Problem with Veblen's proof for the transcendence of $\pi$

Question

I'm trying to understand the following proof, (no need to read all of it), but there's one point where I'm stuck. The proof is by contradiction, so they use the definition of an algebraic number (a is algebraic if there exists a polynomial in the field extension such that $f(a)=0$:

"$x_1,...,x_n$ are algebraic, so they are the roots of an equation $$f(x) = a_0+a_1x+...+a_nx^n$$

with integral coeffiecients, $a_0\neq0, a_n\neq0$."

Yet, on the last page the Girard–Newton formulas are used:

"But from Newton's formulas $$S_1+a_1=0, S_2+a_1S_1+2a_2+0,...$$ it follows that $S_1,S_2,...,S_{s-p}$ are whole numbers.

Where from my own calculations based on Wikipedia (so not sure if this is correct), these formulas are given by

$$\sum^k_{i=1} (-1)^{i-1}a_{n-k+i}S_i = ka_{n-k}$$ for $1 \leq k \leq n$.

Meaning that we would get $a_nS_1 + a_1 = 0$ etc. So if $a_n$ isn't 1, if $f$ isn't monic, $S_i$ wouldn't be an integer. Why do we suddenly assume $f$ to be monic, since this doesn't follow from the definition of an algebraic number. Is the proof wrong? What am I missing? This has been messing with me for days, thank you in advance.

Edit: This is guessing work, but I think we may assume $f$ is monic from the start. By looking at the proof for the transcendence of $e$, this is obvious.But I think there might be a way to choose $x_1,...,x_n$ as roots for $f$ such that it is monic and

$$c+\sum^{n}_{i=1} e^{x_i} = 0.$$

Edit²: The formulas should be as follow: $a_nS_1 + a_{n-1} = 0, S_2 = S_1^2-2\frac{a_{n-2}}{a_n}$ etc

Answer 1

The statement is wrong.

The author states that if $f(x)=\sum_{0\le i\le n}a_ix^i$ is such that $a_0,a_n\ne0$ then \begin{align}S_1+a_1&=0\\S_2+a_1S_1+2a_2&=0.\end{align}

This is not true; for instance the monic polynomial $1-5x-4x^2+x^3$ has $a_1=-5$ but the sum of roots is $S_1=4$.

How did this error arise?

At the bottom of the page, there is the reference "*Cf. Burnside and Panton, Theory of Equations, Chapter VIII, or any book on higher algebra."

I found the PDF version of the book and Article 126, Chapter XII (note that Chapter VIII discusses limits instead) states

Let the equation be \begin{align}f(x)&=x^n+p_1x^{n-1}+p_2x^{n-2}+\cdots+p_n\\&=(x-a_1)(x-a_2)\cdots(x-a_n).\end{align}

[next page] whence, comparing this value of $f'(x)$ with the former, we have the following relations: \begin{align}s_1+p_1&=0\\s_2+p_1s_1+2p_2&=0\\s_3+p_1s_2+p_2s_1+3p_3&=0\\\cdots\end{align}

There are two things to note here. First is that this polynomial is monic, so the following discussion only applies to this case. Secondly, the indexing is the other way around; that is, $f(x)=\sum_{0\le i\le n}p_{n-i}x^i$.

Therefore, it is likely that these two points combined constituted to the incorrect equalities in the original paper.