When affine variety is complete? How to prove that an affine variety $X$is complete only if $\dim X=0$? It is clear that in this case $X$ must be a single point. But I don't known why its dimension should be zero. Could anyone help me? Thanks a lot!
If an affine variety $X\subset \mathbb A^n_k$ has more than one point (which is certainly the case if  $dim (X)\gt 0$) some coordinate function  $x_i$ is not constant on $X$ and we thus get a non constant regular function $x_i\in \mathcal O(X)\setminus k$.
On the other hand,  every regular function on a complete variety $Y$ is constant (in other words $\mathcal O(Y)= k$) and thus our positive-dimensional affine variety $X$ cannot be complete.  

Edit
In the above answer, set in an elementary context not assuming schemes,  I Assumed $X$ irreducible and $k$ algebraically closed.
At a more advanced level one can only say that $\operatorname {dim} _k \Gamma(X, \mathcal O_X) \lt \infty $ for a scheme $X$ proper over a ( maybe not algebraically closed) arbitrary field $k$  . Thanks to @Alex for his comment which caused  this edit.
If I understand properly you are asking why an affine variety which is a single point has dimension zero? 

It is very clear from the definition of dimension: A variety X has dimension n is there exists an chain of irreducible closed subsets of X $Z_0 \subsetneq Z_1\subsetneq...\subsetneq Z_n \subsetneq X$. 

If X has to have dimension 1 then  u need at least  two points because you need two irreducible closed subsets of  X one containing the other properly.