?

No. If $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ were a Galois extension over $\mathbb Q$, then every irreducible polynomial $f(x) \in \mathbb Q[x]$ with a root in $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ would split completely in $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$. But $x^3 - 5$ is irreducible in $\mathbb Q[x]$ by Eisenstein's criterion with $p = 5$, and $x^3 - 5$ does not split completely in $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ since the roots $\sqrt[3]{5}e^{2\pi i / 3}$ and $\sqrt[3]{5}e^{-2\pi i / 3}$ are non-real whereas all elements in $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ are real.

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Edit: Addressing the additional questions you asked in the comments.

How would I find an extension of $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ that is Galois over $\mathbb Q$?

$\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3}, \sqrt{2})$ is a Galois extension of $\mathbb Q$ that contains $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$. $\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3}, \sqrt{2})$ is the splitting field for the polynomial $(x^3 - 5)(x^2 - 2)$ over $\mathbb Q$.

How would I find an extension of $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ that is Galois over $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$?

Again, $\mathbb Q(\sqrt[3]{5}, e^{2\pi i / 3}, \sqrt{2})$ works. This is the splitting field for the polynomial $x^3 - 1$ over $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$.