Finding the Degree of \sqrt{3} + \sqrt[3]{4} Over Q: A Polynomial Search

[T-O7]

Does anyone know how to find the degree over Q of this number:
$$\sqrt{3} + \sqrt[3]{4}$$

In fact I'm having trouble finding any generic polynomial that this satisfies! Please help

 

[Zurtex]

I don't know what the question is that you're asking so I'm probably being really stupid here, but I assume it would help if you had that expressed as the rote of some polynomial equation with integer coefficients:

$$x = \sqrt{3} + \sqrt[3]{4}$$

$$x - \sqrt{3} = \sqrt[3]{4}$$

$$x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 4$$

$$x^3 + 9x - 4 = 3\sqrt{3}x^2 + 3\sqrt{3}$$

$$x^6 + 6x^4 - 8x^3 + 81x^2 - 72x - 16 = 27x^4 + 54x^2 + 27$$

$$x^6 - 21x^4 - 8x^3 + 27x^2 - 72x - 43 = 0$$

That help at all? (still probably worth checking my steps haha)

 

[Hurkyl]

Or, if you'd like to do it a different way, just keep raising your number to powers until you get a linearly dependant set. (Treating each distinct irrational number as a basis vector)

For example, for a = √2 + √3:

a^0 = 1
a^1 = √2 + √3
a^2 = 5 + 2 √6
a^3 = 11 √2 + 9 √3
a^4 = 49 + 20 √6

So, Q(a) is a vector space over Q, with basis vectors 1, √2, √3, and √6. We have 5 different vectors, so 1, a, a^2, a^3, and a^4 form a linearly dependant set...

 

[HallsofIvy]

Of course, the fact that you have to go to 5th power to get a dependent set means that the degree is 4?

(And notice that Hurkyl was using $\sqrt{2}+\sqrt{3}$, NOT the number of the original question. I like Zurtex's method: $\sqrt{2}+ ^3\sqrt{3}$ is algebraic of order 6 and so must have degree 6 over Q.

 

[Hurkyl]

Zurtex's approach is certainly easier to execute, but sometimes it's nonobvious how to manipulate things so that radicals don't proliferate. For example, if the cube roots of both 2 and 4 are in the number, when you cube to get rid of the cube root of 2, the cube root of 4 will just introduce more cube roots of 2.

 

[T-O7]

Great, thanks a lot guys!
Following Zurtex's method, i calculated:
$$x^6-9x^4-8x^3+27x^2-72x-11=0$$ for $$x=\sqrt{3}+\sqrt[3]{4}$$.
(There was a little error with zurtex's calculation i think)
I managed to prove that this was irreducible tediously...but (this might be a dumb question) how did you know x was algebraic of order 6 to begin with? That would save me a lot of trouble heh

 

[Hurkyl]

Because $[\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]{4}}) : \mathbb{Q}] = 3$. Therefore, both 2 and 3 must divide $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{4}}) : \mathbb{Q}]$.

This strongly suggests that the degree of your number must be 6... but more work would be needed to prove it.

Oh, and a bit of shamelessness. If you used my approach to come up with a polynomial your number satisfies, you can also use it directly to prove that it is the minimum polynomial.

For example, for √2 + √3, I just have to prove that {1, a, a^2, a^3} is linearly independent, which is a fairly straightforward task.