# Talk:Integral domain

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Field:  Algebra

## Noncommutative integral domains

I want an example about integral domain is not field

Do you mean in the article, or in general? ${\displaystyle \mathbb {Z} }$ is the obvious example.--Clipdude 07:01, 19 Nov 2004 (UTC)

So how about adding that an integral domain isn't always assumed to be commutative and what happens when it's not? http://www.ams.org/msc/16Uxx.html --mindspank 08:07, 15 Feb 2005 (UTC)

I am interested in such generalizations, but I've never seen an integral domain defined as such without the commutative condition. A section on "noncommutative generalizations" could provide links to the appropriate articles. - Gauge 18:20, 15 March 2006 (UTC)
Over here, a "noncommuative integral domain" is called "domain", and integral domains are always commutative. I find it curious that the AMS classification disagrees in this regard. Anyway, a search through all the algebra books I have at hand agrees with my interpretation. However, a search on mathscinet *did* reveal a few (14) articles referring to "noncommutative integral domains". Well, in my oppinion, this is just yet another case where we mathematicians can't agree on a single common terminology :-). So for now, I added a note that some people talk about noncommutative integral domains, but that we call those simply "domains". The article on the latter certainly should be extended. Of course a lot of things break in the noncommutative case. - BlackFingolfin 00:03, 3 May 2006 (UTC)

## Is there some kind of requirement on the order of the integral domain?

For example, is there an integral domain with exactly 6 elements? -unsigned

There is no requirement on the order. The question for which numbers there exist integral domains with that order is I think very hard to answer. I don't know if there exist integral domains with six elements. Oleg Alexandrov (talk) 03:54, 31 October 2006 (UTC)
Er, a finite integral domain is a field, and finite fields are of prime power order - or have I got that wrong? Septentrionalis 22:02, 1 November 2006 (UTC)

No, you had it absolutly right. The order of a finite field (therefore a finite integral domain) must be a prime power. Conversly, there exists (up to isomorphism) exactly one finite filed of order a prime power.

## rating?

Do you think this article is really "top" importance? E.g. group is considered to be only "high" importance and to me a group is way more an important gadget than an integral domain. I would change it to "mid", if no one objects. Jakob.scholbach 01:02, 13 May 2007 (UTC)

I changed the importance to mid. Jakob.scholbach 21:24, 13 May 2007 (UTC)

## Entire ring

In his classic Algebra text, Serge Lang calls integral domains entire rings. From the discussion on pages 91–92 (3rd ed.) it is clear that this terminology is his own. Does anyone else use this terminology? Is it worth mentioning here (and providing a redirct)? -- Fropuff (talk) 07:11, 20 January 2008 (UTC)

This synonym also appears in the entry Glossary of ring theory. It was added (without citing sources) in 2005, but the editor who added it hasn't edited before or since under the same name. Michael Slone (talk) 18:21, 20 January 2008 (UTC)
I did notice that there. I suspect they took the definition from Lang. I think if we are to include the term in Wikipedia it should really enjoy wider usage then in a single textbook. -- Fropuff (talk) 18:27, 20 January 2008 (UTC)

## uncomfortable with terminology

I am a working algebraic number theorist, and I am not so comfortable with the convention that "domain" means a ring without zero divisors and "integral" connotes commutativity. The adjective "integral" already has a meaning in algebra: it describes a certain kind of element of an overring, or of an extension of rings. I think these two quite different usages will be confusing to many.

As a general rule of thumb, one knows in advance whether one is entertaining possibly non-commutative rings or not. I would call a noncommutative ring without zero divisors a "noncommutative domain", or, if the noncommutativity of the rings is understood, just a "domain."

