# [ilmath]\mathbb{K} [/ilmath] (field)

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Created to define what [ilmath]\mathbb{K} [/ilmath] means when encountered as a field (eg if [ilmath](X,\mathbb{K})[/ilmath] is a vector space - Alec (talk) 03:54, 8 April 2017 (UTC)

- Also make sure that the doctrine page lists references that this page does, as they're the same thing!

## Contents

## Meaning

As per the Doctrine:K (field) we use [ilmath]\mathbb{K} [/ilmath] to denote a field which is either the field of real numbers, [ilmath]\mathbb{R} [/ilmath], or the field of complex numbers, [ilmath]\mathbb{C} [/ilmath].

This only applies if we do not explicitly define what [ilmath]\mathbb{K} [/ilmath] is^{[Example 1]}, in the absence of any mention of what [ilmath]\mathbb{K} [/ilmath] is the doctrine applies and it is a stand in for either [ilmath]\mathbb{R} [/ilmath] or [ilmath]\mathbb{C} [/ilmath].

This convention is not unusual^{[1]}.

### Caveats

- If we say "let [ilmath](X,\mathbb{K})[/ilmath] and [ilmath](Y,\mathbb{K})[/ilmath] be vector spaces" it is unclear whether [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are both over the same field or not.
- We will qualify this with "over the same field" or "over potentially distinct fields"; or use [ilmath]\mathbb{K}_1[/ilmath] and [ilmath]\mathbb{K}_2[/ilmath] if the fields will play a part in what is to come.

- If we say "let [ilmath](X,\mathbb{K})[/ilmath] be a vector space over the field [ilmath]\mathbb{K} [/ilmath]" it suggests that [ilmath]\mathbb{K} [/ilmath] is just being used to represent the field, the doctrine doesn't apply then. It'd be as if the more conventional [ilmath]\mathbb{F} [/ilmath] were used instead.

## Examples

- ↑ Take:
- "
*Let [ilmath]\mathbb{K}\in\mathbb{N} [/ilmath] be given and suppose [ilmath]X[/ilmath] is a set, define [ilmath]z:\eq (X,\mathbb{K})[/ilmath]*"

- "