A problem about an algebraic number

Asked Apr 15 '15 at 05:45

Active Apr 15 '15 at 09:53

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Show that $2^{\frac{1}{2}}+5^{\frac{1}{3}}$ is algebraic over $\mathbb{Q}$ of degree $6$.

Can I just construct $x=2^{\frac{1}{2}}+5^{\frac{1}{3}}$, $x-2^{\frac{1}{2}}=5^{\frac{1}{3}}$, $x^3-3*2^{\frac{1}{2}}x^2+6x-2\frac{1}{2}=5$, $x^3+6x-5=\sqrt{2}(3x^2+2)$, $(x^3+6x-5)^2=2(3x^2+2)^2$, $x^6+12x^4-10x^3+36x^2-60x+25=18x^4+24x^2+8$, $x^6-6x^4-10x^3+12x^2-60x+17=0$.

Does it show what is required?

Then, $2^{\frac{1}{2}}+5^{\frac{1}{3}}$ is algebraic over $\mathbb{Q}$ of degree 6.

edited Apr 15 '15 at 09:53

user26857

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asked Apr 15 '15 at 05:45

Richard

2

yes it is right – Learnmore Apr 15 '15 at 05:48
3

@learnmore correct me if I'm wrong, but this proof only shows that $x$ is algebraic, not necessarily that the degree is 6? I believe this only shows that the degree is at most 6. – Nico Apr 15 '15 at 05:58
I believe "an element is said algebraic of degree 6" if there exists a polynomial of degree 6 of which it is a root @Nico – Learnmore Apr 15 '15 at 06:25
you can ask for clarification of the above definition from OP @Nico – Learnmore Apr 15 '15 at 06:26
@learnmore An element is algebraic of degree 6 if there is an irreducible polynomial of degree 6 of which it is a root. – ajotatxe Apr 15 '15 at 08:23
Hi everyone, so what can I do for the degree 6? – Richard Apr 15 '15 at 09:50

A problem about an algebraic number

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