2
$\begingroup$

We could talk about the formal smoothness of an algebra. See for example Ginzburg's lecture notes For an associative algebra $A$ over a field $k$ we define $$ D(A)=T(A+\bar{A})/(\bar{ab}=a\bar{b}+\bar{a}b, a\otimes a^{\prime}=aa^{\prime}\otimes 1). $$ And Theorem 19.4.1 of Ginzburg's notes claims that: An associative algebra $A$ is formally smooth if and only if the natural map $A\to \bar{A}$ can be extended to a derivation of $D(A)$ of degree $+1$.

Ginzburg also points out that the commutative polynomial ring $k[x_1\ldots x_n]$ is not formally smooth for any $n>1$.

My question is why $k[x,y]$ is not formally smooth in the sense of the above theorem, i.e, what is the difficulty to extend the morphism $x\to \bar{x}, y\to \bar{y}$ to a derivation on $D(k[x,y])$?

$\endgroup$
1
  • 1
    $\begingroup$ This notion of smoothness is also known as (Cuntz-)Quillen smoothness (they have the formal lifting property in the category of all algebras), and for a (finitely generated) commutative algebra it implies being of global dimension $\leq 1$, in particular $k[x,y]$ is not smooth in this sense. $\endgroup$ – pbelmans Jun 15 '16 at 4:51
2
$\begingroup$

I took a look at the Ginzburg lecture notes, and my claim from the comment is proven there as lemma 19.1.6. You can also take a look at the surrounding discussion, and it might be interesting to reverse-engineer lemma 19.1.7 to show that $k[x,y]$ is not formally smooth by exhibiting some obstruction to lifting (I haven't done this exercise, probably I should).

Also, section 5 of the original Cuntz--Quillen article (see below for the precise reference) is an interesting read.

Joachim Cuntz and Daniel Quillen, MR 1303029 Algebra extensions and nonsingularity, J. Amer. Math. Soc. 8 (1995), no. 2, 251--289.

$\endgroup$
2
  • $\begingroup$ I want to make sure: the derivation $D$ in Ginzburg Theorem 19.4.1 means a derivation of ordinary algebra or a derivation of graded algebra? $\endgroup$ – Zhaoting Wei Jun 30 '16 at 1:47
  • $\begingroup$ Sorry, I misinterpreted your question. You consider $D(A)$ as a graded algebra if I'm not mistaken. $\endgroup$ – pbelmans Jun 30 '16 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.