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Results tagged with ra.rings-and-algebras
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user 22989
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.
2
votes
Invertibility of all left multiplication maps in non-unital rings
Another example. Let $R=\mathbb{C}$ as an additive group, with multiplication
$$(w,z)\mapsto\overline{wz}.$$
- 35.2k
2
votes
On the socle of rings
The example given by Mariano in comments (or rather, its opposite ring) is a counterexample if "socle" means "right socle". I think the following example works for both the right and left socle.
Let …
- 35.2k
8
votes
Direct product of rings
Barbara Osofsky proves this is not possible (assuming the Axiom of Choice, I think) in "Noninjective cyclic modules", Proc. Amer. Math. Soc. 19 (1968), 1383-1384.
- 35.2k
4
votes
Countable Maximal Ideals
Eric's result for commutative rings, that an example must have the ring no bigger than the continuum and the maximal ideal infinitely generated, extends to left ideals for non-commutative rings, assum …
- 35.2k
5
votes
Accepted
Tensor decomposition under derived equivalence
Not in general.
For an easy example, let $C$ and $D$ be derived equivalent algebras. Then $A=C\times C$ and $B=C\times D$ are derived equivalent, and $A=C\otimes_K(K\times K)$ has a tensor decompositi …
- 35.2k
13
votes
Accepted
Finite Dimensional Simple nonunital associative Algebras
I suspect there's a simpler argument that doesn't involve adjoining a unit, but ...
Adjoin a unit to get a unital algebra $B=K\oplus A$.
Since $B$ is a finite dimensional algebra, the Jacobson radic …
- 35.2k
5
votes
Accepted
Can you test flatness on $FP_3$-modules?
Answering my own question in comments:
Let $k$ be a field, and $A=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $A$ has no non-projective $FP_3$-modules (using your …
- 35.2k
2
votes
Accepted
Under what assumptions can endomorphisms of $M/IM$ be realized as a subquotient of endomorph...
Let $A=\mathbb{C}[x,y]$, $I=(x,y)$, and $M$ the $3$-dimensional $A$-module $(x,y)/(x^2,y^2)$ (with basis $\{x,y,xy\}$).
Then $M/IM$ is isomorphic to a direct sum of two copies of $\mathbb{C}=A/I$, so …
- 35.2k
5
votes
Accepted
Modules "projective in a subcategory"
More generally, the following is true:
Let $A$ be a finite-dimensional $k$-algebra ($k$ a field), $M$ a finite-dimensional indecomposable (right) $A$-module, and $S$ the category of coproducts of cop …
- 35.2k
10
votes
Accepted
Bass' stable range for Bezout rings
In "Rings of continuous functions in which every finitely generated ideal is principal" by L. Gillman and M. Henriksen (Trans. Amer. Math. Soc. 82 (1956), 366-391 link), Example 3.4 is of a topologica …
- 35.2k
5
votes
Accepted
$D(A) \otimes D(A)= A$?
If $A\cong D(A)^{\otimes i}$as $A$-bimodules, for some $i\geq1$, then $-\otimes_AD(A)$ is a self-equivalence of the module category, and so takes projectives to projectives. But it takes $A$ to $A\oti …
- 35.2k
5
votes
When for every module $M$, $|E(M)| = |M|$
If $R$ is a finite-dimensional algebra (of dimension $d$, say) over a field $k$, and $M$ is a left $R$-module, then
$$\dim(M)\leq\dim\left(E(M)\right)\leq d.\dim(M),$$
and so if $k$ is infinite then $ …
- 35.2k
4
votes
Accepted
Dual of a module
As @S.Carnahan shows. this is not true.
Perhaps the statement you want is that the dual of $M[p^n]$ is $X/p^nX$?
Take the exact sequence
$$0\to M[p^n] \to M \to M \to M/p^nM \to 0,$$
where the mid …
- 35.2k
12
votes
Accepted
For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring?
I think the answer to (2) is yes (which also answers (1) and (4), of course).
First, note that $R=\mathbb{Z}/4\mathbb{Z}$ and $S=\mathbb{F}_2[x]/(x^2)$ have isomorphic multiplicative monoids (via the …
- 35.2k
13
votes
Accepted
Where is my mistake in calculating duals?
If $C$ is a right module then $C^*$ is a left module.
In Landrock, $\text{soc}(R)$ is the right socle. As he proves, it is a two-sided ideal, but it may not be semisimple as a left module (it is not …
- 35.2k