# The irreducibility of $X^q - 3$ in the field extension $\mathbb{Q}(\zeta_q, 2^{1/q})$

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 $L$.

• The cleanest way to do this problem is perhaps Kummer theory. Have you covered this? – knsam Apr 20 '16 at 12:12
• 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 $q=27$. – Lubin Apr 20 '16 at 12:18
• I know the statement of the theorem, although I am not sure how to apply it here. @knsam – Ege Erdil Apr 20 '16 at 12:47
• 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. – Lubin Apr 20 '16 at 17:49

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 $X^{12}-3=(X^6-\sqrt3\,)(X^6+\sqrt3\,)$.
• 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. – Lubin Apr 20 '16 at 19:59
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 .