How to prove directly that, if $A$ is Noetherian and $X=\operatorname{Spec} A,$ then $\mathscr O_X$ is a coherent $\mathscr O_X$-module? I use the following definition:
Definition
Let $(X,\mathscr O_X)$ be a locally ringed space.
An $\mathscr O_X$-module $\mathscr F$ is coherent if
(i) it is locally finitely generated.
(ii) for every $n,$ for every open $U\subset X,$ and for every $u:\mathscr O_X^n|_U\rightarrow \mathscr F|_U,$ the kernel of $u$ is locally finitely generated.  

Then how can I show directly that, if $A$ is a Noetherian ring, $X=\operatorname{Spec} A,$ then $\mathscr O_X$ is a coherent $\mathscr O_X$-module?  

P.S. Here is a related question. 
Edit
An $\mathscr O_X$-module is locally finitely generated if, for each $x\in X,$ there exists a nbd. $U$ of $x$ such that $\mathscr O_X|_U$ is generated by a finite family of sections of $\mathscr O_X$ over $U.$
 A: Let $X = \operatorname{Spec} A$, where $A$ is a Noetherian ring.
Condition (i) is clear.  For condition (ii), let $U$ be an open subset of $X$, and let $n$ be an integer.  Consider the kernel of a map $u$,
$$u: \mathcal{O}_X^n |_U \rightarrow \mathcal{O}_X |_U.$$
We want to show that $\ker u$ is locally finitely generated.  Let $x \in U$ and let $U' \subseteq U$ be an open affine neighborhood of $x$, so that $\mathcal{O}_X |_{U'} \cong \operatorname{Spec} B$, for some ring $B$.  Then $B$ must be Noetherian (see Hartshorne Prop 3.2).  Thus $\Gamma(U',\mathcal{O}_X |_U)] \cong B$  and $\Gamma(U',\mathcal{O}_X^n |_U)] \cong B^n$ is a finitely generated Noetherian module.
Now we want to show $(\ker u)|_{U'}$ is generated by a finite family of sections of $\mathcal{O}_X$ over $U'$.
Since $$\Gamma(U',\ker u) = \ker[\Gamma(U',\mathcal{O}_X^n |_U) \rightarrow \Gamma(U',\mathcal{O}_X |_U)]$$
is a submodule of a finitely generated Noetherian module, $\Gamma(U',\ker u)$ itself is finitely generated, and we are done.
