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The title is the question. Neither cubic root $\sqrt[3]{2}$ not $\sqrt[3]{3}$ is constructible from $\mathbb Q$, because the minimal polynomials $X^3-2$ and $X^3-3$ are irreducible over $\mathbb Q$. I am not able to see just by computation whether $X^3-3$ is irreducible over $\mathbb Q(\sqrt[3]{2})$.

I guess that this is an easy question for experts in algebra but I am an analyst with very little knowledge of Galois theory so that I would most appreciate elementary answers.

Jochen
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  • From How to ask a good question: "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." – jjagmath Feb 03 '25 at 12:25
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    The question seems fine to me. The title and body seem sufficiently clear on their own. – Joshua P. Swanson Feb 03 '25 at 12:30
  • If $X^3-3$ were reducible over $\mathbb Q(\sqrt[3]2)$, it would have a root, i.e. $\sqrt[3]3\in\mathbb Q(\sqrt[3]2)$. – anankElpis Feb 03 '25 at 12:58
  • @anankElpis Yes, but how to show that $\sqrt[3]{3}\notin \mathbb Q(\sqrt[3]{2})$? – Jochen Feb 03 '25 at 13:05
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    See here. – anankElpis Feb 03 '25 at 13:15
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    Somehow the title and the asked question do not really cover the same information. For instance, $\sqrt[6]2$ is constructible from $a=\sqrt[3]2$ in geometric sense, it is in a tower of extensions of degree two from $\Bbb Q(a)$. (In fact it is exactly $\sqrt a$.) But the text translates constructibility as being in the field (not in the upper field of a tower of $2$-extensions). – dan_fulea Feb 03 '25 at 13:20
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    There seems to be no actual confusion regarding the question, despite the title using a somewhat unusual notion of "constructible". In any case, anankElpis' link obviously answers the question, so it's moot at this point. I'd like to note that Lubin's clever answer there is exactly the sort of "ad-hoc approach to handle the literal question" I was talking about in another comment. As far as I'm concerned it's better to simply ask the computer and move on unless you have an infinite family you're genuinely interested in. If so, it's time to look for another ad-hoc technique! – Joshua P. Swanson Feb 03 '25 at 13:25
  • I agree with Dan. Particularly in field theory constructible has a different meaning. Not the least because that is a classical application of the theory of field extensions, used for motivation. – Jyrki Lahtonen Feb 04 '25 at 05:09

1 Answers1

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No. Here's SageMath code to do the calculation:

R.<x> = PolynomialRing(QQ)
K = (x^3-2).splitting_field('x0')
S.<y> = PolynomialRing(K)
print((y^3-3).is_irreducible()) # returns True

You can run it here on the SageCell server.

Strictly speaking the above gives the slightly stronger result that $X^3-3$ is irreducible even over the splitting field of $X^3-2$, not just the "rupture field" $\mathbb{Q}(\sqrt[3]{2})$.

Joshua P. Swanson
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    SageMath is a cool tool -- but I still don't know how to prove the desired result. – Jochen Feb 03 '25 at 12:35
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    @Jochen This is a proof. It's an instance of factoring univariate polynomials over number fields, which is a messy algorithm in general, see e.g. section 2.7 in this report. There are many tricks to handle special cases, from Eisenstein's criterion on up. Here's an MO thread using some algebraic number theory which shows that your particular problem follows if $3$ is unramified in $\mathbb{Q}(\sqrt[3]{2})$. [...] – Joshua P. Swanson Feb 03 '25 at 12:55
  • [...] Unfortunately it doesn't help for you since the discriminant works out to be $-2^2 \cdot 3^3$. Undoubtedly there's some ad-hoc approach to handle the literal question you've asked in a human-readable way, but what would be the use? Do you really want an infinite sequence of statements? – Joshua P. Swanson Feb 03 '25 at 13:02
  • Okay, I am convinced. – Jochen Feb 03 '25 at 13:52