Admittedly the use of integral in abstract algebra is ripe for being overhauled -- more consistent with the geometric terminology would be just to call a commutative ring "integral" if it has no zero divisors: this would make it true that an affine scheme is integral iff it is Spec of an integral ring. But then one needs to come up with some "geometric" terminology for an integral extension. To the best of my knowledge, there is -- strangely -- no geometric word for this. 72.152.146.78 (talk) 07:49, 18 March 2008 (UTC)Plclark

The above comment seems to me to be an attempt to start a discussion on the suitability of changing practice in use of mathematical terminology, 'not a suggestion about editing this Wikipedia article, which is what this talk page is for. Or have I misunderstood? JamesBWatson (talk) 15:39, 5 March 2009 (UTC)

## Variation of definition

Ozob has deleted the statement that there is some variation in the definition of an Integral Domain, giving as edit summary: "Correct definition. All the books I looked at (I gave cites) assume 0 != 1". Clearly this depends on what books Ozob has looked at, but Hartley & Hawkes ("Rings, Modules, and Linear Algebra", Chapman & Hall (1970), page 49) after giving the usual definition, go on to say "It is worth pointing out, however, that there is no universal convention about the definition of an integral domain; some authors leave out the requirement that 1 ≠ 0, and others drop the requirement of commutativity". While it is quite right that the article use the usually recognised definition, I see no reason for not mentioning the occasional variation. JamesBWatson (talk) 16:52, 5 March 2009 (UTC)

The variation doesn't seem to occur in practice. I've never heard of anyone adopting such a definition, and I looked at all the books that were close at hand to confirm: Everyone seems to assume 1 ≠ 0. If you can provide a source that actually uses that definition, then I'll agree that we should mention it. Otherwise, I'd say that it's an error or misunderstanding in Hartley and Hawkes. Ozob (talk) 16:58, 5 March 2009 (UTC)
For the moment I've left the mention of the unusual definition and the citation of Hartley and Hawkes in the article. But I'm going to take it out unless you can give me notable sources that actually use this definition. All of the five books I've looked at support me—and they're standard references, not obscure stuff. If nobody uses this definition, or if it's used only by a very, very small number of people, then it shouldn't be mentioned here. Ozob (talk) 17:19, 5 March 2009 (UTC)
There are two things I don't understand: (1) Why it shouldn't be mentioned, even though it is rare, and (2) why on earth Ozob feels so strongly about so unimportant a point. Since there is some variation in definition I thought it worth mentioning, and I still think so, but I don't think it's worth spending any more time or effort on it. JamesBWatson (talk) 17:26, 5 March 2009 (UTC)
I think that it is wrong and that Wikipedia should not include what is wrong. It's not wrong for any moral or philosophical reason, and when I condemn it as wrong I'm not condemning it as bad or evil. But integral domains have a standard definition, and that definition includes the axiom 1 ≠ 0. It feels like saying "Some authors define yellow to be the color of the ocean".
As I keep saying, if there are notable references that use this definition, then I'll admit to my own wrongness and to some variation in the literature. But I suspect the literature is totally consistent on this one. Ozob (talk) 17:42, 5 March 2009 (UTC)
I agree that it may be worth pointing out to readers that there s/he may possibly encounter different definitions (in particular the noncommutative version) but I don't think it is very useful to spread these rarely used alternative definitions to other articles as was done here. —Ruud 22:04, 5 March 2009 (UTC)
What I'm still not convinced of is that there is an alternative definition, and that it's not just a misunderstanding. JamesBWatson quoted Hartley and Hawkes above as saying that sometimes people drop the commutativity assumption or the assumption that 0 ≠ 1. I agree that sometimes the commutativity assumption may be dropped. I've never seen the non-triviality assumption dropped, anywhere. Even Hartley and Hawkes are not dropping that restriction themselves; they're just saying that some others do. It makes me want to put in a [who?] template.
I suppose it may not be obvious why I care so much about what must look like a trivial point. So let me explain: An ideal P of a (commutative) ring R should be prime if and only if R/P is an integral domain. If the zero ring is a domain, then R is a prime ideal of itself. In particular, every ring has a prime ideal which contains all other prime ideals. In terms of the geometry of the prime spectrum, this means that every point of any affine algebraic variety specializes to this strange point which I'll call the empty point since it contains no ordinary points. So (ordinary) points are no longer the smallest things in an algebraic variety; still smaller is the empty point. Every ring would have an empty point, so there would be no affine scheme corresponding to the empty set; the smallest you could get would be the spectrum of the zero ring, which would contain only the empty point. But the empty set would still be a (non-affine) scheme, so there would be an even smaller scheme than the scheme containing only the empty point! This is undesirable, and this is why I object so strongly.
Since nobody has put forward any references, I've removed the unusual definition from the article as well as the citation to Hartley and Hawkes. Ozob (talk) 16:56, 6 March 2009 (UTC)
J.C. McConnel and J.C. Robson use (obviously) the noncommutative definition in their book on "Noncommutative Noetherian Rings" (Garduate studies in Mathematics Vol. 30, AMS).

