# Prime Element in a Ring ...

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## Main Question or Discussion Point

On page 284 Dummit and Foote in their book Abstract Algebra define a prime element in an integral domain ... as follows: My question is as follows:

What is the definition of a prime element in a ring that is not an integral domain ... does D&F's definition imply that prime elements cannot exist in a ring that is not an integral domain ... but why not ...?

Can someone please clarify this situation ...

Peter

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andrewkirk
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I think the reason it excludes rings that are not integral domains is that it makes the definition of prime a little messier, because such rings will have zero divisors and every element of R divides zero. So a ring element $p$ that is a 'genuine' prime would be disqualified because for any nonzero pair $a,b$ such that $ab=0$ we would have $p|ab$ but $p\not|a$ and $p\not|b$.

I think it might be able to be resolved fairly painlessly by adding the requirement that $ab\neq 0$.

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fresh_42
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I think it might be able to be resolved fairly painlessly by adding the requirement that $ab\neq 0$.
This won't work. Take $6 \in \mathbb{Z}_{24}$. Then one has $6 \, | \, 12 = 3 \cdot 4 ≠ 0$ and $6 \nmid 3 \wedge 6 \nmid 4$. Therefore $6$ would be prime in $\mathbb{Z}_{24}$. On the other hand $6 \, | \, 12 = 1 \cdot 12 ≠ 0$ and $6 \, | \, 12,$ so $6$ is not prime.
To resolve these situations one has to define:

$c \in R \; \text{prime} ⇔ c \neq 0 \wedge c \nmid 1 \wedge (∀ \{a,b\} ⊆ R: c | ab ⇒ c | a ∨ c | b)$.

This is indeed the general definition for prime elements of unitary (i.e. with 1), commutative rings $R$.

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andrewkirk
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Take $6 \in \mathbb{Z}_{24}$. Then one has $6 \, | \, 12 = 3 \cdot 4 ≠ 0$ and $6 \nmid 3 \wedge 6 \nmid 4$. Therefore $6$ would be prime in $\mathbb{Z}_{24}$.
Am I missing something? Isn't the conclusion from this that 6 is not prime in $\mathbb Z_{24}$ - which is what we would expect?

fresh_42
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Am I missing something? Isn't the conclusion from this that 6 is not prime in $\mathbb Z_{24}$ - which is what we would expect?
No, you're right. I simply wanted to show that $ab ≠ 0$ alone doesn't help. One has to add the for-all-quantor.

Edit: My bad. The quantor is there anyway.

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mathwonk
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in general in a (unitary) commutative ring, the primes are defined just as the non zero elements which generate a (proper) prime ideal*. It follows that the prime elements cannot be units. they can be zero divisors however. intuitively think of the algebraic curve XY = 0, the union of the x and y axes in the plane. The ring of function on this set is the quotient ring k[X,Y]/(XY). Both X and Y are zero divisors in this ring. However both are also prime since modding out by eiother one gives a polynomial ring k[X] or k[Y]. Geometrically an element is prime (essentially) if its vanishing set has only one component (and counted without multiplicity), so the vanishing set of X e.g. is just one component, namely the Y axis. it is a zero divisor if its vanishing set takes up an entire component of the whole, since then multiplying by an element vanishing on the other components will give zero. (*this is exactly the same as fresh42's definition.)

fresh_42
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Nice example.

andrewkirk
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define [primeness of c as]:

$c \in R \; \text{prime} ⇔ c \neq 0 \wedge c \nmid 1 \wedge (∀ \{a,b\} ⊆ R: c | ab ⇒ c | a ∨ c | b)$.
We can take this and add the $ab\neq 0$ requirement to get:

$$c \in R \; \text{prime} ⇔ c \neq 0 \wedge c \nmid 1 \wedge (∀ \{a,b\} ⊆ R: (c | ab \wedge ab\neq 0) ⇒ c | a ∨ c | b)$$

Without that extra requirement many elements we would wish to call primes would be disqualified in a ring that has zero divisors. For instance in $\mathbb Z_{24}$ the element 5 would not be prime because 5 divides $8\cdot 3=0$ (ie 0 = 5\cdot 0) but doesn't divide 8 or 3.

