# Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

I have to find the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. The suggested way of doing it is to prove that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ first.

I can prove that. It's enough to prove that $\sqrt 2,\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ and that $\sqrt 2 + \sqrt[3] 2\in \mathbb Q[\sqrt 2,\sqrt[3] 2]$. The latter is obvious and $\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ will follow from $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$ It remains to prove $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$

For $\alpha=\sqrt 2 + \sqrt[3] 2,$ we have

$$\begin{eqnarray} &\alpha^0&=1,\\ &\alpha^1&=\sqrt 2 + \sqrt[3] 2,\\ &\alpha^2&=2+\sqrt[3]4+\sqrt 2\sqrt[3]2\\ &\alpha^3&=2+2\sqrt 2+6\sqrt[3]2+3\sqrt 2\sqrt[3]4\\ &\alpha^4&=4+4\sqrt 2+12\sqrt[3]4+8\sqrt 2\sqrt[3]2\\ &\alpha^5&=32+4\sqrt 2+20\sqrt[3]2+4\sqrt 2\sqrt[3]2 \end{eqnarray}$$

If I can express $\sqrt 2$ as a linear combination of $\{\alpha^0,\alpha^1,\cdots,\alpha^5\}$, I'm done. I don't know how to find out whether $$\{1,\sqrt 2,\sqrt[3]2,\sqrt[3]4,\sqrt 2\sqrt[3]2,\sqrt 2\sqrt[3]4\}$$ are linearly independent over $\mathbb Q$ but I think I don't have to care. I can use Gaussian elimination anyway. I write the matrix

$$\left[ \begin{array}{rrrrrr|r} 1 & 0 & 2 & 2 & 1 & 8 & 0\\ 0 & 1 & 0 & 2 & 1 & 1 & 1\\ 0 & 1 & 0 & 6 & 0 & 5 & 0\\ 0 & 0 & 1 & 0 & 3 & 0 & 0\\ 0 & 0 & 1 & 0 & 2 & 1 & 0\\ 0 & 0 & 0 & 3 & 0 & 5 & 0 \end{array}\right]$$

and by performing row operations I obtain a row echelon form, which proves that I'll obtain the reduced row echelon form if I want, which will give me the desired coefficients $a_0,\cdots,a_5$ such that

$$\sqrt 2=\sum_{i=0}^5a_i\alpha^i.$$

I don't have to find the coefficients -- I'm just satisfied with their existence.

If this is correct, I have proven that

$$\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2].$$

But I don't know how to proceed further. Of course, I can find the minimal polynomials of $\sqrt 2$ and $\sqrt[3] 2$ over $\mathbb Q$ but I don't know how to find the minimal polynomial of $\sqrt 2$ over $\mathbb Q[\sqrt[3]2]$ or the minimal polynomial of $\sqrt [3]2$ over $\mathbb Q[\sqrt 2].$

What I did is this. I wrote the equation

$$x=\sqrt 2+\sqrt [3]2$$

and by squaring and cubing obtained

$$W(x):=x^6-6x^4-4x^3+12x^2-24x-4=0.$$

I belive this is the required polynomial but I would need to prove that it's irreducible over $\mathbb Q.$ Eisenstein's criterion doesn't work, which means I'm lost.

1) Is my proof of $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ correct?

2) How can I use this fact? I believe it's supposed to make the solution easier to find.

3) How can I tell whether $W(x)$ is reducible over $\mathbb Q$ or not?

EDIT: 1) has been answered by André Nicolas in a comment. All of this answers 3) but I would like to ask whether we could know that the polynomial is irreducible without knowing that $\sqrt 2+\sqrt[3]2$ is its zero. Also, it would be to know whether

4) there is a theorem that would give me $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ without the need of performing the Gaussian elimination?

For these numbers it was possible to do, but a bit higher dimension and it would turn out impossible without a computer.

