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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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar