Let $L/K$ be extension of number field. Let $p$ be a prime ideal of $K$. Let $P$ be a prime ideal above $p$.
Then we can define local field extension $L_P/K_p$,where $L_P$ and $K_p$ is completion at $P$ and $p$.
My question is ; For fixed $p$, there are many choice for $P$ in general, take different $P_1$ and $P_2$ for $p$.Is $L_{P_1}/K_p$ and $L_{P_2}/K_p$ the same field extension ? Or they differ ?
Thank you in advance.
On
i dont know if u mean by "the same" that they are equal ? if $L/K$ is Galois, we have $[L_{P_{1}}:K_{p}]=[L_{P_{2}}:K_{p}]=ef$, with $e$ is ramification indice of both $P_{1}$ and $P_{2}$ under $p$ and $f$ is the residue field degrees, so we can conclude that $L_{P_{1}}$ and $L_{P_{2}}$ have the same dimension over $K_{p}$ means they are isomorphic as $K_{p}$-vector space ( they are the same). but when $L/K$ is not Galois we dont have that always because of the difference of ramification indices of $P_{1}$ and $P_{2}$ and residue field degrees.
When $L/k$ is Galois then that is true, but there are many counterexamples when $L/K$ is not Galois: if $\mathfrak p$ is a (nonzero) prime ideal in $\mathcal O_K$, perhaps there are prime ideals $\mathfrak P_1$ and $\mathfrak P_2$ lying over $\mathfrak p$ with different ramification indices or different residue field degrees. Then the completions of $L$ at the two prime ideals are not isomorphic as extensions of $K_\mathfrak p$.