My past research interests included differential algebraic equations, nonlinear analysis and relations between symmetries and structural properties.

I recently investigated hierarchical structures, starting from group cohomology, continuing with semi-group theory and ending with lattices and universal algebra.

Nov
17
awarded Critic
Nov
17
awarded Citizen Patrol
Nov
12
revised Which associative and commutative operations are defined for any commutative ring?
Added an appendix with a more intuitive explanation/description of the found operations
Nov
10
comment Is the equational theory of commutative vN regular rings decidable?
@EmilJeřábek Yes, they are the same thing as meadows. I didn't know "meadows" before, but I have the impression that Bergstra’s papers will answer most of my open questions with respect to "meadows" (and commutative vN regular rings).
Nov
10
comment Is the equational theory of commutative vN regular rings decidable?
@Ycor The equations $x\phi(x)x=x$ and $\phi(\phi(x))=x$ in addition to the normal commutative ring equations are sufficient. (I remember that the equation $\phi(\phi(x))=x$ was derivable from the other equations, but at least for an inverse semigroup, it was an awfully long derivation.)
Nov
10
comment Is the equational theory of commutative vN regular rings decidable?
@YCor Note that $x_0=0$, i.e. $\phi(0)=0$ can be deduced from the available identities. The notion is basically equivalent to a subdirect product of fields. One reason why this notion might be useful is that the free commutative vN regular ring with $n$-generators exists. If the equational theory should turn out to be decidable, then this would give one more reason to suspect that this notion is interesting. If not, then maybe it's not that interesting after all.
Nov
10
revised Is the equational theory of commutative vN regular rings decidable?
Removed wrong statement about residual finiteness of free vN regular group with 0-generators
Nov
10
comment Is the equational theory of commutative vN regular rings decidable?
@BenjaminSteinberg I wanted to map $x=\frac{z}{n}$ to $z_pn_p^{-1}$. You are right that this mapping is not a homomorphism, because $p$ and $1/p$ would be mapped to $0$, but $1=p*1/p$ gets mapped to $1$. I will remove the corresponding part from the question.
Nov
10
revised Is the equational theory of commutative vN regular rings decidable?
add link and explanation for "well known"
Nov
10
comment Is the equational theory of commutative vN regular rings decidable?
@YCor I added an appendix to the question trying to clarify these things. The vN regular rings don't form a variety, but for different reasons than you state. The commutative vN regular rings do form a variety, or at least can be treated as a variety. The vN regular rings with unique inverses can be treated as a variety, here the identity $xx^{-1}yy^{-1}=yy^{-1}xx^{-1}$ holds because of the uniqueness and is sufficient to ensure uniqueness.
Nov
10
revised Is the equational theory of commutative vN regular rings decidable?
Let's hope that adding concrete examples and explanations will reduce the confusion
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
@Ycor en.wikipedia.org/wiki/Free_object, i.e. via a universal property. It exists, because a commutative vN regular ring is a en.wikipedia.org/wiki/Variety_(universal_algebra). Free objects on $n$-generators exists for a variety (equational class) and can be represented as the quotient of the term algebra (on the generators) over the congruence relation generated by the equations. This is proved in universal algebra texts in connection with Birkhoff's theorem.
Nov
9
revised Is the equational theory of commutative vN regular rings decidable?
Let's hope that adding vN in front of regular will reduce the confusion
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
@AntonKlyachko The free commutative von Neumann regular ring of $n$-generators is different from the free commutative ring, because the generalized inverse operation $()^{-1}$ allows to form elements otherwise not available. Hence it is not so obvious whether it is residually finite. Maybe the terminology here is a bit unhappy.
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
@Ycor The statement that the equation will fail in some finitely generated free commutative regular ring is a bit trivial, because one just needs to use the free variables from the two terms to get the required finite set of generators. So the statement "and the latter is residually a finite field" implies that my first question actually asks whether the free commutative regular ring of $n$-generators is residually finite, i.e. that for every of its elements I can find a homomorphism to a finite ring.
Nov
9
comment Is the equational theory of commutative vN regular rings decidable?
@YCor The generalized inverse operation $()^{-1}$ in a field is the normal inverse operation for non-zero elements, and maps zero to zero. Hence it satisfies $xx^{-1}x=x$ as required.
Nov
9
asked Is the equational theory of commutative vN regular rings decidable?
Nov
8
revised Can Schwartz-Zippel be formulated for commutative rings instead of fields?
Added answer to question 1 based on the observations from the comments
Nov
7
comment Simply typed lambda calculus and higher order logic
As I said, you use equality to get HOL. For STT ~= PROP (via Curry-Howard), you are not allowed to use equality in STT. And think about it, equality is simply undecidable, so it has to increase the expressive power if you allow it.
Nov
7
comment Simply typed lambda calculus and higher order logic
With respect to Curry-Howard, it don't think it will be HOL. Maybe it's the multiplicative fragment of intuitionistic PROP, i.e. intuitionistic PROP without "or". But that was for CCC (cartesian closed category), and I'm a bit tired at the moment. Lambda will probably be translated as "implication", which was the "exponential" in CCC. The "product" from CCC was "and", so you would need a "pair" in STT for that. And "or" would be a "sum" type in STT then, i.e. a disjoint union, maybe an if "a" then "b" else "c" does that.
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