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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:

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

  2. 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.


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You can compute the module of differentials in your example of the node using Macaulay2. – Mariano Suárez-Alvarez Jan 28 '11 at 21:30

Let's do your example (1).

There is a short exact sequence $$0\to A\to A\mathrm dx\oplus A\mathrm dy\to\Omega^1\to 0$$ in which the first map maps $1$ to $2x\,\mathrm dx+3y\,\mathrm dy$. It follows that $\Omega^1$ is isomorphic to the module $$\frac{A\oplus A}{(2x,3y)}.$$

If the characteristic is $2$ or $3$, this looks like $A\oplus A/(y)$ or $A\oplus A/(y)$. If the characteristic is not one of those two, then it is clear, from this description, that if you localize at a prime not containing $x$ or $y$, then you get a free module. Now localize at one of the other ideals, and see what you get!

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Thank you for your answer. I now see that I should do these computations using the concrete definition of a module of differentials. However, I do not completely understand your calculation. Since in our ring we have that $y^2 = x^3$, it follows that $d(y^2) = 2ydy = d(x^3) = 3x^2dx$, so why do you claim that $2xdx+3ydy = 0$? – the L Jan 29 '11 at 10:29

As for your question (2), here is an example. It requires some backgrounds on the theory of divisors on algebraic curves.

Let $C$ be a (projective) smooth curve of genus $2$ over a field $k$ (say of characteristic different from $2$), defined by a hyperelliptique equation $$y^2=x^6+a_5x^5+\cdots +a_0, \quad a_i\in k.$$ Let $\sigma : C\to C$ be the hyperelliptique involution (it takes $(x,y)$ to $(x,-y)$). There are two points $\infty_1, \infty_2=\sigma(\infty_1)$ of $C$ above $x=\infty$. Consider $U=C\setminus \{\infty_2 \}$ and let $A=O_C(U)$. As $U$ is smooth, $\Omega_{A/k}^1$ is projective of rank $1$ as you know.

Now let us prove that $\Omega^1_{A/k}$ is not free. Suppose it is free. Let $\Omega_{C/k}^1$ be the sheaf of differential forms over $C$. Then $\Omega_{C/k}^1|_U=\Omega_{U/k}^1$ is free. This implies that the canonical divisor $K$ on $C$ is linearly equivalent to a divisor supported only in $C\setminus U=\{ \infty_2 \}$: $K\sim d \infty_2$ for some integer $d$. As $K$ has degree $2g(C)-2=2$, we have $d=2$. On the other hand, it is known that $K\sim P+\sigma(P)$ for any rational point $P$ of $C$. Therefore $$ 2\infty_2 \sim \infty_1 + \infty_2 $$ hence $\infty_1\sim \infty_2$. This implies that $C$ has genus $0$. Contradiction.

Please tell me if you can't understand all the details, I could try some direct computations.

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"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):

  1. $\Omega^1_{k[x_1,\dotsc,x_n]}$ is a free $k[x_1,\dotsc,x_n]$-module with basis $d(x_1),\dotsc,d(x_n)$
  2. 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 $\Omega^1_{A}$.

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I think there is a typo. The relation should be $2yd(y) - 3x^2d(x) = 0$. – user44400 Oct 12 '15 at 21:48
In algebraic topology, the homology modules are used to distinguish nonhomotopic spaces. How can HH (via Hochschild-Kostant-Rosenberg thm) be used to distinguish nonisomorphic varieties in $k^n$? For instance, how can $\Omega^1_A$ tell us that $Z(x^2-y^2-1)\ncong Z(x^2+y^2-1)$? – Leon Oct 20 '15 at 22:13
If $A=K[x,y|x^2+y^2-1]$ and $B=K[x,y|x^2-y^2-1]$, I don't know how to compare the $A$-module $\mathrm{Coker}[\begin{smallmatrix} x\\y\end{smallmatrix}]$ and $B$-module $\mathrm{Coker}[\begin{smallmatrix} x\\-y\end{smallmatrix}]$. – Leon Oct 20 '15 at 22:34

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