It is known that the fact that $\zeta(1+it)\ne 0$ for each $t$ is equivalent to the Prime Number Theorem. I want to understand succinctly
(a) Why $\zeta(1+it)\ne 0$
I know the very long proof of this, but I also think it might be ok to note that $$\zeta(1+it)\gg \frac{1}{(\log t)^7}$$ and so does not vanish?
(b) Why this is equivalent to the Prime Number Theorem? Especially in the form $\psi(x)\sim x$.
Can anyone elaborate on these?
Your questions are answered in the standard literature (and there are also shorter proofs). Let me give a few hints:
Question $(a)$: Assume that $\zeta(1+it)=0$. Then $t\neq 0$. By a standard lemma we have for all $\sigma>1$, \begin{align*} \left((\sigma-1)\zeta(\sigma)\right)^3 \left\lvert\frac{\zeta(\sigma+it)}{\sigma-1}\right\rvert^4|\zeta(\sigma+2it)| & \ge \frac{1}{\sigma-1}. \\ \end{align*} Consider the limit $\sigma\to 1+$. The first factor tends to $1$, because $\zeta(s)$ has residue $1$ at the pole $s=1$. The third factor tends to $|\zeta(1+2it)|$. Then the second one satisfies \begin{align*} \left\lvert\frac{\zeta(\sigma+it)}{\sigma-1}\right\rvert^4 & = \left\lvert\frac{\zeta(\sigma+it)-\zeta(1+it)}{\sigma-1}\right\rvert^4 \to |\zeta'(1+it)|^4 \\ \end{align*} for $\sigma\to 1+$. So the left side tends to $|\zeta'(1+it)|^4 \cdot |\zeta(1+2it)| $, but the right side tends to $\infty$, for $\sigma\to 1+$. This is a contradiction.
Question $(b)$: See here: Why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?
