# Showing that the ring of $n\times n$ matrices has exactly two 2 sided ideals, even though it is not a division ring

Show the ring $A=\mathrm{Mat}(F)$ has exactly two 2-sided ideals, even though it is not a division ring.

$F$ is any field, and $\mathrm{Mat}(F)$ is all $n\times n$ matrices with elements of $F$ as entries. I don't understand how this is true. If you have a $2 \times 2$ matrix, and you have only 1 entry being nonzero (say $e_{11}$, the top left entry), wouldn't the ideal generated by this element be only this matrix multiplied by some coefficient? Also, why is it not a division ring even though it has two 2-sided ideals (i know its not a division ring since not all matrices are invertible, but what condition is not met?)

• the ring $\,A\,$ is simple in this case, and the only two-sided ideals are the trivial ones: the zero ideal and the whole ring. – DonAntonio Apr 20 '13 at 0:55

Also, I'm not sure what you're confused by in your second question - there are non-zero elements of $\mathrm{Mat}(F)$ that are not invertible; a division ring is a ring in which every non-zero element is invertible; therefore $\mathrm{Mat}(F)$ is not a division ring.