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

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

• 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. – pbelmans Jun 15 '16 at 4:51

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).
• 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? – Zhaoting Wei Jun 30 '16 at 1:47
• Sorry, I misinterpreted your question. You consider $D(A)$ as a graded algebra if I'm not mistaken. – pbelmans Jun 30 '16 at 14:36