Decomposition of coherent module sheaf. Lex $X$ be an irreducible noetherian seperated scheme all of whose stalks $\mathcal{O}_x$ at closed points are discrete valuation rings. If $\mathcal{F}$ is a coherent $\mathcal{O}_X$-module, then it's said that 
$\mathcal{F}\cong \mathcal{F}_1 \oplus \mathcal{F}_2$
where $\mathcal{F}_1$ is a locally free $\mathcal{O}_X$-module, and $\mathcal{F}_2$ has support at a finite number of closed points $x_1,x_2,...,x_n$.
Here is my attempt. Suppose we have an open affine cover {$U_i$} of $X$ with $U_i=Spec(R_i)$. Let's look at one open affine $U_0=Spec(R_0)$, then the stalk of closed point $x_1$ in $U_0$ is a discrete valuation ring, i.e, a regular local ring of dimension 1. This means that the maximal ideal $m_{x1}$ is generated by one element $a$. If $R_0$ contains another maximal ideal $m_{x_2}$, then the generater $b$ of it must equal to $a$, otherwise, $(a,b)$ will be an ideal contain maximal ideal $aR$, a contradiction. Therefore, $R_0$ itself must be a regular local ring of $dim=1$. In other words, each $U_i$ contains at most one closed point.
Since $\mathcal{F}$ is coherent module. We have surjections 
$R^{n_i}_i\to \mathcal{F} |_{U_i}\to 0$ for some integer $n_i$. If we denote field $k=R_i/m_i$, where $m_i$ is a maximal ideal of $R_i$, then the stalk at closed point can be identified with $k[X_1]$ with some indeterminent $X_1$. Thus, localization at closed point $x_1$ of the surjection can be written down as:
$k[X_1,...,X_{n_i}]\to k[X_1,...,X_{n_i}]/I\to 0$
with some prime ideal $I$. And the middle is the module of $\mathcal{F}_1$.
Now, if $I$ just invole, after permutation, $X_{r+1},...,X_{n_1}$, we let $\mathcal{F}_{1}$ to be $k[X_1,...,X_r]$. Under this construction, $\mathcal{F}$ is locally free. And let $\mathcal{F}_2$ to be $k[X_{r+1},...,X_{n_1}]/I$, it's not empty at closed point, but it will be empty at generic point, which is the only point besides closed point in a regular local ring. Also, notice that $X$ is notherian, thus {$U_i$} is finite and support of $\mathcal{F}_2$ is finite.
My question is: wether every open set in a cover must contain a closed point of $X$? If it does, the argument works through; if it doesn't, then all the work is in vain. Also, I am not certianly sure about the argument above. So, point out the mistakes if there were any. Thank in advance!
 A: There are some mistakes I can point out. First, it is completely false that the ring $R$ of any open affine $U = \text{Spec}(R)$ inside $X$ is local (as a counterexample, take $X$ to be the affine line). Second, the stalk at a closed point cannot have the form $k[X]$ as this ring is not local. I must confess I don't understand much of your argument after these mistakes.
I would suggest another approach. Take $\mathcal{T}$ to be the maximal torsion subsheaf of $\mathcal{F}$, i.e. if locally $\mathcal{F}|_U = \widetilde{M}$ then $\mathcal{T}|_U = \widetilde{T(M)}$ where $T(M) = \{ m \in M \mid 
\exists \:a \in A \setminus \{0\} \: am=0 \}$. Then show that $\mathcal{F}/\mathcal{T}$ is locally free and that the exact sequence $0 \to \mathcal{T} \to \mathcal{F} \to \mathcal{F}/\mathcal{T} \to 0$ splits.
To answer to your first question, if $X = \text{Spec}(R)$ with $R$ a DVR, then the open affine $\{(0)\}$ does not contain the closed point. However, if the scheme is of finite type over field, then it's true as the closed points are dense.
