Intuition/Heuristic behind I/I^2 definition of Kähler differentials Hello,
this one has always been mysterious to me. The Kähler differentials $\Omega_{A/k}$ are definined, by the universal property
$$Der_k(A,M)=A-Mod(\Omega_{A/k},M)$$
so for $M=A$ we get that $\Omega_{A/k}$ is the cotangent space of $spec(A)$.
(or a relative version of it if k is no field).
There are two constructions of Kähler Differentials I know.
The first one is $$\Omega_{A/k}=\langle df  : \text{relations satisfied by any derivation} \rangle$$ 
I think I sort of understand this one, it says that the differential of a function just contains enough information to extract the derivation of the function out of it. 
And this is what a section into cotangent space should be. Something that contains just enough information to pair it with a vector-field into a function.
The other construction is
$$\Omega_{A/k}=I/I^2$$
Where $I$ is the Ideal of functions vanishing on the diagonal in $spec(A)\times_{spec(k)} spec(A)$.
More geometrically it says 
sections into cotangent space=functions vanishing on the diagonal mod higher order.
But still I don't think I understand this equality on an intuitive level. Can someone explain the heuristic behind this equality? Or maybe explain $\Omega_{A/k}=I/I^2$ from another intuitive viewpoint?
 A: Think of the Taylor expansion of a smooth function on ${\mathbb R}^n$.
Let $I$ be the ideal of smooth functions vanishing at a given point $x_0$, then the zero order part of the Taylor series of a smooth function $f$ gives just the value of $f$ at the point.
If we subtract this constant from $f$, we land in the ideal $I$.
Now the first order derivatives of $f$ correspond to the first order terms in the Taylor series abd these are given by the image of $f$ in $I/I^2$.
A: Start with $Spec(A)$ being a vector space $V$, and let $V'$ be the subspace $V\times 0$ complementary to the diagonal $V_\Delta$. Then the ideal $I$ is functions on $V\times V$ vanishing on the diagonal, which we can think of as functions on $V'$ vanishing at $0$, indexed by points of $V_\Delta$. As Anton Deitmar explained, $I/I^2$ is then linear functions on $V'$, indexed by points of $V_\Delta$. Identifying $V' = V = V_\Delta$ we get $V \times V^*$.
Then this analysis works locally at smooth points of a general $Spec(A)$.
A: There are a number of definitions of the module of Kähler differentials that are proven to be equivalent in these notes.  Your question is discussed on page 20 near the bottom.
A: In calculus we teach that if $x = x_0 + \Delta x,$ then $f(x) = f(x_0) + f'(x_0) \Delta x.$
In other words, the derivative of $f$ at $x_0$ tells us the scaling factor for the change
in $f$ when $x_0$ moves by an infinitesimal amount $\Delta x$. What does infinitesimal mean? Well, we are ignoring higher order terms, i.e. terms in $\Delta x^n$ for $n \geq 2,$ i.e. we are working not at the level $\Delta x =0$, which says $x = x_0$, i.e. that we are on the diagonal, but at the next level $\Delta x^2 = 0,$ which is working modulo $I^2$.
(Working modulo $I$ is the same as setting $\Delta x = 0$.)
This is written in one variable, but works for any number of variables.  
