On pi being transcendental [closed]

Question

There are probably 101 reasons why this argument is plain wrong.

At the same time however there probably is a subtle truth to it I imagine:

We note that $\sin(x) \in \mathbb{Q}[[x]]$

Suppose $\pi$ is algebraic. Let $f(x)$ be the minimal polynomial with one root being $\pi$.

Then $f(x) | \sin(x)$. Note all the roots of $\sin(x)$ are of the form $k\cdot \pi$ where $k \in \mathbb{Q}$ .

So lets us consider the field $K/\mathbb{Q}$ which is the Galois Closure of $\mathbb{Q}(\pi)/\mathbb{Q}$

Let $G=Gal(K/\mathbb{Q})$ and for all $\sigma \in G$ we have $\sigma(\pi) = k \cdot \pi$ for some $k \in \mathbb{Q}$.

Now from this it follows that $N_{K/\mathbb{Q}}(\pi)=Q_1 \cdot \pi^{n} = Q_2$

Where quite obviously $n=[K:\mathbb{Q}]$ and $Q_1, Q_2 \in \mathbb{Q}$.

So you get then that if $\pi$ is algebraic then $\pi = \sqrt[n]{Q}$ where $Q \in \mathbb{Q}$

So it follows then that $K=\mathbb{Q}(\pi, \zeta_n)$

Also note then that $\sigma(\pi)={\zeta_n}^a \cdot \pi$ for some $a \in \mathbb{Z}$

By the argument before we have that ${\zeta_n}^a \in \mathbb{Q}$

So we must conclude that $\mathbb{Q}(\pi) = \mathbb{Q}$

Conclusion: $\pi$ is algebraic $\Leftrightarrow$ $\pi \in \mathbb{Q}$

Love to know what you all think of it, criticisms et all.

Thank You!!!

Answer 1

Your argument breaks down at the point where you claim that if $\alpha$ is an algebraic number which is a root of an entire function with rational coefficients, then the algebraic conjugates of $\alpha$ are also roots (as this is the step that allows you to claim that the algebraic conjugates of $\pi$ must be among the $k \pi$ which is the claim "$f(x) | \sin x$")).

For example let $\alpha =\sqrt 2 +1$ so its conjugate is $\beta=1-\sqrt 2, |\beta|<|\alpha|$ and one can easily find by induction positive rational numbers $c_n, n \ge 1$ st $\sum c_n\alpha^n=3$ and $f(z)=-3+\sum c_nz^n$ is entire; then clearly $f(\beta) < 0$ and $f(\alpha)=0$ and $f$ has rational coefficients

(By induction one can easily find $0<a_k <1/k!, (z-\alpha)\sum a_kz^k$ has positive rational coefficients except for the first one which is $-3$ so with $a_0=3/\alpha <1$ starting the induction so one needs to choose $0<a_1<1$ st $3/\alpha-\alpha a_1$ is positive rational etc; then $g(z)=\sum a_kz^k$ is entire hence $f(z)=(z-\alpha)g(z)$ is too and its coefficients are the ones above)