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Results tagged with ra.rings-and-algebras
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user 17734
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.
9
votes
Accepted
On rings $R$ such that $xR\cap yR$ is non zero whenever $x$ and $y$ are non zero
These rings are called right uniform rings.
Generally, a right ideal $I$ is called (right) uniform if all nonzero right subideals $J,K\subseteq I$ have nonzero intersection: $J\cap K \neq 0$.
Note: …
- 2,019
5
votes
About the Wedderburn-Malcev Theorem
$S \cong A/rad(A)$ as $F$-algebra.
Proof: The composition $S \hookrightarrow A \twoheadrightarrow A/rad(A)$ is a surjective $F$-algebra homomorphism with kernel $S \cap rad(A)=0$.
By Wedderbur …
- 2,019
4
votes
0
answers
167
views
For $P$ $\mathbb{Z}G$-projective, $\mathbb{Q}\otimes P$ is $\mathbb{Q}G$-free
I'm looking for a proof of a theorem of Swan [1, Theorem 3]:
If $G$ is a finite group and $P$ a finitely generated projective $\mathbb{Z}G$-module, then $\mathbb{Q}\otimes_\mathbb{Z}P$ is a free …
- 2,019
3
votes
Accepted
Subring of ring
No, this isn't true in general:
$$R_0:= \mathbb{Z}[X] \le \mathbb{Z}[X,Y]/(X^2-pY)=: R$$
is a counter-example because
$R/pR = \mathbb{F}_p[X,Y]/(X^2)$ has non-trivial radical.
$R$ is a domain sinc …
- 2,019
1
vote
The center of a(n endomorphism) ring is a PID
Two classes of torsion-free abelian groups having the desired property are
free abelian groups
torsion-free divisible groups (here I use the axiom of choice)
By noting that a torsion-free divis …
- 2,019