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Old question: I am not sure whether this statement is even true, although it "feels" as if it should be. The statement is as stated in the title: $ X^q - 3 $ is irreducible over the splitting field $ L $ of $ X^q - 2 $ over $\mathbb{Q}$ for all $ q \geq 2 $, in other words, it is irreducible in $ L[X] $ where $ L = \mathbb{Q}(\zeta_q, 2^{1/q}) $. ($\zeta_q$ is a primitive qth root of unity.)

This follows if we can show that no number $ 3^{n/q}$ where $1 \leq n < q $ is in the field , so I tried to show that $ 3^{1/q} $ is not in $ L $, but I could not find a general method which works for all values of $ q $. Any ideas?

Edit: An answer by Lubin supplies the counterexample $ q = 12 $ which contains $ \sqrt{3} $, and therefore $ X^{12} - 3 $ splits into two factors of sixth degree (why didn't I think of that?). Weakening the statement, I would like a way to show that $ X^q - 3 $ has no roots in $ L $.

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  • $\begingroup$ The cleanest way to do this problem is perhaps Kummer theory. Have you covered this? $\endgroup$ – knsam Apr 20 '16 at 12:12
  • $\begingroup$ Certainly if $q$ is relatively prime to $3$. Then, the field $\Bbb Q(\zeta_q,2^{1/q})$ is unramified above $3$, but you have $q$ of ramification at $3$ when you adjoin $3^{1/q}$. To get insight, you might try $q=9$ and $q=27$. $\endgroup$ – Lubin Apr 20 '16 at 12:18
  • $\begingroup$ I know the statement of the theorem, although I am not sure how to apply it here. @knsam $\endgroup$ – Ege Erdil Apr 20 '16 at 12:47
  • $\begingroup$ For using Kummer Theory, you’d need to show that $3$ remains square-free in $L$, or some such thing that looks to me very difficult. $\endgroup$ – Lubin Apr 20 '16 at 17:49
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No, not irreducible.

For, $\Bbb Q(\zeta_{12})$ contains the fourth roots of unity, thus $i$, and the cube roots of unity, thus $\sqrt{-3}$, so that $\sqrt3\in\Bbb Q(\zeta_{12})$. And $X^{12}-3=(X^6-\sqrt3\,)(X^6+\sqrt3\,)$.

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  • $\begingroup$ Nice, I suppose the feeling that Kummer might be involved is only misleading. I just got back from an exam in Differential topology and began writing out various facts from Kummer to see what the situation is. (I'd also like to say that your posts here are very insightful, so thank you.) $\endgroup$ – knsam Apr 20 '16 at 19:55
  • $\begingroup$ It seems to me that when the base is not $\Bbb Q$ or some other field with unique factorization, Kummer is much harder to use. We know so little in general about the arithmetic in OP’s field $L$ that I for one wouldn’t know where to start when kummerizing. $\endgroup$ – Lubin Apr 20 '16 at 19:59
  • $\begingroup$ That's very true -- that's why it is a misleading suggestion, on my part. Thank you again. $\endgroup$ – knsam Apr 20 '16 at 20:07
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    $\begingroup$ I contest your claim of having misled, @knsam. We never know which route will take us to the end; reaching it, we think the successful route was obvious. It wasn’t. Specifically, I thought the conjecture was true, and was about to spend a lot of time trying to prove it. $\endgroup$ – Lubin Apr 21 '16 at 13:20
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Let k = ($\zeta_q$) and L = k($\sqrt[3] 2$), k/$\mathbf Q$ of degree m. We want to show that $X^q – 3$ has no root in L. Suppose there is such a root $\sqrt[q]3$ in L . Then k($\sqrt[q]3$) would be contained in k($\sqrt[3] 2$), and by Kummer theory, we would have a relation of the form $3 = 2^i$.$x^q$, with $x$ $\in$ k*. Norming down from k to $\mathbf Q $, we would get $b^q. 3^m = a^q.2^{im}$, where $a$ and $b$ are two coprime integers. This implies that 3 divides $a$ and not $b$, hence $3^q$ appears in the RHS and $3^m$ in the LHS; but $m = \phi (q)$ is strictly smaller than $q$ ($\neq 2$) : impossible . Needless to say, this argument needs the base field to be $\mathbf Q$ , or the fraction fild of a UFD .

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