How can we find an example of derivation on commutative rings? $R$ is an commutative ring and an additive map $d:R \to R$ such that $d(xy)=d(x)y+xd(y)$. I need examples of derivations on basic commutative rings, for example $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{Z}_p$, $\mathbb{Q}$. If we take $a \in R$ and all $ x\in R$, then $[a,x]=ax-xa=0$ because $R$ is commutative, so all inner derivation is zero. So I don't want to find zero derivation as a example.
 A: Let me expand on the comment Emrah Korkmaz made:

There are no nonzero derivations on $\mathbb Z, \mathbb Z_p, \mathbb Q$.

First of all, for any derivation $d$, $d(1) = 0$. This follows from
\begin{equation}
d(1) = d(1 * 1) = d(1) + d(1).
\end{equation}
Since $d$ is additive we have therefore on any ring that the value of $d$ on the integers vanishes.
Let us take a look on fractions:
Suppose $b$ is a non-zero divisor and invertible. We have
\begin{equation}
0 = d(1) = d(\frac{b}{b}) = b d( \frac{1}{b} ) + d(b) \frac{1}{b}
\end{equation}
therefore $d(\frac{1}{b}) = - \frac{1}{b}^2 d(b)$.
From $ d( \frac{a}{b} ) = a d( \frac{1}{b} ) + d(a) \frac{1}{b} = - \frac{a}{b^2} d(b) + d(a) \frac{1}{b}$ we get that $d$ even vanishes on all fractions of integers.
This settles the case for the above rings.
Now, there could still be a nonzero derivation on $\mathbb R$, but since its value on the rationals must vanish it has to be discontinuous. I assume proving the existence of one (if it can be done) is highly nontrivial, and requires the axiom of choice.
