Consider the linear action of of $SL(n)$ on $n$ by $m$ matrices $F^{n\times m}$ by left-multiplication. Equivalently, $SL(n)$ acts on a set of $m$ vectors from $F^n$ by simultaneous left-multiplication. Where can I find some results on the polynomial invariants of this action? Or, probably, those follow immediately from some other well-known results?
The case $m=1, n>1$ is trivial: there are no invariants, since any vector can be mapped to any vector by an element of $SL(n)$.
In the case $m=1, n=1$ the only invariant is the single vector coordinate, since $SL(1)$ is the trivial group.
In case $m=2, n>2$ we could rotate & scale the space so that the two vectors coincide. Thus, again, no invariants.
In case $m=2, n=2$ the determinant would be a non-trivial invariant.
In general, I suspect that for $n>m$ there are no invariants (intuitively, we can use the extra dimension to rotate & scale everything). For $n=m$ I suspect the determinant to be the only invariant (any matrix $A$ can be put to a diagonal form $\operatorname{diag}(\det A, 1, 1, \dots, 1)$ using left-multiplication by $SL(n)$).
However, I am completely lost in the most interesting case of $n<m$, e.g. $n=2, m=3$.
The answer is contained in CLASSICAL INVARIANT THEORY A Primer by Hanspeter Kraft and Claudio Procesi, section 8.4 The First Fundamental Theorem for $SL_n$. As Omnomnomnom correctly suspected in the comments, the invariant ring is generated by all $n \times n$ minors, i.e. determinants of any $n$ out of $m$ vectors. In particular, there are no invariants in case $n > m$.