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Background:

Exercise:

If $R$ is a Euclidean domain, is it true that $\delta(ab)=\delta(a)\delta(b)$ for all nonzero $a,b\in R?$

Questions:

For the above exercise, I don't think the question is an affirmative yes. For the ring of Gaussian integers, yes. But what if I am given a constant function for a Euclidean domain $R$ defined as $f(a)=k$ for some nonzero constant $k\in R$ and for all $a\in R,$

Then for any $x,y$ nonzero in $R,$ $f(xy)=k$, but $f(x)f(y)=k^2,$ hence $f(xy)\neq f(a)f(b).$ Am I correct in my guess?

Thank you in advance

Arturo Magidin
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Seth
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    What is $\delta$ ? – Digitallis Jan 28 '24 at 21:25
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    Seconding the query from @Digitallis. If it stands for the function we need to drive Euclid's algorithm, then the obvious counterexample is the degree of a polynomial, when we have $\delta(ab)=\delta(a)+\delta(b)$ instead. – Jyrki Lahtonen Jan 28 '24 at 21:27
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    @JyrkiLahtonen ok ok, so it is not universsally true. Thank you for confirming my suspicious. – Seth Jan 28 '24 at 21:29
  • @Digitallis $\delta$ is the function defined in the definition of Euclidean domain. Sory, I should have been more clear in my post. – Seth Jan 28 '24 at 21:30
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    Note that we can normalize any Euclidean function to a "least" version satisfying $,\delta (ab)\ge \delta(a)\ \ $ – Bill Dubuque Jan 28 '24 at 21:47
  • @BillDubuque what does the term "normaliize" mean, and is it standard terminology? I look at your answer at the other post, I don't see the word being defined? – Seth Jan 29 '24 at 20:35
  • For this purpose just think of it as meaning transforming it to satisfy a property we normally encounter in familiar Euclidean domains. Please delete your duplicate comment on the answer. – Bill Dubuque Jan 30 '24 at 02:05

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This fails even for well-known Euclidean functions such as the degree in a polynomial ring. More generally we can define a new Euclidean function by changing the value of $\,\delta\,$ at some point to any greater value, making it trivial to destroy this $\delta$'s multiplicative property. Obviously the same method works to twiddle Euclidean functions to destroy many other properties they may possess.

Bill Dubuque
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  • can you explain what you mean by "making it trivial to destroy"? – Seth Jan 30 '24 at 01:28
  • @Seth e.g. in $\Bbb Z$ change the value of $,\delta(n) = |n|,$ at $,n=2,$ so that $,\bar\delta(2)>\delta(2) = 2,,$ e.g. put $,\bar\delta(2) = 3.,$ Then $,4 = \bar \delta (4) \neq \bar\delta(2)^2 = 9,,$ which destroys the multiplicative property of $,\delta(n) = |n|.\ \ $ – Bill Dubuque Jan 30 '24 at 01:50