Algebraic/Transcendence Theory

« first day (2 days earlier) ← previous day next day → last day (151 days later) »

DHMO

00:40

@BalarkaSen Wikipedia claims that Lindemann-Weierstrass theorem is equivalent to the following:
If $a_1,\ldots,a_n$ are distinct algebraic numbers, then $e^{a_1},\ldots,e^{a_n}$ are linearly independent over the algebraic numbers.

Balarka Sen

@DHMO That's literally the statement of L-W

DHMO

2 days ago, by DHMO

Hermite-Lindemann-Weierstrass theorem (HLW):

2 days ago, by DHMO

If $a_1, \ldots, a_n$ are linearly independent over the rational numbers, then $e^{a_1}, \ldots, e^{a_n}$ are algebraically independent over the rational numbers.

Balarka Sen

huh, never seen that formulation before

DHMO

Lindemann–Weierstrass theorem

In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α1, ..., αn are algebraic numbers which are linearly independent over the rational numbers ℚ, then eα1, ..., eαn are algebraically independent over ℚ; in other words the extension field ℚ(eα1, ..., eαn) has transcendence degree n over ℚ. An equivalent formulation (Baker 1975, Chapter 1, Theorem 1.4), is the following: If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over t...

That's basically the first line

And the equivalent formation is the second paragraph

Balarka Sen

@DHMO Oh, one is over \bar Q, another is over Q

Then that's clear.

DHMO

00:47

I don't think "algebraically independent over Q" means the same thing as "linearly independent over Q"

Balarka Sen

No, but "algebraically independent over Q" does mean the same thing as "linearly independent over \bar Q".

\bar Q being algebraic numbers

DHMO

but why?

the premise is changed as well

from "linearly independent over Q" to "distinct algebraic numbers"

Balarka Sen

@DHMO Why don't you try to prove it?

DHMO

I don't have any idea

Balarka Sen

Prove that "algebraic dependent over Q" implies "linearly dependent over \bar Q", and it's converse.

That's easier.

DHMO

00:52

did you swap the two sets? right.

Balarka Sen

Right, fixed.

DHMO

huh

still no idea

Balarka Sen

Recall definitions

And give it more than 2 seconds :)

DHMO

So $e$ and $e^2$ are algebraically dependent over $\Bbb Q$.

and how on earth are they linearly dependent over $\bar {\Bbb Q}$?

In fact, if $ae+be^2=0$ where $a,b \in \bar{\Bbb Q}$

Then $e=-\dfrac a b$

which is a contradiction, since $e$ is transcendental

so I don't see how they are equivalent

@BalarkaSen

Balarka Sen

@DHMO Yeah, you're right. I'm a little confused. Oops.

Alright, suppose $a_1, \cdots, a_n$ are distinct alg. ints. L-W2 says $e^{a_1}, \cdots, e^{a_n}$ are $\bar{\Bbb Q}$-linearly independent.

If $a_1, \cdots, a_n$ are $\Bbb Q$-linearly independent, L-W1 says $e^{a_1}, \cdots, e^{a_n}$ are $\Bbb Q$-algebraically independent.

To go from LW2 to LW1, assume $a_i$ are $\Bbb Q$-linearly ind. Then $e^{a_i}$ are $\bar{\Bbb Q}$-linearly ind. Suppose for a contradiction $e^{a_1}, \cdots, e^{a_n}$ are $\Bbb Q$-algebraically dependent. Then they satisfy a polynomial relation $P(e^{a_1}, \cdots, e^{a_n}) = 0$, $P$ a poly. with coeffs in $\Bbb Q$.

That's like $\sum p_i e^{q_i} = 0$ where $q_i$ are linear combinations of $a_i$'s, isn't it?

Oh, so there you go. $q_i$'s are all distinct because $a_i$'s are $\Bbb Q$-linearly ind by assumption. So apply L-W2 to $e^{p_1}, e^{p_2}, \cdots$, @DHMO. You break it because you're saying they are $\Bbb Q$-linearly dependent, whereas they are not even $\bar{\Bbb Q}$-linearly dependent.

Ok, this was harder than it looked.

DHMO

01:20

@BalarkaSen did you mean $e^{q_1}\cdots$?

@BalarkaSen wonderful proof

Balarka Sen

Yeah, sorry. Thanks for pointing out.

