For the quotient of polynomial rings over complex number field,
its global dimension is finite is equivalent to it is domain.
is this true?
For the quotient of polynomial rings over complex number field, its global dimension is finite is equivalent to it is domain. is this true? 


No, but Serre proved that for noetherian local rings having finite global dimension is the same as being regular. So, choose any nonregular local ring which at the same time an integral domain such as the localization of the cuspidal curve at the origin: $$ k[x,y]_{(x,y)}/(y^2x^3)k[x,y]_{(x,y)} $$ 


It is true. If global dimension of $R$ is finite say $ = r$ then for any maximal ideal $\mathfrak{m}$ we have homological dimension of $R_\mathfrak{m}$ is finite and $\leq r$. By Serre's result $\dim R_\mathfrak{m}$ = global dimension of $R_\mathfrak{m}$. It remains to note that there exists at least one maximal ideal $\mathfrak{m}$ with global dimension of $R_\mathfrak{m}$ equal to $r$. This is so since $$ r = \max \{ hdim \ R_\mathfrak{m} \ \mathfrak{m} \ \text{a maximal ideal of $R$} \}.$$ 

