I'm trying to understand the exercise in the title, which should be an alternative way to construct the sheaf of differentials:
" Let $B$ be an $A$-module. Let $\rho$: $B \otimes_A B$ $\longrightarrow$ $B$ be the 'diagonal' homomorphism $\rho(b_1 \otimes b_2)=b_1b_2$. Let $I=Ker(\rho)$. We can consider $B \otimes_A B$ as a $B$-module. Let $d$: $B$ $\longrightarrow$ $I/I^2$ be defined by $d(b)=b \otimes 1 – 1 \otimes b$.
(a) Show that d is an $A$-derivation.
(b) Show that the $B$-module $I/I^2$ endowed with the $A$-derivation $d$ verifies the universal property of the module of differentials of $B$ over $A$.
(c) Let $f$: $X$ $\longrightarrow$ $Y$ be a morphism of schemes. Let $\Delta$: $X$ $\longrightarrow$ $X \times_Y X$ be the diagonal morphism. Let $\mathcal{I}$ denote the kernel of $\Delta^\text{#}$: $\mathcal{O}_{X \times_Y X}$ $\longrightarrow$ $\Delta_∗$ $\mathcal{O}_X$.
Show that $\Omega_{X/Y}$ $\simeq$ $\Delta$*$(\mathcal{I}/\mathcal{I}^2)$ "
(a):
For what concerns this point, after observing that $d(a)=0$ with $a \in A$, I just verified that $d$ follows Leibnitz rule:
$d(b_1 b_2)=b_1 b_2 \otimes 1 – 1 \otimes b_1 b_2 = (b_1 b_2 \otimes 1 – b_1 \otimes b_2) + (b_1 \otimes b_2 – 1 \otimes b_1 b_2)$
then just note that
$(b_1 b_2 \otimes 1 – b_1 \otimes b_2) – (1 \otimes b_1 b_2 – b_2 \otimes b_1) = (b_2 \otimes 1 – 1 \otimes b_2)(b_1 \otimes 1 – 1 \otimes b_1) \in I^2$
so conclude that $d(b_1 b_2) = b_1d(b_2) + b_2d(b_1)$
(b):
Here I'm already stuck: I tried in vain, making a few attempts, to find the homomorphism $\Phi$ that satisfies the universal property.
(c):
I don't know if solving point (b) gives hints on how to solve this point, but up to now I have no Idea of how to proceed
Best Answer
About point (b) I proved that $I/I^2$ satisfies the universal property of $\Omega_{B/A}$:
I need to define a homomorphism $\Phi$: $I/I^2$ $\longrightarrow$ $M$ for every $M$ B-module and $d':B \longrightarrow M$ A-derivation and then prove the uniqueness of $\Phi$
$\Phi\big( \big[ \displaystyle \sum_{i \in I} c_i b_i \otimes c_i \big] \big) := \displaystyle \sum_{i \in I} d'(b_i)c_i $ which is a homomorphism and satisfies $\Phi \circ d = d'$ and is well defined as:
$[\displaystyle \sum_{i \in I} b_i \otimes c_i] = [\displaystyle \sum_{i \in I} \tilde{b}_i \otimes \tilde{c}_i]$ $\Longrightarrow$ $\displaystyle \sum_{i \in I} \tilde{b}_i \otimes \tilde{c}_i = \displaystyle \sum_{i \in I} b_i \otimes c_i + (\displaystyle \sum_{i \in I} h_i \otimes e_i)(\displaystyle \sum_{j \in I}f_j \otimes g_j)$ with $h_ie_i = f_jg_j = 0$
But $\Phi([\displaystyle \sum_{i \in I} h_i \otimes e_i)(\displaystyle \sum_{j \in I}f_j \otimes g_j]) = 0$ indeed:
$\Phi([\displaystyle \sum_{i \in I} h_i \otimes e_i)(\displaystyle \sum_{j \in I}f_j \otimes g_j]) = \displaystyle \sum_{i,j} \Phi([h_if_j\otimes e_i g_j])$
But $\forall i,j$ $\Phi([h_if_j\otimes e_i g_j])= e_ig_jd'(h_if_j)=0$ where the last equality follows very easily by Leibinitz.
So we just need to prove the uniqueness of $\Phi$ but this follows from the fact that it is the inverse homomorphism of $\tilde{\Phi}: \Omega_{B/A}\longrightarrow I/I^2$ which sends $b_1d_{B/A}b_2 \longrightarrow [b_2 \otimes b_1 - 1 \otimes b_1b_2]$.
So we proved that $\Omega_{B/A}$ and $I/I^2$ are isomorphic.
Point (c): Point c of exercise 6.1.1 in Qing Liu's book