In Gelfand's book Discriminants, resultants, and multidimensional determinants he gives a formula:
$$J_{1}(L)\cong J_{1}(\mathcal{O}_{X})\otimes L$$
Here $L$ is a line bundle on some complex variety.
I am confused about this formula. The explanation of the book is the following:
Take regular function $h$ and regular section $f$ of $L$ around some point $x$. Then we send $j(h)\otimes f$ to $j(hf) - hj(f)$.
I am confused. At the point $x$, if we think in Taylor expansion, $j(hf)-hj(f)$ should have value $0$ on the first term, that is to say, the value in $L$. So it seems to me that we could not get all the elements in $J_{1}(L)$.
Question: "Here L is a line bundle on some complex variety. I am confused about this formula."
Answer: If $k$ is a field and $k \rightarrow A$ a $k$-algebra and $L$ a projective $A$-module of rank 1 (or any $A$-module), we may define
$$J1.\text{ }J^1(L):=A\otimes_k A/I^2 \otimes_A L$$
and a map
$$d_L: L \rightarrow J^1(L)$$
by $d_L(s):=1\otimes s \in J^1(L)$.
The map $d_L$ is a differential operator of order $1$:
$$d_L \in Diff_k(L, J^1(L)).$$
There is an isomorphism
$$Hom_A(J^1(L), L) \cong Diff_k^1(L,L),$$
hence the first order jet bundle $J^1(L)$ "generalize" the module of Kahler differentials.
Example: In general let $J^n_A:=A\otimes_k A/I^{n+1}$. Let $k$ be the complex numbers and $A:=k[x]$ with $A\otimes_k A \cong k[x,y]$ and $I=(y-x) \subseteq A$. Define $dx:=y-x$. It follows
$$d: A \rightarrow J^n_A$$
is the map
$$d(f(x)):=f(y)=f(x+y-x):=f(x+dx) \cong \sum_{i \geq 0}\frac{f^{(i)}(x)}{i!}dx^i.$$
Hence the differential operator $d$ calculates the "Taylor expansion" of the polynomial $f(x)$.
The module $J^1_A:=A\otimes_k A/I^2$ has a left and right $A$-module structure and there is a relation
$$a(\omega b)=(a\omega)b$$
for all $a,b \in A, \omega \in J^1_A$. When you tensor you use the right $A$-module structure on $J^1_A $ and the left $A$-module structure on $L$. With this definition you get the isomorphism
$$J^1_A(A) \otimes_A L \cong A\otimes_k A/I^2 \otimes_A A \otimes_A L \cong A\otimes_k A/I^2 \otimes_A L :=J^1(L).$$
Note: If you are to be "rigorous", for the general definition of the jet bundle you should use the projection maps $p,q: X \times X \rightarrow X$ and make the following definition: Let $I \subseteq \mathcal{O}_{X \times X}$ be the "ideal of the diagonal" and define
$$J^1_X:=p_*(\mathcal{O}_{X \times X}/I^2)$$
and
$$J^1_X(L):=p_*(\mathcal{O}_{X \times X}/I^2 \otimes_{\mathcal{O}_{X\times X}} q^*L).$$
Example: If $X \subseteq \mathbb{P}^n_k$ is a quasi projective variety with $L^* \in Pic(X)$ a line bundle with $U:=Spec(A) \subseteq X$ an open affine subscheme and $L^*(U):=L$ with $L\in Pic(A)$, it follows the restriction to $U$
$$J^1_X(L^*)_U \cong J^1(L) \cong A\otimes_k A/I^2 \otimes_A L$$
is the jet bundle from Definition J1 above.
Since $q^*(\mathcal{O}_X) \cong \mathcal{O}_{X \times X}$ it follows
$$J^1_X(\mathcal{O}_X) \cong p_*(\mathcal{O}_{X \times X}/I^2 \otimes_{\mathcal{O}_{X \times X}} q^*(\mathcal{O}_{X})) \cong p_*(\mathcal{O}_{X \times X}/I^2 \otimes _{\mathcal{O}_{X \times X} }\mathcal{O}_{X\times X}) \cong J^1_X.$$
Question: "I am confused about this formula."
Response: You get an isomorphism
$$J^1_X(\mathcal{O}_X) \otimes_{\mathcal{O}_{X \times X}} q^*L \cong J^1_X \otimes_{\mathcal{O}_{X \times X}} q^*L \cong J^1_X(L).$$
Note: If $\mathfrak{m} \subseteq A$ is a $k$-rational point (with $k$ a field), there is an isomorphism
$$J^1_A(\mathfrak{m}):=A/\mathfrak{m}\otimes_A J^1_A \cong A/\mathfrak{m}\otimes_A (A\oplus I/I^2) \cong k \oplus \mathfrak{m}/\mathfrak{m}^2 \cong A/\mathfrak{m}^2.$$
More generally
$$F1.\text{ }J^1(L)(\mathfrak{m}) \cong L/\mathfrak{m}^2L.$$
Usually they define the jetbundle "fiberwise" using the formula $F1$. The above definition is a "global" definition.
Example: In the book on discriminants they use the jet bundle to construct the classical discriminant. If $C$ is the complex projective line and $L:=\mathcal{O}(d)$ the tautological linebundle with $d\geq 2$, it follows $H^0(C,L)$ is the vector space of homogeneous polynomials $f(x_0,x_1)$ of degree $d$. The classical discriminant $D_d \subseteq H^0(C,L)$ is the "set" of such polynomials with "multiple roots". A polynomial $g(t)$ in one variable $t$ has a multiple root iff $g(t)=g'(t)=0$, hence you may use the derivative to tell if $g(t)$ has a repeated root. In the book they use the jet bundle $J^1(L)$ to construct $D_d$ using "Taylor expansion" of sections as described above.