$M_n(I)\subseteq J$ for ideal $J$ in $M_n(R)$ Let $R$ be a unital ring, $J$ an ideal in $M_n(R)$ and $I= \{x\in R\colon x\text{ is an entry in } A\in J\}$.
I'm trying to understand the proof here of correspondence between two-sided ideals of $R$ and $M_n(R)$, but I'm stuck on the detail $M_n(I)\subseteq J$. Everything else is clear. I believe I'm missing something obvious.
Let $(a_{i,j}) \in M_n(I)$ then $(a_{i,j})$ is an $n\times n$ matrix with entries in $I$, i.e. $a_{i,j} \in I \implies$ there exists a matrix $A^{i,j}\in J$ which has $a_{i,j}$ as an entry somewhere. It is not a priori clear to me that $(a_{i,j}) \in J$. I've tried to construct $(a_{i,j})$ from the "entry associated matrices" $A^{i,j}$ by doing operations together with $E_{p,q}$ matrices but with no luck. I'm staring myself blind on this little problem, it feels like it should be obvious. What am I missing?
 A: Starting from your assertion that each $a_{ij}$ is an entry of some $A^{ij}\in J$, let's explicitly write
$$
  a_{ij}=A^{ij}_{p_{ij}q_{ij}},
$$
for some indices $p_{ij},q_{ij}$. Since $J$ is a two-sided ideal,
$$
  (a_{ij})=\sum_{i,j}E_{ip_{ij}}A^{ij}E_{q_{ij}j}\in J.
$$
A: Let $E_{ij}$ be the standard basis for $M_n(R)$. These multiply by the rule $E_{ab}E_{cd} = \delta_{bc}E_{ad}$ where $\delta_{bc}$ is Kronecker's delta function. Let $A = (a_{ij}) = \sum_{i,j} a_{ij}E_{ij} \in J$. Then
$$ E_{pq} A E_{rs} = \sum_{ij} a_{ij} E_{pq}E_{ij}E_{rs} = \sum_{ij} a_{ij} \delta_{qi}\delta_{jr} E_{ps} = a_{qr}E_{ps}. $$
The point is that we can isolate any entry of a given matrix in $J$. We can also move any single-entry $a E_{ij}$ anywhere we want in the matrix by the same trick:
$$ a E_{kl} = E_{ki} (a E_{ij}) E_{jl}. $$
Using these two facts, it follows that $M_n(I) \subseteq J$.
A: T.Gunn's answer is very concise and cannot be improved. however I hope it is not intrusive to add a few remarks which attempt to address your specific difficulty regarding the inclusion $M_n(I) \subseteq J$ 
since $R$ is unital $M_n(R)$ contains a multiplicative subgroup $P$ isomorphic to the symmetric group $S_n$. elements of $S$ are called permutation matrices.
an element of  P acts on any matrix $X \in M_n(R)$ by left multiplication (permuting the rows of $X$). csll this group of actions $P_L$
there is a corresponding action by right multiplication by (permuting the columns of $X$). call this group of actions $P_R$.
note that the left and right actions commute. so that the full set of actions is isomorphic to $P_L\times P_R$
for $i,j \in \{1,2,\dots,n\}$ let $e_{ij}$ be the matrix with only a single non-zero entry - a $1$ in the $ij^{\text{th}}$ position, then $P_L\times P_R$ acts transitively on the set $\{e_{ij}\}$.
and since $J$ is a two-sided ideal, we must have $P_L\times P_R(J) \subseteq J$
thus if a matrix in $J$ has an element $r \in R$ in the $ij^{\text{th}}$ position, then $J$ must also contain a matrix with $r$ in the $kl^{\text{th}}$ position, for any $k,l \in \{1,2,\dots,n\}$
let $A \in M_n(I)$. by the definition of $I$ each entry $a_{ij}$ of $A$ occurs as an entry in some matrix $M(i,j)\in J$ and by the foregoing remarks we may choose each $M(i,j)$ to have $a_{ij}$ as its $ij^{\text{th}}$ entry.
now using elementary matrices as selectors we have in $J$ the following $n^2$ matrices, each having at most one non-zero entry:
$$
e_{ii}M(i,j)e_{jj} = a_{ij}e_{ij}
$$
since $J$ is an ideal the sum of all these matrices is in $J$. but then
$\sum_{ij} a_{ij}e_{ij} = A$
hence $A \in J$. and as $A$ was any element of $M_n(I)$ this gives:
$$
M_n(I) \subseteq J
$$
