Ring of matrices has no nontrivial ideals It is a theorem that a commutative ring is a field if and only if it has no nontrivial ideals. Clearly this does not hold in the noncommutative case. I am trying to show for instance that the ring of $n $ x $n$ matrices in a given field has no nontrivial ideals despite it not being a field. Now this is the part I am not completely sure about. My understanding is that the kernel of a ring homomorphsim is an ideal and viceversa, every ideal of a ring is the kernel of some ring homomorphsim. So my approach has been to show that there are no homomorphisms in the ring of matrices that do not have a nontrivial kernel. However I am stuck on this step. Any thoughts?
 A: A theorem of linear algebra reads:

A matrix over a field $A\in\mathcal M_{n\times n}(\Bbb K)$ has rank $k$ if and only if there exist $C,D\in \operatorname{GL}(n,\Bbb K)$ such that $$C\cdot A\cdot D=\left(\begin{array}{c|c} I_k & 0\\\hline 0&0\end{array}\right)$$

If $\mathfrak a\subseteq \mathcal M_{n\times n}(\Bbb K)$ is a two-sided ideal and $0\ne A\in\mathfrak a$, then $\operatorname{rk}A=k\ne0$.
Using the result above, $\mathfrak a$ must contain all the matrices of rank $k$, hence $\left(\begin{array}{c|c} I_k & 0\\\hline 0&0\end{array}\right)\in\mathfrak a$.
But then, observing that $$\left(\begin{array}{c|c|c} 1&0&0\\\hline 0&0 & 0\\\hline 0&0&0\end{array}\right)\left(\begin{array}{c|c|c} 1&0&0\\\hline 0&I_{k-1} & 0\\\hline 0&0&0\end{array}\right)=\left(\begin{array}{c|c|c} 1&0&0\\\hline 0&0 & 0\\\hline 0&0&0\end{array}\right)$$
You get that $\mathfrak a$ contains a matrix of rank $1$. By the above lemma, it must contain all the matrices of rank $1$. Since every $n\times n$ matrix can be written as a linear combination of matrices of rank $1$, it must hold $\mathfrak a=\mathcal M_{n\times n}(\Bbb K)$.
A: An elementary proof could go like this.
Suppose $I$ is a nonzero ideal, and let $0 \ne a \in I$. Thus there are indices $s, t$ such that $a_{st} \ne 0$. Let $e_{ij}$ be the matrix with $1$ in the $i, j$ position, and $0$ elsewhere. Then for each $i$
$$
a_{st}^{-1} e_{is} a e_{ti} = e_{ii} \in I,
$$
so that the identity $\sum_{i=1}^{n} e_{ii}$ belongs to $I$.