It'd have been more wonderful if I hadn't said the nonsense I told you above :3

DHMO

heh, let me understand the second theorem (pi is transcendental) before we move on to the third theorem

planetmath.org/sites/default/files/texpdf/39438.pdf

Balarka Sen

Yeah, totally

mick

Intresting

DHMO

@mick welcome

mick

01:32

Hi. Working on other things now srr

If you wanna know for instance math.stackexchange.com/questions/2075456/…

13 hours later…

Balarka Sen

14:34

@DHMO Any new breakthroughs?

DHMO

@BalarkaSen still reading the proof that pi is transcendental

Balarka Sen

great.

DHMO

got quite stuck in the "symmetrical polynomial" thing

Balarka Sen

if you want to discuss i can try to help

DHMO

basically the top of P.5

I don't understand a word

Balarka Sen

14:39

@DHMO Ok, so you want to understand the fundamental theorem of symmetric polynomials, I suppose? That says that if $P(x_1, \cdots, x_n)$ is a polynomial which remains invariant under any permutation of $\{x_1, \cdots, x_n\}$, then $P(x_1, \cdots, x_n) = Q(e_1, \cdots, e_m)$ where $Q$ is also a polynomial, and $e_i$ are the elementary symmetric polynomials on $x_1, \cdots, x_n$.

DHMO

I looked that up

Balarka Sen

yup. it's a neat theorem (the general mathematical framework is known as invariant theory)

DHMO

I don't understand how this theorem applies here at all

Balarka Sen

let me read what they're doing then

Balarka Sen

14:59

@DHMO Sorry I got disconnected. Why's $\sum_{k= 1}^n f^j(\alpha_k)$ an integer polynomial in $a_d\alpha_i$? Shouldn't that be $\alpha_i$?

Oh there's a factor of $a_d^{stuff}$ on $f$.

@DHMO Once you agree that's symmetric, the rest seems straightforward. It's a polynomial on the symmetric polynomials in $a_d\theta_i$ - those are coefficients of the min. poly. of $\pi$, which are integers.

DHMO

I don't get it

Balarka Sen

can you be explicit? which part do you not get?

DHMO

What does "inner sum over k" refer to?

Balarka Sen

15:15

$\sum_{k =1}^n f^j(\alpha_k)$.

DHMO

why is it a polynomial?

Is f(x) a polynomial with integer coefficients?

Balarka Sen

It's a polynomial; not in integers coefficients.

DHMO

I can't wrap my head around f(x)

Balarka Sen

The sum, however, is a poly. with integer coefficients - that's what they are proving

DHMO

but the sum isn't even a polynomial

Balarka Sen

15:18

Polynomial in $\alpha_1, \alpha_2, \cdots, \alpha_n$, no?

DHMO

but they aren't variables

Balarka Sen

So what? A polynomial need not be on "variables".

It's a linear combination - with integers coefficients - of terms of the form $\alpha_1^{p_1} \cdots \alpha_n^{p_n}$. That's it.

We are trying to prove that the whole sum is an integer.

DHMO

"It is clear that this is a symmetric polynomial with integer coefficients..."

why are the coefficients integer?

I think I get it

@BalarkaSen why do we need to prove that they are algebraic integers?

Balarka Sen

@DHMO $a_d\alpha_1, \cdots, a_d\alpha_n$? Those are algebraic integers by assumption.

DHMO

@BalarkaSen I mean, the proof specifically mentioned that $a_d \alpha_i$ are algebraic integers

yes, I understand that they are algebraic integers

but why do we need to point that out?

Balarka Sen

15:51

@DHMO My internet is terrible today. Interesting point; it doesn't seem relevant.

Well, not directly, anyway. The elem. sym. polys on $a_d\alpha_i$ are coefficients of the min poly of $a_di\pi$, and that's the relevant bit

DHMO

but even if the min poly isn't monic

the coefficients are still integers

Balarka Sen

the elementary symmetric polynomials on the roots of a non-monic polynomial are not in general integers, @DHMO

DHMO

oh, thanks

2 hours later…

DHMO

18:13

@BalarkaSen so are we done with pi?

Balarka Sen

@DHMO I didn't make an effort to understand the proof yet. Can you explain it in a couple of sentences?

DHMO

ah, tomorrow.

you know my time i know your time

Balarka Sen

@DHMO Sure.

Transcript for

Dec28

Dec '1629

Dec30

$Mathematics$ Algebraic/Transcendence Theory

Discussions about algebraic numbers and transcendent numbers.

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$