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Noncommutative geometry in the sense of Connes and beyond: noncommutative algebras viewed as functions on a noncommutative space.
8
votes
Accepted
Path algebras are formally smooth
Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.
I …
3
votes
Elementary linear algebra over a (possibly skew) field $K$
If I understand correctly what Question 1 is asking, then there are easy counterexamples even using commutative fields.
Let $K=\mathbb{R}$ and $L=\mathbb{C}$. Then $\begin{pmatrix}1&i\\1&i\end{pmatri …
11
votes
Accepted
Is a "smooth" finite-dimensional algebra separable modulo its radical?
Let $K$ be an algebraic closure of $k$.
The following lemma must surely be well-known, but I haven't found an explicit reference, so I'll include a proof at the end of this post.
Lemma. If $S$ is …
25
votes
Accepted
Is this ring isomorphic to a quotient of a group algebra?
If $A$ is a $\mathbb{Q}$-algebra, then there is a group $G$ such that $A$ is a quotient of $\mathbb{Q}[G]$ if and only if $A$ is generated by units. For the "if" direction, take $G$ to be the group of …