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While thinking about this question I noticed that if $\alpha$ is a root of $f(x) = x^9 + 3 x^6 + 165 x^3 + 1$ then $$\left(\frac{\alpha^3+1}{3 \alpha}\right)^3 = -6$$ so $\mathbb{Q}(\alpha)$ contains a cube root of $6$. This shows that $3 \mid [\mathbb{Q}(\alpha) : \mathbb{Q}]$. Moreover, $f(x)$ factors over $\mathbb{Q}(\zeta, \sqrt[3]6)$ as $$f(x) = (x^3 + 3 \sqrt[3]6\,x + 1)(x^3 + 3 \zeta \sqrt[3]6\,x + 1)(x^3 + 3 \zeta^{-1} \sqrt[3]6\,x + 1)$$ where $\zeta$ is a primitive third root of unity. Is there a "simple" way to use these facts to show that in fact $f(x)$ must be irreducible over $\mathbb{Q}$, e.g. by relating this to some finite field that contains a cube root of $6$?

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  • $\begingroup$ Do we really need to go through finite field? I think the factorization you have is very good! I think we can consider the tower of fields $\mathbb{Q}(\sqrt[3]{6})$, $\mathbb{Q}(\sqrt[3]{6},\zeta)$, $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\alpha,\zeta)$ $\endgroup$ – i707107 Jul 20 '18 at 5:24
  • $\begingroup$ Great stuff with the factorization. I just noticed that $\alpha=\root3\of2-\root3\of3$ is a zero of your first cubic factor. Looks like the nonic is the minimal polynomial of this difference of cube roots! Need to think whether that leads to a simple irreducibility proof? $\endgroup$ – Jyrki Lahtonen Jul 30 '18 at 21:11
  • $\begingroup$ Anyway, looks like $L=\Bbb{Q}(\root3\of2,\root3\of3,\zeta)$ is the degree 18 splitting field of $f(x)$ over $\Bbb{Q}$. $\endgroup$ – Jyrki Lahtonen Jul 30 '18 at 21:16
  • $\begingroup$ Basically "all" we need to do is to show that $x^3-3$ is irreducible over $\Bbb{Q}(\root3\of2)$. Then we are done. The usual proof of the existence of a primitive element tells us that $\alpha$ generates $\Bbb{Q}(\root3\of2,\root3\of3)$. Therefore its minimal polynomial has degree nine. $\endgroup$ – Jyrki Lahtonen Jul 30 '18 at 21:21
  • $\begingroup$ In other words, this or this will settle the question. Approach0 gives more related threads. $\endgroup$ – Jyrki Lahtonen Jul 31 '18 at 5:20
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It suffices to show that $g(x)=x^3 + 3 \sqrt[3]6\,x + 1$ is irreducible over $K=\mathbb{Q}(\sqrt[3]6)$ because $$[\mathbb{Q}(\alpha):K]=3 \implies [\mathbb{Q}(\alpha):\mathbb{Q}] = 9$$ There are general algorithms to factor polynomials over number fields. However, I present two ad hoc ways for this particular polynomial.


First method

If $g$ is reducible, then it has a root in $K$. The only real root of $g$ is $\gamma= -0.182329$, this number is an algebraic unit. Hence $-1/\gamma = 5.48461$ is also an algebraic unit.

Let $u>1$ be the fundamental unit of a real cubic field with one real embedding, then $$u^3 > \frac{|\delta|-27}{4}$$ where $\delta$ is the discriminant of the field.

An instructive proof is given in Number Fields by Daniel Marcus (chapter 5). Introductory algebraic number theory by Alaca (chapter 13) contains an excessively verbose proof.

Now apply this proposition to ring of integer of $K$, with $\delta = -972$, we found its fundamental unit is $>6.1819$. Therefore $-1/\gamma \notin K$ so $\gamma \notin K$.


Second method

Replace $x$ by $x/\sqrt[3]6$ in $g(x)$ gives $x^3+18x+6$, irreducible over $\mathbb{Q}$. Denote a real root of $x^3+18x+6$ by $\gamma$. It suffices to show $\gamma \notin K$. Assume $\gamma \in K$, then $\mathbb{Q}(\sqrt[3]6) = \mathbb{Q}(\gamma)$. Both $$\{1,\sqrt[3]6, \sqrt[3]6^2\} \qquad \{1,\gamma, \gamma^2\}$$ are basis of $K$ over $\mathbb{Q}$. Let $\text{tr}$ denote the trace of $K$ over $\mathbb{Q}$. There exists $a,b,c\in \mathbb{Q}$ such that $$\gamma = a + b\sqrt[3]6+c\sqrt[3]6^2$$ Taking trace both sides gives $$0=\text{tr}(\gamma) = 3a \implies a=0$$ So $$\gamma^2 = b^2\sqrt[3]6^2+c^2\sqrt[3]6^4+12bc$$, taking trace again gives $$\tag{1} -36=\text{tr}(\gamma^2) = 36bc \implies bc=-1$$ Consider $\gamma^3$ gives $$\tag{2}-18=\text{tr}(\gamma^3)=18(b^3+6c^3) \implies b^3+6c^3=-1$$ It is easy to see that $(1),(2)$ have no rational solutions, this contradiction shows $\gamma \notin K$.

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  • $\begingroup$ Isn’t $\gamma/b = \sqrt[3]6 -\sqrt[3]6^2/b^2$? $\endgroup$ – WimC Jul 20 '18 at 12:06
  • $\begingroup$ @WimC Yes, I corrected it. $\endgroup$ – pisco Jul 20 '18 at 15:42
  • $\begingroup$ (+1) Second method is one of my favorite argument. $\endgroup$ – i707107 Jul 20 '18 at 18:42
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    $\begingroup$ Nice. Not unlike the answers here. Those became relevant after I realized that $\root3\of2-\root3\of3$ is a zero of the first cubic in WimC's factorization. All the nine zeros are $\zeta^k\root3\of2-\zeta^\ell\root3\of3$. I don't know whether I should be embarrassed or not about missing this earlier :-) $\endgroup$ – Jyrki Lahtonen Jul 31 '18 at 5:24
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This answer elaborates on the observations made by @JyrkiLahtonen in the comments. Let $\alpha$ be a root of $x^9+3x^6+165x^3+1$ then $$\tag1 (\alpha^3+1)^3+6(3\alpha)^3=0$$ and $$\tag2 3(\alpha^3+7)^3+2(\alpha^3-8)^3=0.$$ Equation $(1)$ is already mentioned in the question and $(2)$ was found by observing that $$(1-b)(\alpha^3+a)^3+(a-1)(\alpha^3+b)^3$$ is linear in $\alpha^3$ for any $a,b$ and then picking suitable values. From this it follows that $\mathbb{Q}(\alpha)$ contains both a cube root of $3$ and a cube root of $2$. As mentioned in the comments, this suffices to show that $\alpha$ has degree $9$ over $\mathbb{Q}$.

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  • $\begingroup$ Nice that you worked out how to get the individual cube roots with $\alpha$. $\endgroup$ – Jyrki Lahtonen Jul 31 '18 at 20:15

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