An alternative would be to alter the definition of the 'divides' operator to exclude the case where one of the factors is zero. But I suspect that may require too many changes with the many theorems that use the fact that any element 'divides zero'.

fresh_42
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Without that extra requirement many elements we would wish to call primes would be disqualified in a ring that has zero divisors. For instance in $\mathbb Z_{24}$ the element 5 would not be prime because 5 divides $8\cdot 3=0$ (ie 0 = 5\cdot 0) but doesn't divide 8 or 3.
But $5 \cdot 16 = 8$.

andrewkirk
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But $5 \cdot 16 = 8$.
Very nice.

More generally, if $p$ is a prime in $\mathbb N$ and we write $p_n$ for its counterpart in $\mathbb Z_n$ then the ideal it generates in $\mathbb Z_n$ is all of $\mathbb Z_n$, as long as $p\nmid n$.

So if $p\nmid n$ then $p_n$ will satisfy the post #3 requirement for primeness regardless of the presence of divisors of zero (ie whether $n$ is prime or not).

If $p$ is prime in $\mathbb Z$ and $p\mid n$ then the ideal of $\mathbb Z_n$ generated by $p_n$ will be proper but again we can show that for any $ab$ in the ideal generated by $p_n$ we will have $p_n\mid a$ or $p_n\mid b$.

OK, you have officially convinced me that we don't need the requirement $ab\neq 0$ in general in a (unitary) commutative ring, the primes are defined just as the non zero elements which generate a (proper) prime ideal*...... (*this is exactly the same as fresh42's definition.)
I'm having trouble seeing the equivalence where the ring is $\mathbb Z_n$, $p$ is prime in $\mathbb N$ and $p$ does not divide $n$ in $\mathbb N$. Then if we label by $p_n$ the counterpart of $p$ in $\mathbb Z_n$, won't the ideal generated by $p_n$ be all of $\mathbb Z_n$, in which case the ideal is not prime (since it's not proper) and so $p_n$ is not prime. Yet $p_n$ divides every element of $\mathbb Z_n$ and hence satisfies the definition of prime in post #3.

fresh_42
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I'm having trouble seeing the equivalence where the ring is $\mathbb Z_n$, $p$ is prime in $\mathbb N$ and $p$ does not divide $n$ in $\mathbb N$. Then if we label by $p_n$ the counterpart of $p$ in $\mathbb Z_n$, won't the ideal generated by $p_n$ be all of $\mathbb Z_n$, in which case the ideal is not prime (since it's not proper) and so $p_n$ is not prime. Yet $p_n$ divides every element of $\mathbb Z_n$ and hence satisfies the definition of prime in post #3.
No, in this case it doesn't satisfy the definition because $p_n \in \mathbb{Z}_n$ is a unit if $p_n \nmid n$, i.e. $p_n \, | \, 1$.

The claimed equivalence, however, is in my opinion a bit sloppy. E.g. the ideal $4R$ of $R = \mathbb{Z}_{24}$ isn't a prime ideal for $2 \cdot 2 = 4$, although it is proper and $4$ is neither zero nor a unit.

An ideal $I$ is a prime ideal in a ring $R$ iff its quotient $R / I$ is an integral domain.

For the principal ideal, euclidean and integral domain $\mathbb{Z}$ the equivalence of prime numbers $p$ and the above definition of prime ideals is obviously true: $\mathbb{Z}_p$ is even a field and $p\mathbb{Z}$ prime and maximal. But in general one has to handle the zero divisors as my example has shown and the ideal $\{0\}$ as well, which is surely proper and not prime if $R$ isn't an integral domain.

