equivalent statements of Lindemann-Weierstrass theorem

equivalent statements of Lindemann-Weierstrass theorem

Proposition 1

.

The following versions of the

Lindemann-Weierstrass Theorem

are

equivalent

:

1.

If

α

1

,

…

,

α

n

are

linearly independent

algebraic numbers

over

ℚ

, then

e

α

1

,

…

,

e

α

n

are

algebraically independent

over

ℚ

.

2.

If

α

1

,

…

,

α

n

are distinct algebraic numbers over

ℚ

, then

e

α

1

,

…

,

e

α

n

are linearly independent over

ℚ

.

Proof.

(

1

⟹

2

)

.

Write

A

i

=

e

α

i

for each

i

=

1

,

…

,

n

. Suppose

0

=

r

1

⁢

A

1

+

⋯

+

r

n

⁢

A

n

,
where

r

i

∈

ℚ

. Moving

r

1

⁢

A

to the LHS and by multiplying a common

denominator

we can assume that

r

1

⁢

A

1

=

r

2

⁢

A

2

+

⋯

+

r

n

⁢

A

n

where

r

i

∈

ℤ

. We want to show that

r

1

=

⋯

=

r

n

=

0

. We induct on

n

. The case when

n

=

1

is

trivial

because

A

1

is never 0 and therefore

0

=

r

1

⁢

A

1

forces

r

1

=

0

.

By

induction hypothesis

, suppose the statement is true when

n

<

k

. Now suppose

n

=

k

. If

α

1

,

…

,

α

k

are linearly independent over

ℚ

,

A

1

,

…

,

A

k

are algebraically independent and certainly
linearly independent over

ℚ

. So suppose

α

1

,

…

,

α

k

are not linearly independent over

ℚ

.
Without loss of generality, we can assume

s

1

⁢

α

1

=

s

2

⁢

α

2

+

⋯

+

s

k

⁢

α

k

, where

s

i

∈

ℚ

and

s

1

≠

0

. By multiplying a common denominator we can further assume that

s

i

∈

ℤ

and

s

1

>

0

. Then

A

1

s

1

=

e

s

1

⁢

α

1

=

e

s

2

⁢

α

2

+

⋯

+

s

k

⁢

α

k

=

e

s

2

⁢

α

2

⁢

⋯

⁢

e

s

k

⁢

α

k

=

A

2

s

2

⁢

⋯

⁢

A

k

s

k

.

Since

r

1

⁢

A

1

=

r

2

⁢

A

2

+

⋯

+

r

n

⁢

A

n

, we get

(

r

1

s

1

)

⁢

(

A

2

s

2

⁢

⋯

⁢

A

k

s

k

)

=

(

r

1

s

1

)

⁢

A

1

s

1

=

(

r

1

⁢

A

1

)

s

1

=

(

r

2

⁢

A

2

+

⋯

+

r

k

⁢

A

k

)

s

1

=

g

⁢

(

A

2

,

…

,

A

k

)

,

where

g

⁢

(

x

2

,

…

,

x

k

)

=

(

r

2

⁢

x

2

+

⋯

+

r

k

⁢

x

k

)

s

1

∈

ℚ

⁢

[

x

2

,

…

,

x

k

]

.

Partition

the numbers

s

i

′

⁢

s

into non-negative and

negative

ones, so that, say,

s

c

⁢

(

1

)

,

…

,

s

c

⁢

(

l

)

are non-negative and

s

d

⁢

(

1

)

,

…

,

s

d

⁢

(

m

)

are negative, then

(

r

1

s

1

)

⁢

A

c

⁢

(

1

)

s

c

⁢

(

1

)

⁢

⋯

⁢

A

c

⁢

(

i

)

s

c

⁢

(

i

)

=

g

⁢

(

A

2

,

…

,

A

k

)

⁢

A

d

⁢

(

1

)

s

d

⁢

(

1

)

⁢

⋯

⁢

A

d

⁢

(

j

)

s

d

⁢

(

j

)

.

