Talk:Irreducible component

Real Numbers Reducible

What is this part of the article saying. I neither believe the statement nor understand the example.

Every open set is dense in an irreducible space, which is clearly false for the real numbers. Thus R is reducible. Father Chaos (talk) 23:13, 23 August 2012 (UTC)[reply]

February 2014

It is essential to refer local irreducibility. Coffeebrake60 (talk) 11:37, 8 February 2014 (UTC)[reply]

I took out the comment saying that R is reducible with the usual topology. This is false. I am a expert in the field. Coffeebrake60 (talk) 11:42, 8 February 2014 (UTC)[reply]

This is true that R is reducible for the usual topology and this definition of irreducibility: it is the union of two closed subsets, for example the nonnegative numbers and the non-positive numbers. Maybe you have another definition in mind, or you are confusing irreduciblr and connex. Therefore I'll revert your edit. Nevertheless this article, although correct, is very badly written. The main point is that it emphasizes the topological definition, when the notion originates in algebraic geometry, and from the topological point of view, it makes sense only for Zariski topology. D.Lazard (talk) 12:25, 8 February 2014 (UTC)[reply]

The empty space

The article (contradicts its own reference) is missing clarifying the status of the irreducibility of the empty space. Namely, that there is (currently) a unique definition.

It is true that currently are in existence texts that both include it as irreducible (as does the definition currently in the article), and those that explicitly exclude it. The article should reflect this. Moreover, the definition that exclude it (roughly) appears in more modern texts. It appears in the single reference that the article has. I would dare to say that with time it is likely that everyone will move to opt to exclude, for the same reasons as 1 is excluded from being a prime. Thatwhichislearnt (talk) 14:10, 23 January 2024 (UTC)[reply]

There is no ambiguity in this article (after my recent edit): the empty set is an irreducible variety, and the irreducible components of a non-empty algebraic set are all non-empty. The only question that is left open by this article is the characterization of the irreducible components of the empty set. One may consider that the empty set has a unique irreducible component that is itself; one may also consider that it has no irreducible component; in this case, it is indeed the empty union of its irreducible components. The latter choice is the only one that is compatible with the other common conventions, namely that the zero ring is not an integral domain, that a prime ideal in a ring is a proper subset of the ring, and that 1 is not a prime number.

As the case of the empty algebraic variety is totally trivial, it is not useful to spent much time on it, if all stated assertions are correct. D.Lazard (talk) 15:17, 23 January 2024 (UTC)[reply]

I didn't say that there was any ambiguity, there wasn't any, nor talked about the case of varieties. The article currently defines the empty topological space as irreducible, as it resorted to simply negate the condition on being reducible. This is done in some books in topology, while some others explicitly exclude it. An example of the latter is the single reference that the article cites. For the first, I would need to check, but I know some good books do (although they are getting older). Since both positions are currently in use, and even more things are moving towards not calling the empty space irreducible, the Wikipedia article should mention both definitions, instead of mentioning only one (the one that is becoming older). Thatwhichislearnt (talk) 15:38, 23 January 2024 (UTC)[reply]

And regarding "the empty set is an irreducible variety", that very much depends on who you ask. In particular, Algebraic Geometers are more likely to the opposite. In topology texts is where is still possible to find such definition. Thatwhichislearnt (talk) 15:56, 23 January 2024 (UTC)[reply]

As @D.Lazard explained, there is no contradiction in assuming the empty set is irreducible, and there is a logical way to talk about the irreducible components of the empty set; and also there in nothing very interesting about the empty space. Algebraic geometry is not my expertise. But in general topology at least the simplest way to define a hyperconnected space is to not make a special case for the empty set. And then the empty set (with its unique topology) happens to be hyperconnected according to that definition. It is just what it is. But there is nothing particularly interesting about the empty set, so nobody should really care one way or the other about that convention. On the other hand, in algebraic geometry does assuming the empty set is not irreducible make some theorems easier to state without special cases maybe? PatrickR2 (talk) 07:20, 25 January 2024 (UTC)[reply]

No need to explain well known facts. Yes, it is just a definition, and theorems can be written correctly any way one chooses to fix the terminology. It is exactly the same as whether one considers 1 a prime or not. All theorems can be adjusted accordingly. What I am pointing out is regarding current use. While almost no one calls 1 a prime nowadays, the empty set is called irreducible still. But it is also called not irreducible, as well. So, Wikipedia should inform. Thatwhichislearnt (talk) 13:16, 25 January 2024 (UTC)[reply]

References:

Empty space is not irreducible:
- http://alpha.math.uga.edu/~pete/pointset.pdf See page 126, exercise 3.46
- https://stacks.math.columbia.edu/tag/004U The only current citation in the article, which contradicts the definition currently in the article.
Empty space is irreducible:
- Bredon's Topology and Geometry. Page 10, Problem 5.
- I don't have my Kelley at hand, but I think it falls in this case. — Preceding unsigned comment added by Thatwhichislearnt (talk • contribs) 19:01, 23 January 2024 (UTC)[reply]

