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Recently, while undertaking a study of commutative algebra, I learned three concepts: (i) a local ring, (ii) a regular local ring and (iii) a regular ring.

At the end, I found myself asking this seemingly naïve question: Are regular local rings the same objects as local rings that are regular? At first, I thought, "My mind must be acting stupid again." However, upon further analysis, it turned out that the answer to my question was non-trivial after all.

One direction, namely proving that a local ring that is regular is actually a regular local ring, is not very hard to establish. Indeed, it can be assigned as a homework problem in an undergraduate abstract algebra course. The key observation is that for a local ring $ (R,{\frak{m}}) $, upon localization at $ {\frak{m}} $, we obtain $ R_{\frak{m}} = R $. This is because $ R \setminus {\frak{m}} $ is precisely the set of units of $ R $. Hence, by the definition of regular ring, we see that $ (R,{\frak{m}}) $ is a regular local ring.

The other direction is a well-known (in my opinion, highly) non-trivial result in homological algebra, which states that the localization of a regular local ring at any prime ideal is still a regular local ring. By the definition of regular ring once again, regular local rings are therefore local rings that are regular.

I am wondering, are there any pairs of concepts in other areas of mathematics that look so similar that their similarity may be mistaken for tautology but, in reality, can only be established by a hard proof?

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An anti-example could be: "A function that coincides almost everywhere with a continuous function is not necessarily almost everywhere continuous" ...Or are you looking for something considerably less trivial? –  Qfwfq Dec 17 '12 at 22:24
Shouldn't this be community wiki? Also, you could theoretically put a "big list" tag on it if there end up being several quick good answers and you expect more. –  Neil Epstein Dec 17 '12 at 22:46
The fact that a localization of a regular local is regular is, in fact, a deep result. The fact that this can be stated as "a regular local ring is regular" is mildly amusing; but this is just word play, and, in my opinion, of no interest whatsoever. I voted to close as "not a real question". –  Angelo Dec 18 '12 at 5:55
I think the "regular" example is poor, as Angelo says, but I think there may be merit in other examples - e.g. analogies between results for groups and results for von Neumann algebras, some of which are even true rather than just "advertising"... –  Yemon Choi Dec 19 '12 at 20:05
To expand on what Angelo says -- we only define "regular ring" the way we do because of this property of regular local rings. Or, in other words, the fact that we have the term "regular ring" at all encapsulates this fact. –  Harry Altman Dec 20 '12 at 1:16

4 Answers 4

up vote 10 down vote accepted
  1. $f:\mathbb R^2\to \mathbb R$ is $C^\infty$.

  2. $f:\mathbb R^2\to \mathbb R$ is $C^\infty$ along each $C^\infty$-curve $c:\mathbb R\to \mathbb R^2$; i.e., $f\circ c$ is $C^\infty$ for each such $c$.

Equivalence was proved only in 1979 by Jan Boman. EDIT: It was 1967, sorry for being careless.

EDIT: Using "general abstract nonsense" and functional analysis, one push this result from $\mathbb R^2$ to Frechet spaces. Beyond Frechet spaces, the notions start to divergence. Analysis based on (2) is called convenient analysis, since it leads to a diffeomorphism $$C^\infty(U,C^\infty(V,W)) \cong C^\infty(U\times V, W)$$ and a monoidally closed category.


A.Frölicher and A.Kriegl: Linear spaces and differentiation theory. John Wiley & Sons Ltd., Chichester, 1988.


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Wow! That's interesting. –  Todd Trimble Dec 18 '12 at 21:30
Very interesting indeed! Looks trivial, but after looking at the actual proof, I am deeply impressed by its depth. –  Leonard Dec 19 '12 at 2:13
I can't believe the result is really that recent: I've seen a version of (2) as a definition of smoothness in old differential geometry books. Aside from that, I remember this being assigned as a homework problem in an analysis class; it's harder than it seems at first, but it just came down to compactness of the circle if I recall correctly. –  Paul Siegel Dec 19 '12 at 3:47
If the nLab gives a correct reference, then Boman proved it in 1967. Reference: Jan Boman, Differentiability of a function and of its compositions with functions of one variable, Math. Scand. 20 1967 249–268. –  Todd Trimble Dec 19 '12 at 18:17

In my example, the similarity did not require a hard proof but it was not seen for many years for the reasons which I would call "social".

In 1928 Weil (and simultaneously Siegel) defined and studied heights in algebraic number theory. In 1933 Henry Cartan introduced the Nevanlinna characteristic of a holomorphic curve in projective space. From certain view point these two things are the same:-)

Only in 1987, Paul Vojta pursued this analogy quite far. This became famous as "Vojta's analogy", and many new results were proved inspired by this observation. S. Lang, was very much excited and widely popularized this discovery. Since then, Vojta's analogy led to substantial progress in both areas. The story is described in Lang's book Introduction to Complex hyperbolic spaces, on page 185.

Why I called the reasons of this almost 60 years gap social?

Because on my opinion the reason is that complex analysts and algebraic geometers do not communicate sufficiently with each other.