If we define

f

⁢

(

x

2

,

…

,

x

k

)

=

r

1

s

1

⁢

x

c

⁢

(

1

)

s

c

⁢

(

1

)

⁢

⋯

⁢

x

c

⁢

(

i

)

s

c

⁢

(

i

)

-

g

⁢

(

x

2

,

…

,

x

k

)

⁢

x

d

⁢

(

1

)

s

d

⁢

(

1

)

⁢

⋯

⁢

x

d

⁢

(

j

)

s

d

⁢

(

j

)

,

then

f

⁢

(

x

2

,

…

,

x

k

)

∈

ℚ

⁢

[

x

2

,

…

,

x

k

]

and

f

⁢

(

A

2

,

…

,

A

k

)

=

0

.

By the induction hypothesis,

f

=

0

. It is not hard to see that

r

1

=

⋯

=

r

k

=

0

and
therefore

A

1

,

…

,

A

k

are linearly independent.

(

1

⟸

2

)

.

We first need two lemmas:

Lemma 1.

Given 2., if

α

≠

0

is algebraic over

ℚ

, then

e

α

is transcendental
over

ℚ

.

Proof.

Suppose

f

⁢

(

e

α

)

=

0

where

f

⁢

(

x

)

=

r

0

+

r

1

⁢

x

+

⋯

+

r

n

⁢

x

n

∈

ℚ

⁢

[

x

]

. Then we have

0

=

r

0

+

r

1

⁢

e

α

+

⋯

+

r

n

⁢

(

e

α

)

n

=

r

0

⁢

e

0

+

r

1

⁢

e

α

+

⋯

+

r

n

⁢

e

n

⁢

α

.

Since

α

≠

0

,

0

,

α

,

…

,

n

⁢

α

are all distinct,

1

,

e

α

,

…

,

e

n

⁢

α

are
linearly independent by the

hypothesis

. Thus,

r

0

=

r

1

=

…

=

r

n

=

0

and we have

f

⁢

(

x

)

=

0

, which means that

e

α

is transcendental over

ℚ

.
∎

Lemma 2.

Given 2., if

α

and

β

are linearly independent and algebraic over

ℚ

, then

e

α

is transcendental over

ℚ

⁢

(

e

β

)

.

Proof.

Let

A

=

e

α

and

B

=

e

β

. Suppose

f

⁢

(

A

)

=

0

where

f

⁢

(

x

)

∈

ℚ

⁢

(

B

)

⁢

[

x

]

. We want to show that

f

⁢

(

x

)

=

0

.
Write

f

⁢

(

x

)

=

r

0

⁢

(

B

)

+

r

1

⁢

(

B

)

⁢

x

+

⋯

+

r

n

⁢

(

B

)

⁢

x

n

,

where each

r

i

⁢

(

x

)

=

p

i

⁢

(

x

)

/

q

i

⁢

(

x

)

with

p

i

⁢

(

x

)

,

q

i

⁢

(

x

)

≠

0

∈

ℚ

⁢

[

x

]

. Let

Q

⁢

(

x

)

=

q

1

⁢

(

x

)

⁢

⋯

⁢

q

n

⁢

(

x

)

. So

Q

⁢

(

B

)

, being the

product

of the denominators

q

i

⁢

(

B

)

≠

0

, is non-zero.
Multiply

f

⁢

(

x

)

by

Q

⁢

(

B

)

we get a new

polynomial

g

⁢

(

x

)

such that

g

⁢

(

x

)

=

R

0

⁢

(

B

)

+

R

1

⁢

(

B

)

⁢

x

+

⋯

+

R

n

⁢

(

B

)

⁢

x

n

,

where each

R

i

⁢

(

x

)

=

r

i

⁢

(

x

)

⁢

Q

⁢

(

x

)

=

p

i

⁢

(

x

)

⁢

Q

⁢

(

x

)

/

q

i

⁢

(

x

)

∈

ℚ

⁢

[

x

]

. Now,

g

⁢

(

A

)

=

f

⁢

(

A

)

⁢

Q

⁢

(

B

)

