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What is the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ ?

I proved that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)\supset \mathbb Q(\sqrt 5)$, and thus that $\mathbb Q(\sqrt 3+\sqrt5 )/\mathbb Q(\sqrt 5)$ has degree $2$. Since the minimal polynomial of $\sqrt 3+\sqrt 5$ is $X^4-16X^2+4$, the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt 5)$ must be a polynomial of order 2 that divide $X^4-16X^2+4$. It's roots are $\pm\sqrt{8\pm 2\sqrt{15}}$. But I tried to make all combinaisons $(X\pm\sqrt{8\pm2\sqrt{15}})(X\mp\sqrt{8\mp2\sqrt{15}})$, but I didn't get any polynomial of order $2$ in $\mathbb Q(\sqrt 5)$ that canceled $\sqrt 3+\sqrt 5$. Where is my mistakes ?

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  • $\begingroup$ Hint: its conjugate is $\sqrt5-\sqrt3$. $\endgroup$ – DHMO Apr 19 '17 at 16:28
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Since $\sqrt3\notin k=\Bbb Q(\sqrt5\,)$, the minimal polynomial for $\sqrt3$ over $k$ is $x^2-3=f(x)$. Clearly, then, the minimal polynomial for $\sqrt3+\sqrt5$ is $f(x-\sqrt5\,)=(x-\sqrt5\,)^2-3=x^2-2\sqrt5x+2$.

I know this wasn’t your strategy, but it is the quickest way to get the minimal polynomial.

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  • $\begingroup$ Thanks for your answer. Is it possible anyway to conclude with my method ? $\endgroup$ – MSE Apr 19 '17 at 17:25
  • $\begingroup$ I think that, following @DHMO’s lead, you should have written the various conjugates of $\sqrt3+\sqrt5$ in the form $\pm\sqrt3\pm\sqrt5$ $\endgroup$ – Lubin Apr 19 '17 at 17:30
  • $\begingroup$ It's strange since we know that the minimal polynomial is unique and if $p(X)$ is an annihilator polynomial the the minimal polynomial divide $p(X)$. Therefore, using my technic we should be able to conclude... no ? $\endgroup$ – MSE Apr 19 '17 at 17:33
  • $\begingroup$ Yes, yes, your strategy was sound, but: yet again: $\bigl(8+2\sqrt{15}\bigr)^{1/2}=\pm(\sqrt3+\sqrt5)$. You made things difficult for yourself by squaring $\sqrt3+\sqrt5$ $\endgroup$ – Lubin Apr 19 '17 at 17:37
  • $\begingroup$ Fantastic ! Thanks a lot . $\endgroup$ – MSE Apr 19 '17 at 19:09
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Its conjugate is $\sqrt5-\sqrt3$.

Therefore, the minimal polynomial is:

$$\begin{array}{rcl} (x-(\sqrt3+\sqrt5))(x-(\sqrt5-\sqrt3)) &=& (x-\sqrt5)^2 - (\sqrt3)^2 \\ &=& x^2-2\sqrt5x+5-3 \\ &=& x^2-2\sqrt5x+2 \end{array}$$

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