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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.

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 …
Todd Leason 's user avatar
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 …
Todd Leason 's user avatar
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 …
Todd Leason 's user avatar
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: …
Todd Leason 's user avatar
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 …
Todd Leason 's user avatar