Talk:Division ring

Strictly noncommutative

I tagged the "strictly" in "strictly noncommutative" as needing clarification. I can't figure out what's meant here, nor can I even find any good sources that explain it. --Deacon Vorbis (talk) 14:08, 27 March 2017 (UTC)

I'm slightly guessing, or half-remembering discussions that may or may not have taken place, but I think the point is that some people might use "noncommutative" to mean "not assumed/required to be commutative". Probably not for a particular division ring, but for division rings in general. So "strictly" noncommutative would mean that, not only are we not specifying that they're commutative, but there is at least one example of elements that in fact do not commute.

If my guess is right, then I would just remove "strictly", which I don't think really clarifies anything. --Trovatore (talk) 16:46, 27 March 2017 (UTC)

I suspect that the intent is to say a "division ring which is not a field". The problem here seems to stem from the fact that some people (and they should have their licenses to define things revoked!) permit noncommutative rings to include commutative rings, so "strictly" is important if you are working with that definition. So I wouldn't just remove strictly, I'd ditch the whole construct and replace it with "division ring which is not a field." --Bill Cherowitzo (talk) 04:14, 28 March 2017 (UTC)

Well, in context, it's about two specific division rings (the quaternions, and the rational quaternions), and it doesn't say "noncommutative ring" but rather "noncommutative division ring", which I think is clear on its own without further elaboration. I assume that anyone who uses "noncommutative ring" to include rings that may be commutative does so because he/she takes "ring" to mean "commutative ring" by default. That's actually a reasonable choice if you use that convention (which I don't). But no one (AFAIK) uses "division ring" to mean "field" by default, so the issue doesn't come up. --Trovatore (talk) 04:54, 28 March 2017 (UTC)

French terminology

It is slightly misleading to say that French "corps" stands for English field. First "corps" is not necessarily commutative, second, French "champ" (which literally means field) is used by French speakers to indicate a "corps commutatif", i.e., a field.

cerniagigante (talk) 10:24, 15 April 2024 (UTC)

I am French, and professional mathematician since more tham 60 years. I never heard "champ" used for "commutative field". If "champ" has had this meaning, this was before the second world war. D.Lazard (talk) 15:11, 15 April 2024 (UTC)

One-sided invertibility of matrices

in the "Relation to fields and linear algebra" section, the sentence "However, a matrix that is left invertible need not to be right invertible, and if it is, its right inverse can differ from its left inverse. (See Generalized inverse § One-sided inverse.)" seems highly dubitable. First: the link discusses one-sided inverses of non-square matrices over commutative fields and thus seems irrelevant. Second: if inverses exist on both sides, they have to be identical by associativity. Third: for the claim about one-sided inverses not implying the other side to be nontrivial, it has to be about square matrices. If this is true, we should give an example. Gnampfissimo (talk) 11:54, 8 December 2025 (UTC)

@Gnampfissimo: It is certainly true that if inverses exist on both sides, they have to be identical by associativity. I also agree that the link is irrelevant, because the one-sided inverses it refers to differ from the more usual inverse matrices not because they are over division rings that are not fields, but because they are not square matrices. As for the possibility of there being a one sided inverse on one side but not the other, that doesn't seem right to me. As far as I can see, the linear independence of the rows of the matrix is a necessary and sufficient condition for the existence of an inverse on either side, just as it is for fields, but maybe I've got that wrong? The statement was added to the article by D.Lazard, so perhaps he can clarify the issues. JBW (talk) 01:06, 11 December 2025 (UTC)

@JBW : I've played around with this a bit more and I think I get it now. Somewhere on stackexchange I saw someone claim that the rank theorem generalizes from fields to division rings in the sense that the left row rank agrees with the right column rank (and vice versa). If I've thought this through correctly, for right invertibility the left row rank is the relevant one and for left invertibility it's the right column rank, so invertibility on either side follows from the other. What doesn't follow is invertibility of the transpose, because that would rely on the right row rank and left column rank. Consider for example the quaternion matrix A = 1/sqrt(2) ((1, j),(i,k)). This fulfills A_{2n} = A_{1n}j, i.e. the columns aren't right-linearly independent, i.e. the matrix isn't left invertible. Analogously A_{m2} = iA_{m1}, i.e. the rows aren't left-linearly independent, i.e the matrix isn't right invertible. No such thing holds for its transpose though, in fact that's a unitary matrix (idk if you use that term in this context, but ykwim). Gnampfissimo (talk) 21:00, 11 December 2025 (UTC)

@Gnampfissimo: Yes, I suppose it's possible that this results from a confusion concerning what happens with transposes. (Incidentally, my failure to specify left or right linear independence was an oversight, no doubt due to being more used to dealing with commutative rings, but my working actually used right independence in both cases.) The failure to cite a source for the content is not helpful. I was intending to wait for D.Lazard to respond. He's usually very prompt in doing so, but I have discovered that recently he has not edited for a while, so waiting might be slow. I'm therefore going to go ahead with the following:

