Is X^M-N the Minimal Polynomial of Irrational Root \sqrt[M]{N} in \mathbb{Q}[X]?

[jostpuur]

Assume that $\sqrt[M]{N}$ is irrational where $N,M$ are positive integers. I'm under belief that

$$X^M-N$$

is the minimal polynomial of $\sqrt[M]{N}$ in $\mathbb{Q}[X]$, but I cannot figure out the proof. Assume as an antithesis that $p(X)\in\mathbb{Q}[X]$ is the minimal polynomial such that $\partial p < M$.

We can divide $X^M-N$ by $p(X)$ and obtain

$$X^M-N = q(X)p(X) + r(X)$$

where $\partial r < \partial p$. According to the assumption, $r=0$ must hold.

So I'm somehow supposed to use the information $X^M-N=q(X)p(X)$ and $r=0$ and find a contradiction.

Doesn't look very obvious to me.

 

[mfb]

$\sqrt[4]{4}$?

$X^2-2=0$

 

[wisvuze]

I had never really thought about this. All I know is that it will be true when M is a prime number, or if N is a number with a square-free prime factorization ( use eisenstein's ). I don't if there are easier conditions

edit: also, I think it will be true if M is not a power of 2. Over Q, you can factorize your polynomial into irreducibles that are either linear factors of x^2 + n terms, where n is a positive integer. If M is not a power of 2, then you must have a linear factor in such a decomposition. Then, the polynomial must have a rational root. But it is easy to see that it cannot, since this is a monic polynomial with coefficients in Z. Any rational root of these polynomials must themselves be in Z

 

[HallsofIvy]

No, being a power of 2 is not relevant. For example, the ninth root of 9, $$\sqrt[9]{9}$$ is a zero of $$x^3- 3$$. This will be false whenever M is a composite number and N is an n power where n divides M.

 

[jostpuur]

some linear combinations of irrationals

Well at least I was careful enough to mention that I was "under belief"...

I'll hijack my own thread to a new topic. I have so many questions with similar themes, so I guess I might as well concentrate them here...

I want to prove that there does not exist coefficients $a_1,a_2,a_3,a_4\in\mathbb{Q}$ such that

$$a_1 + a_2\sqrt{2} + a_3\sqrt{3} + a_4\sqrt{6} = 0$$

without all coefficients being zero. How does that happen?

I have a feeling that the proof might have something to do with polynomials

$$p(X_1,X_2) = (X_1^2-2)^2 + (X_2^2-3)^2$$

and

$$a(X_1,X_2) = a_1 + a_2X_1 + a_3X_2 + a_4X_1X_2$$

which would both satisfy $p(\sqrt{2},\sqrt{3})=0$ and $a(\sqrt{2},\sqrt{3})=0$, but I'm not sure what's the final trick.

 

[mfb]

I think I would consider $(b_1+b_2\sqrt{2}+b_3\sqrt{3})^2$. It gives all 4 different terms, and if the bracket cannot be rational the square of it cannot be rational either.

 

[jostpuur]

I think I would consider $(b_1+b_2\sqrt{2}+b_3\sqrt{3})^2$. It gives all 4 different terms, and if the bracket cannot be rational the square of it cannot be rational either.

If I first fix arbitrary $a_1,a_2,a_3,a_4\in\mathbb{Q}$, you are not going to find $b_1,b_2,b_3\in\mathbb{Q}$ such that they would map to the $a_1,a_2,a_3,a_4$ through that square.

 

[jostpuur]

My question is equivalent to asking that how do I prove that $\mathbb{Q}(\sqrt{2})(\sqrt{3})$ is two dimensional vector space with respect to the scalar field $\mathbb{Q}(\sqrt{2})$. Does this remark help those who know the theory?

$$\mathbb{Q}(\sqrt{2})(\sqrt{3}) = \big\{x_1+x_2\sqrt{3}\;\big|\;x_1,x_2\in\mathbb{Q}(\sqrt{2})\big\}$$

It looks like two dimensional, but it's not obvious that non-zero $x_1,x_2\in\mathbb{Q}(\sqrt{2})$ wouldn't exist so that $x_1+x_2\sqrt{3}=0$.

 

[mfb]

If I first fix arbitrary $a_1,a_2,a_3,a_4\in\mathbb{Q}$, you are not going to find $b_1,b_2,b_3\in\mathbb{Q}$ such that they would map to the $a_1,a_2,a_3,a_4$ through that square.

a_1 does not have to match, as my expression is not supposed to be 0. You can write it on the RHS.

 

[jostpuur]

I invented one proof on my own now. I can assume that $a_1,a_2,a_3,a_4\in\mathbb{Z}$. Equations

$$(a_1 + a_2\sqrt{2})^2 = (a_3\sqrt{3} + a_4\sqrt{6})^2$$

and

$$(a_1 + a_3\sqrt{3})^2 = (a_2\sqrt{2} + a_4\sqrt{6})^2$$

eventually imply

$$2a_2^2 = 3a_3^2$$

which then implies $a_2=\pm\infty$ and $a_3=\pm\infty$ when you work out how $2^n$ divides these coeffients with arbitrarily large $n$.

 

[jostpuur]

No, no, mfb, you'r idea aint working.

$$(b_1 + b_2\sqrt{2} + b_3\sqrt{3})^2 = b_1^2 + 2b_2^2 + 3b_3^2 + 2b_1b_2\sqrt{2} + 2b_1b_3\sqrt{3} + 2b_2b_3\sqrt{6}$$

So you were thinking, that once $a_2,a_3,a_4\in\mathbb{Q}$ were fixed, you could find $b_1,b_2,b_3\in\mathbb{Q}$ such that

$$a_2 = 2b_1b_2$$
$$a_3 = 2b_1b_3$$
$$a_4 = 2b_2b_3$$

These equations actually fix $b_1,b_2,b_3$ uniquely, and they can be solved. The answer is

$$b_1=\sqrt{\frac{a_2a_3}{2a_4}}$$
$$b_2=\sqrt{\frac{a_2a_4}{2a_3}}$$
$$b_3=\sqrt{\frac{a_3a_4}{2a_2}}$$

We see they can be irrational too.

 

[jostpuur]

No, being a power of 2 is not relevant. For example, the ninth root of 9, $$\sqrt[9]{9}$$ is a zero of $$x^3- 3$$. This will be false whenever M is a composite number and N is an n power where n divides M.

Actually

$$(\sqrt[9]{9})^3 = 3^{\frac{2}{3}} \neq 3$$



$\sqrt[9]{27}$ would be a zero of $X^3-3$.

Anyway, it's clear that $\sqrt[M]{N}=\sqrt[M']{N'}$ often holds while $M\neq M'$ and $N\neq N'$. It still seems reasonable, that $X^M-N$ will turn out to be the minimal polynomial of $\sqrt[M]{N}$ if there does not exist $M'<M$ and $N'<N$ such that $\sqrt[M]{N}=\sqrt[M']{N'}$.

 

[mfb]

No, no, mfb, you'r idea aint working.

Okay. I did not check it in detail, it was just an idea.

Your proof is nice.