For a modern reference in general topology, Encyclopedia of General Topology (2003): https://books.google.com/books/about/Encyclopedia_of_General_Topology.html?id=JWyoCRkLFAkC&source=kp_book_description defines hyperconnected without special case for the empty set. See here: https://books.google.com/books?id=JWyoCRkLFAkC&q=hyperconnected#v=snippet&q=hyperconnected. PatrickR2 (talk) 07:29, 25 January 2024 (UTC)[reply]

Let's see how many times I need to repeat it. I never said that there was any contradiction, or error. Once again. What I said is that there was an omission of a version of the definition that is very much in use. Well, the very single reference that the article has/had used the definition excluding the empty set. An encyclopedia should reflect this state of affairs. A bit of speculation on my part, but I think it is likely that modern publication will tend to favor not calling the empty irreducible. Like everything with language, it takes time. Thatwhichislearnt (talk) 13:09, 25 January 2024 (UTC)[reply]

You need more care in your affirmations. I suspect that most sources exclude the empty set from the definition of irreducibility as well as from the definition of reducibility. So the two possible definitions are not those you suggest: nobody defines the empty set as being reducible. The empty set may be said irreducible; otherwise the concept of irreducibility does not apply to it; that is, it is neither irreducible nor reducible. Similarly, 1, 0 and

π

are neither prime nor composite.

However, there is, here, a big difference with the case of the primality of 1. There is no theorem whose statement depends on the definition chosen for empty set, while for the fundamental theorem of arithmetic, uniqueness is wrong if 1 is supposed to be prime.

In summary, irreducibility was originally defined for non-empty sets only. The extension of the definition to the empty set is a minor generalization that has no significant consequences. As an encyclopedia cannot take into account every minor development, there is no reason to include this in Wikipedia. D.Lazard (talk) 14:16, 25 January 2024 (UTC)[reply]

"There is no theorem whose statement depends on the definition chosen for empty set". Bold, bold statement. One has to have an overinflated opinion on one's own knowledge to make such a claim. And it is in fact, wrong. Thatwhichislearnt (talk) 14:33, 25 January 2024 (UTC)[reply]

"uniqueness is wrong if 1 is supposed to be prime" is a high school teacher's argument. The FTA's uniqueness already has a caveat with 1 not a prime, regarding the ordering of the factors. In UFD even more caveats, regarding associated elements. It is but a minor inconvenience to add to FTA the presence of a unit to any power, like in UFD. It is not about one theorem being wrong. Every single theorem can be adjusted accordingly. The real inconvenience is to have to add exceptions to all other theorems on primes, connected sets, and irreducible sets. Of which there are many. Any way. The discussion here is not about the merits of the definitions. Opinions are irrelevant. Both definitions are in use. Wikipedia should mention both. Thatwhichislearnt (talk) 14:50, 25 January 2024 (UTC)[reply]

The same clarification is done for connected space, as well as for cases in which the alternative is not in use: For example, for fields clarifying that the trivial ring is not one. Even for prime number, there is the mention of the now defunct definition that includes 1. In the case of irreducible there are two in use. It is fine if a Wikipedia's article chooses to mainly use the one that is more likely to go away in the future. But mentioning the two, when there are two in use, it should. Thatwhichislearnt (talk) 15:10, 25 January 2024 (UTC)[reply]

Forget about prime numbers, connected sets, etc not relevant here. Both of you make some valid point. We should mention that both conventions are used by different authors. But as @D.Lazard explained well, the probable reason many algebraic geometers use the convention that the empty set is not irreducible is historical: people who cared about the notion of ireducible set originally defined it for nonempty sets, because they did not care about the empty set. But there was no intrinsic reason to do so. And on the other hand, in general topology, as already explained, hyperconnected is defined with no special condition for the empty set.

So saying things like modern publication will tend to favor not calling the empty irreducible and ... the one that is more likely to go away in the future makes little sense. PatrickR2 (talk) 00:11, 26 January 2024 (UTC)[reply]

It is the exact opposite. Lazard is just plain wrong. The dated definition is that which ignored the empty set as special and thus included it as irreducible. It is once you start considering its role in statements about decomposition into components, functors, theory of species, existence of generic points, etc. - most of these appearing relatively more recently - that the tacit classification of the empty set as irreducible becomes inadequate. It is the more modern publication the ones that do take care of explicitly exclude the empty set from being irreducible. Thatwhichislearnt (talk) 13:52, 26 January 2024 (UTC)[reply]

I have slightly streamlined the section about the empty set, but preserved all your contents. (Also putting a term in bold should only be used the first time it is defined.) Hope you don't mind. Cheers. PatrickR2 (talk) 06:39, 26 January 2024 (UTC)[reply]

It is not "my content" and as everything in Wikipedia it is expected to change. So, feel free to change it. If anything it is the content of the (currently) only reference that this Wikipedia article has. Thatwhichislearnt (talk) 13:54, 26 January 2024 (UTC)[reply]