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Since $\mathbb Q[\sqrt 2,\sqrt[3] 2]=\mathbb Q[\sqrt 2 + \sqrt[3] 2]$, the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ has degree equal to $[\mathbb Q[\sqrt 2,\sqrt[3] 2]:\mathbb{Q}]$. Can you compute this? –  Chris Eagle Feb 25 '12 at 19:26
@ChrisEagle Thank you for the comment. I can't! I think I would have to find either the minimal polynomial of $\sqrt 2$ over $\mathbb Q[\sqrt[3]2]$ or the minimal polynomial of $\sqrt[3]2$ over $\mathbb Q[\sqrt 2]$. Or at least their degrees, but I don't know how to find their degrees without computing them, either... –  user23211 Feb 25 '12 at 19:29
Your big field has $\mathbb{Q}(\sqrt{2})$ as a subfield. Therefore the degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ divides the degree of your big field over $\mathbb{Q}$. So $2$ divides the degree. Similarly, so does $3$. The end. –  André Nicolas Feb 25 '12 at 19:42
Since we seem to be in a degree-6 extension anyway, I wonder whether it would be easier to work in $\mathbb Q[\sqrt[6]{2}]$ instead? –  Henning Makholm Feb 25 '12 at 20:08
Let $\beta$ be a sixth root of $2$. Then $\sqrt 2=\beta^3$ and $\sqrt[3]2=\beta^2$, so you just have to find the minimal polynomial for $\beta^3+\beta^2$ given $\beta^6=2$. Granted, this is not much different from what you're already doing -- but the polynomial arithmetic modulo $\beta^6-2$ is slightly easier than keeping track of two different generator polynomials at once. –  Henning Makholm Feb 25 '12 at 20:31

Here is an extremely easy way of showing that $\sqrt{2} \in \Bbb{Q}(\sqrt{2} + \sqrt[3]{2})$. Write $\alpha = \sqrt{2} + \sqrt[3]{2}$. Then $(\alpha-\sqrt{2})^3 = 2$, so that

$$\alpha^3 - 3\alpha^2(\sqrt{2}) + 6\alpha -2\sqrt{2} = 2.$$

It follows that $\alpha^3+ 6\alpha - 2 = \sqrt{2}(3\alpha^2+ 2)$, so that

$$\sqrt{2} = \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}.$$

Since $\Bbb{Q}(\alpha)$ is a field, the expression on the right hand side is in here, so that $\sqrt{2}$ is in here. No using horrible row reduction to calculate anything! It follows that $\sqrt[3]{2}$ is in here. Hence $\Bbb{Q}(\alpha)$ contains the fields $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}(\sqrt[3]{2})$ which means that $[\Bbb{Q}(\alpha) : \Bbb{Q}]$ is a multiple of 3 and 2. Since we already know that it is at most $6$, it follows now that since $2$ and $3$ are coprime that $[\Bbb{Q}(\alpha) : \Bbb{Q}] = 6$. Whatever monic polynomial of degree 6 that you find for $\alpha$ over $\Bbb{Q}$ will then be irreducible, and hence will be the minimal polynomial of $x$ over $\Bbb{Q}$.

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I approve! This is better. Note that $3\alpha^2 + 2 \neq 0$ since $[\mathbb Q(\alpha) : \mathbb Q] > 2$. –  Patrick Da Silva May 8 '12 at 2:41
@PatrickDaSilva Whoops forgot to check that, sort of tacitly assumed that was true. –  user38268 May 8 '12 at 3:03
@Benjamin : Well, you are dividing by some element you know nothing about. You could be fooled by assuming that this number "is real", but algebraists don't like those kind of assumptions. ;) –  Patrick Da Silva May 8 '12 at 3:28

