Why $k[x,y]$ is not a formally smooth algebra?

Question

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 $D(k[x,y])$?

Answer 1

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.