Edit: A prime ideal $P$ can also be defined as: $P \neq (1)$ and $ab \in P$ implies $a \in P$ or $b \in P$. This shows the similarity to prime numbers.

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mathwonk
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indeed i was careless in equating the definition i gave with fresh42's. i think the example of 4R with R = Z/24 may not illustrate this however since it seems non prime with both definitions, but in general the requirement that ab ≠ 0 must be omitted from fresh's definition. i.e. c is prime iff c is non zero and a non unit and also if whenever c divides ab then c divides one of a or b, without assuming ab≠0. this seems equivalent to c generating a prime ideal. please correct me if wrong. i.e., in my opinion, the "standard" definition (i.e. that used in a book written by experts such as auslander and buchsbaum) of a prime element is that it generates a proper prime ideal. but there is always room for discussion.

i was indeed wrong not to notice that ab≠0 was a condition in fresh's definition. this apparently makes it different from saying c generates a prime ideal. but at the moment i cannot think of an example to illustrate the difference. i.e. i would like to see an element which is prime in fresh's defn but does not generate a prime ideal. it may be obvious but i don't see it just now.

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fresh_42
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The requirement $ab \neq 0$ is indeed not necessary (and I didn't make it). Of course a prime ideal $P$ in rings $R$ with unit has to be proper for otherwise $R / P = \{0\}$ would be an integral domain, which it is not by definition. $\{0\}$ however, is a prime ideal iff the ring is an integral domain, e.g. $\mathbb{Z}$, whereas we don't call $0$ a prime number.
I'm not sure whether one can connect prime ideals to prime elements that easy in case one hasn't a principal ideal domain, i.e. I have no example in mind. The definition via integral domains for prime ideals (and fields in case of maximal ideals) always works, as well as $(ab \in P ⇒ a \in P ∨ b \in P)$ does. In case $P$ is generated by a single element $p$ it is the same as to say $p$ is a prime element. I don't know what it could be, if $P$ is generated by more than one element.
The definition with the integral domain and equivalently the "ab"-form mentioned above are usual and "Introduction to Commutative Algebra" by M.F.Atiyah and I.G.Macdonald (Addison-Wesley, p. 3) a book written by experts, too.

andrewkirk
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Based on that illuminating exploration of how definitions of prime operate in various types of ring s that are not integral domains, it seems that the answer to MA's question in the OP:
What is the definition of a prime element in a ring that is not an integral domain
is 'it's the same as for rings that are integral domains', on the basis that fresh's definition in post #3 which, it seems to have been agreed, works for all commutative rings, is the same as the one at item (2) in the OP.

If that's correct then we can surmise that the authors restricted their definition to operating in integral domains for reasons not related to the difficulty of defining primes.

mathwonk
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well i have been doubly careless in attributing that extra condition to fresh, i see where it appears now. My "expert" comment was in reference to the history of auslander and buchsbaum's famous result that all regular local rings are ufd's. I.e. they got that adjective from me because they have done famous research specifically in the area of ufd's.

do we all agree that the definition by fresh in post #3 is entirely equivalent to saying that c is non zero and generates a (proper) prime ideal? i.e. apparently all the experts agree on this definition of prime element, although some prefer to restrict themselves to discussing domains. (I cannot find the definition of prime element in my copy of atiyah macdonald, however. i.e. on p.3 of my copy only the notion of prime ideal appears.)

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mathwonk
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i have been searching my books for the notion of prime element. the (to me) classical experts, like zariski-samuel, and matsumura, do not use the term at all, but eisenbud does so. the little book of miles reid seems to make an error on p. 2 and omit to require the element to be non zero, which seems to make zero (in a domain) a counterexample to his claim that all prime elements are "trivially" seen to be irreducible. of course he has required his element to be non zero in the previous sentence when giving another definition, so we might grant him the assumption he was still holding to that hypothesis, although it is not at all clear.

it seems DF are concerned in that chapter with ufd's and perhaps it is hard for a ring to have unique factorization if it is not a domain. e.g. a prime element would have to divide one of the factors a,b for every equation ab=0, which seems to reduce the number of primes seriously.

