Off hand, does anyone know of some useful conditions for checking if a ring (or more generally a semiring) has non-trivial derivations? (By non-trivial, I mean they do not squish everything down to the additive identity.) Part of the motivation for this is that I was thinking about it the other day, and had trouble finding any good example of a semiring with an interesting derivation. 


For example, the multiplicative Banach algebra of positive functions is an algebra of the semifield of nonnegative reals. However, the usual definition for derivative breaks down due to the fact that you can have positive functions with negative slope. So, this leads me to wonder if there are any semirings with derivations at all?


As a related question, is there a known classification of all the derivations for an algebra? It feels like this should be a pretty standard thing, but I don't think I've ever encountered it in one of my courses and my initial googling around was not too successful at finding references.



Necessary and sufficient criteria for non-trivial derivations to exist?
There is a notion of a universal derivation for an algebra. I'll assume
everything is commutative for simiplcitity. If $A$ is a $k$-algebra
($k$ a commutative ring) then there is an $A$-module $\Omega\_{A/k}$,
the module of *Kahler differentials* of $A$ over $k$ and a $k$-derivation
$d:A\to\Omega\_{A/k}$ which is universal for derviations of $A$.
That is, if $\delta:A\to M$ is a $k$-derivation from $A$ to an $A$ module,
then $\delta=f\circ d$ for a unique $A$-module homomorphism $f$.


There is an explicit desciption of $\Omega\_{A/k}$ as $I/I^2$
where $I$ is the ideal in the ring $B=A\otimes\_k A$ which is the kernel
of the map $\mu:B\to A$ with $\mu(x\otimes y)=xy$. Then $d$ maps $x\in A$
to $1\otimes x-x\otimes 1$. So to find a derivation from $A$ with values in
your favourite $A$-module $M$ all one has to do is to find an
$A$-homomorphism from $\Omega\_{A/k}$ to $M$. Of course $\Omega\_{A/k}$
may be hard to determine concretely, and even if that is possible, perhaps
it may not be easy to find a homomorphism from that into $M$. Indeed
using this method may be no easier than finding a derivation directly :-)


For details see the commutative algebra texts by Eisenbud or Matsumura.


