Let $R$ be a unitary noncommutative Artinian ring, how can I prove that $M_n(R)$ is Artinian? I found a topic here Matrix Rings over Artinian commutative Rings, and know that Artinian is a Morita invariant condition. But I'm looking for a direct proof of it.
Now let $\rho_1 \supset \rho_2 \supset...$ be a descending chain of right ideals of $M_n(R)$. I'll try to show that it'll finally terminate.
Since I need to make some connections betweeb $\rho_k$, and the ideals of $R$, I just kind of projecting everything into $R$. Let $\rho_{k_{i, j}} = \pi_{i, j}(\rho_k)$, where $\pi_{i, j}$ is the projection of the $(i, j)-$coordinate into $R$.
For each pair of $(i, j)$, I know that $\rho_{1_{i, j}} \supset \rho_{2_{i, j}} \supset ...$ is a descending chain of right ideal in $R$, hence terminates.
However, having all chains $\{ p_{k_{i, j}} \}_k$ terminate for all $(i, j)$ doesn't mean that $\{ \rho_k \}$ terminates. Indeed, let's consider $M_2(\mathbb{R})$, and let $\rho = \left\langle \left[ \begin{array}{cc} 1 & 2 \\ 2 & 4 \end{array} \right] \right\rangle$, of course $\pi_{i, j}(\rho) = \mathbb{R}, \forall i, j$, but $\rho \neq M_2(\mathbb{R})$; i.e, we can make 'holes' inside $\rho$, such that all projections of it remain the same, put in other words, the chain $\rho_k$ can still continue to descend even though all of its projections terminate.
I can also show that, given any $\rho_k$, then $\pi_{i, 1}(\rho_k) = \pi_{i, 2}(\rho_k) = ... = \pi_{i, n}(\rho_k), \forall i$, i.e if we project each row in $\rho_k$ into $R^n$, then we'll have $I^n$, for some right ideal $I$ of $R$. But that doesn't seem to help. :(
Can I fix this proof in anyway, or is there some other ways to prove the claim?
Thank you guys very much,
And have a good day.
 A: Here is a sketch:
Any right ideal in $M_n(R)$ is in particular a right $R$-module.  So it suffices to prove something stronger: $M_n(R)$ is Artinian as a right $R$-module.
Now $M_n(R)$ is free as an $R$-module, of rank $n^2$.  So it suffices to prove something stonger still: if $R$ is Artinian, then any finite rank free module over $R$ is Artinian.   
(In fact, one has the yet stronger statement that any finitely generated $R$-module is Artinian.  This follows from the case of free modules, because any quotient of an Artinian module is Artininian, and any f.g. module is a quotient of a finite rank free module.)
There are various approaches to this.  One is to use the concept of length: an Artinian ring has finite length as a module over itself, and the length of a direct sum is the sum of the lengths.  Finally, a finite length module is Artinian.  

A little more detail:
Recall that finite length means that the module $M$ admits a Jordan--Holder series (equivalently, composition series)  of finite length, and the value of the length is just the number of Jordan--Holder factors.  So all these statements are simple applications of the basic theory of Jordan--Holder series.
The key to these argments is the following easy fact: if $M$ is a module and $N$ a submodule,
and $A \subseteq B$ are two other submodules such that 


*

*the images of $A$ and $B$
in $M/N$ coincide;

*the intersections $A \cap N$ and $B \cap N$ coincide;
then $A = B$.  
In your attempted argument, you were implictly using the "projection to $M/N$" part of this argument (by applying the $\pi_{i,j}$), but not the "intersection with $N$" part.  This is why you had trouble with "holes".