There's a really easier way to do this. Notice that over $\mathbb Q[\sqrt 2]$, $\sqrt 2 + \sqrt[3]2$ is a root of $(x-\sqrt 2)^3 - 2$. The roots of this polynomial are $\sqrt 2 + \omega \sqrt[3]2$ with $\omega^3 = 1$, and clearly none of these are in $\mathbb Q[\sqrt 2]$ because if they would, we would have $\omega \sqrt[3]2 \in \mathbb Q[\sqrt 2]$ and the minimal polynomial of those three over $\mathbb Q$ is $x^3-2$, which is irreducible over $\mathbb Q$, hence $[\mathbb Q[\omega \sqrt[3]2] : \mathbb Q] = 3$, which means they can't be in $\mathbb Q[\sqrt 2]$ since $[\mathbb Q[\sqrt 2] : \mathbb Q] = 2$. Now knowing this, since a polynomial of degree $3$ is irreducible if and only if it has no roots, $(x-\sqrt 2)^3 - 2$ is also irreducible over $\mathbb Q[\sqrt 2]$, hence it is the minimal polynomial of $\sqrt 2 + \sqrt[3]2$ over $\mathbb Q[\sqrt 2]$ and $$[\mathbb Q[\sqrt 2 + \sqrt[3]2] : \mathbb Q] = [\mathbb Q[\sqrt 2 + \sqrt[3]2] : \mathbb Q[\sqrt 2]][\mathbb Q[\sqrt 2] : \mathbb Q] = 3 \cdot 2 = 6.$$ Therefore you know that the minimal polynomial of $\sqrt 2 + \sqrt[3]2$ over $\mathbb Q$ has degree $6$. It now suffices to compute a polynomial of degree $6$ of which this guy is a root. That's computations : compute all powers of $\sqrt 2 + \sqrt[3]2$ up to the sixth power and use linear algebra. I see that you did, so I guess you're done.

And 1) is correct. You can use this fact to show that $2$ and $3$ divides the order of your degree extension, which means $6$ divides it. Since you found a poylnomial of degree $6$ of which your element is a root, the polynomial must be irreducible.

Hope that helps,

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Thank you for the answer. I know that the fact that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ is true. I would like to know whether my proof of this fact is correct. I'm not sure whether you are saying that the statement is correct or that the proof is correct. –  user23211 Feb 25 '12 at 19:52
@ymar : The proof is. –  Patrick Da Silva Feb 25 '12 at 21:33
@PatrickDaSilva When you use the dimension formula aren't you assuming that $\Bbb{Q}(\alpha) \supset \Bbb{Q}(\sqrt{2})$? How do you know that? –  user38268 Mar 21 '12 at 12:10
@Benjamin Lim : Good question. I am in a hurry but I'll answer it later. –  Patrick Da Silva Mar 21 '12 at 16:21

As noted in Patrick's answer, the degree of the minimal polynomial must be six. This given, it suffices to construct a polynomial of degree six with your number as a root. One way is brute force linear algebra. Here is a more efficient and systematic way; it is a part of a constructive proof that, given commutative rings $R \subseteq S$, the set of elements of $S$ which are integral over $R$ is a subring.

Let $\alpha$ and $\beta$ be algebraic numbers with conjugates $\alpha=\alpha_1,\alpha_2,\dots,\alpha_n$ and $\beta=\beta_1,\beta_2,\dots,\beta_m$. Then the polynomial

$$f(x)=\prod (x-\alpha_i-\beta_j)$$

has $\alpha+\beta$ as a root, and thanks to a symmetry argument, has coefficients in $\mathbb{Q}$. Evidently the degree of $f(x)$ is $mn$, which forces it to be the minimal polynomial if, as in your situation, you know for other reasons that $\alpha+\beta$ has degree $mn$. This gives a quick way to get your answer.

In regards to your question 4) above, it is a theorem that, given algebraic numbers $\alpha$ and $\beta$, "most" linear combinations $\alpha+\mu \beta$ (for $\mu \in \mathbb{Q}$) generate $\mathbb{Q}[\alpha,\beta]$. But to check whether this definitely happens for a particular choice may be time consuming (though you have to get "unlucky" for it to fail).

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Thank you. Could you please say more about the theorem? What is the meaning of "most" here? –  user23211 Feb 25 '12 at 23:08
All but finitely many. See the wikipedia page on the theorem of the primitive element, en.wikipedia.org/wiki/Primitive_element_theorem –  Stephen Feb 25 '12 at 23:41

To answer (3), one straightforward way to show an integer polynomial irreducible is to try and factor it over finite fields. If you find a finite field where it's irreducible then it's irreducible as an integer polynomial. With some care, you can speed this up; e.g. if:

• You find a finite field where it factors into degree 2 and degree 4
• You find another finite field where it factors into degree 3 and degree 3

Then it has to be irreducible. This, of course, assumes you know how to factor in finite fields....

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