Prove that $M_{nn}(\mathbb{F})$ is a simple algebra, that is, prove that the only ideals in $M_{nn}(\mathbb{F})$ are $\{$o$_{nn}\}$ and $M_{nn}(\mathbb{F})$
There is a definition says that " An ideal in an algebra A with identity is a vector subspace $I$ of $A$ which further stisfies:if $r\in A$ and $b\in I$, then $rb\in I$ and $br \in I$. An algebra A is said to be simple if the only idelas in A are A and $\{$0$_{A}\}$
would someone please write me up a proof or help me out ?
Let $E_{ij}$ be the matrix which has a 1 in position $i,j$ and zeros everywhere else. To show that an ideal $I$ is equal to $M_n(\mathbb{F})$, it's enough to show that $I$ contains $E_{ij}$ for some $i,j$, because then we can multiply on the left and right by permutation matrices to show that $I$ contains $E_{ij}$ for all indices $i$ and $j$.
So let $I$ be an ideal of $M_n(\mathbb{F})$, and suppose that $I$ contains a non-zero matrix $A=(a_{ij})$. Then there are indices $i,j$ for which $a_{ij}\neq 0$.
If we multiply $A$ on the left by $E_{1i}$ and on the right by $E_{j1}$, the result is the matrix $a_{ij}E_{11}$. And multiplying this matrix by the scalar matrix $a_{ij}^{-1}I_n$ shows that $I$ contains $E_{11}$.
Let us prove that if $\mathscr{E}$ is a non-zero ideal of $M_{nn}(\mathbb{F})$ then $\mathscr{E}=M_{nn}(\mathbb{F})$:
Let us first show that $\mathscr{E}$ contains at least one invertible matrix: Since $\mathscr{E}\neq \{0\}$, then $\mathscr{E}$ contains at least one non-zero matrix. Let it be $A$, and let $x\in\mathbb{F}$, $x\neq 0$ a non-zero element of $A$ in the $(i,j)$ entry. If we denote by $e_{kl}$ the matrix having $1$ in the $(k,l)$ entry and $0$ everywhere else, then it is easy to show that the matrix $$e_{ki}Ae_{jk}$$ has $x$ in the $(k,k)$ entry and zero everywhere else. By definition $e_{ki}Ae_{jk}\in\mathscr{E}$. Let $k=1,2,...,n$, where $n$ is the dimension of the v.s. $V$. Since $\mathscr{E}$ is closed under addition, we can sum $$E=\Sigma_{k=1}^{n}e_{ki}Ae_{jk}=xI\in\mathscr{E}$$ But $x$ is invertible (since it is an element of a field) and let $x^{-1}$ its inverse. $E\in\mathscr{E}$ and thus $x^{-1}E=I\in\mathscr{E}$. But if an ideal $\mathscr{E}$ contains the unit matrix $I$ then it is equal to the whole ring since for any $S \in M_{nn}(\mathbb{F})$ $$I*S=S*I=S \in \mathscr{E}$$ Thus $$\mathscr{E}=M_{nn}(\mathbb{F})$$ Consequently there are no non-trivial ideals in $M_{nn}(\mathbb{F})$ or equivalently: $M_{nn}(\mathbb{F})$ is a simple ring.
