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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.

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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.

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    $\begingroup$ Thank you for your answer. I got it. I leave two lemmas for who read this answer: (1) $E_{ij}XE_{kl} = x_{jk}E_{il}$, (2) For a ideal $\mathcal{I}$ of $n \times n$ matrices over $R$, there is the ideal $I$ of $R$ such that $\mathcal{I} = \{\, a_{ij} \mid a_{ij} \in I \,\}$. $\endgroup$ – Orat Apr 19 '13 at 17:29
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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.)

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