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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.
57
votes
Accepted
$R$ is isomorphic to $R[X,Y]$, but not to $R[X]$
The answer to this quite beautiful question is that there does exist a commutative ring $R$ with $R\cong R[X,Y]$ but $R\not\cong R[X]$.
Let $F$ be a field, and take
$$
R=F[x_i,y_i,r_i\ (i\geq 0)]
$$
…
- 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
24
votes
2
answers
1k
views
What do you do if you believe a problem is undecidable?
While the title of this question is subjective, I hope to make what I'm looking for quite concrete. The first, and main question is this: If you believe that a problem you are working on is formally …
- 18.7k
13
votes
Uncountable counterexamples in algebra
In rings: Let $R$ be a ring where idempotents lift modulo the Jacobson radical $J(R)$. Any countable set of orthogonal idempotents in $R/J(R)$ lifts to an orthogonal set of idempotents; but this fai …
11
votes
4
answers
2k
views
When is it okay to intersect infinite families of proper classes?
For experts who work in ZFC, it is common knowledge that one cannot in general define a countable intersection/union of proper classes. However, in my work as a ring theorist I intersect infinite col …
- 18.7k
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
10
votes
2
answers
544
views
A back and forth Euclidean algorithm over the integers--does it have bounded length?
cLet $a,b,c,d\in \mathbb{Z}$ and suppose we have the equation $ac+bd=1$. One way of thinking about this equation is it expresses the fact $\gcd(c,d)=1$. It is well-known that all other similar equat …
- 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
9
votes
4
answers
489
views
Basic Algebraic Applications of Stationary Sets?
Background: I've been working my way through Thomas Jech's "Set Theory" because I'm working on some problems that have the potential to be logically independent of the usual axioms, or at least invol …
- 18.7k
8
votes
3
answers
820
views
Does a left basis imply a right basis, without AC?
If $_DV_D$ is a $D$-$D$-bimodule, and we have a $D$-basis for $V_D$, do we still need AC to get a $D$-basis for $_DV$?
(The original question appears below. But this shorter question gets at the hea …
- 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
7
votes
Monoids in which every prime is an atom
Fact 1: If $M$ is a monoid where primes are atoms, then $M$ is Dedekind-finite.
Proof. Working contrapositively, assume $ab=1$ with $a,b\in M\setminus M^{\times}$. Now $a$ is prime since it divides …
- 18.7k
7
votes
1
answer
419
views
Sequences without long arithmetic progressions
First, a bit of notation. If we have an arithmetic progression $a, a+k, a+2k, \ldots, a+(n-1)k$ we will call $k$ the distance, and $n$ the length.
While trying to find an example for a paper I'm wri …
- 18.7k
7
votes
Polynomial roots in the ring extension
In the noncommutative case, your condition for a "root" is called a "right root". I remember that T.Y. Lam worked with this condition a bit (you might search through his papers, or look in his "First …
- 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