I think this should be mentioned in the article. —Preceding unsigned comment added by 2009 92.192.40.92 (talkcontribs) 14:05, 14 December

## Subring of field?

The article offers one definition of an integral domain "as a subring of a field." How is this reconciled with the theorem that every commutative ring with no zero divisors extends to its field of fractions? --Vaughan Pratt (talk) 04:14, 10 June 2009 (UTC)

By "extends to" you probably mean "admits a natural injection into". The existence of the field of fractions shows that every integral domain is a subring of a field; and conversely, every subring of a field has no zero divisors (and has 0 not equal to 1), so it is an integral domain. Ozob (talk) 18:48, 10 June 2009 (UTC)
Are you assuming that rings (and hence subrings) must contain a unit? If so then why the distinction between an integral domain and a commutative ring with no zero divisors in the article on field of fractions? If not then why should a subring of a field contain the unit? --Vaughan Pratt (talk) 02:19, 27 June 2009 (UTC)
Well, I suppose that in the greatest generality, one should not assume that subrings have a unit. This is usually done when working with operator algebras, I understand. But in the context of commutative algebra, it's most common to assume that all rings have a unit and that all subrings preserve the unit; that is, all rings are non-trivial Z-algebras and all subrings are Z-subalgebras. I assumed this in my above comment; in particular the zero ring is not a subring of anything other than itself. The field of fractions article, on the other hand, is trying to be very general, so it does not assume the existence of a unit. Does that help?
To answer your question in the more general setup where rings need not have unit: The existence of the field of fractions again shows that every integral domain is a subring of a field, and we can check easily that the multiplicative identity of the integral domain maps to the multiplicative identity of the field. Conversely, every subring of a field is a commutative ring with no zero divisors, so if it also has a non-zero multiplicative identity, then it is an integral domain. If it has a non-zero multiplicative identity u, then that multiplicative identity must be the same as the multiplicative identity 1 of the original field, because the identity u2 = u in the subring implies u = 1 after multiplication by u−1 in the field. Therefore we see that a subring of a field is an integral domain if and only if it contains the multiplicative identity of the field. Ozob (talk) 01:47, 3 July 2009 (UTC)

## Why is "${\displaystyle a^{2}+5b^{2}=3}$ has no integer solution" a proof that 3 is irreducible?

Hi,

in the section Divisibility, prime and irreducible elements one finds the following argument:

The numbers 3 and ${\displaystyle \left(2\pm {\sqrt {-5}}\right)}$ are irreducible as there is no ${\displaystyle \pi =a+b{\sqrt {-5}}}$ where ${\displaystyle \pi |3}$ or ${\displaystyle \pi |\left(2\pm {\sqrt {-5}}\right)}$ as ${\displaystyle a^{2}+5b^{2}=3}$ has no integer solution.

Yet, it doesn't make sense to me. If 3 were reducible, this would mean that it could be written as a product: ${\displaystyle (a+b{\sqrt {-5}})(c+d{\sqrt {-5}})}$, but not necessarily as the square of ${\displaystyle (a+b{\sqrt {-5}})}$, which is ${\displaystyle a^{2}+5b^{2}}$. Am I not right?

