I'm stuck with the following problem.
If $R$ is an unitary ring and $n\in \mathbb{N}$, there is a bijective correspondence which preserves the order given by the inclusion between $\mathcal{A}$ and $\mathcal{B}$, where $$\mathcal{A}:=\big\{I \trianglelefteq R : I \mathrm{\, \,ideal\,of\, \,} R \big\} \rightleftarrows \big\{ \mathfrak{a}\trianglelefteq\mathcal{M}_{n}(R): \mathfrak a \mathrm{\,\,ideal\,of\,} \mathcal{M}_n(R)\big\}=:\mathcal{B}. $$
I was thinking in using the following "generalized" correspondence theorem, but I can't think of a suitable ring morphism from $R$ onto $\mathcal{M}_n(R)$
Also, my first attempt was to give an explicit function from $\mathcal A$ to $\mathcal B$, namely $F: I\mapsto \mathcal{M}_n(I):=\big\{ (a_{ij})\in \mathcal{M}_n(R): a_{ij}\in I\quad \forall i, j \big\}.$ This seems to work, but I can't find an explicit function from $G: \mathcal B \to \mathcal A$ which also serves as an inverse for $F$ ($GF=1_{\mathcal A}$ and $FG=1_{\mathcal B}$ ).
What do you think? Is it possible to use the theorem above? Is there a "painless" way to give such $G: \mathcal B \to \mathcal A$?
Thanks in advance!
The map $I\mapsto M_n(I)$ is clearly well defined and injective.
It remains to show that every ideal $J$ of $M_n(R)$ is of the form $M_n(I)$, for some ideal $I$ in $R$.
Consider the matrix unit $E_{ij}$ defined as the matrix having $1$ at the place $(i,j)$ and $0$ elsewhere. By suitably multiplying of a matrix $A=(a_{ij})\in J$ with matrix units on the left and on the right, show you can express $$ A=\sum_{i,j}a_{ij}E_{ij} $$ and that $a_{ij}E_{ij}\in J$. Now consider the subset $I$ of $R$ consisting of those elements that appear in at least one matrix in $J$ and prove $J=M_n(I)$.