This is from the Wikipedia page on Local Fields:
"A field $K$ is called a (non-Archimedean) local field if it is complete with respect to a topology induced by a discrete valuation $v$ and if is residue field $k$ is finite".
My question is I am confused on what is meant be a field being complete with respect to a topology. Here is my best guess so far –
I'm assuming the field $K$ can be made into a metric space under the discrete valuation $v$, and if $\tau$ is he topology induced by the metric then we consider the metric space $(K, \tau)$. We can form the completion of $(K, \tau)$ by considering an isometric embedding $h:K \rightarrow L$ where $L$ is any complete metric space. The completion of the image of $K$ under the isometric embedding $h$ is a subspace of $L$ and is the completion of $K$. In other words, $\overline{h(K)}$ is the completion of $K$.
So, my guess is that saying a field $K$ is complete with respect to a topology is the same thing as saying that $K$ is the same as its completion, i.e., $K = \overline{h(K)}$. Is this the correct understanding of what is being said?
Best Answer
The comment has basically answered your question but I'll expand a little bit: your definition is a bit overly complicated because you're trying to define what it is to be complete and the way you're doing this is to embed into a complete space $L$ then ask whether $K$ maps isomorphically onto its image. But you can't even make sense of what it means for $L$ to be complete if you don't have a notion of completeness already. So more generally recall that any metric space $(X,d)$ is complete if every cauchy sequence converges; now your valuation $v$ induces a metric (e.g. $d_v(x,y)=e^{-v(x-y)}$) and then $(K,v)$ is complete if $(K,d_v)$ is a complete metric space.
In more generality one can define what it means for any topological group to be complete: there is a natural notion of cauchy sequence/net (in the most general case one should use cauchy nets) in a topological group, and then a topological group $G$ is complete if every cauchy net in $G$ converges.