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I'm trying to find the minimal polynomial of $\alpha = \sqrt{\sqrt[3]{7}-5}$. Rather, I know that it will be $p(x) = x^6 +15x^4+75x^2+118$ by just squaring and then cubing appropriately, but I would like to verify that this is the minimal polynomial.

The usual method I use (if for example, we have a square root inside the square root) is to verify that $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$ by noting that $$ \mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{7}) \subset \mathbb{Q}(\alpha) $$ and then showing that $\sqrt[3]{7} - 5$ is not a square inside $\mathbb{Q}(\sqrt[3]{7})$. But I can't get the usual method of supposing it is a square and then arriving at a contradiction to work. I.e. writing $\sqrt[3]{7} - 5 = (a + b \sqrt[3]{7})^2$ and conjugating with $-\sqrt[3]{7} - 5 = (a - b \sqrt[3]{7})^2$ and then multiplying the two doesn't really help us (because we have a cube root instead of a square root). So my question is whether there is a simple way to do this kind of question when we have a cube root inside the main root?


Also, as a side question, I'd like to ask if we really have that, say $p(x)$, is a minimal polynomial for $\theta$ over $K$ a field is equivalent to $p(\theta)=0$ and $p(x)$ is irreducible? Or is this not enough for minimality? Do we also have to show that the degree is minimal? So if we could show that $p(x)$ in the question above is irreducible, would that be enough to show it's the minimal polynomial or would we still have to do some stuff for the degree?

Many thanks for any help.

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    $\begingroup$ Yes, the minimal polynomial must be irreducible (over the field of base). In particular if you can show that your sextic polynomial is irreducible over $\mathbb Q$ then you are done. $\endgroup$ – Piquito May 31 '18 at 11:39
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$\sqrt[3]7-5<0$ so $\Bbb Q(\sqrt{\sqrt[3]7-5})\not\subseteq\Bbb R$. Therefore $|\Bbb Q(\sqrt{\sqrt[3]7-5}):\Bbb Q(\sqrt[3]7)|=2$

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  • $\begingroup$ Thanks. What was the norm function you were using in the earlier version of you answer? $\endgroup$ – mathphys May 31 '18 at 11:31
  • $\begingroup$ Just the usual norm for an extension of fields, but in this example, it was overkill.... $\endgroup$ – Lord Shark the Unknown May 31 '18 at 11:48
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Putting $x^2=z$ the only real root, $z_0$, of $z^3+15z^2+75z+118=0$ is $z_0\approx-3.0871$ so $x^2\lt0$. Consequently $$x^6 +15x^4+75x^2+118=(x^2-z_0)(x^2-(a+bi))(x^2-(a-bi))\quad a,b\in\mathbb R$$ This is sufficient to show that your $p(x)$ is irreducible over $\mathbb Q$.

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