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I am trying to answer the following question

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For (a) I have said that a and ab are in the ring R, by the definition of a ring. Therefore, by the definition of a Euclidean domain a=abq+r. As we are told that r is non-zero, we know that $\delta(r)<\delta(ab)$.

For (b) I am less sure how to prove this. We know that $\delta(r)<\delta(ab)$ and $\delta(ab)\ge\delta(b)$, but I'm not sure how to show what is required.

user26857
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Mark
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1 Answers1

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For a), you are not given that $r \neq 0$, you need to show this. By definition of an Euclidean domain, you can find $q,r \in R$ but either $r=0$ or $\delta(r) < \delta(ab)$. Suppose $r=0$, then $a=abq$ which implies since $a \neq 0$ and you are in an integral domain that $1=bq$, but that means $b$ is invertible (a contradiction). So $r \neq 0$.

For b), we have $r=a-abq=a(1-bq)$ so $\delta(a) \leq \delta(r)$.

ET93
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  • Can you expand on why a(1-bq) implies result? – Mark Mar 08 '15 at 16:11
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    For an Euclidean valuation/function, $\delta(x)\leq \delta(xy)$ for any nonzero $x,y \in R$. So $\delta(a) \leq \delta(a(1-bq))=\delta(r)$. – ET93 Mar 08 '15 at 16:12
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    @user43290 The property used in the prior comment need not necessarily hold true, so check you definitions. One can however always choose some $\delta$ for which it is true, e.g. see here. – Bill Dubuque Mar 08 '15 at 22:35
  • I wonder how you would show b) without assuming the valuation had that property. – ET93 Mar 09 '15 at 05:09