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I was studying Algebra (partially from I.N. Hertstein book). In it, an element is defined as prime $\pi$, if $\pi = ab \implies \text{a or b is unit.}$ However, this is the definition for irreducible elements, isn't it? The definition for primes is as follows: $\text{if}\ p |ab, \text{then}\ p|a \ \text{or}\ p|b$

Further, one of the theorems states that every non-zero element in Euclidean domain R is either unit or a finite product of primes. I understand the proof, which is done in the standard manner, by first showing that dab>da, if b is non-unit, then proceeding by induction on $d(r),\ r \in R.$ However, the proof uses this strange definition for primes, which is really proving that every non-zero element in Euclidean domain R is either unit or a finite product of irreducible elements.

Also, I understand that in a UFD, primes and irreducibles are the same. However, the book uses this previous statement in proving that Euclidean domains are UFD. Only after we have proved that R is a UFD, we can use this fact, can we not? Why does the book use this strange definition? Or is there a misunderstanding by me? Please clear this.

IamThat
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  • I agree, that is quite strange. On first glance, it seems to me like an error, but I'll defer to someone who knows more than I do. – Rushabh Mehta Jun 13 '21 at 20:21
  • Why do you think there is a problem? Exactly the same terminology is used in $\Bbb Z$ in most elementary number theory textbooks. Simply replace his "prime" by "irreducible" (or atom) to get the standard algebraic language. Note that he proves early on that "primes"(atoms) are prime, i.e. Euclid's Lemma (and the converse is trivially true in every domain). – Bill Dubuque Jun 13 '21 at 20:29
  • @BillDubuque But, we should prove that irreducibles and primes are the same, first, shouldn't we? In a general Integral domain, they need not be the same. For example, 5 in $Z[\sqrt{-5}]$. 5 is irreducible, but not prime, since $5|\sqrt{-5}\sqrt{-5}$ – IamThat Jun 13 '21 at 20:34
  • Do you think there is a logical flaw, or do you simply not like the terminology? If the former, what is the flaw? – Bill Dubuque Jun 13 '21 at 20:37
  • @BillDubuque I got confused because in most other algebra books, it is irreducibles that are defined that way, not primes. I doubt it would be quite easy to prove the statement that every non-zero element is unit or a finite product of primes. If it is simply a matter of convention, then I don't have any problem. I thought that there must be a deeper philosophical reason for calling irreducibles as primes, as the book by Mr. Herstein is quite reputed and develops the content in a logical manner. – IamThat Jun 13 '21 at 20:44
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    It seems Herstein desires to proceed in exact analogy as in $\Bbb Z$, e.g. he states "We wish to show that prime elements in a Euclidean ring play the same role as prime numbers play in the integers". So presumably he thinks it is pedagogically ok to delay introducing the distinction between atoms and primes in general domains till he actually discusses non-UFDs (but I didn't verify that by reading further) – Bill Dubuque Jun 13 '21 at 20:56
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    One can view Herstein's exposition as simply showing that the common proofs that $\Bbb Z$ is a UFD via the Euclidean algorithm generalize verbatim to every Euclidean domain (with very minor tweaks). If that was the goal then using the same terminology is natural. – Bill Dubuque Jun 13 '21 at 21:04
  • @BillDubuque I understand. Thank you for your patience. I haven't read the entire book, so there was a confusion. I see now that UFD is introduced much later in the book, and in the section titled Polynomial rings over commutative rings, the previously defined 'prime' separates into primitives and irreducibles. – IamThat Jun 13 '21 at 21:04
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    Beware that Herstein's "Euclidean rings" are domains by definition, i.e. they are what most authors call Euclidean domains (for true Euclidean rings see the paper cited here). – Bill Dubuque Jun 13 '21 at 21:07

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