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Let $p=n^2+1$ be prime, $a=\sqrt{p+\sqrt{p}},b=\sqrt{p-\sqrt{p}}$ and ${\rm Irr}\left(a,\mathbb{Q}\right)=x^4-2px^2+p\left(p-1\right).$

How do I check if $\;b\in \mathbb{Q}(a)$?

I tried by claiming that if $b\in \mathbb{Q}(a)$, then $b$ could be written as a linear format of $\mathbb{Q}(a)$, using its base, which is $\{1,a,a^2,a^3\}$.

That really got me nowhere or at least nowhere somewhat ''elegant''.

Is there another way?

Some of my guesses are using $b^{-1}$ or maybe $ab$, since we know a for sure belongs in $\mathbb{Q}(a)$.

pieq3
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    Hint: there are at least two different ways to show that $\sqrt p\in \Bbb Q(a)$: consider either $ab$ (using the fact that $p=n^2+1$) or $a^2$. Can you combine these observations to show that $b\in \Bbb Q(a)$? – Greg Martin Mar 20 '22 at 18:29
  • @GregMartin $ab=n\sqrt{p}\in \mathbb{Q}\left(\sqrt{p}\right)$(?) and if we thing $:\mathbb{Q}\left(a\right)$ as an extension of $\mathbb{Q}\left(\sqrt{p}\right)$,doesnt that mean $n\sqrt{p}\in \mathbb{Q}\left(a\right)$?And thus we can conclude,with a little more elaboration,that $b\in \mathbb{Q}\left(a\right)$ – pieq3 Mar 20 '22 at 21:21

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One way is to show $b = a^3/n-a(p+1)/n$ by simplifying the expression on the right side of the equation. Another way is realize that the four roots of $f(x) = x^4-2px^2+p(p-1)$ are $\pm a$, $\pm b$. Combine this with the fact that the Galois group of $f(x)$ is cyclic of order 4 to deduce that $\mathbb{Q}(a)$ contains all four roots of $f(x)$, hence contains $b$. Note: Galois groups of polynomials of the form $x^4+ax^2+b$ are straightforward to determine; using Galois group of a biquadratic quartic for example.

lm6612
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