Discrepancy in Proof of π's Transcendence

Question

Consider the following:

"From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{iπ}$ is transcendental.

However, from Euler's Identity: $$e^{iπ} = −1$$

which is the root of $h(z) = z+1$

and so is algebraic. This contradicts the conclusion that $e^{iπ}$ is transcendental. Hence by Proof by Contradiction it must follow that $π$ is transcendental."

Monic, minimal polynomial expressions are of degree two. When no such polynomial exists for an irrational number it is deemed transcendental. Note that $e^{iπ} = −1$ is the root of a linear equation: $h(z) = z+1$. How is this possible? Logically, $e^{iπ} = −1$ should be the root of a quadratic equation in order for the contradiction to be warranted. Doesn't this call into question the proof of $π$ transcendence?

Answer 1

Not really. It is not true that all minimal monic polynomials are of degree $2$. Precisely, $x+1$, the minimal polynomial of $-1$, has degree $1$. There are minimal polynomials of any degree.

Answer 2

As it was pointed out already, it is allowed in the theory for the minimal polynomial to be linear.

If you do not like this, you can reach a contradiction by observing that $$ e^{ i \frac{\pi}{2}}=i $$ is algebraic with minimal polynomial $x^2+1$. This implies that $\frac{\pi}{2}$ is transcendental, and it is easy to show that implies that $\pi$ must also be transcendental.