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

<br /> X^M-N<br />

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 &lt; M.

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

<br /> X^M-N = q(X)p(X) + r(X)<br />

where \partial r &lt; \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

<br /> a_1 + a_2\sqrt{2} + a_3\sqrt{3} + a_4\sqrt{6} = 0<br />

without all coefficients being zero. How does that happen?

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

<br /> p(X_1,X_2) = (X_1^2-2)^2 + (X_2^2-3)^2<br />

and

<br /> a(X_1,X_2) = a_1 + a_2X_1 + a_3X_2 + a_4X_1X_2<br />

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?

<br /> \mathbb{Q}(\sqrt{2})(\sqrt{3}) = \big\{x_1+x_2\sqrt{3}\;\big|\;x_1,x_2\in\mathbb{Q}(\sqrt{2})\big\}<br />

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

<br /> (a_1 + a_2\sqrt{2})^2 = (a_3\sqrt{3} + a_4\sqrt{6})^2<br />

and

<br /> (a_1 + a_3\sqrt{3})^2 = (a_2\sqrt{2} + a_4\sqrt{6})^2<br />

eventually imply

<br /> 2a_2^2 = 3a_3^2<br />

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.

<br /> (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}<br />

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

<br /> a_2 = 2b_1b_2<br />
<br /> a_3 = 2b_1b_3<br />
<br /> a_4 = 2b_2b_3<br />

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

<br /> b_1=\sqrt{\frac{a_2a_3}{2a_4}}<br />
<br /> b_2=\sqrt{\frac{a_2a_4}{2a_3}}<br />
<br /> b_3=\sqrt{\frac{a_3a_4}{2a_2}}<br />

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

<br /> (\sqrt[9]{9})^3 = 3^{\frac{2}{3}} \neq 3<br />



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

Anyway, it's clear that \sqrt[M]{N}=\sqrt[M&#039;]{N&#039;} often holds while M\neq M&#039; and N\neq N&#039;. 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&#039;&lt;M and N&#039;&lt;N such that \sqrt[M]{N}=\sqrt[M&#039;]{N&#039;}.

 

[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.