MHB Euclidean Rings - Rotman Example 3.76

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.5 From Numbers to Polynomials ...

I need help with Example 3.76 ... ... the example concerns Euclidean rings and their defining characteristics so I am including the definition of a Euclidean ring in the relevant text shown below ... ...

The relevant text from Rotman's book is as follows:https://www.physicsforums.com/attachments/4649

View attachment 4648
I am trying to understand Example 3.76 which indicates that every field is a Euclidean ring ... ...

I can see that point (i) of the definition is satisfied with $$\partial$$ set identically to zero ...... BUT ... I fail to understand what Rotman is saying about how point (ii) is satisfied ... ...

In order for (ii) to be satisfied, for every $$g \in R$$ and every $$f \in R^{\times}$$ we have to find $$q, r \in R$$ such that:

$$g = qf + r$$ ... ... ... (*)

... BUT ...

Rotman says to set $$q = f^{-1}$$ and $$r = 0$$

but if we do this (*) above becomes

$$g = f f^{-1} + 0 = 1$$ ...

but $$g$$ may be any element of $$R$$ ... ?Can someone please explain what is going on ... that is, what Rotman means in this example ...

Hope someone can help ...

Peter

 

[Fernando Revilla]

Rotman says to set $$q = f^{-1}$$ and $$r = 0$$ but if we do this (*) above becomes $$g = f f^{-1} + 0 = 1$$ ... but $$g$$ may be any element of $$R$$ ... ?

You are right. The book should say: if $g\in R$ and $f\in R^{\times}$ set $q=gf^{-1}$ and $r=0.$ So, for all $g\in R$ and for all $f\in R^{\times}$ we verify $g=(\underbrace{gf^{-1}}_{q})\;f+\underbrace{0}_{r}.$

 

[Math Amateur]

You are right. The book should say: if $g\in R$ and $f\in R^{\times}$ set $q=gf^{-1}$ and $r=0.$ So, for all $g\in R$ and for all $f\in R^{\times}$ we verify $g=(\underbrace{gf^{-1}}_{q})\;f+\underbrace{0}_{r}.$

Thanks for that clarification Fernando ... I appreciate your help ...

Peter