question on proof in Frohlich/Taylor regarding extension of a discrete absolute value on a complete field

Question

I am having some trouble understanding a proof in Fröhlich and Taylor, pages 105-106.

There $K$ is a complete field with a discrete absolute value $u$ $\mathfrak o,\mathfrak p$ is the valuation ring, prime ideal, respectively, associated with $u$. $E/K$ is a finite separable extension and $\mathfrak o_E$ is the integral closure of $\mathfrak o$ in $E$.

They are trying to show that there is a unique prime ideal $\mathfrak P$ of $\mathfrak o_E$ over $\mathfrak p$ and that $\mathfrak o_E=\varprojlim \mathfrak o_E/\mathfrak P^n$. The latter statement implies that $E$ is complete.

Indeed, if there were two primes above the prime ideal $\mathfrak p$ of $\mathfrak o$, then the $B:=\mathfrak o_E/\mathfrak p\mathfrak o_E\big/$Rad$(\mathfrak o_E/\mathfrak p\mathfrak o_E)$ has a non-trivial idempotent. By (I.1.40) this lifts to a non-trivial idempotent $e$, say, of $\mathfrak o_E/\mathfrak p\mathfrak o_E$. Since $\mathfrak o$ is a PID, by Theorem 5(a) we know that $\mathfrak o_E$ is a free $\mathfrak o$-module of rank $(E:K)$; hence by Theorem 12, $\mathfrak o_E=\varprojlim \mathfrak o_E/\mathfrak p^n\mathfrak o_E$.

Questions/doubts:

1. Why the existence of two prime ideals above $\mathfrak p$ implies the existence of non-trivial idempotents in the $\mathfrak o/\mathfrak p$-algebra $B$? I feel I am missing some commutative algebra knowledge here.

2. I assume the argument about freeness comes from the fact that $\mathfrak o_E$ is f.g., clearly torsion-free over $\mathfrak o$ which is a PID (being a valuation ring), and f.g. torsion-free modules over PIDs are free. Correct?

3. I don't understand the argument after that, where Theorem 12 is used.

4. Using $e$, the proof then constructs an idempotent element $e'\in\mathfrak o_E$ with certain properties. I understand the construction of $e'$ and its properties but I am not sure why/where we need it. Is it because $\mathfrak o_E$ is a Dedekind ring, in particular an integral domain, so it shouldn't have idempotents other than 0 and 1? So this would contradict the existence of idempotents in $B$, and hence would imply there is only one prime ideal above $\mathfrak p$?