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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.

Kenny Wong
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12_18
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  • You are being a bit inaccurate: the splitting field of $x^3-5$ is not $\mathbb{Q}({\root 3 \of 5})$... But, yes, the "Galois closure" of a given (not-Galois) field extension is certainly obtained by adjoining a few further things. :) – paul garrett Apr 24 '23 at 21:15
  • Can you give an example element of Q cube root 5 root 2? Is it arbitrary products and sums of integer numbers of 5^1/3 and 2^1/2? What do you mean exactly by that, and whether or not something "Is Galois". Galois is a mathematician who has created many concepts – Snared Apr 24 '23 at 21:16
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    @Snared, well, a field extension "being Galois" is a pretty standard notion, as silly as the syntax and/or semantics may be... :) – paul garrett Apr 24 '23 at 21:18
  • Been awhile since I looked at this but I believe so, since the rationals have char 0, the splitting field should also be separable, and your extension is an intermediate field (that is, the extension of your field to the s.f. for that polynomial should be galois over the rationals. I don't know about adjoining the square root of two as that is not a root of the equation. You likely need to adjoin a cubic root of unity. – David Reed Apr 24 '23 at 21:23
  • So how would I find a galois extension of Q(3root5, root2)? I have a feeling it's galois group would be of degree 4 but i am unsure how to show this. – 12_18 Apr 24 '23 at 21:24
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    Does this answer your question? Separability and normal closure – Anne Bauval Apr 24 '23 at 21:25
  • You need to find an irreducible polynomial (over the rationals) that has both the square root of two and the cubed root of five as roots. The splitting field of this polynomial will then be galois over the rationals. – David Reed Apr 24 '23 at 21:29

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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.

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 $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$.

Kenny Wong
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  • I don't think I am asking my question correctly. The question is to see if Q(3root5,root2) is a Galois extension of Q. If it is a Galois extension, then I must find the Galois group for Q(3root5,root2) over Q. – 12_18 Apr 24 '23 at 21:37
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    Okay. And the answer is that $\mathbb Q(\sqrt[3]{5}, \sqrt{2})$ is not a Galois extension over $\mathbb Q$. – Kenny Wong Apr 24 '23 at 21:38
  • Thank you, that is what I was thinking. – 12_18 Apr 24 '23 at 21:42