Does a simple $k$-algebra has finite dimension? Let $k$ be a field, $A$ a non-commutative ring over $k$. Suppose $A$ has no two sided ideals. Is there an example for such an $A$ with infinite dimension over $k$? 
 A: There are plenty. If $k$ is a field of characteristic $0$, the (first) Weyl algebra
$$
A = A_1(k) = k\langle x,y \mid yx-xy=1 \rangle
$$
is the perhaps simplest example. An equivalent definitition of $A$ is as ring of skew polynomials: $A=k[x][y;\frac{d}{dx}]$ where $yf = \frac{df}{dx} + fy$ for all $f \in k[x]$.
Note that every element of $A$ can (uniquely) be written in the form $\sum_{i,j\ge 0} a_{i,j} x^i y^j$ with $a_{i,j} \in k$. So $\dim(A)=\infty$.
Now one can check that $yf-fy=\frac{df}{dx}$ and $xf-fx=-\frac{df}{dy}$ for all $f \in A$. Suppose $I$ is a nonzero ideal of $A$ and $f \in I$. By the rules just mentioned, then also all derivatives of $f$ are contained in $I$. In particular, differentiating in a way that only the lead term of the original element remains, $I$ contains a nonzero constant, hence a unit. Thus $A$ is simple.
A similar argument works for the $n$-th Weyl algebra. There are also characterizations of when a skew polynomial ring of the form $R[x;\delta]$ with a derivation $\delta$ where $R$ is a $\mathbb Q$-algebra, respectively a skew Laurent polynomial ring of the form $R[x,x^{-1};\sigma]$ with an automorphism $\sigma$, is a simple ring. See McConnell, Robson, Noncommutative Noetherian Rings, Thms. 1.8.4 and 1.8.5. This way one can construct further examples.
A: Another example:
Let be the ring of linear transformations $R=End(V_k)$ where $V$ is a countably infinite dimensional $k$-vector space. It is known that this ring has exactly one nontrivial twosided ideal, namely the transformations with finite dimensional image.
$S=R/I$ is obviously a simple ring. It is also known that $R\cong R\times R$ as $R$ modules. 
Now given matrices $A\in M_{1,2}$, $B\in M_{2,1}$ realizing the isomorphism (that is, $AB=I_1$ and $BA=I_2$) you can apply the quotient map $R\to R/I$ to get an $S$ linear isomorphism of $S\cong S\times S$. But this is impossible if $S$ is finite dimensional as a $k$ algebra. If the left side had dimension $n$, then the right side would have to have dimension $2n$, but then they wouldn't be isomorphic.