Actually I noticed the analogy in 1982 (I am sure that I was not alone) and I told about my observation to a famous algebraic geometer in. He was not excited. But when the news of the Vojta's analogy reached him few years later, he run into my office, and said: "Alex, can you quickly tell me what's Nevanlinna theory about?"

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Vojta's advisor was Mazur, not Lang; Vojta had a postdoc position at Yale. –  KConrad Dec 18 '12 at 0:51
Thanks. I corrected. –  Alexandre Eremenko Dec 18 '12 at 14:53
Did Jean-Pierre Serre's GAGA not change the way that complex analysts and algebraic geometers look at each other's field? –  Leonard Dec 19 '12 at 2:16

Here is a possible example that I have been speculating about, on and off, since around 2006: it might not fit the OP's question, and I may be talking rubbish, so I'd welcome corrections from the true connoisseurs (as opposed to this dilletant).

Call a short exact sequence of Banach spaces and bounded linear maps naively exact if it is exact as a s.e.s. of vector spaces (this is just to get round the fact that category-theoretic cokernels and epimorphisms in Ban don't behave as they do in Vect). Call a Banach space $E$ homologically flat if $E\widehat{\otimes}\underline{\quad}$ preserves the short naively exact sequences of Banach spaces, where the tensor is the projective tensor product a.k.a. the "right one from the POV of closed monoidal categories".

Now if I understand the literature correctly, the homologically flat Banach spaces are known: they are precisely the ${\mathcal L}^1$-spaces of Lindenstrauss and Pelczynski, which are roughly speaking those where each fin-dim subspace $E$ is contained in a fin-dim subspace $F$ which is uniformly isomorphic to $\ell_1^{|F|}$.

As $\ell_1^n$ is the "free Banach space" on an $n$-element set, this result could be airily if imprecisely captured by saying "the flat Banach spaces are those built locally from finitely generated free Banach spaces". But now this sounds awfully like the Govorov-Lazard theorem from ring theory: a module over a fixed commutative ring $R$ is flat iff it is a colimit of finitely generated free $R$-modules.

As far as I know the Banach-space result is proved using duality (duals of flat spaces are injective, and the injective Banach spaces admit a description as ${\mathcal L}^\infty$-spaces) and doesn't seem to have a proof along the lines of the GL-theorem. But one might hope to find a proof that made use of a meaningful analogy between the category of Banach spaces and bounded linear maps, and categories of modules...

(Of course the module-categories are abelian, while Ban is not, but there may be some way to exploit the "embedding" of Ban into an abelian envelope, cf. work of Noël, Waelbroeck and in greater generality Schneiders. Calling Theo Bühler... calling Theo Bühler...)

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You rang...? I'm sorry to disappoint, I don't have all that much to say on this matter right now. We apparently noticed this analogy around the same time, and I agree that it is not totally trivial to make it precise. I will contact you in the first few weeks of January with some more specifics. Meanwhile, happy holidays and a happy new year to you! –  Theo Buehler Dec 26 '12 at 1:49
Thanks Theo! Happy holidays and New Year to you too. –  Yemon Choi Dec 27 '12 at 0:49

(1) $f:\mathbb R\to \mathbb R$ is $C^\infty$.

(2) $f^2:\mathbb R\to \mathbb R$ is $C^\infty$.
EDIT: The right formulation is: If $f\ge 0$ is $C^\infty$, then one can choose a square root of $f$ which is twice differentiable, but not better in general.

(3) $f^2$ and $f^3:\mathbb R\to \mathbb R$ are both $C^\infty$.

(4) $f^p$ and $f^q:\mathbb R\to \mathbb R$ are both $C^\infty$. EDIT: Where $p$, $q$ are relatively prime.

Obviously, (1) $\implies$ (3), (4).

(2) See:
Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Choosing roots of polynomials smoothly, Israel J. Math 105 (1998), p. 203-233.(pdf)

(3) $\implies$ (1). See:
MR0682456 Joris, Henri: Une $C^\infty$-application non-immersive qui possède la propriété universelle des immersions. Arch. Math. (Basel) 39 (1982), no. 3, 269–277.
MR2179865 Myers, Robert: An elementary proof of Joris's theorem. Amer. Math. Monthly 112 (2005), no. 9, 829–831.

(4) $\implies$ (1). See:
MR0833407 Duncan, John; Krantz, Steven G.; Parks, Harold R.: Nonlinear conditions for differentiability of functions. J. Analyse Math. 45 (1985), 46–68.

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Thank you very much for the references! –  Leonard Dec 19 '12 at 11:21
Set $f(x) = |x|$. $f$ is not even once differentiable at $0$, but $f^2(x) = x^2$ is $C^\infty$. Am I missing something? –  Paul Siegel Dec 19 '12 at 11:48
Paul: The linked article seems to be about the \emph{possibility} of a smooth choice of a root of what is here called $f^2$. In your example, $|x|$ is a bad choice, $x$ is a good one. –  Philip van Reeuwijk Dec 19 '12 at 12:39
What are the conditions on $p$ and $q$ in (4)? –  rem Dec 19 '12 at 13:59

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