Best regards, Wisapi (talk) 13:54, 29 May 2011 (UTC)

The idea is as follows: If 3 were reducible, we could write ${\displaystyle (a+b{\sqrt {-5}})(c+d{\sqrt {-5}})=3}$. Taking norms on both sides results in ${\displaystyle (a^{2}+5b^{2})(c^{2}+5d^{2})=9}$. The only non-trivial factorisation of 9 is 3x3, so we should have ${\displaystyle a^{2}+5b^{2}=3}$. The same goes for ${\displaystyle \left(2\pm {\sqrt {-5}}\right)}$, because the norm is again 9.
Best, 128.214.200.220 (talk) 15:22, 14 September 2011 (UTC)

## Why does the ideal (3) equal the ideals ((2+sqrt(-5)) and ((2-sqrt(-5))?

I don't understand this remark (section "Divisibility, prime and irreducible elements"). Why are these ideals identical? --Thn2010 (talk) 13:27, 25 December 2011 (UTC)Thn2010 (talk) 09:08, 27 December 2011 (UTC)

I'm not sure that they are, although I believe the statement about the product of prime ideals. Not sure whether the double bracketing is some piece of notation I'm not familiar with, but I would guess that ${\displaystyle ((2+{\sqrt {-5}}))}$ at least contains 2+√(-5), which isn't in (3). I would just change it in the article, but it's been there for almost 2 years and I'm a little worried that I've made a silly error. Raelthelamb (talk) 13:41, 11 January 2012 (UTC)
I've removed the example. It's false.
To see that (3) isn't equal to (2 + sqrt(-5)), notice that 2 + sqrt(-5) times a + b sqrt(-5) is (2a - 5b) + (a + 2b)sqrt(-5). We want the first coordinate to equal 3 and the second to equal zero, and that's just solving a linear system. We have a = -2b, so -9b = 3, so b = -1/3 and a = 2/3. But that's not in our ring. Now, in general, an element of an ideal is a sum of linear combinations of the generators; in this case we have a single generator, so an element of (2 + sqrt(-5)) consists of the multiples of 2 + sqrt(-5). We just showed that 3 isn't one of those elements, so (3) is not contained in (2 + sqrt(-5)). Ozob (talk) 23:57, 11 January 2012 (UTC)

## Domain and integral domain: Merge proposal

There is a page domain (ring theory). On the other hand "Integral domain" is a pleonasm, as for most mathematicians "domain" is a synonym of "integral ring". Thus I propose to merge this page into domain (ring theory). If for some rare authors there is a slight difference between these notions, it deserve to be quoted in a single page. D.Lazard (talk) 22:29, 26 February 2012 (UTC)

As far as I understand, this article is about commutative rings, while domain (ring theory) is centered on non-commutative examples, with a suitable bibliography. This splitting seems to make sense, does not it ? French Tourist (talk) 18:03, 29 February 2012 (UTC)
I'm with the last comment that the split is OK. In a perfect world they would be in the same article, but keeping commutative domains apart seems like a good idea because of the depth of analysis on them that is not paralleled in the noncommutative case. I think a lot of mathematicians will interpret "domain" as "commutative domain", but I've never seen "integral domain" used for a noncommutative domain, so the commutative case is well placed here and "domain" takes care of the noncommutative case. Rschwieb (talk) 19:11, 9 April 2012 (UTC)
No, leave Integral Domain as a separate article. I just arrived there from Discriminant, following a link within the sentence, "As the discriminant is a polynomial function of the coefficients, it is defined as soon as the coefficients belong to an integral domain R and, in this case, the discriminant is in R." The aptness of that link would really suffer if Integral Domain were merged with anything else. Dratman (talk) 10:04, 15 July 2012 (UTC)
This is not a comment on the merge, but, if the merge were done, integral domain would redirect to domain (ring theory), or a section of it, so the link wouldn't suffer. I think commutative vs. non-commutative might be an adequate reason to keep them separate, if that is used in the "real world" (i.e., reliable sources), but I'm not sure. — Arthur Rubin (talk) 23:40, 10 August 2012 (UTC)
No. To merge integral domain to domain is not just wrong but really stupid. These two are can be very different topics. I am not much aware about domain. But integral domain is a frequently used terminology in abstract algebra. Aravind V R (talk) 08:57, 12 September 2012 (UTC)
It's been seven months, and I think the consensus has been pretty clearly against a merge. I'm going to take down the template and hope I'm not breaching some time limit I don't know about. Rschwieb (talk) 12:53, 18 September 2012 (UTC)

## Equivalent definitions of an integral domain

Reddyuday has detected that one of the item was wrong in the equivalent definition of an integral domain. This item is

• An integral domain is a ring for which the set of nonzero elements is a commutative monoid under multiplication.