=

0

. So

0

=

R

0

⁢

(

B

)

+

R

1

⁢

(

B

)

⁢

A

+

⋯

+

r

n

⁢

(

B

)

⁢

A

n

=

∑

j

=

0

m

0

a

0

⁢

j

⁢

B

j

+

∑

j

=

0

m

1

a

1

⁢

j

⁢

B

j

⁢

A

+

⋯

+

∑

j

=

0

m

n

a

n

⁢

j

⁢

B

j

⁢

A

n

=

∑

j

=

0

m

0

a

0

⁢

j

⁢

e

j

⁢

β

+

∑

j

=

0

m

1

a

1

⁢

j

⁢

e

j

⁢

β

+

α

+

⋯

+

∑

j

=

0

m

n

a

n

⁢

j

⁢

e

j

⁢

β

+

n

⁢

α

,

where each

a

i

⁢

j

∈

ℚ

. Now, the exponents in the above equation are all distinct, or else we would end up with

α

and

β

being linearly dependent, contrary to the

assumption

. Therefore, by 2 (Lindemann-Weierstrass
Version 2), all

e

i

⁢

β

+

j

⁢

α

are linearly independent, which means each

a

i

⁢

j

=

0

. This implies that

g

⁢

(

x

)

=

0

.
But

g

⁢

(

x

)

=

f

⁢

(

x

)

⁢

Q

⁢

(

B

)

and

Q

⁢

(

B

)

≠

0

, we must have

f

⁢

(

x

)

=

0

.
∎

Now onto the main problem. We proceed by

induction

on the number of linearly independent algebraic elements over

ℚ

. The case when

n

=

1

is covered in Lemma 1, since a linearly independent

singleton

is necessarily non-zero.
So suppose

α

1

,

…

,

α

k

are linearly independent and algebraic over

ℚ

. Then each pair

α

k

,

α

i

are independent and algebraic over

ℚ

,

i

≠

k

. Thus

e

α

k

is transcendental over

ℚ

⁢

(

e

α

i

)

for all

i

≠

k

. This means that

e

α

k

is transcendental over

ℚ

⁢

(

e

α

1

,

…

,

e

α

k

-

1

)

.

Now let

A

i

=

e

α

i

for all

i

=

1

,

…

,

k

. Suppose

f

⁢

(

A

1

,

…

,

A

k

)

=

0

where

f

∈

ℚ

⁢

[

x

1

,

…

,

x

k

]

.
To show the algebraic independence of the

A

i

′

⁢

s

, we need to show that

f

=

0

. Rearranging terms of

f

and we have

0

=

f

⁢

(

A

1

,

…

,

A

k

)

=

∑

j

=

0

m

g

j

⁢

(

A

1

,

…

,

A

k

-

1

)

⁢

(

A

k

)

j

.

If we let

g

⁢

(

x

)

=

f

⁢

(

A

1

,

…

,

A

k

-

1

,

x

)

, we see that

g

⁢

(

x

)

∈

ℚ

⁢

(

A

1

,

…

,

A

k

-

1

)

⁢

[

x

]

and

g

⁢

(

A

k

)

=

f

⁢

(

A

1

,

…

,

A

k

-

1

,

A

k

)

=

0

. Since

e

α

k

is transcendental over

ℚ

⁢

(

A

1

,

…

,

A

k

-

1

)

, we must have

g

⁢

(

x

)

=

0

. This implies that each

g

j

⁢

(

A

1

,

…

,

A

k

-

1

)

=

0

. But then

A

1

,

…

,

A

k

-

1

are algebraically
independent by the induction hypothesis, we must have each

g

j

=

0

. This means that

f

=

0

.
∎

Title

equivalent statements of Lindemann-Weierstrass theorem

Canonical

name

EquivalentStatementsOfLindemannWeierstrassTheorem

Date of creation

2013-03-22 18:05:18

Last modified on

2013-03-22 18:05:18

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

4

Author

CWoo (3771)

Entry type

Result

Classification

msc 12D99

Classification

msc 11J85

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