Give $a=\sqrt{3}+\sqrt{5}$, I need to find the minimal polynomial of $a$ in $\mathbb{Q}$ and the polynomial $f\in\mathbb{Q}[X]$ with $f(a)=a^{-1}$.

Now the first part is rather easy, but how do I calculate the second part?

  • $\begingroup$ Just play with the powers of $a$. You have $a^3=18\sqrt 3+14\sqrt 5$ so $ \sqrt 5=\frac 14(a-a^3)$. Get a similar expression for $\sqrt 3$. Use those two expressions to get $a^{-1}$. $\endgroup$ – lulu Feb 14 at 13:51

Give $a=\sqrt3+\sqrt5$, I need to find the minimal polynomial of $a$.

$$\begin{align} a=\sqrt3+\sqrt5 &\quad\Rightarrow\quad a^2=(\sqrt3+\sqrt5)^2 = 8+2\sqrt{15}\\ &\quad\Rightarrow\quad (a^2-8)^2 = 4\cdot15\\ &\quad\Rightarrow\quad 0=a^4-16a^2+4\\ \end{align}$$


$$0=a^4-16a^2+4 \;\Leftrightarrow\; 0=a^3-16a+4a^{-1}$$ Thus $$a^{-1}=-\frac14a^3+4a$$

  • $\begingroup$ Is it really that simple? Thank you! $\endgroup$ – KingDingeling Feb 14 at 14:39
  • $\begingroup$ For $a^{-1}$, yes. For the minimal polynomial you'd also have to show that no polynomial of smaller degree will do. $\endgroup$ – emacs drives me nuts Feb 14 at 14:48
  • $\begingroup$ thank you but the polynomial is the same I found as well, it is normed, has a root in $a$ and is irreducible with Eisenstein with $p=2$. So it is the minimal polynomial. $\endgroup$ – KingDingeling Feb 14 at 15:29

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