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Hello I am a novice in fields and was asked this question in an assignment: I need to find the minimal polynomial of the expression

$\sqrt[3]{7-\sqrt{2}}$ over the rationals Q Here is where I am stuck: I think that $\sqrt[3]{7-\sqrt{2}}$ is not in the field

Q($ 7 - \sqrt{2} $) but how to show this formally?

Thank you to all helpers

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First, let's note that the field $\mathbb{Q}(7 - \sqrt{2})$ is the same as the field $\mathbb{Q}(\sqrt{2})$ which is maybe a little easier to write. So an element of that field is of the form $\alpha = a + b\sqrt{2}$ where $a, b\in \mathbb{Q}$. We will define the "norm" of such an element $\alpha$ to be $\mathbf{N}(\alpha) = a^2 - 2b^2$. Note that for $\alpha, \beta \in \mathbb{Q}(\sqrt{2})$, one has $\mathbf{N}(\alpha\beta) = \mathbf{N}(\alpha)\mathbf{N}(\beta)$.

Suppose we could write $$ \sqrt[3]{(7-\sqrt{2})} = \alpha $$ for some $\alpha \in \mathbb{Q}(\sqrt{2})$. Then $\alpha^3 = 7 - \sqrt{2}$, so $\mathbf{N}(\alpha)^3 = 47.$ But $\mathbf{N}(\alpha)$ is rational and $47$ is not a cube of a rational number.

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  • $\begingroup$ That's perfect thanks a million $\endgroup$ – kroner Apr 4 '15 at 14:28
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Wahat about this reasoning, suppose there exist an integer of $x\in \mathbb{Z}[\sqrt 2]$ and $a\in \mathbb{Z}$ such that $x^3=a^3(7-\sqrt 2)$ then: $$N(x)^3=a^6N(7-\sqrt 2) $$

which $N$ is the usual norm in $\mathbb{Z}[\sqrt 2]$ and here you get a contradiction

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