ABSTRACT ALGEBRA. Euclidean Norm. Question will be posted on the comment box. Thanks

Question

darkprince14 · Answer

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darkprince14 · Answer

v is a euclidean valuation. it is not defined because the problem was meant to be proved in general.

darkprince14 · Answer

v has a property that v(a) $\leq$ v(ab) so it doesn't actually follow

loser66 · Answer

One more question: what does (a) mean? is it an ideal generated by a?

darkprince14 · Answer

Can I prove it by contradiction?
Assume$$v(a)\ge v(b)$$
since a divides b, there exists m in D s.t. b=am,
then we'll have
$$v(a)\ge v(b) =v(am)$$ but since we have $$v(a)\le v(am)$$ it must be that $$v(a)= v(b) =v(am)$$  which implies that $$v(a)= v(am) =v(a*1)$$ 
but $$b=a*m=a*1, m=1$$ . Since m=1, is a unit. a and b are associates, which is a contradiction to our assumption that a and b are not associates.
Hence v(a)< v(b)

darkprince14 · Answer

please correct the proof if there are any problems with this proof.
@Loser66 v is function mapping and v(a) is a nonnegative integer

loser66 · Answer

I dare not to say anything because to me, it is TRIVIAl, just one line proof to get the result. Why do we have to use contradiction?? a | b, D is Euclidean domain , a is not unit, hence a* m =b . If v is function mapping, $v: a\mapsto  v(a)\v:am\mapsto v(am) = v(b) $

loser66 · Answer

but, ignore me. I am not good. hihihi. gotta go.

anonymous · Answer

degree function right?  i mean $v(a)$ is the degree of $a$?

anonymous · Answer

not familiar with that notation 
in any case if it is the degree function then it is always the case that the degree of a unit is zero, and $v(ab)=v(a)+v(b)$ 
that should work for you i think

darkprince14 · Answer

the euclidean valuation is a function that could differ from a euclidean domain to another so it's not necessarily the degree function

anonymous · Answer

i think it is still the case that $v(ab)=v(a)+v(b)$ no matter what you call it