Linear & Abstract algebra

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11:53 AM

Herstein , presents the definition of an Euclidean ring as , :

A ring , such that for every element , $a , in it , there exists , an integer d(a) , such that
for all $a, $b in the ring , R, d(a) < d(ab)
and for all $a , $b in R , there exists $c and $r such that $a = $b$c + $r , such that $r is either 0 or d(r) <= d(b) .

But then a theorem is presented stating that if $b is not a unit in R , then d(a) < d(ab) .

Is the only difference between the definition and the theorem is the greater than or equal to sign ?

Martin Sleziak

It seems that you have unpaired dollars after "$a in it...". (Also in "$b is not a unit...", but in that line there are no closing dollars, so that does not really matter.)

> A ring , such that for every element $a$ in it , there exists an integer $d(a)$ such that, for all $a$, $b$ in the ring $R$, $d(a) < d(ab)$ and for all $a , b \in R$ , there exists $c$ and $r$ such that $a = bc + r$ , such that $r$ is either $0$ or $d(r) \le d(b)$.

Isn't there some condition such as $d(a)>0$ in the definition?

In the third edition the definition says that $d$ goes from $R$ into nonnegative integers.

However, I do not see the theorem about units that you mention. (I am looking at the 3rd edition of Herstein's Abstract Algebra.)

beerzil charlemagne

12:11 PM

@MartinSleziak : I have the one from 1975

Martin Sleziak

Hm... this one seems to be from 1996.

beerzil charlemagne

@MartinSleziak : Do you know of any such theorem or lemma ? I was basically confused about as to why there is another theorem which is same as the first property of the euclidian ring . Then I saw the difference was the greater than and greater than or equal to sign

Martin Sleziak

This is what I have in the book right after the definition: i.stack.imgur.com/iFYis.png

It is relatively easy to see that if $b$ is unit then $d(ab)=d(a)$.

We have $c$ such that $bc=1$, and thus $a=abc$, therefore $d(ab)\le d(abc) = d(a)$.

beerzil charlemagne

@MartinSleziak . Yes , thanks a lot

Martin Sleziak

But you're asking about the other case - when $b$ is not a unit.

beerzil charlemagne

12:18 PM

yes

Martin Sleziak

I have to admit that I do not see immediately how can it be shown.

Still, it that's the case, it is probably worth mentioning it - even in a separate theorem or lemma.

beerzil charlemagne

@MartinSleziak : Not a problem , I will go through the proof given here and will post here the specifuc parts I find difficult to understand

Martin Sleziak

@beerzilcharlemagne Is it possible that the book you're talking about is Herstein's Topics in Algebra and not Abstract Algebra?

beerzil charlemagne

@MartinSleziak Yes that is the case , thanks

Martin Sleziak

So the result you've mentioned is Lemma 3.7.3, right? i.stack.imgur.com/po8vH.png

beerzil charlemagne

12:26 PM

@MartinSleziak : Yes exactly this part , getting a hang of the proof , strategized with the ideal of multiples (loose terminology though :P) , and coming to a contradiction with the assumption of equality .

beerzil charlemagne

1:02 PM

@MartinSleziak : On a different note , are undergrads in maths expected to complete the whole of a book like , Dummit and foote , in one semester of algebra course ?

Martin Sleziak

TBH I'm not able to answer that.

beerzil charlemagne

@MartinSleziak : Ok

Martin Sleziak

I assume there is big difference between what can be expected from undergraduates in one university from another one. And it also depends on how much of the stuff people already know before starting this course.

Anyway, the whole Dummit-Foote in one semester seems quite a lot to me. Perhaps in one year would be doable - but I'd still consider that rather ambitious.

beerzil charlemagne

@MartinSleziak : Thanks for your answer

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