Matrices with ideals Let $R$ be a ring with identity, let $n\in \mathbb N^*$, and let $S=M_n(R)$ be the ring of $n\times n$ matrices with entries in $R$. Let $J$ be an ideal of $S$. Prove that the ideal $K := \{a \in R \mid a$ is the $(1,1)$ entry of some element of $J\}$  coincides with the set of elements of $R$ that occur as entries of elements of $J$.  First, I need to show that $M_n(K)=J$. I see it is kind of ambiguous to show that for any $A \in J$, $A \in M_n(k).$ Also, if $B \in M_n(K)$ it has entries in $K$, but how can we know that this entries are the same entries for a matrix in $A \in J$. I would apprecitate any help with that.
Proving $M_n(I)\lhd M_n(R)$ and proofs about ideal rings. I'm trying to do a few proofs about rings: $(1)$ I want to show that for a ring $R$ and the ring of matrices $M_n(R)$, if $I\lhd R\implies M_n(I)\lhd M_n(R)$ and that all ideals of $M_n(R)$ are of the form $M_n(I)$. $(2)$ Prove that if $R$ is a ring with division then $M_n(R)$ is a simple ring.  The relation $I\lhd R$ implies that with $r\in R$ and $x\in I$ we have $rx\in I, xr\in I$. But how can I prove the statement for matrices?, here's what I would do: Consider $\psi \in M_n(R)$, if is given any matrix $\phi\in M_n(I)$ it must be $\psi \phi \in M_n(I),\phi \psi \in M_n(I)$, if we denote the element $(i,j)$ of each matrix by $\phi_{ij}$ and $\psi_{ij}$, then $(\psi \phi)_{ij}=\sum_{k=1}^n \psi_{ik}\phi_{kj}$ and $(\phi \psi)_{ij}=\sum_{k=1}^n \phi_{ik}\psi_{kj}$.  Since $I\lhd R$ is $\psi_{ik}\phi_{kj}\in I$ and $\phi_{ik}\psi_{kj}\in I$, which means that $I\lhd R\implies M_n(I)\lhd M_n(R)$.  To show that every ideal of $M_n(R)$ are of the form $M_n(I)$ with $I$ ideal: Suppose $I$ is not an ideal of $R$, then exists $r\in R$ such that $rx\notin I$ or $xr\notin I$, then with taking a matrix $\phi\in M_n(I)$ such that $x$ is an entry of $\phi$ and a matrix $\psi$ such that $r$ is an entry of $\psi$ we get that the matrices $\phi$ and $\psi$ have an entry $xr$ or $rx$, in either case this is not an element of $I$ then the matrix is not in $M_n(I)$. This concludes my proof (is it?) for $(1)$.  For $(2)$ I'd like to proceed by absurdum . Let's say that $M_n(R)$ is not a simple ring, then it has at least one ideal. Denote by $\varphi$ this ideal in $I$, then for any matrix $\psi$ we have $\varphi\psi \in M_n(I)$ and $\psi \varphi \in M_n(I)$.  Now, can $R$ be a ring with division?.I'm not sure how to prove that it can't. I believe I could use this theorem: "If $I\lhd R$ and $1\in I$ then $I=R$. The case $I\neq R$ implies $1\neq R$ and since $\varphi \psi$ and $\psi \varphi$ are in $M_n(I)$ then there are at entries in the product matrix which belong to $I$ (using $I \lhd R$) but are not $1$ (1\notin I); then there are elements in $I$ (and therefore in $R$ by $I\subset R$) that are not invertible, follows that $R$ is not a ring with division. But if $I=R$?, here I can't see a why $R$ can't be a ring with division.
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