In my fresh look at Galois Theory that I’ve had a year ago, I was content when I arrived at the proofs for the impossibility of the classical geometry problems (squaring the circle and the like) and for the Fundamental Theorem of Algebra (which I presented here). A gap in the business about squaring the circle was the proof that is actually not contained in some finite field extension of having a degree of a power of . Actually, as Lindemann proved in 1882, there is a stronger result: is even transcendental. I have taken some time to look at the proof of this, and as a by-product, we can also prove that is transcendental.

Again, quite a stronger result holds which is known as the Lindemann-Weierstrass-Theorem: If the equation

holds for some , algebraic numbers , and pairwise distinct algebraic numbers – then all the must be equal to . We won’t prove that, even though the ideas are roughly the same as presented here. The results that follow are special cases of this theorem.

By usual cardinality arguments, it is easy to see that there are very many transcendental numbers. Actually putting your finger on one of these numbers is quite hard. Liouville was the first to give a criterion for this in 1851. Hermite proved in 1873 that is transcendental, and 9 years later, Lindemann followed up about .

I have learned these proofs from an exposition I found on the internet, and hence the notation will be very close to this exposition. I also took some clues from a German book for undergraduate students. In this book, the complete Lindemann-Weierstrass-Theorem is proved.

We will first be concerned with the following statement:

**Theorem**: The number is transcendental.

For the *proof* we will apply the function

Let be a polynomial of degree .

If the coefficients of are , we will use the abbreviation for the polynomial which has as its coefficients. Then, it is immediate:

Integration by parts and a little induction will show

**The Integral-Lemma**: Let and be defined as above. Then,

.

*Proof*: Let us integrate by parts:

For the inductive step, let’s assume we have done this integration by parts times already and do it now once again:

which means that this equation holds for all . On the other hand, is identical to , since is the degree of . So, the integral will vanish after such steps, and the assertion follows. **q.e.d.**

Note that we have vitally used the properties of the number here, to be precise we used that the function is it’s own derivative. This is the one moment in the proof that hinges specifically on the number and this is why we will have proved that is transcendental in the end.

We wish to prove that is not algebraic. So, let’s assume otherwise: we assume that there is some such that . Without loss of generality, we can take (or else, take out and use the new polynomial instead of ).

Consider , which equals (by the Integral-Lemma):

where we used the assumption that . Note, that the polynomial itself has not mattered yet. Let a sufficiently large prime number with and and let us specify as follows:

The degree of is then . We will show that with this choice of is divisible by , but not divisible by . Each term in is divisible by , except for .

Let us first deal with the first summand of . The following is true:

**The First -Lemma**: is divisible by for any , and not divisible by for .

**Proof**: By the product rule and the binomial theorem, we have

Now,

and hence

Therefore, only the case is relevant (which immediately implies to get a non-trivial case), and we still have to take care of the term

The case is particularly easy, we arrive at

In particular, this expression is not divisible by , since we chose a prime from the start.

For the cases , we use the product rule again, and the multinomial theorem for the first time:

where is an integer of which the exact value is not important (note: may vanish, for instance if is sufficiently large). This implies

Altogether, we have shown

That’s our claim. **q.e.d.**

The second summand of works similarly, but the details are a little more cumbersome.

**The second -Lemma**: is divisible by for any and any .

**Proof**: By the product rule and the binomial theorem, we have

Of course,

and so

We now have left to take care of the term

For , we find that this is equal to

where is an integer (note that is prime, hence and hence there will be no fraction in this term). In particular, is divisible by .

For , we can apply the multinomial theorem once again to find

This means, that we have proved

where is some integer which is of no importance in itself (note: may vanish, if is sufficiently large for instance – in the pathological cases, the binomial and multinomial coefficients vanish by definition). This is what we claimed: is divisible by for any and hence for any . **q.e.d.**

Where do we stand? We consider the expression

where is a certain polynomial that has multiple roots in . As in the proof of both -Lemmas, we find that for any and for any and . Hence

.

The second -Lemma showed, that the second summand is divisible by . The first -Lemma showed that is only divisible by , not by . Now, since we chose in the start, the whole first summand of is not divisible by . On top of all that, is an integer, since the are by assumption and the other terms are by construction and by the Lemmas. Altogether, we conclude is a multiple of , but not divisible by , and hence

.

Now, we give an upper bound for for any :

From the immediate bound we gave at the very beginning,

with the unimportant constants and that do not depend on .

So, on the one hand, is large, (larger than ), on the other hand it’s small (smaller than ). This is impossible, because eventually gets larger than — this can be seen by considering the exponential sequence that is summable and hence converges to . In particular, eventually, which means . There’s the contradiction: cannot satisfy both inequalities that we have shown. The contradiction has its roots in our assumption — must be transcendental. **q.e.d.**

We shall now prove

**Theorem**: is transcendental.

In this *proof* we will use many of the ideas from the proof about , and therefore we will not be as detailed as above. Besides, the proof requires some results from algebra courses. We will briefly review those here.

The number is called an algebraic integer, if its minimal polynomial has coefficients in . The minimal polynomial is the (unique) polynomial of lowest degree which has as a root and whose leading coefficient is equal to .