This is a blatant error, as every commutative ring satisfies this definition. Reddyuday has corrected this item as

• An integral domain is a commutative ring for which the set of nonzero elements is a cancellative monoid under multiplication.

Ozob has reverted twice this correction despite it is perfectly correct. In fact, cancellation property means that ab = ac implies b = c for nonzero a, b and c. This may be rewritten as a (b-c) = 0 implies b - c = 0. Every element different of 1 has the form d = b - 1. with b non zero. This shows that the condition may be rewritten ad = 0 implies d =0 for non zero a, showing the equivalence with the usual definition.

Reddyuday edit being correct, I will restore it. D.Lazard (talk) 18:38, 16 February 2014 (UTC)

It is correct, but it is also needlessly complicated. What I put on the page was both simpler and also correct. It stated:
• An integral domain is a ring for which the set of nonzero elements is a commutative monoid under multiplication.
We are in no dispute as to the forward implication. For the other, suppose that R is a ring with the stated property. To see that the ring is commutative, let a and b be elements of R. Their product ab in R is by definition their product in the semigroup consisting of R and the operation of multiplication. The monoid of the statement is obtained by restricting this multiplication to the non-zero elements of R, so if this monoid is commutative and if a and b are nonzero, then ab = ba. If either a or b is zero, then ab = 0 = ba. Therefore ab = ba always, so the ring is commutative. There is therefore no distinction between the ring being commutative and the monoid being commutative. I suspect that we agree on this and that we disagree only about the domain property.
To see that the ring is an integral domain, first recall that the multiplication map on R is a function R × RR. To determine whether or not R is an integral domain, we may restrict it to a function (R \ {0}) \times (R \ {0}) → R. The domain here is exactly the domain of the monoid, and the function is the operation of the monoid. The stated hypothesis that R \ {0} is a monoid means the range of this function is contained in R \ {0}. Therefore, for any two non-zero elements a and b, it must be that ab is non-zero, and hence R is a domain.
In particular this implies that the cancellative hypothesis is unnecessary. I have therefore reverted the article to the simpler statement, but added a short explanation as to why that statement is correct. Ozob (talk) 20:06, 16 February 2014 (UTC)
I think that there is still an interesting statement to be made regarding cancellative monoids and integral domains. I believe that the following is true:
• An integral domain is a commutative ring for which every non-zero element is cancellable under multiplication.
Ozob (talk) 20:10, 16 February 2014 (UTC)

## Equivalent definitions: need of a third party intervention

An IP user has removed four time the first of the equivalent the equivalent definitions of an integral domain

An integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero,

with the fallacious argument that it is exactly the same as the second definition:

An integral domain is a nonzero commutative ring with no nonzero zero divisors.

In fact there are not the same for several reasons: firstly, the first definition is less technical than the second one, as the latter requires to know what is a zero divisor for being understood. Secondly the first definition involves two elements the ring, while the second one involves only one element. Moreover, if they would be the same, they would be formalized by the same logical formula, which is not the case. The formalization of the first definition is

${\displaystyle (\forall x)(\forall y)\;x\neq 0\;\operatorname {and} \;y\neq 0\implies xy\neq 0,}$

while the second definition is

${\displaystyle (\not \exists x)\;x\neq 0\;\operatorname {and} \;((\not \exists y)\;xy=0\;\operatorname {and} \;y\neq 0).}$

Clearly these are not the same formula, and showing that they are equivalent requires several logical operations.

Thus the first definition must restored. Because of WP:3RR, this should be done either by the IP himself. if his math expertise is sufficient for understanding these explanations, or by a watcher of this article. D.Lazard (talk) 20:55, 3 November 2017 (UTC)

I have restored the disputed definition outside the list, because it is the basic definition, and the others are variants that use more WP:TECHNICAL terminology. D.Lazard (talk) 13:04, 5 November 2017 (UTC)