Homotopy Type Theory UMyn8W7b (Rev #39)

On foundations

The natural numbers are characterized by their induction principle (in second-order logic/in a higher universe/as an inductive type). If one only has a first order theory, then one cannot have an induction principle, and instead one has a entire category of models. Thus, the first order models of arithmetic typically found in classical logic and model theory do not define the natural numbers, and this is true even of first-order Peano arithmetic.

Euclidean rings

Given a commutative ring RR, a term e:Re:R is left cancellative if for all a:Ra:R and b:Rb:R, ea=ebe \cdot a = e \cdot b implies a=ba = b.

isLeftCancellative(e) a:R b:R(ea=eb)(a=b)\mathrm{isLeftCancellative}(e) \coloneqq \prod_{a:R} \prod_{b :R}(e \cdot a = e \cdot b) \to (a = b)

A term e:Re:R is right cancellative if for all a:Ra:R and b:Rb:R, ae=bea \cdot e = b \cdot e implies a=ba = b.

isRightCancellative(e) a:R b:R(ae=be)(a=b)\mathrm{isRightCancellative}(e) \coloneqq \prod_{a:R} \prod_{b :R}(a \cdot e = b \cdot e) \to (a = b)

An term e:Re:R is cancellative if it is both left cancellative and right cancellative.

isCancellative(e)isLeftCancellative(e)×isRightCancellative(e)\mathrm{isCancellative}(e) \coloneqq \mathrm{isLeftCancellative}(e) \times \mathrm{isRightCancellative}(e)

The multiplicative submonoid of cancellative elements in RR is the subset of all cancellative elements in RR

Can(R) e:RisCancellative(e)\mathrm{Can}(R) \coloneqq \sum_{e:R} \mathrm{isCancellative}(e)

A Euclidean ring is a commutative ring RR for which there exists a function? d:Can(R)d \colon \mathrm{Can}(R) \to \mathbb{N} from the multiplicative submonoid of cancellative elements in RR to the natural numbers, often called a degree function, a function ()÷():R×Can(R)R(-)\div(-):R \times \mathrm{Can}(R) \to R called the division function, and a function ()%():R×Can(R)R(-)\ \%\ (-):R \times \mathrm{Can}(R) \to R called the remainder function, such that for all aRa \in R and bCan(R)b \in \mathrm{Can}(R), a=(a÷b)b+(a%b)a = (a \div b) \cdot b + (a\ \%\ b) and either a%b=0a\ \%\ b = 0 or d(a%b)<d(g)d(a\ \%\ b) \lt d(g).

Revision on May 12, 2022 at 23:27:37 by Anonymous?. See the history of this page for a list of all contributions to it.