Thread: Rings - definition of prime element

I am trying to tie together the various elements of the definition of a prime element in a ring/integral domain. [why integral domain and not ring??]

On page 284 Dummit and foote define a prime element as follows:

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"Let R be an integral domain.

The non-zero element p R is called prime if the ideal (p) generated by p is a prime ideal. In other words, a non-zero element p is a prime if it is not a unit and whenever p | ab for any a, b R, then either p | a or p | b."

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I need help to understand how the definition of a prime in terms of a prime ideal ties up with the definition given after "In other words".

In particular why does (p) being a prime ideal imply that p cannot be a unit.

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Further if you check D&F's definition of a prime ideal it seems to specify things in terms of inclusion rather than p dividing elements.

Specifically the definition of a prime ideal on page 255 is as follows:

"Assume R is commutative. An ideal P is called a prime ideal if P R and whenever the product ab of two elements a, b R is an element of P, then at least one of a and b is an element of P.

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Can someone please clarify the above for me. In particular why is the definition of prime ideal couched in terms of set inclusion while the definition of a prime element is couched in terms of divisors?

Peter

ok, suppose in a ring R a|b. what does this mean? it means there is some c in R such that b = ca (we are going to assume commutativity, the non-commutative case gets too involved).

well if b = ca, this means b is in the ideal generated by a, (a), and hence (b) ⊆ (a).

on the other hand suppose that for a,b in R, (b) ⊆ (a). this means that b = ra, for some r in R, so that a|b.

that is: "b divisible by a" is equivalent to (b) ⊆ (a).

now suppose p in R is a prime element. this means that if p|(ab), then either p|a, or p|b. equivalently if ab is in (p), then either a is in (p), or b is in (p).

this means a prime element generates a prime ideal.

the proviso that p not be a unit, is to ensure that (p) ≠ R, because trivially ab in R implies a in R AND b in R (in other words, only PROPER ideals can be prime, or else they wouldn't tell us anything).

for commutative rings R, (a,b,...,k) can be thought of as R-linear combinations of {a,b,...,k}. if we have just a 1-element set, like {a}, then (a) is "R-multiples of a".

saying a|b is the same as saying: b is an "R-multiple" of a. that's why we study IDEALS, for any element x in an ideal I, it also contains all "R-multiples" of x.

this situation is particularly handy with principal ideal domains, because then there is a 1-1 correspondence between "prime elements" and "prime ideals".

now: why integral domains?

well, suppose R has a zero divisor a. then there is some b ≠ 0, with ab = 0. now: consider (ab) = (0) = {0}. clearly neither a nor b is in (0) = {0}, so {0} is not a prime ideal. why is this bad?

because 0 is an element of EVERY IDEAL so for an ideal to be prime in a ring with zero divisors it needs to contain a lot of zero divisors.

for example, since 2*3 = 0 in Z₆, any prime ideal of Z₆ has to either contain ALL multiples of 2 or ALL multiples of 3, or else we wouldn't have one at all.

in general, in rings, we don't have a linear order, so we tend to order elements of rings by divisibility. zero divisors sort of mess with this (but it IS possible to have prime ideals in rings with zero divisors).

Thanks so much for the help!

Will work through this now.

Peter

We all known that the notion of the Prime Element generalizes the ordinary definition of prime number in the Ring, expect it allows for negative prime elements while every prime is irreducible, the converse is not a general true. The Integral domain is a commutative ring with identity in which the product of any two non zero elements is not equal to zero and is a prime deal when zero is identity is equal to zero. The integral domain is a ring that is a sub ring of the field and means it is also a commutative ring with identity. For each integer n>1, the set of real numbers of the from a+ b√n with and b integers is a sub ring of R and hence an integral domain. A regular local ring is an integral domain.


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