Let $(A,\mathfrak m)$ be a regular, Noetherian, local, domain of dimension $2$ and consider a prime ideal $\mathfrak p\subset A$ of height $1$. Moreover let $\hat{A}$ be the completion of $A$ with respect to the maximal ideal $\mathfrak m$ and consider a prime ideal $\mathfrak P\subset \hat{A}$ such that $\mathfrak P\cap A=\mathfrak p$
The localizations $A_\mathfrak p$ and $\hat{A}_{\mathfrak P}$ are discrete valuation rings, and the canonical immersion $A\to \hat{A}$ induces a local injective homomorphism:
$$f:A_\mathfrak p\to\hat{A}_{\mathfrak P}$$
The following result is true but I don't understand the reason, please help me:
A uniformizer element $\pi$ of $A_\mathfrak p$ is also a uniformizer of $\hat{A}_{\mathfrak P}$
I think that in this case one should prove that the extension of the maximal ideal $\mathfrak p A_{\mathfrak p}$ through $f$ is equal to $\mathfrak P A_{\mathfrak P}$...
Note: The result is quoted without proof in a paper, I don't mention the source because it doesn't matter. It is a technical detail in the proof of a bigger theorem.
I do not know about general Noetherian rings, but at least in the geometric case, this follows from the fact that the completion of a reduced ring is reduced (one of the proofs is due to Zariski). Assuming this, in your case $\mathfrak{p}=\pi A$ and then $\pi\in \widehat{A}$ is product of distinct prime elements. Then $\mathfrak{P}$ is generated by one of these primes and the rest should be clear.