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I need to show that:

a) v(u) = 1 <=> v is in E(R), where E(R) is the group of elements in R with an inverse
b) a|b => v(a)|v(b) => v(a) ≤ v(b) oder b=0
c) a~b <=> v(a)=v(b) and a|b

I think I have managed to solve part b. Here is my proof:
a|b => there exists x in R s.t. b=ax
=> v(b)=v(ax)
(v euclidean) => v(b) =v(a)v(x) let v(x) = k v(b)=v(a)*k => v(a)|v(b)
I just need to solve the last part.

I'm really not sure where to begin with the other parts. I know that a euclidean ring has no zero divider but apart from that am a bit stumped. Google hasn't thrown up many good explanations of any of these points - so any clues would be greatly appreciated. Thanks

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In (a) one of the implications should be clear (if $a$ is a unit in $R$, there is $b$ in $R$ with $ab = 1$; apply $v$ to both sides and use the defining properties of a norm to conclude that $v(a) = 1$). For the other, use the "division with remainder" property (i.e., the one thing that makes a "Euclidean ring" special). If $a$ in $R$ satisfies $v(a) = 1$, by the division with remainder property (dividing $1$ by $a$), either there is $q$ in $R$ with $1 = aq$, in which case $a$ is a unit, or there are $q$ and $r$ in $R$ with $r$ nonzero, $1 = aq + r$ and $v(r) < v(1)$. But nothing can be smaller than $v(1)$ (because $1$ divides $r$ and (b)) so this case does not occur.

In (c) you do not explain what the relation $\sim$ is but I imagine that the definition of the statement $a \sim b$ is that there is a unit $u$ with $a = bu$ (or something equivalent to this) and that you know that $\sim$ defined in this way is an equivalence relation (so in particular, if $a \sim b$ then $b \sim a$). So if $a \sim b$, there is a unit $u$ with $b = au$, and hence $a$ divides $b$, and the equality $v(b) = v(a)$ follows from applying $v$ to both sides of $b = au$ and using the defining properties of $v$ (and $v(u) = 1$, from (a)). Conversely, if $a$ divides $b$ and $v(a) = v(b)$, from the first of these hypotheses there is $c$ in $R$ with $b = ca$; applying $v$ to both sides of this equation and using the defining properties of $v$ we conclude that $v(b) = v(c) v(a)$; from the hypothesis that $v(a) = v(b)$ we deduce that $v(c) = 1$ and hence that $c$ is a unit by (a); this shows that $a \sim b$.

The only trick to proving statements like these is close study of the definitions of the things that appear in the statements. If you are getting stuck, it is likely because you are not making full use of these definitions. I do not think it is possible to prove (a), for example, using only the ring axioms and the properties of the norm that do not refer to the "division with remainder" property; by this I mean that it is likely you can find an integral domain $R$, with a function $v$ from it to the integers satisfying $v(ab) = v(a) v(b)$ and other identities of this type, for which the statement (a) is not true.

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    thanks alot for your help. i will have to have a careful look and see if any other preconditions have been given to prove part a.2011-01-10