Kernel of a morphism between coherent sheaves. Throughout this book, http://books.google.co.kr/books/about/Algebraic_Geometry_and_Arithmetic_Curves.html?id=uaLKdA0PxS4C&redir_esc=y, kernel of a morphism between coherent sheaves on a locally Noetherian scheme is also coherent.
More precisely, let $X$ be a locally Noetherian scheme and  $u:\mathcal{F}\to\mathcal{G}$ be a morphism of coherent sheaves on $X$. Then $\textrm{Ker }u$ is also coherent.
The author says, this is trivially true. However, I cannot prove it. Moreover, I do not know whether it is really true or not.
Even though I do not have any explicit example, I think this would be false in following sense; when $X$ is locally Noetherian scheme and $\mathcal{F}$ is a coherent sheaf on $X$, it is known that $\mathcal{F}(U)$ is finitely generated $\mathcal{O}_X(U)$-module for any affine open subset $U$ of $X$. If $\textrm{Ker }u$ is coherent, then the sequence $\textrm{Ker }u(U)\to\mathcal{F}(U)\to\mathcal{G}(U)$ of $\mathcal{O}_X(U)$-modules is an exact for any affine open $U$. However, in general, a submodule of a finitely generated module is not finitely generated. So, maybe there would be an example that arises form the sequence $\textrm{ker} f \to L\xrightarrow{f}M$ where $L,M$ are finitely generated modules and $\textrm{ker }f$ cannot be generated by any finite set. Is this reasonable?
Actually, we can find the same statement for $X$ is Noetherian case in Hartshorne's book. If the above statement is true, how can we expand the case?
 A: Since $X$ is locally noetherian, and since submodules of f.g. modules over a noetherian ring are finitely generated, we can see that this is locally true: that is for any affine open spec$R$, there is a cover $U_i=D(f_i)$ of Spec$R$, $f_i\in R$ generating the unit ideal, such that ker$(f)(U_i)$ is finitely generated. 
Thus, all we have to prove is the following gluing result: Let $f_1,...,f_k$ generate the ideal $(1)$ in a ring $R$. Let $M$ be a module on $R$. We want to show that if $M_{f_i}$ are all finitely generated $R_{f_i}$-modules, then $M$ is a finitely generated $R$-module. Indeed, let $\frac{m_{ij}}{1}$ generate $M_{f_i}$, where $m_{ij}\in M$. 
Let $m\in M$. By our assumption, we can fix a number $N$ and elements $r_{ij}\in R$ such that 
$f_i^Nm=\sum_j f_i^Nr_{ij}m_{ij}$ for all $i$. 
But now we use the fact that $(f_1,...,f_r)=1$, so we may write $1=\sum_{i=1}^k s_if_i$, $s_i\in R$. Taking this equation to the $kN$th power, we see that, in fact $(f_1^N,...,f_r^N)=(1)$ as well. 
Thus we may write $1$ as a linear combination of the $f_i^N$, so we may write $m$ as a linear combination of the elements $f_i^Nm$, which, by our earlier equation, means we can write $m$ as a linear combination of the $m_{ij}$.
So the $m_{ij}$ generate $M$ and we have proven $M$ is finitely generated.
A: Hint: A ring $R$ is noetherian iff submodules of f.g. $R$-modules are again f.g.
