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Show that $\sqrt{2} + \sqrt[3]{5}$ is algebraic of degree $6$ over $\mathbb{Q}$

What is the degree of a root? Is it the smallest polynomial that gives this thing as root?

What I tried:

$x = \sqrt{2} + \sqrt[3]{5} \implies x^2 = 2 + 2\sqrt{2}\sqrt[3]{5} + \sqrt[3]{5^2}$

I don't see it going anywhere.

Maybe if I try to elevate to the power of something that is common to both $2$ and $3$, as $6$, I'll get something. But elevating the root itself won't help: http://www.wolframalpha.com/input/?i=(%5Csqrt%7B2%7D+%2B+%5Csqrt%5B3%5D%7B5%7D)%5E6

Do I really need to find the polynomial or just show that it is algebraic using another technique?

I couldn't find anything helpful here The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$

José Carlos Santos
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Paprika
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  • Show $\sqrt{2}$ inside the field extension and similarly for $\sqrt[3]{5}$ as well. Then deduce the degree from relative primeness subfield of extensions. – user45765 Sep 11 '17 at 00:03
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    To find a polynomial of degree $6$ with integer coefficients having $x$ as a root, expand $(x-\sqrt{2})^3=5$ then collect the terms in $\sqrt{2}$ on one side and square the equation. To prove that it is the minimal polynomial, see the answers to the question you linked. – dxiv Sep 11 '17 at 00:06

3 Answers3

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It is easily seen that a sum of two algebraic numbers $\alpha$ and $\beta$ is algebraic since one has: $$\mathbb{Q}(\alpha+\beta)\subseteq\mathbb{Q}(\alpha,\beta).$$ Hence, $\mathbb{Q}(\alpha+\beta)$ is finite-dimensional since $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)(\beta)$ is, indeed recall that: $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}].$$

If $p$ is an annihilator polynomial of $\alpha$ and $q$ one of $\beta$, then $\textrm{res}_Y(p(Y),q(X-Y))$ is an annihilator polynomial of $\alpha+\beta$.

In your case, use $p=X^2-2$ and $q=X^3-5$ to find that $X^6-6X^4-10X^3+12X^2-60X+17$ is an annihilator polynomial of $\sqrt{2}+\sqrt[3]{5}$. You are left to show that it is indeed irreducible over $\mathbb{Q}$.

C. Falcon
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Hint: Use the formula for the degree:

$[\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$

Cornman
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  • Thank you, that should work. But how should I find the degree of $\sqrt{2}+\sqrt[3]{5}$ over $\sqrt{2}$ for example? – Paprika Sep 11 '17 at 00:06
  • By the polynomial $X^3-5\in\mathbb{Q}(\sqrt{2})[X]$, which has the root $\sqrt[3]{5}$. You have to show, that it is irreducible. If it is reducible it is $(X^3-5)=(X^2+aX+b)(X+c)$. With comparison of the coefficents you get a contradiction. There might be a more efficient way. – Cornman Sep 11 '17 at 00:12
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    @Cornman It's enough to show that $x^3 - 5$ doesn't have a root over $\mathbb{Q}(\sqrt{2})$ and that should be easier, as it reduces to showing that $\sqrt[3]{5} \not \in \mathbb{Q}(\sqrt{2})$ – Stefan4024 Sep 11 '17 at 00:33
  • You are right. :) @Paprika: I would still recomment you to go the way i have given above. It is a commen strategy and worth looking at. (Obviously not needed) – Cornman Sep 11 '17 at 00:35
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$$\sqrt[3]5+\sqrt2=q\implies$$ $$q-\sqrt2=\sqrt[3]5\implies$$ $$q^3-3\sqrt2q^2+6q-2\sqrt2=5\implies$$ $$q^3+6q+5=(3q^2+2)\sqrt2.$$ Square the last identity.