Proving e^(n/m) is transcendental from e being transcendental

[Dick]

I'm not sure now if that applies. I can see if a is in F then a finit extension would contain a^n, but maybe not the other way.
I see my exponent error.

Yes, it doesn't apply. If I extend Q by sqrt(2), I have an extension field of dimension 2. But it does not include 2^(1/4).

 

[happyg1]

Ok, so I can go from a being algebraic to a^n being algebraic. I get that, but does it HAVE to go the other way? I've confused myself here.
I'm taking a break and getting a Dr. Pepper.
CC

 

[Dick]

Ok, so I can go from a being algebraic to a^n being algebraic. I get that, but does it HAVE to go the other way? I've confused myself here.
I'm taking a break and getting a Dr. Pepper.
CC

Yeah, take a break. And forget about finite field extensions. Just consider that a^n algebraic means there is a polynomial satisfied by a^n e.g.:
a_0+a_1*a^n+a_2*(a^n)^2+...a_m*(a^n)^m=0

Can you think of a polynomial satisfied by 'a'?

 

[HallsofIvy]

Do you mean "if aⁿ is algebraic, then a is algegraic"? Certainly that's true. If aⁿ is algebraic, of order m, then there exist a polynomial equation of degree m, with integer coefficients, such that
$$k_m(a^n)^m+ k_{m-1}(a^n)^{m-1}+ ...+ k_0= 0$$
but then
$$k_m a^{n+m}+ k_{m-1}a^{n+m-1}+ ...+ k_0= 0$$
a is algebraic of order (less than or equal to) n+m.

 

[Dick]

Do you mean "if aⁿ is algebraic, then a is algegraic"? Certainly that's true. If aⁿ is algebraic, of order m, then there exist a polynomial equation of degree m, with integer coefficients, such that
$$k_m(a^n)^m+ k_{m-1}(a^n)^{m-1}+ ...+ k_0= 0$$
but then
$$k_m a^{n+m}+ k_{m-1}a^{n+m-1}+ ...+ k_0= 0$$
a is algebraic of order (less than or equal to) n+m.

You gave it away! But your exponent arithmetic could use some work too.

 

[happyg1]

Ahhhhhh...
A little caffeine and sugar help my brain a lot sometimes.
I Think I understand this now. It's almost trivial, the a^n and a algebraic thing, I'm trying to make it harder than it is.
So now I have that a^(1\m) algebraic implies a algebraic and that a^n algebraic implies a algebraic, so then a^((1\m)^n) algebraic implies a algebraic. I just need to fill in the reasoning for each step and then e^(n\m) CAN'T be algebraic.

Am I finally on target here?
CC

 

[Dick]

Ahhhhhh...
A little caffeine and sugar help my brain a lot sometimes.
I Think I understand this now. It's almost trivial, the a^n and a algebraic thing, I'm trying to make it harder than it is.
So now I have that a^(1\m) algebraic implies a algebraic and that a^n algebraic implies a algebraic, so then a^((1\m)^n) algebraic implies a algebraic. I just need to fill in the reasoning for each step and then e^(n\m) CAN'T be algebraic.

Am I finally on target here?
CC

Yes. Dead on target.

 

[Dick]

Ahhhhhh...
A little caffeine and sugar help my brain a lot sometimes.
I Think I understand this now. It's almost trivial, the a^n and a algebraic thing, I'm trying to make it harder than it is.
So now I have that a^(1\m) algebraic implies a algebraic and that a^n algebraic implies a algebraic, so then a^((1\m)^n) algebraic implies a algebraic. I just need to fill in the reasoning for each step and then e^(n\m) CAN'T be algebraic.

Am I finally on target here?
CC

Except you are still being careless with exponents. Seems to be a big day for that.
a^((1/m)^n) is not equal to (a^(1/m))^n.

 

[happyg1]

Thanks for all of your patience and help on this. I learned A LOT about algebraic numbers from this problem. This stuff is SO COOL! (math nerds everywhere are cheering!)
I look back at my initial try UGGGHHHHH!
Thanks again.
CC

 

[happyg1]

So what if m is negative? Then I get $$\frac{1}{c^{1/m}}$$
Can I just refer this case to the previous case where m is not negative?
CC

 

[Dick]

You'll want to prove if c is algebraic then 1/c is. You can do it pretty directly - just solve a polynomial in c to get an expression for 1/c and argue that it is algebraic. Or if you've proved the algebraic numbers are a field, isn't it immediate?