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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.
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
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
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 …
3
votes
Product lattice
As I explain in this MO answer, there is a choice of product structures to place on the product of two lattices.
Please click through and read the discussion there. But to summarize, one choice is t …
- 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
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
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 …
- 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
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 …
- 236k
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 …
- 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
2
votes
Decidability of matrix algebra
If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expr …
- 236k
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
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 …
- 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