# Proof $\pi$ is transcendental without symmetric function theory

Recently for a bonus homework assignment in my algebra class, I was asked to review the literature and write up a proof that $$\pi$$ is transcendental. Essentially every source I found ("The Transcendence of $$\pi$$" by Steve Mayer for example) presents the classic proof of Lindemann, which heavily relies on symmetric function theory and in particular the fundamental theorem of elementary symmetric functions.

Before this assignment, I did not know symmetric function theory, and by far the biggest difficulty in my solution and write up was understanding this theory and the argument used in the proof (which in my opinion, was not spelled out enough for a beginner to the theory to easily understand the argument in the sources I consulted).

Now, symmetric function theory was useful to learn, and I am aware it is very useful in the proof of the Lindemann-Weierstrass theorem, but it begs the question: Is there any proof (preferably understandable to approximately a beginning graduate student) that $$\pi$$ is transcendental, without using symmetric function theory, and if not, is there a theoretical explanation for why?

Edit: Crossposted to MO after 2.5 weeks https://mathoverflow.net/questions/466288/proof-pi-is-transcendental-without-symmetric-function-theory

• @user2661923 Defining $e^z$ for complex $z$ as a Taylor series is not "artificial", it's a very natural generalization from the real case. And once that's done, $e^{i\pi} = \sum \frac{(i\pi)^n}{n!}$ doesn't look like gibberish to me.
– Sam
Commented Feb 15 at 16:54
• To settle this, the problem is not in the definition of the complex exponential, but rather (roughly as m-stgt points out) that the proof that goes: $e$ is transcendental, therefore $e^{i\pi}$ is algebraic only if $i\pi$ is transcendental, is essentially the entire fact to be proven (or a stronger form of it like Lindemann-Weierstrass). Commented Feb 15 at 16:57
• You should understand the need of symmetric function theory here. It is needed to ensure that certain polynomial used in the proof has integer coefficients (this also applies for derivatives of the polynomial). There may be an alternative in Galois theory but I guess that is even more complicated compared to symmetric function theory. Commented Feb 16 at 0:13
• Most irrationality/transcendence proofs use an expression which is a non-zero integer and hence has absolute value greater than or equal to $1$ and the same expression can be shown to be arbitrarily small in absolute value thereby giving a contradiction. Proving the expression to be a non-zero integer is where one needs the coefficients of the polynomial referred in previous comment. Commented Feb 16 at 0:18
• I disagree with the implication that "$e$ is transcendental, therefore $e^{i\pi}$ is algebraic only if $i\pi$ is transcendental". Indeed, let $a\not\in\{0,1\}$ and $b\not\in\mathbb Q$ be any algebraic numbers. Then, by Gelfond–Schneider theorem, answering Hilbert's seventh problem, $a^b$ is transcendental. But the numbers $1/b$ and $(a^b)^{1/b}=a$ are algebraic. Commented Feb 16 at 4:08

Matt Baker presents a $$p$$-adic argument to prove the Lindemann-Weierstrass theorem (and thus the transcendence of $$\pi$$) here: https://mattbaker.blog/2015/03/20/a-p-adic-proof-that-pi-is-transcendental/