Every simple ring is a division ring and vice versa.

Question

My definition of simple ring is:

A ring $R$ where $\{0\}$ is a maximal ideal, is called a simple ring.

Now, assume $R$ is not trivial. Wherever I write $I \subset R$, I mean that $I$ is an ideal.

$R$ simple ring $\iff \{0_R\}$ is a maximal ideal of $R$

$ \iff\forall I \subset R: I \neq R: (\{0_R\} = I \lor \{0_R\} \not\subset I)$

$\iff \forall I \subset R: I \neq R: \{0_R\} = I$

$\iff$ the only ideals in $R$ are $\{0\}$ and $R$

$\iff R$ is a division ring

Where is my mistake? I just 'proved' that $R$ simple ring $\iff R$ division ring and this result is false.

Answer 1

The mistake is in the last step. $R$ is a division ring if and only if the only one-sided (left or right) ideals of $R$ are $\{0\}$ and $R$.

If $D$ is a division ring, it is not too difficult to show that the matrix ring $M_n(D)$ is simple. You can easily see that any non-zero ideal will have matrices with a single non-zero entry. Matrix products from left/right then move that non-zero entry to any location you want, and you are done.

Observe that such matrix rings have non-trivial one-sided ideals. The set consisting of matrices with certain rows (resp. columns) all zeros form a right (resp. left) ideal.