TITLE: Diagonal map and "infinitesimal points"
QUESTION [9 upvotes]: Let $f:X\to Y$ be a morphism between schemes. To construct the relative sheaf of differentials on $X$ (relative to $Y$), we first consider the diagonal map $\Delta: X \to X\times_Y X$ and then define $\Omega_{X/Y} = \Delta^{-1} \mathscr{I}/\mathscr{I}^2$ where $\mathscr{I}$ is the sheaf representing the immersion $X\to X\times_Y X$ (it's the kernel of $\mathscr{O}_{X\times_Y X} \to \Delta_* \mathscr{O}_X$.
Algebraically, this works out fine, due to the theory of abstract Kahler derivation defined on algebras.  Is there a way to actually see the motivation behind this? 
Moreover, what's the analog in higher infinitesimal approximation (instead of just 1st order one given by the differentials)?  What's the (say, "analytic") insight behind the relationship between higher infinitesimal and higher diagonal?

REPLY [6 votes]: I would like to add another answer to this old question. Consider
the case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$:
$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$
Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in fact
we can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the universal point $S = R[T_0]$, $a_0 = T_0$, we see that the ``universal Taylor expansion'' of $f$ is
$$
f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T].
$$
If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as
$$
1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i
$$
Looking mod $(1\otimes T-T\otimes 1)^2$ we get:
$$
1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2}
$$
Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by
$1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear terms
of Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function
$f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$.
This also explains nicely what happens in higher degree. We can introduce the algebras
$P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the ring of Taylor expansions of degree $\leq n$ where
the terms of degree at most $n$ of the Taylor expansion live. There is a natural map
$d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion.
This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), because
locally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and this
map induces an isomorphism on $I/I^2$ and $P^n$ more generally.