Let $F$ be a field and $A$ an $F$-algebra. (And assume that $A$ is finite dimension over $F$ if necessary.) A textbook says that $A$ is simple if it has no proper two-sided ideals.
To understand this definition well, I'm looking for an example of $F$-algebra which has no proper two-sided ideals but has one-sided ideal(s). But I cannot find out such one. Is there such example?
Thank you.
On
Tobias's example of matrix rings over fields is definitely the most straightforward example! The only way I can contribute is to provide a different and exotic one.
There exists a simple domain which is not a division ring: Let $F$ be a field of characteristic zero, and let $F(x)$ be the ring of rational functions with indeterminate $x$, and take $F[t,x]$ to be the ring of differential polynomials in $t$ over $F(x)$, that is, the $y\in F(x)$ such that $yt=ty+\frac{dy}{dx}$.
I originally learned about this in Lam's Lectures on modules and rings in the introductory examples section, and I know it appears in the article Some aspects of ring theory by Goldie. (Bull. London Math. Soc. (1969) 1 (2): 129-154.)
Consider the algebra of $n\times n$ matrices over $F$. This clearly has one-sided ideals (since the left or right ideal generated by an element is the entire ring iff that element is left or right invertible).
On the other hand, it is a classical exercise to prove that this algebra has no non-trivial proper twosided ideals.