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Results tagged with ra.rings-and-algebras
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user 3199
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.
11
votes
Accepted
Does every cancellative duo semigroup embed into a group?
Let $S$ be a cancellative duo semigroup. Let $a,b\in S$ be arbitrary, and consider the element $x:=ba$. Clearly, $x\in Sa$ and by the duo property $x\in bS=Sb$. Thus, $S$ is right reversible, and s …
- 18.7k
3
votes
Accepted
Lemma of Harada and Sai on sums of modules with a "chain" of monomorphisms between them
Note that $M_i'\oplus M_{i+1}=M_i\oplus M_{i+1}$. Thus,
$$
\left(\bigoplus_{i=1}^{n}M_i'\right)\oplus M_{n+1}=\bigoplus_{i=1}^{n+1}M_i.
$$
Case 1: Suppose that for every integer $n\geq 1$ that $\psi …
- 18.7k
4
votes
0
answers
109
views
Adjoining new factors for primes in UFDs
It is well-known that if we pass from a UFD to a new ring where we have factored one of the primes, it does not need to stay a UFD. The classic example is passing from $\mathbb{Z}$ to $\mathbb{Z}[\sq …
- 18.7k
10
votes
1
answer
215
views
Matrix ring isomorphisms of different sizes
Do there exist (unital, associative, noncommutative) rings $R$ and $S$, where $\mathbb{M}_2(R)\cong \mathbb{M}_3(S)$, but these matrix rings are not isomorphic to $\mathbb{M}_6(T)$ for any ring $T$?
- 18.7k
4
votes
Left quasi-inverse elements: motivation
Consider an equation like $(1-b)(1-a)=1$, which says that $1-a$ has a left inverse $1-b$. Multiplying through, and cancelling the $1$'s, we are left with $a+b=ba$. This expresses essentially the sam …
- 18.7k
7
votes
2
answers
575
views
Deriving consequences of identities
Suppose we are given a variety in the universal algebra sense.
For concreteness, suppose that we have two binary operations $+,\cdot$, three unary operations $-,\ast,'$, and two zeroary operations $0, …
- 18.7k
4
votes
Accepted
Do you know of any indecomposable ring that has no isolated elements and is neither reversib...
Let $F$ be a field, and let
$$
R:=F\langle x,y\, :\, x^2=xy=y^2=0\rangle.
$$
Notice that $\{1,x,y,yx\}$ is an $F$-basis for $R$ as an $F$-vector space.
This ring is not reversible since $xy=0$ but $yx …
- 18.7k
7
votes
Accepted
Symmetry of unique generator property
Lam informed me that, as far as he knew, this problem was still open. However, the example below shows that the condition is not left-right symmetric.
Let
$$
R=\mathbb{F}_2\langle a,b,c\, :\, a^2=ab= …
- 18.7k
28
votes
2
answers
849
views
$A^2$ is isomorphic to $A^{(\omega)}$, but not $A$
Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)?
Edited to add: As no answers are forthcoming, does anyone …
- 18.7k
1
vote
Accepted
Automorphisms of special egg-box diagrams
It turns out (surprisingly) that the answer to my question is yes, and there are even finite examples. I came up with the following diagram
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\circ & \circ & \bul …
- 18.7k
6
votes
1
answer
135
views
Automorphisms of special egg-box diagrams
By a egg-box diagram I will simply mean a (possibly infinite) rectangular array of holes, with some of the holes containing an egg (denoted by a filled-in circle) and the rest of the holes are empty ( …
- 18.7k
1
vote
$R$ is a UFD iff $R_{\frak{m}}$ is a UFD?
The following are just some partial thoughts about the new question which are too long for comments. Throughout, assume that $R_{\mathfrak{m}}$ is a UFD, $R_0$ is a field, etc...
First, let $x\in \ma …
- 18.7k
5
votes
$R$ is a UFD iff $R_{\frak{m}}$ is a UFD?
(Note: The original question had an incorrect premise. The ideal generated by elements of positive degree in an arbitrary graded ring is not generally maximal.)
Here is the solution to the original p …
- 18.7k
5
votes
Accepted
Existence of a finite extension of ℤ providing a finite extension of the primes
This post started as some minor observations, but I believe it now contains a full proof that there is no such ring. Throughout, we let $R$ be an example of the kind wanted.
Observation 1: $R$ is in …
- 18.7k
8
votes
1
answer
1k
views
First isomorphism theorem for sets?
Let $f\colon S\to T$ be any function. There is the obvious refinement of $f$, by replacing the codomain $T$ with the image. Thus, every function factors into a surjection followed by an injection (a …
- 18.7k