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Let A be a unital associative algebra over a field k.

Then A is smooth if and only if X:=Spec(A) is smooth. That is $\Omega_{X|Spec(k)}$ is locally-free. The later module is isomorphic to $\Omega_{A|k}$ the Khaler differentials on A over k.

Now a result equates local freeness with projectiveness of modules, therefore $\Omega_{A|k}$ is projective.

Now, the HKR theorem gives an isomorphism $\Omega_{A|k} \cong \Omega (A)$, where $\Omega (A)$ are the non-commutative (or abstract) differential forms on A.

Finally, a quasi-free algebra is one for which $\Omega$ is projective.

However, $k[X,Y]$ is smooth but can be seen to not be quasi-free, this leads me to believe there is a flaw somewhere in my reasoning, but I'm uncertain where?

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The flaw is that the an algebra $A$ is quasi-free if and only if its module of non-commutative forms $\Omega (A)$ is projective not as an $A$-module but as an $A$-*bi*module.
The second condition is much much stronger, as it implies $A^e$-projective dimension of atmost $1$, and so $A$-global dimension of atmost $1$, which is shown via the isomorphism $Hom_{A}(N,M)\cong Hom_{A^e}(A,Hom_k(N,M))$.

So for example if $A$ is a commutative Cohen-Macaulay then its $A^e$-projective dimension is bounded below by its Krull dimension. In such a rather common case, $A$ can only be quasi-free if it is of Krull dimension at most $1$. Which is a rather stringent statement.
Particularly, this is the case for the affine plane $k[X,Y]$.

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