Locally compact field

In algebra, a locally compact field is a topological field whose topology forms a locally compact Hausdorff space.^[1] These kinds of fields were originally introduced in p-adic analysis since the fields ${\displaystyle \mathbb {Q} _{p}}$ are locally compact topological spaces constructed from the norm ${\displaystyle |\cdot |_{p}}$ on ${\displaystyle \mathbb {Q} }$. The topology (and metric space structure) is essential because it allows one to construct analogues of algebraic number fields in the p-adic context.

One of the useful structure theorems for vector spaces over locally compact fields is that the finite dimensional vector spaces have only an equivalence class of norm: the sup norm^{[2] pg. 58-59}.

Given a finite field extension ${\displaystyle K/F}$ over a locally compact field ${\displaystyle F}$, there is at most one unique field norm ${\displaystyle |\cdot |_{K}}$ on ${\displaystyle K}$ extending the field norm ${\displaystyle |\cdot |_{F}}$; that is,

${\displaystyle |f|_{K}=|f|_{F}}$

for all ${\displaystyle f\in K}$ which is in the image of ${\displaystyle F\hookrightarrow K}$. Note this follows from the previous theorem and the following trick: if ${\displaystyle ||\cdot ||_{1},||\cdot ||_{2}}$ are two equivalent norms, and

${\displaystyle ||x||_{1}<||x||_{2}}$

then for a fixed constant ${\displaystyle c_{1}}$ there exists an ${\displaystyle N_{0}\in \mathbb {N} }$ such that

${\displaystyle \left({\frac {||x||_{1}}{||x||_{2}}}\right)^{N}<{\frac {1}{c_{1}}}}$

for all ${\displaystyle N\geq N_{0}}$ since the sequence generated from the powers of ${\displaystyle N}$ converge to ${\displaystyle 0}$.

If the index of the extension is of degree ${\displaystyle n=[K:F]}$ and ${\displaystyle K/F}$ is a Galois extension, (so all solutions to the minimal polynomial of any ${\displaystyle a\in K}$ is also contained in ${\displaystyle K}$) then the unique field norm ${\displaystyle |\cdot |_{K}}$ can be constructed using the field norm^{[2] pg. 61}. This is defined as

${\displaystyle |a|_{K}=|N_{K/F}(a)|^{1/n}}$

Note the n-th root is required in order to have a well-defined field norm extending the one over ${\displaystyle F}$ since given any ${\displaystyle f\in K}$ in the image of ${\displaystyle F\hookrightarrow K}$ its norm is

${\displaystyle N_{K/F}(f)=\det m_{f}=f^{n}}$

since it acts as scalar multiplication on the ${\displaystyle F}$-vector space ${\displaystyle K}$.

All finite fields are locally compact since they can be equipped with the discrete topology. In particular, any field with the discrete topology is locally compact since every point is the neighborhood of itself, and also the closure of the neighborhood, hence is compact.

The main examples of locally compact fields are the p-adic rationals ${\displaystyle \mathbb {Q} _{p}}$ and finite extensions ${\displaystyle K/\mathbb {Q} _{p}}$. Each of these are examples of local fields. Note the algebraic closure ${\displaystyle {\overline {\mathbb {Q} }}_{p}}$ and its completion ${\displaystyle \mathbb {C} _{p}}$ are not locally compact fields^{[2] pg. 72} with their standard topology.

Field extensions ${\displaystyle K/\mathbb {Q} _{p}}$ can be found by using Hensel's lemma. For example, ${\displaystyle f(x)=x^{2}-7=x^{2}-(2+1\cdot 5)}$ has no solutions in ${\displaystyle \mathbb {Q} _{5}}$ since

${\displaystyle {\frac {d}{dx}}(x^{2}-5)=2x}$

only equals zero mod ${\displaystyle p}$ if ${\displaystyle x\equiv 0{\text{ }}(p)}$, but ${\displaystyle x^{2}-7}$ has no solutions mod ${\displaystyle 5}$. Hence ${\displaystyle \mathbb {Q} _{5}({\sqrt {7}})/\mathbb {Q} _{5}}$ is a quadratic field extension.

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• Inequality trick https://math.stackexchange.com/a/2252625