Proving pi^2 is transcendental over Q

[PsychonautQQ]

Homework Statement

If we know pi is transcendental over Q, how could we show pi^2 is also transcendental?

Homework Equations

The Attempt at a Solution

Yeah, I'm a little confused. My homework is asking 'true or false' for if pi^2 is transcendental over Q, and I'm quite sure we can assume pi is transcendental. Anyone have any tips? I'm really quite lost.

 

[Ssnow]

Hi,
assume that $\pi$ is transcendental (you said you can...) and that $\pi^2$ is a root of some polynomial $P(x)$, can you see a contradiction?

 

[PsychonautQQ]

Hi,
assume that $\pi$ is transcendental (you said you can...) and that $\pi^2$ is a root of some polynomial $P(x)$, can you see a contradiction?

Is it because polynomials have the ability to 'extract roots' in a way (I'm sorry math people, this is the clearest way I could think of expressing my thoughts!) What I mean is since 2 is algebraic, 2^1/2 is also algebraic. So for any algebraic element, all it's roots are also algebraic?

 

[fresh_42]

Is it because polynomials have the ability to 'extract roots' in a way (I'm sorry math people, this is the clearest way I could think of expressing my thoughts!) What I mean is since 2 is algebraic, 2^1/2 is also algebraic. So for any algebraic element, all it's roots are also algebraic?

What does it mean for a number to be algebraic?

 

[PsychonautQQ]

What does it mean for a number to be algebraic?

A number is algebraic over a field K if there is some polynomial in K[x] for which that number is a root of. So if pi^2 is a root of f(x), somehow that means that pi is a root of some polynomial with coefficients in the same field, which would be a contradiction because pi is transcendental. I don't think we could just say f(x)^1/2.

If an element is algebraic, then the degree of it's minimal polynomial will be the degree of the extension between the ground field and the field that the element is in. So if pi^2 is algebraic, say or degree n, then perhaps pi would be algebraic of degree n^1/2 which is a contradiction?

 

[Ssnow]

Is it because polynomials have the ability to 'extract roots' in a way

this can be part of the idea, the fact is that by assumption you have $\pi$ is transcendental, you can assuming also that $\pi^2$ is a solution of certain polynomial $P(x)$ (so assuming $\pi^2$ that is algebraic in fact) and be able to arrive to a contradiction so conclude that $\pi^2$ is transcendental ...

 

[fresh_42]

So if pi^2 is a root of f(x), somehow that means that pi is a root of some polynomial with coefficients in the same field, which would be a contradiction because pi is transcendental.

Correct. And done. Write it in formulas:

If $\pi^2$ is a root of $f(x)$, then $f(\pi^2)=0$.
Now let $f(x) = a_nx^n+\dots+a_1x+a_0$. Then $f(\pi^2) = a_n\pi^{2n}+\dots+a_1\pi^2+a_0=0$.

This is what @Ssnow has meant in post #2: Is $f(\pi^2) = g(\pi)$ for another polynomial $g(x)$? And if so, what does this mean?

 

[PsychonautQQ]

Correct. And done. Write it in formulas:

If $\pi^2$ is a root of $f(x)$, then $f(\pi^2)=0$.
Now let $f(x) = a_nx^n+\dots+a_1x+a_0$. Then $f(\pi^2) = a_n\pi^{2n}+\dots+a_1\pi^2+a_0=0$.

This is what @Ssnow has meant in post #2: Is $f(\pi^2) = g(\pi)$ for another polynomial $g(x)$? And if so, what does this mean?

Thank you guys so much, this community is amazing.

 

[Ssnow]

This is what @Ssnow has meant

yes exactly...