mathwonk
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in geometry the rings are rings of polynomial functions on a space. intuitively a prime function will define a zero set that consists of just one piece. so a product of prime functions will define a zero set that is the union of the sets defined by each factor. It can happen however that zero set can have more than one piece, but none of the individual pieces can be defined by a single equation. That would be a set defined by an irreducible function which is not prime. i.e. irreducible just means not a product of iother functions, but prime means the zero set has only one piece. thus a prime function is irreducible, since if it were a product of other factors, each factor wopuld define a separate piece, and there is only one piece. But irreducible does not always mean prime. e.g. on a cone, any function will cut more than one piece, one from each "side" of the cone, but neither piece can be defined by just one function. (a tangent plane to the cone appears to cut out just one piece, but it cuts it twice, which does not count.)

mathwonk
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well this is an interesting topic.

notice in the ring k[X,Y]/(XY), that modding out by Y gives k[X], a domain, so Y is prime. however modding out by 1-cX where c ≠0 is in k, I claim gives k, also a domain. I.e. Y(1-cX) = Y - cXY = Y, since XY = 0. So modding out by 1-cX also mods out Y, and we have the quotient ring k[X,Y]/(Y,1-cX) ≈ k[X]/(1-cX) ≈ k.

Thus 1-cX is prime, and moreover 1-cX divides Y, also a prime. So we have a prime as a proper divisor of another prime. Thus the prime Y is not irreducible! It also can have infinitely many non associate prime divisors, namely 1-cX for infinitely many choices of c, (if our field is infinite).

These oddball phenomena suggest we have trouble doing factorization theory in a non domain, or at least that we should probably avoid trying to factor zero divisors, like the Y in our example. So if we are doing factorization theory in such a ring we might try factoring only non zero - divisors, and we might want to use as factors only primes that are also irreducible.

note the geometric intuition still holds that primes are elements that define irreducible sets, since Y in our example defines the X axis, one piece, and 1-cX defines the point X=1/c on the X axis.

still it is hard to resist thinking that Y is the best prime factorization of Y, not Y.(1-cX). I.e. Y seems like a good prime element, even if not irreducible.

mathwonk
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this is interesting. a ring R is a domain iff the ideal (0) is prime. suppose the ring R is noetherian and has no nilpotents. then at least (0) is an intersection of a unique finite collection of distinct prime ideals. suppose e.g. that (0) = ImeetJ where I and J are prime ideals. Then we can map R injectively into the product of the two domains R/I x R/J. So at least R is a subring of a product of domains, and we can try to relate the prime elements of R to the prime elements of those domains.

Moreover we can recognize the image of the map, i.e. an element (x,y) of the product comes from an element of R if and only if the elements x and y have the same image in the quotient R/(I+J). Then a prime element f of R will generate an ideal (f) in R that contains either I or J, hence will define a prime element in either R/I or R/J.

But I don’t quite see when a pair (x,y) in the product, such that either x or y is prime in one of the factor rings, and x and y have same image in R/(I+J), comes from a prime element of R. e.g. i can imagine cases where say x is prime and y is also prime but the element they come from is not prime in R, and also cases where x is prime and y is a unit and they do come from a prime elment, but no general statement.

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mathwonk
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fresh 42. this is the statement i have been trying to understand; "The claimed equivalence, however, is in my opinion a bit sloppy."

I.e. it seems to me that a non zero element that generates a proper prime ideal is exactly one that satisfies your definition in post #3. i.e. the ideal is prime, which is equivalent to your divisibility condition, and the ideal is proper, which is equivalent to it not being a unit. what am i missing?

your example explaining the non equivalence does not seem to work, since it satisfies neither of our definitions, i.e. although the ideal 4R in R = Z/24 is proper, it is not prime. but maybe we do understand each other and are just spinning our wheels at this point.

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mathwonk