Construct nonzero Derivation of Commutative Ring Given any ring $R$ and an additive map $d:R \to R$ such that $d(xy)=d(x)y+xd(y)$, then $d$ is called derivation in ring $R$. For any $r\in R$, take $d_{r}(x)=[x,r]=xr-rx$ is a derivation and named inner derivation.
As we know, the inner derivation of any commutative ring is zero derivation, hence the only thing we could get nonzero derivation is by outter derivation. But we know too for $\mathbb{Z}, \mathbb{Q}, \mathbb{R} ,\mathbb{Z}n$, they don't have such derivation (nonzero derivation). 
Taking polynomial ring $\mathbb{R}[x]$ and $\mathbb{Z}[x]$, and define mapping $d_{1}:\mathbb{Z}[x]\rightarrow \mathbb{Z}[x]$ and $d_{2}:\mathbb{R}[x]\rightarrow \mathbb{R}[x]$ by $d_1(p(x))=p'(x)$ and $d_{2}(q(x))=q'(x)$ for all $p(x)\in\mathbb{Z}[x]$ and $q(x)\in\mathbb{R}[x]$, it can be show that they are derivation. Is there any other derivation on them? 
and how about any other commutative ring (not polynomial form) that have nonzero derivation? (please give me example) 
 A: For the derivations of a polynomial ring $S[x_1,\ldots ,x_n]$ we have the following result; ${\rm Der}_S(S[x_1,\ldots ,x_n])$ is a free $S[x_1,\ldots ,x_n]$-module on the basis
$$
\frac{\partial}{\partial x_1},\ldots , \frac{\partial}{\partial x_n}.
$$
So for every $f_1,\ldots ,f_n\in S[x_1,\ldots ,x_n]$ there exists a unique derivation
$d$ such that $d(x_i)=f_i$.
Besides polynomial rings, also power series rings $S[[x_1,\ldots ,x_n]]$ are commutative rings with nonzero derivations (among many other examples).
A: Dietrich describes actually the "recipe" for building derivations on a polynomial ring. So for a commutative ring $K$ and a polynomial $f \in K[x]$ you define $d=f\frac{\partial}{\partial x}$, which means that $d(h):=f\frac{\partial h}{\partial x}$, for all $h\in K[x]$. Conversely, any K-linear derivation $d$ of $K[x]$ is of that form, as you easily can show by induction and where $f=d(x)$. Dietrich also told you that this more generally is possible for a polynomial (or power series) ring in multiple variables.
So for an easy example, define $d(x)=x$, which means that $d\left(\sum_{i=0}^n \lambda_i x^i\right)=\sum_{i=1}^n i\lambda_i x^i$ for any $\sum_{i=0}^n \lambda_i x^i\in K[x]$. This is a derivation, that you might not find interesting, but if you construct the "Ore extension" $R=K[x][y;d]$, where $yx=xy+d(x)=xy+x$, then you see that $[y,x]=x$ and you can easily show that $R$ is the enveloping algebra of the $2$-dimensional non-abelian Lie algebra.
Of course, if you have a derivation $d$ on a ring $R$, then you can try to extend $d$ to a possible quotient ring $R/I$. The canonical way would be to define $d(r+I)=d(r)+I$, but to be well-defined, you need that $d(I)\subseteq I$. So here is a concrete example: let $R=K[x,y,z]$, with $K$ a field and $I=\langle x^2+y^2+z^2-1\rangle$. If we define a derivation $d$ on $R$ by
$$ d = yz\frac{\partial}{\partial x} + xz\frac{\partial}{\partial y} -2xy\frac{\partial}{\partial z},$$ then $d(x^2+y^2+z^2-1)=0$ and $d$ extends to a derivation of $K[x,y,z]/\langle x^2+y^2+z^2-1\rangle$, which is the coordinate ring of the sphere. Derivations on the coordinate ring of an algebraic variety correspond in general to vector fields on that variety (see here).
Derivations on a commutative domain also extend to its field of fractions, by the usual formula, i.e. if $d$ is a derivation on an integral domain $R$ with field of fraction $Q=\{ab^{-1} \mid a,b\in R, b\neq 0\}$, then define $d(ab^{-1}):=(d(a)b-ad(b))b^{-2}$. Hence if $d$ is the derivation on $K[x]$ defined by $d(x)=x$, with $K$ a field, then $d$ extends to a derivation of the field $K(x)$, where $d(1/x)=-1/x$. The derivation $d$ is $K$-linear, but of course not $K(x)$-linear.
The book "Polynomial derivations and their rings of constants" by Nowicki is a nice book about derivations on polynomial rings (see here).