**The Integer Lemma**:

- Any algebraic integer is already in .
- If is algebraic then there is some such that is an algebraic integer.

*Proof*: 1. Any is algebraic, because its minimal polynomial is . By assumption, is an algebraic integer, hence its minimal polynomial is in . Hence, , as claimed.

2. Let us designate the minimal polynomial of by . Let us put

.

Any such coefficient can be written as for appropriate and . Let so we can get

.

This polynomial is not necessarily a minimal polynomial, since its leading coefficient is no longer equal to , in general. But, on the other hand, we find

$latex \begin{aligned}

0 = g(z) = c_n^{n-1}qg(z) &=\displaystyle\sum_{k=0}^n c_n^{n-1}c_kz^k\\

&= \displaystyle \sum_{k=0}^{n-1} c_n^{n-1-k} c_k c_n^kz^k + c_n^{n-1}c_nz^n\\&= \displaystyle \sum_{k=0}^{n-1} c_n^{n-1-k} c_k (c_nz)^k + (c_nz)^n\\&= \displaystyle \sum_{k=0}^n d_k (c_nz)^k\\&=: \tilde g(c_nz).\end{aligned}$

Thus, we have found a polynomial which has as a root and which has leading coefficient . This means is an algebraic integer. A closer inspection shows that is equal to which, in turn, was the lcm of several positive integers, itself is positive. Hence we can choose in the claim and we are done. **q.e.d.**

**The Lemma on Algebraic Numbers** (aka the **LAN**): If and are algebraic, then so are , and .

*Proof*: This follows from a theorem in field theory stating that in a field extension , is algebraic if and only if its degree is finite. This can easily be seen as follows:

In a field extension of degree , the elements are linear dependent, so we find . This is a polynomial which has as its root, so is algebraic. On the other hand, if is algebraic and is the degree of its minimal polynomial, any has a representation as – keep in mind that is a vector space over and the set must be linear dependent. In particular, the degree of the extension is finite.

Now the usual theorem on degrees of extensions tells us that any finite field extension is algebraic, because if is a finite extension of degree , and :

,

and so is finite, hence is algebraic. Note that the converse of this is false ().

Now consider the algebraic numbers and and the field extension . By the arguments above, the extension is finite (it’s a tower of two finite extensions) and hence every element in is algebraic. The lemma is proved. **q.e.d.**

Now we can turn to the proof that is transcendental. But we won’t prove that directly, instead we show that is transcendental. Because of the LAN, if is algebraic, then so is .

Let us assume that is algebraic with minimal polynomial and degree . We will designate with the leading coefficient of that arises from getting rid of the denominators in . Let , be the roots of . Then, by the Integer Lemma, is an algebraic integer. It is immediate that

because of . By multiplication of this product, we get terms that look like where and (this comes from the fact that you have to choose either or from each factor, and each of the possible choices will be made once).

Let us denote the non-zero summands by and the number of non-zero summands by . So, there are summands that show up as . Altogether,

.

Now we re-use some ideas from the proof that is transcendental. Let be a large prime number and let

Here, consider the expression

and note that it is symmetric in the : when you re-number the , you will also re-number the , but the polynomial given here will not change. The Fundamental Theorem of Elementary Symmetric Polynomials says that this symmetric polynomial can be written as a polynomial in the elementary symmetric polynomials with coefficients in the underlying field . In particular, the polynomial

has coefficients in , the remaining factor does not affect any of the coefficients. Besides, the coefficients are algebraic by assumption and by the LAN (the are algebraic). By the Integer Lemma (closely inspect the proof for the possible choice of the number in that Lemma!) and by the definition of as a leading coefficient here, the numbers are algebraic integers. The second part of the Integer Lemma now says that the polynomial has coefficients in – each of the factors of the symmetric polynomial is multiplied by to make the coefficients in algebraic integers.

Now, remember the definition of and its representation from the Integral Lemma:

,

where is the degree of as defined above. Then, we set

Similar reasoning now shows that the term

is a symmetric polynomial: when you re-number the and hence the , will not change, and by the product rule, neither will the derivatives . By the Fundamental Theorem of Elementary Symmetric Polynomials, this sum takes a value in the underlying field . Again, the Integer Lemma implies that the sum is even in (the helpful factor is of course present in the relevant derivatives as well).

Now, remember the first -Lemma to see that for any and any . This hinges on the fact that none of the factors can have vanished in the -th derivative, and then setting gives a zero of this polynomial. By the same argument, for and by the second -Lemma, is divisible by for . Finally, one can show by explicit calculation

Since this expression is symmetric, it is in , and from the arguments we already used, . The symmetry arguments also show that is an integer in . If we choose the prime sufficiently large, is not divisible by . This is a condition depending on and on the absolute value of the , hence on the degree of . In particular, the sum is divisible by .

.

Finally, the bound from the Integral Lemma tells us that

with two constants that do not depend on . The exponent comes from the definition of the polynomial .

Similar to the proof of our first theorem, we have a contradiction: is small on the one hand and large on the other hand. This shows that cannot be algebraic. **q.e.d.**