# 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.

Your proof of the second part of $(1)$ shows that if $I$ is not an ideal of $R$, then $M_n(I)$ is not an ideal of $M_n(R)$. Although this is true, it isn't quite what we're hoping to prove. We want to show that if $I$ is any ideal of $M_n(R)$, then there must be some ideal $J\subset R$ such that $I=M_n(J)$.
A brief hint on how to do this: let $S$ be the subset of $R$ consisting of all elements which appear as an entry in some matrix in $I$, and let $J$ be the ideal generated by $S$. I claim that $I=M_n(J)$. First, by definition of $J$ we have that $I\subset M_n(J)$. To show the reverse containment, notice that if $A=(a_{ij})\in I$, then multiplying $A$ by appropriate matrices on the left and right, we can show that for any $a_{ij}$, the matrix whose only nonzero entry is $a_{ij}$ in position $(i,j)$ is also in $I$. Then, we can multiply by an appropriate permutation matrix to move the nonzero entry to any position. In essence, we are using the rather strong condition that $I$ is a two-sided ideal to show that any matrix in $I$ leads to the existence of a lot of other, simpler matrices in $I$. These simpler matrices can then be added appropriately to show that $M_n(J)\subset I$.
For $(2)$, you are definitely on the right track. Suppose $I$ is an ideal in $M_n(R)$, so that there is some ideal $J\subset R$ such that $I=M_n(J)$ by what we just proved. However, division rings are simple (why?), so it follows that... (see if you can finish the argument).