Are there ideals in $M_n(P)$ that are not of the form $M_n(I)$? Let $P$ be a noncommutative ring with units, $M_n(P)$ be a ring of matrices with coefficients in $P$. Is there an two-sided ideal $J$ in $M_n(P)$ that is not of the form $J=M_n(I)$, where $I$ is some two-sided ideal in $P$?
Thanks. 
 A: You said $P$ has units, so I'll assume it has a $1$ (the proper way of saying that $P$ has units is by saying that it is a unitary ring). 
Consider the matrix $E_{ij}$ to be the matrix with zeros everywhere except at the $ij^{\text{th}}$ position, where it is worth $1$. It is a quick calculation exercise to show that $E_{ij}AE_{k\ell}$ is a matrix whose $i\ell^{\text{th}}$ coefficient is $A_{jk}$. 
Assume $J$ is a two-sided-ideal of $M_n(P)$ that is non-zero. Let 
$$I = \{ x \in P \, | \, \exists A \in J \text{ such that } x = A_{ij} \text{ for some } i,j\in\{1,\cdots,n\}\},$$
i.e. $I$ is the set of all possible coefficients of elements in $J$. We show that $J = M_n(I)$. One containment is clear : $J \subseteq M_n(I)$. 
Next consider an element $A \in M_n(I).$ For each $a_{ij}$ coefficient of $A$, there exist a matrix $J_{ij} \in J$ whose $k\ell^{\text{th}}$ coefficient is precisely $a_{ij}$, for some $k,\ell\in\{1,\cdots,n\}.$ But then $E_{ik} J_{ij} E_{\ell j}$ is a matrix with zeros everywhere except at $(i,j)$, where it is worth $(J_{ij})_{k\ell} = a_{ij}$. Therefore
$$
A = \sum_{i=1}^n \sum_{j=1}^n E_{ik} J_{ij} E_{\ell j} \in J
$$
which gives you the reverse containment. In conclusion, there is no two-sided ideal of the form you request.
ADDED : I thought Matt's answer was too particular and not very understandable by everyone so I thought of another one. The set of matrices of the form
$$
\begin{bmatrix}
a_1 & \dots & a_n \\
0 & \dots & 0 \\
\vdots & & \vdots \\
0 & \dots & 0 \\
\end{bmatrix}
$$
form a right ideal of $M_n(P)$ but not a left ideal. Similarly, the set of all square matrices of the form
$$
\begin{bmatrix}
a_1 & 0 & \dots & 0  \\
 \vdots & \vdots &  & \vdots \\
a_n & 0 & \dots & 0 \\
\end{bmatrix}
$$
form a left ideal but not a right ideal. This only works for the cases $n > 1$, but for $n=1$, everything depends on $P$, so there's not much to say there.
Hope that helps,
A: Yes, there is. There's probably a simpler answer than this, but the first one that I thought of was if you take $P=KQ$ to be the path algebra of a quiver $Q$, and take:
$$J=\begin{pmatrix}KQ&KQ\\e_1KQ&e_1KQ\end{pmatrix}$$
where $e_1$ is one of the vertices. Then elements of $KQ$ are linear combinations of paths in the quiver (directed graph) $Q$ with coefficients in $K$, and $e_1KQ$ consists of those paths that start from the vertex $e_1$. The multiplication is by concatenation of the paths where possible, so unless I've made a silly error, this is a right ideal of $M_n(KQ)$, but not of the form you describe for any $Q$ with at least one path not starting at $e_1$.
A: If by ideal you mean a two-sided ideal then no. Every ideal $J$ of $M_n(P)$ is of the form $M_n(I)$ for some ideal $I$ of $P$.
A: Let $E_{i,j}$ be the matrix with $1$ at the $(i,j)$ place and $0$ elsewhere.
Note that for any matrix $A$, we have that $E_{(i,j)}A E_{(k,l)}$ is a matrix with $0$ everywhere, except at the $(i,l)$ place, where it is equal to $A_{(j,k)}$. It follows that if $I$ is any (Two-sided!) ideal in $M_n(R)$, and $J$ is the set of elements of $R$ which are part of matrices in $I$, then $I$ contains $M_n(J)$. Of course, $I$ is contained in $M_n(J)$, so there is an equality.
