Let $K$ be a number field. If there exists a prime $l$ such that $l$ spilts in $K$, why does that imply there exists a place of $K$, which satisfies $K_v \cong \Bbb{Q}_l$ ?
Here, $K_v$ denotes completion of $K$ at $v$. If $K/\Bbb{Q}$ is Galois, I try to embed local version of Galois group int ogalobal version, but here we don't assume $K$ is Galois over $\Bbb{Q}$.
Thank you for your help.
The meaning of a prime $p$ splitting completely in $K$, of degree $n$ over $\mathbf Q$, is that $(p) = \mathfrak p_1\cdots \mathfrak p_n$ where each $\mathfrak p_i$ has residue field degree $1$. Each $\mathfrak p_i$ defines an absolute value on $K$, and its restriction to $\mathbf Q$ is the $p$-adic absolute value on $\mathbf Q$, with $e(\mathfrak p_i|p) = 1$ and $f(\mathfrak p_i|p) = 1$. Completing a field with respect to a non-Archimedean absolute value does not change the value group or the residue field. Then $[K_{\mathfrak p_i}:\mathbf Q_p] = e(\mathfrak p_i|p)f(\mathfrak p_i|p) = 1$, so $K_{\mathfrak p_i} = \mathbf Q_p$.
There is no grand mystery about where the places on $K$ come from with completion isomorphic to $\mathbf Q_p$: they're the $\mathfrak p$-adic absolute values at primes lying over $p$.
And the argument in the first paragraph does not need $p$ to split compeltely in $K$: what you need is at least one prime $\mathfrak p$ over $p$ in $K$ with $e(\mathfrak p|p) = 1$ and $f(\mathfrak p|p) = 1$.
Example. Let $K = \mathbf Q(\sqrt[3]{2})$, so $\mathcal O_K = \mathbf Z[\sqrt[3]{2}]$. Since $x^3 - 2 \equiv (x-3)(x^2+3x+4) \bmod 5$, we have $(5) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ both have $e = 1$ and $f(\mathfrak p|5) = 1$ while $f(\mathfrak q|5) = 2$. Thus $[K_{\mathfrak p}:\mathbf Q_5] = 1$, so $K_{\mathfrak p} \cong \mathbf Q_5$ even though $5$ doesn't split completely in $K$.