Why is directed limit of simple rings simple?

Question

This is a statement mentioned in Exercise 1.2.7 of Rosenberg, Algebraic K-theory.

"Note that since matrix rings over fields are simple, directed limit of simple rings is simple."

$\textbf{Q:}$ Why is directed limit simple ring simple? Let $J$ be a 2 sided ideal. Take $a\in J$. Now $a$ must be in some matrix ring say $M_n(k)$. Hence $M_n(k)\subset J$. But why is $J$ containing arbitrary $M_n(k)$?

Answer 1

Let $\{A_i\}$ be a directed system of simple rings, and $A=\varinjlim A_i$. Let $\phi_i:A_i\to A$ be the structure maps for the direct limit. Now let $I$ be a nonzero ideal of $A$ and let $0\ne x\in I$. By definition of direct limit there $i$ and $0\ne x_i\in A_i$ such that $\phi_i(x_i)=x$. Now since $A_i$ is simple, $A_i=\langle x_i\rangle $, the two side ideal generated by $x_i$. Hence, $A=\langle 1_A\rangle =\langle \phi_i(1_{A_i})\rangle =\langle \phi_i(\langle x_i\rangle )\rangle =\langle x\rangle \subseteq I$. Thus $I=A$.

Answer 2

Let $\{A_i\}$ be a directed system of simple rings, and $A=\operatorname{colim}_i A_i$. Let $\phi_i:A_i\to A$ be the structure maps for the colimit. Finally, let $I\subseteq A$ be a nonzero two sided ideal. Then it follows that $\phi_i^{-1}(I)$ is a two sided ideal of $A_i$, and hence is either zero or the whole ring. There is either a cofinal subcategory where $\phi_i^{-1}(I)$ is zero, or a cofinal subcategory where it is the whole ring, so the ideal itself is either the whole ring or zero.