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Meadow theory (see here) allows us to apply the results and concepts of universal algebra to the study of fields. Obviously, this is very, very nice.

However, I have the following issue with the meadow-theoretic approach: since every meadow satisfies $0^{-1}=0,$ and since this makes reciprocation in both $\mathbb{R}$ and $\mathbb{C}$ discontinuous (at $0$), thus these number systems cannot be viewed as models of the theory of meadows in the category $\mathrm{Top}$ unless we endow them with a non-standard topology.

Questions.

  1. Should this be viewed as a serious issue with the meadow-theoretic approach?

  2. Does anyone know of a good solution?

goblin GONE
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1 Answers1

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It really depends on what you want to study! When you are interested in topological questions, then obviously topological meadows don't include topological fields. In my opinion the natural definition of a topological field should be a (commutative) topological ring $K$ whose underlying ring is also a field such that the map $K^* \to K^*, x \mapsto x^{-1}$ is continuous.

Probably a more natural approach would be to replace $K$ by $K \cup \{\infty\}$ and define $0^{-1} := \infty$, $\infty^{-1} := 0$, and built up an axiomatization around this idea. In other words, work with the projective line $\mathbb{P}^1_K$ instead of the affine line $\mathbb{A}^1_K$.

PS: I wonder why the well-known Von Neumann regular commutative rings are called meadows as if this was a new concept.

Martin Brandenburg
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  • Every meadow is a regular ring in the sense of von Neumann, but the homomorphisms are (in principle) different. – Zhen Lin Mar 09 '14 at 14:23
  • @Zhen: What is an example of a homomorphism of von Neumann regular rings which is not a homomorphism of meadows? – Martin Brandenburg Mar 09 '14 at 15:16
  • I don't know, but there should be one. After all, homomorphisms of von Neumann regular rings are just homomorphisms of rings, which may not preserve the "weak" inverses (which are not always unique). – Zhen Lin Mar 09 '14 at 18:52
  • The weak inverses are unique (recall that $b$ is a weak inverse of $a$ if $aba=a$ but also $bab=b$). – Martin Brandenburg Mar 09 '14 at 20:29
  • The definition of von Neumann regular ring only asks for one of those two equations, however. – Zhen Lin Mar 09 '14 at 20:37
  • @ZhenLin : If you know of a commutative von Neumann-regular ring (ᴄᴠNʀʀ) that can be made into a meadow in two different ways, then the identity map is an isomorphism of ᴄᴠNʀʀs that's not an isomorphism between the two meadows. And of course, if there's a ᴄᴠNʀʀ that can't be made into a meadow at all, then the concepts are different. So meadows are the same as ᴄᴠNʀʀs if and only if each ᴄᴠNʀʀ can be made into a meadow in a unique way. (I don't know if that's true or not!) – Toby Bartels Oct 20 '22 at 14:47
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    @TobyBartels Yes, every commutative von Neumann regular ring can be made into a meadow in a unique way, see Lemma 2.11 in Meadows and the equational specification of division. The point is that while there may be multiple elements $y$ such that $xyx = x$, there is a unique element $y$ such that $xyx = x$ and $yxy = y$. The axioms of meadows say $xx^{-1}x = x$ but also $(x^{-1})^{-1} = x$, so we get $x^{-1}xx^{-1} = x^{-1}$ as well, and the value of $x^{-1}$ is uniquely determined. – Alex Kruckman Sep 19 '24 at 15:38
  • Commenting on the original answer instead of the stuff about ᴄᴠNʀʀs … if you have $ 0 ^ { - 1 } = \infty $ and $ \infty ^ { - 1 } = 0 $ literally, then you need $ 0 \infty = 1 $, which is discontinuous. But you can also take the reciprocal as a unary operation in its own right that only sometimes gives a multiplicative inverse. Axiomatizing this can lead the the concept of a wheel. – Toby Bartels Sep 19 '24 at 16:09