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Results tagged with ra.rings-and-algebras
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user 1946
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.
5
votes
Accepted
Countably compact Boolean algebras versus distributivity
There are many countably distributive complete Boolean algebras, and this is an important concept in forcing. For example, the canonical forcing to add a Cohen subset (or any number of Cohen subsets) …
- 236k
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 …
- 236k
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 …
- 236k
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 …
- 236k
6
votes
Basic Algebraic Applications of Stationary Sets?
Stationary sets are exactly the positive-measure sets with respect to the club filter, which is a very natural measure on the subsets of $\omega_1$ or on higher cardinals with uncountable cofinality. …
- 236k
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 …
- 236k
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 …
- 236k
3
votes
Accepted
Two different products of filters
If all the $a_i$ are principal ultrafilters on sets with at least two elements, then $\Pi_1$ will also be principal, since it concentrates on the singleton that picks out the base of each $a_i$. But $ …
- 236k
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 …
- 236k
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 …
- 236k
2
votes
How should one look at the set of compatible ring structures on a given group?
In the case of a countable group, this kind of thing often arises in the subject of Borel equivalence relation theory, which has been considered in a few MO questions (see also links in this answer). …
- 236k
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 …
- 236k
5
votes
Complete De Morgan algebra
I claim that the property is true in every de Morgan algebra,
whenever the expressions in it make sense (on either side). The issue about making
sense is that when $I$ is infinite, the expression $\bi …
- 236k
2
votes
Chain of ideals in a BA
Let $\mathfrak{A}$ be the power set of an uncountable set $X$, which is a complete Boolean algebra. Select disjoint sets $X_n\subset X$ of size $\omega_1$, and let $J_n$ be the ideal generated by $X_0 …
- 236k
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 $ …
- 236k