• Clearly, as you have said, if inverses exist on both sides, they have to be identical by associativity. Therefore that is evidently a mistake, so I shall remove it.
• We both agree that the link about one-sided inverses is irrelevant, so I shall remove it. (Incidentally, that irrelevant link was not given by D.Lazard, but was added later.)
• The claim that invertibility on one side doesn't imply invertibility on the other side seems dubious. I haven't exactly proved that it's wrong in general, but I have done so for small matrices over the quaternions. I am going to remove it on the basis that it's unsourced and has been disputed. Of course if D.Lazard, or anyone else, has an justification to offer they can do so, and the content can be restored.

Here's something which is actually irrelevant to this discussion, because it isn't about division rings, but it's related. I remember my surprise, years ago, when I discovered that over some rings it's possible to have a non-square matrix with a two-sided inverse; that is to say an n × m matrix A and an m × n matrix B, where n ≠ m, such that AB is the n × n identity, and BA is the m × m identity. It went so much against my intuition, which had been built upon the concept of the rank of a matrix, which can be reduced but not increased by multiplication. To put it another way, it seemed obvious that a smaller matrix contains less information than a larger one, so it can't be possible to recover all the information in the larger one from the smaller one. I was unprepared for discovering that those ideas just don't apply to some rings. JBW (talk) 22:19, 11 December 2025 (UTC)

@JBW about the invertible non-square matrices: does this rely on working over a ring without IBN? In that case you can construct an invertible non-square matrix by representing the identity over two bases of different size. I wonder if there's still a way to get around it if we assume IBN though. Gnampfissimo (talk) 11:45, 12 December 2025 (UTC)

@Gnampfissimo: The only way I know of doing it is for a ring without IBN, and it seems to me unlikely that it can work with an IBN. JBW (talk) 12:19, 12 December 2025 (UTC)

Actually I think I may in fact have used left linear independence in one case and right linear independence in the other. I did much of the working in my head, and wrote down only brief notes, so I can't check my working, but I'll do it again, being more careful about writing it all down. JBW (talk) 11:23, 12 December 2025 (UTC)

@JBW I think for the equivalence of left and right invertibility we only need the rank-nullity theorem (besides left row rank = right column rank), and as far as I can see for both the proof transfers just fine from the case over fields. Gnampfissimo (talk) 11:41, 12 December 2025 (UTC)

Unfortunately it's a very long time since I studied this kind of stuff, and I've forgotten much of what I ever knew about matrices, so instead of just using well known theorems I'm having to go through proofs of each step by very basic elementary methods. However, despite what I said about not keeping written notes, I now remember what I did, and it's ridiculously easy. If the rows of a matrix satisfy a linear relatonship (whether a left or a right linear relationship) then the rows of any multiple of that matrix satisfy the same linear relationship; therefore no multiple of that matrix can be the identity matrix, whether the multiplication is on the right or the left. Therefore if one-sided linear dependence of the rows is a necessary & sufficient condition for nonexistence of an inverse on one side, then it's also a necessary & sufficient condition for nonexistence of an inverse on the other side. Therefore I was perfectly OK in using only left linear independence. I am now pretty well convinced that if a matrix over a division ring has a one-sided inverse then it is a two-sided inverse. JBW (talk) 12:19, 12 December 2025 (UTC)

@Gnampfissimo: OK, continuing my process of reinventing the wheel by working through each step by very basic elementary methods, I have reminded myself of something I used to know without having to think about it. If the rows of the matrix are linearly independent then Gaussian elimination works for a division ring, just as it does for a field. Together with what I said above, that establishes that a one-sided inverse must be a two-sided inverse, if I haven't made any mistakes. (And of course if I have then please tell me.) JBW (talk) 14:05, 12 December 2025 (UTC)

Sorry for not having been here for a while, and for having introduced a wrong sentence. As far as I remember I was trying to make understandable obscure sentences, and I ommited to verify everything.

I agree that ssociativity implies immediately that right- and left-invertibility imply invertibility.

I seems also true (I did not checked the details) that the four ranks of a matrix are equal (the ranks of the left- and right- modules spanned by the rows and the columns), and that a square matrix is left- and right-invertible if and only it has a full rank. The simplest proof seems the following (see Rank (linear algebra) § Rank from row echelon forms and Rank (linear algebra) § Proof using row reduction): the elementary row operations implied by Gaussian elimination do not change the left-module spanned by the rows and thransform the three other modules into inomorphic ones. So, these elementary operations do not change any of these ranks. It is immediate that, for the resulting reduced row echelon form, the four ranks are equal.

After such a mistake, I leave to others the task to add this to the article, if useful. D.Lazard (talk) 16:59, 17 December 2025 (UTC)