Let $L/\Bbb{Q}_p$ be ramification index $e$ extension. Let $π$ be prime element of $L$. Then, $p＝π^eu$ ($u$: unit element of ring of integers of $L$)。
But I have met the remark that we can choose prime element $π$ such that $p＝π^e$ (I cannot find the reference now, sorry・・・).
I think this is true in the case $e＝1$, but when $e≧2$, I think we cannot take $π$ in general・・・
Is the remark correct statement ? If not, I want to know counterexample, that is, example of $L/\Bbb{Q}_p$ such that there is no prime element $π$ such that $p＝π^e$.
Let $d$ be a quadratic nonresidue modulo $p$, and take $L = \mathbb{Q}_p(\sqrt{dp})$. Then $e = 2$, so we want to show that $p$ is not square in $L$.
Suppose that $p$ is square in $L$. Then there is some $\alpha \in L$ with $p = \alpha^2$. Clearly there are $a,b\in\mathbb{Q}_p$ such that $\alpha = a + b\sqrt{dp}$, so $$ p = (a + b\sqrt{dp})^2 = a^2 + b^2dp + 2ab\sqrt{dp}, $$ so we have $$ p = a^2 + b^2dp, \quad 2ab = 0. $$ Since $2ab = 0$, either $a = 0$ or $b=0$. Suppose that $a = 0$. Then $b^2dp = p$ so $b^2d = 1$. This implies that $\lvert b^2\rvert_p = 1$, so $b \in \mathbb{Z}_p^\times$ and $d \equiv (b^{-1})^2 \pmod{p}$, which is impossible since $d$ is a quadratic nonresidue. Suppose instead that $b = 0$. Then $a^2 = p$, which is impossible since $a \in \mathbb{Q}_p$.