I have a question on this article proving $\zeta(3)\notin\mathbb{Q}.$ by using modular forms. This is theorem 1 at page 275 (page 5 in the pdf). Most things in the proof are clear but I don't get the penultimate sentence at all, e.g.
Furthermore, it [$E(t)(f(t)-\zeta(3))$] cannot be a polynomial in $t$, since then $f(\tau)-\zeta(3)$ would be a modular form of weight $-2$, which is impossible.
I would say that this has something to do with $E(\tau)$ being modular of weight $2$ but I am really not sure.
Hoping that somebody out there has an idea for me. Thanks!
I think the argument is something like this. Note that $t=t(\tau)$ is a meromorphic modular function for $\Gamma_1(6)$. Therefore, if $E(t)(f(t)−\zeta(3))$ is a polynomial in $t$, then $E(\tau)(f(\tau)-\zeta(3))$ would be a meromorphic modular function for $\Gamma_1(6)$. As you said, $E(\tau)$ is a modular form of weight $2$ for $\Gamma_1(6)$. It follows that $f(\tau)-\zeta(3)$ is a weight $-2$ modular form for $\Gamma_1(6)$. Such a form is necessarily $0$. Contradiction since the $3$rd derivative $(2\pi i)^3F$ of $f$ is nonzero.