Regular sequence of sections of line bundles over a coherent sheaf I am reading the first chapter from the book by Huybrechts and Lehn, where I encountered the following definition.

I have the following doubts regarding this definition :


*

*What is the map $s:E\otimes L^\vee\longrightarrow E$? $E\otimes L^\vee \cong\mathcal{Hom}(L,E)$. So we have $s:\mathcal{Hom}(L,E)\longrightarrow E$. For any open set $U$ of $X$, $s(U):\mathcal{Hom}(L|_U,E|_U)\longrightarrow E(U)$ could be defined as $s(U)(\phi)=\phi(s|_U)$. This is my guess for the map $\phi$. Is this correct?

*They talk about $(s_1,s_2,\cdots,s_{i-1})(E\otimes L^\vee)$. What is this object? In case of a module $M$ over a commutative ring $R$ and an ideal $I$ in $R$, we know what $IM$ is. Is $(s_1,s_2,\cdots,s_{i-1})(E\otimes L^\vee)$ something analogous to that? How is that defined?

*What is the zero set $H$ of a section $s\in H^0(X,L)$. Since this $H\in|L|$, I am guessing that it could be the divisor of zeros of the section $s$ i.e. $(s)_0$ as denoted by Hartshorne. Is this $(s)_0$ the same as the set $\{x\in X|s_x\in\mathcal{m}_xL_x\}$, because this is the set which is generally called the zero set of $s$.

*And finally, what is this definition really saying. I am not able to really understand this definition. For example, what is a regular sequence for $L=\mathcal{O}(1)$ for a projective scheme $X$ over a field $k$. How can I find that out using this definition.
If someone can clarify these doubts, or direct me to some references, I will be grateful! Thanks in advance!
 A: [I can't provide a source other than Huybrechts-Lehn, which is credible very much in my opinion. Therefore I will try to convince you that the definition is a good one.]
Let me quickly address the last question first. Generalising terms from modules over commutative rings to coherent modules on schemes, sometimes, it is worthwhile replacing the structure sheaf with a line bundle. This is done in particular in the cases where one mostly cares about things that happen locally, i.e., at the local rings. This is how the definition should be read: for every point $x\in X$, the germ $s_x\in L_x\cong \mathcal{O}_{X,x}$ is $E_x$-regular. We will prove this below.
Your guess for the first question is correct. Alternatively it can be defined as the morphism obtained from $L^{\vee} \to \mathcal{O}_X$, $L^{\vee}(U)\ni\varphi\mapsto \varphi(s|_U)\in\mathcal{O}_X(U)$ by tensoring with $E$. Note that if $L$ is trivial, then this amounts to multiplication with $s_U$ and in the affine world, say $X = \mathop{Spec}(A)$, $E = \widetilde{M}$ and $L = \widetilde{N}$ for $M$ a finitely generated $A$-module and $N$ a projective rank one $A$-module, we can take an $n\in N$ and look at the map $\mu(n)\colon M\otimes_A N^{\vee}\to M$ mapping $m\otimes_A \varphi$ to $m\varphi(n)$.
This map will be injective if and only if for every prime ideal $\mathfrak{p}\subset A$, the localised map $\mu(n)_p$ is injective, but after localisation we get $N_{\mathfrak{p}} \cong A_{\mathfrak{p}}$ and $\mu(n)_{\mathfrak{p}}$ is injective if and only if multiplication with the corresponding element of $A_{\mathfrak{p}}$ on $E_{\mathfrak{p}}$ is injective, i.e. if and only if this element is $E_{\mathfrak{p}}$-regular.
Concerning the second question, this will certainly be the submodule of $E$ generated by the images of the $s_k$. Note that over an open affine $U\subset X$ such that $L|_U$ is trivial, this is just the coherent sheaf associated with the module $(s_1|_U,s_2|_U,\dots s_{i-1}|_U)E(U)$, where we interpret the former factor as the ideal generated by the sections $s_k|_U\in L(U) \cong \mathcal{O}_X(U)$, $k=1,2,\dots i-1$. So here, again, we generalise the well known notion of a regular sequence. (Some books require from a regular sequence not to span the whole module, but this is possible here.)
For the third, yes, it is the zero divisor of the section. Note that its support is given by the points $x\in X$ where $s_x$ doesn't generate $L_x$. Indeed, this is precisely the locus where $s_x\in \mathfrak{m}_x L_x$. This is because the ideal to the zero divisor of $s$ is locally given by $s$ itself (more precisely when restricted to opens where $L$ becomes trivial of course).
I do not have a nice non-trivial example at hand, so to get a better feeling for the definition we prove:

For a global section $s\in H^0(L)$ of a line bundle on a noetherian scheme $X$ and any coherent $\mathcal{O}_X$-module $E$, the following are equivalent.

