MHB Definition of a Euclidean Domain ....

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In the book "The Basics of Abstract Algebra" Bland defines a Euclidean Domain using two conditions as follows:View attachment 8256
View attachment 8257In the book "Abstract Algebra"by Dummit and Foote we find that a Euclidean Domain is defined using only one of Bland's conditions ... as follows:View attachment 8258What are the consequences of these different definitions ... for example does D&F's definition allow some structures to be Euclidean Domains that are not recognized as such under Bland's definition ...Peter
 
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Let $N$ be the norm of the integral domain $R$ in F&D’s definition. Define another norm $N^\ast$ by
$$N^\ast(a)\ =\ \min_{b\in R\setminus\left\{0_R\right\}}N(ab).$$
Then $R$ is still a Euclidean domain with norm $N^\ast$ and $N^\ast$ is also a Euclidean valuation in Bland’s definition.

The most important property of a Euclidean domain is the second one in Bland; many results about Euclidean domains involve only the second property but not the first. IMHO the importance of the first property only shows up in the study of ideals and algebraic-number theory.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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