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My question is how to determine the minimal polynomial of $\sqrt{2}+\sqrt{5}$ over $\mathbb{Q}$, $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{7})$ and $\mathbb{Q}(\sqrt{10})$.

For the first one, I did: $u=\sqrt{2}+\sqrt{5}$ and I squared in order to obtain that the minimal polynomial is $t^4-10t^2+5$. Is this right? How do I find the other ones?

mathmo
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andrea98
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    Over $\mathbf{Q}(\sqrt{10})$, $\alpha$ is quadratic since $\alpha^2=7+2\sqrt{10}$. By the way, I think the minimal polynomial over $\mathbf{Q}$ is rather $X^4-14X^2+9$. – rae306 May 12 '20 at 14:00
  • See this thread also: https://math.stackexchange.com/questions/1990635/finding-a-minimal-polynomial-over-mathbbq-sqrt5?rq=1 – rae306 May 12 '20 at 14:06

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Let $u=\sqrt{2}+\sqrt{5}$.

  • Over $\mathbb{Q}(\sqrt{10})$, observe that $u^2=7+2 \sqrt{10}$ and so the minimal polynomial is $X^2-(7+2\sqrt{10})$.
  • Over $\mathbb{Q}(\sqrt{2})$, note that $(u-\sqrt{2})^2=5$ so the minimal polynomial is $X^2-2\sqrt{2}-3$.
  • Over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{7})$ you have $u^2=7+2 \sqrt{10}$ so $(u^2-7)^2=4\cdot 10$ and so the minimal polynomial is $X^4-14X^2+9$.

These are obviously polynomials vanishing at $u$. To show that they are minimal indeed, you would need to argue that they are irreducible. In the first two instances, it is enough to see that $u$ doesn't lie in the field. In the final two, verify that the degree of the extension is correct.

mathmo
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