*

*The section $s$ is $E$-regular.

*For each $x\in X$, the element of $\mathcal{O}_{X,x}$ corresponding to $s_x\in L_x\cong \mathcal{O}_{X,x}$ is $E_x$-regular in the sense of commutative algebra (with the possibility that $s_x E_x = E_x$).

*The zero locus of $s$, i.e., the set of $x\in X$ such that $s_x\in\mathfrak{m}_xL_x$, does not contain associated points of $E$.


(I didn't write zero divisor of $s$, because it might happen that $E$ is trivial, on a component of $Y$, say, and then the section would be allowed to be zero on this component too and hence the zero locus would also contain this component.)
I pretty much proved the equivalence of the first two statements in the explanation of the morphism that is required to be injective in $1$. To prove that $1.$ is equivalent to $3.$, we first have to recall some statements from the theory of associated points on noetherian schemes. (See here for example.) First of all, recall that on a noetherian scheme, for all coherent modules $F$ we have $\mathop{Ass}(F) = \mathop{WeakAss}(F)$ and $\mathop{Ass}(F) = \emptyset$ if and only if $F = 0$. Also, if $0\to F_1\to F_2\to F_3\to 0$ is exact, then $\mathop{Ass}(F_1)\subset \mathop{Ass}(F_2)\subset \mathop{Ass}(F_1)\cup\mathop{Ass}(F_2)$.
We will apply this to the kernel of the map $E\otimes L^{\vee}\to E$. This being a local question we may assume that $L$ is trivial. I.e. $E\to E$ is just multiplication with a section $s\in H^0(\mathcal{O}_X)$ and the zero locus of $s$ is the scheme associated with the ideal generated by $s$. If $s$ fulfills 3., then $\mathop{Ass}(\ker(E\xrightarrow{\cdot s} E)) = \emptyset$ and hence the multiplication map is injective, thus $1.$ holds. This is because for all $x\in X$ outside the zero locus of $s$, $s_x$ is a unit of the local ring such that the kernel is trivial at $x$ and for all the other points, we required that $x\not\in\mathop{Ass}(E)\supset\mathop{Ass}(\ker(\cdot s))$.
The other way around, if $s$ is $E$-regular, to get 3. we have to prove that if $s_x\in\mathfrak{m}_x$ (still assuming $L = \mathcal{O}_X$ for simplicity), then $x\not\in\mathop{Ass}(E)$. By assumption, $s$ induces an isomorphism between $E$ and $sE$. If $x$ was an associated point of $E$, then it also would be an associated point of $sE$, meaning we'd have to have an $e\in E_x$ such that $\mathfrak{m}_x$ were the annihilator of $s_x e$. In particular, $s_x^2\in\mathfrak{m}_x$ would have to annihilate $e$ which is impossible, since if $\cdot s\colon E\to E$ is injective, then so is the composition $\cdot s^2\colon E\to E$.
A: (1). A section $s \in H^0(X, L)$ gives a map $\mathcal O_X \to L.$ Taking it's dual and tensoring by $E,$ we get the map $E\otimes L^\vee\longrightarrow E.$
(2). Let $s_1, s_2 \in H^0(X, L).$ Let $L^\vee \oplus L^\vee \to \mathcal O_X$ be the map given by $s_1, s_2.$ So we will have a map $E \otimes L^\vee \oplus E \otimes L^\vee \to E.$ Image of this map is  denoted by $(s_1,s_2)(E \otimes L^\vee).$ Similarly for $(s_1,s_2,\cdots,s_{i-1})(E\otimes L^\vee).$
(3). Here $H$ is the set $\{x\in X|s_x\in\mathcal{m}_xL_x\}.$
