TITLE: Kähler differentials of affine varieties
QUESTION [9 upvotes]: I would like to gain some intuition regarding the modules of Kähler differentials $\Omega^j_{A/k}$ of an affine algebra $A$ over a (say - algebraically closed) field $k$.
Let us recall the definition: let $A^e = A\mathrel{\otimes_k} A$, let $f:A^e\to A$ be the map defined by $f(a\otimes b) = ab$, and let $I = \ker f$. Then $\Omega^1_{A/k} = I/I^2$. And, $\Omega^j_{A/k} = \bigwedge^j \Omega^1_{A/k}$.
An important theorem regarding Kähler differentials says: If $k \to A$ is smooth of relative dimension $n$, then $\Omega^n_{A/k}$ is a projective module of finite rank.
My question:

I was wondering if anyone could provide some examples of:

How does the module of Kähler differentials look for some singular varieties? For example, what is $\Omega^1_{A/k}$ for $A = k[x,y]/(y^2-x^3)$?

Can anyone provide an example of a non-singular affine variety with coordinate ring $A$, such that $k \to A$ is smooth of relative dimension $n$, and $\Omega^n_{A/k}$ is projective but not free?



I would be happy for any concrete example that will help my intuition on the subject.
Thanks!

REPLY [4 votes]: "Let us recall the definition"
$\Omega^1=I/I^2$ is not a definition, it is a construction (and I know at least three other constructions). The definition of the module $\Omega^1$ is its universal property, i.e. that it classifies derivations.
The following two facts follow from the universal property (I will omit the base ring $k$ from the notation):

$\Omega^1_{k[x_1,\dotsc,x_n]}$ is a free $k[x_1,\dotsc,x_n]$-module with basis $d(x_1),\dotsc,d(x_n)$
For an ideal $I \subseteq A$ the sequence $I/I^2 \xrightarrow{d} \Omega^1_{A} \otimes_A A/I \to \Omega^1_{A/I} \to 0$ is exact. 

From this get that $\Omega^1_{k[x_1,\dotsc,x_n]/(f_1,\dotsc,f_r)}$ is the free module over $k[x_1,\dotsc,x_n]/(f_1,\dotsc,f_r)$ generated by symbols $d(x_1),\dotsc,d(x_n)$ subject to the relations $d(f_1)=\dotsc=d(f_n)=0$, where $d(f)$ is expressed using the $d(x_i)$ (as in 1.).
For example, let $A=k[x,y]/(y^2-x^3)$. Then $d(y^2-x^3)=2 y d(y) - 3 x^2 d(x)$. Hence, $\Omega^1_{A}$ is the free $A$-module generated by symbols $d(x),d(y)$ subject to the relation $2 y d(y) = 3 x^2 d(x)$.
For another example, consider $A=k[x,y]/(x^2+y^2-1)$ (the coordinate ring of a "circle"). Let us assume $2 \in k^*$, so that $A$ is smooth over $k$. Then $d(x^2+y^2-1)=2 x d(x) + 2 y d(y)$, so that $\Omega^1_{A}$ is the free $A$-module generated by $d(x),d(y)$ subject to the relation $x d(x) = - y d(y)$. One can show that $x d(y) - y d(x)$ is a basis of $\Omega^1_{A}$.