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Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
76
votes
9
answers
6k
views
Can we unify addition and multiplication into one binary operation? To what extent can we fi...
The question is the extent to which we can unify addition
and multiplication, realizing them as terms in a single
underlying binary operation. I have a number of questions.
Is there a binary operati …
37
votes
What do you do if you believe a problem is undecidable?
The first thing to say is that for a statement to be independent
of some axioms means really two things, namely, that it is
consistent with those axioms, and also that the negation of the
statement is …
32
votes
Accepted
Are the real numbers isomorphic to a nontrivial ultraproduct of fields?
The answer is no, because such ultrapowers are always $\aleph_1$-saturated, but $\mathbb{R}$ is not.
More concretely, the ultraproduct will be an ordered field with uncountable cofinality — every cou …
27
votes
Accepted
Jonsson Boolean algebras?
Boolean algebras are never Jonsson.
Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $ …
24
votes
Does every set admit a ring structure or a field structure?
In ZFC, every nonempty cardinality is the cardinality of a ring. For finite cardinalities, we have $\mathbb{Z}/n\mathbb{Z}$ as you mentioned. For infinite cardinalities, this is an immediate consequen …
20
votes
Accepted
Nice algebraic statements independent from ZF + V=L (constructibility)
Let me address the part of your question seeking algebraic statements independent of ZFC+V=L.
The basic situation is that in set theory our tools are not so flexible for finding statements
independen …
19
votes
Accepted
Is there a version of the Archimedean property which does not presuppose the Naturals?
It is not surprising that some versions of the Archimedean property concern subsets of the order rather than merely elements. The reason is that the Archimedean property is provably not expressible in …
14
votes
Ring with three binary operations
An exponential field is a field with an additional unary operation $x\mapsto E(x)$ extending the usual idea of exponentiation. So it satisfies the usual law of exponents $E(a+b)=E(a)\cdot E(b)$ and al …
13
votes
Accepted
Non-Standard Prime
First, you haven't actually specified a particular field,
since the field $F$ that you have will depend on your
choice of $c$ and of $M$. For example, different
nonstandard models can seriously affect …
12
votes
Accepted
Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| =...
The answer is yes.
Let $\kappa$ be any singular strong limit cardinal of uncountable cofinality, such as the cardinal $\beth_{\omega_1}$ for a specific example, and let $X=\kappa+1$, the ordinals up …
11
votes
LUB and GLB on a lexicographically ordered complete lattice product
There are two natural orders to put on the product of two lattices, the product order and the lexical order.
Product order: (a,b) ≤ (a',b') if and only if a ≤ a' and b ≤ b'
Lexical order: (a …
10
votes
How much are reduced powers different?
Easy differences arise if one allows principal ultrafilters, since the ultrapower of $X$ by a principal filter is canonically isomorphic to $X$, but other ultrapowers are not. Another easy difference …
8
votes
Accepted
Orderings of ultrafilters
I understand your question better now.
First, in your general context of filters the relations
$\leq_1$ and $\leq_2$ are not the same. To see this, let
$G=\{I\}$ be the trivial filter on a set $I$ wi …
8
votes
Examples for "nice" Boolean algebras that are not complete or not atomic
There is up to isomorphism a unique countably infinite atomless Boolean algebra (by a back-and-forth argument), making this algebra highly canonical. But it cannot be complete, since every infinite B …
7
votes
1
answer
490
views
Normal form for terms in language with two ring structures
Suppose I have two different ring structures on the same domain $\langle R,+,\cdot,0,1\rangle$, $\langle R,\oplus,\otimes,\bar 0,\bar 1\rangle$ and I throw the structures together into a single common …