Primitive element extension of discrete valuation ring.

Assume $K$ is a complete discrete valuation ring, with valuation ring $A$ and maximal ideal $\mathfrak p$, $E$ a finite extension of K and $B$ is the integral closure if $A$ in $E$, $\mathfrak B$ the unique prime lying above $\mathfrak p$ in $B$.

Is it always the case that $B/\mathfrak B=A/\mathfrak p(b)$ for some $b$ in $B/\mathfrak B$? Or we need conditions like $A/\mathfrak p$ perfect? (Is $E$ separable over $K$ enough?)

• Hey CYC, if you found Hagen's answer useful, it's usually considered courteous to upvote as well as accept to add to the understanding of total people who agree with it (so that future users can see which approaches are generally favored by the community). – Adam Hughes Feb 28 '17 at 18:59

No it is not always the case: take a finite inseparable extension $l|k$ possessing no primitive element and consider the ring extension $l[[t]]|k[[t]]$ (rings of power series in $t$). The extension $l((t))|k((t))$ of the fields of fractions is inseparable too and the corresponding extension of residue fields is $l|k$.