Let $G$ be an irreducible algebraic group over the field $K$ of characterstic 0. Let $A=K[x_1,...,x_n]/I(G)$ be the coordinate ring and $K(X)=Q(R)$ be the quotient field of $A$. (Since $G$ is irreducible, $A$ is an integral domain). The dimesion of $G$ is defined as the transcendence degree of $K(X)$ over $K$.
Let us suppose $G$ is given to be finite. How do we show that dim$G$=0 in this case ?
Please give some hints.
This has nothing to do with algebraic groups or irreducibility: if $X$ is a (non-empty) finite $k$-scheme, then $\dim(X)=0$. You know that $X$ is affine, so $X=\mathrm{Spec}(A)$ for a non-zero finite $k$-algebra $A$, and $\dim(X)=\dim(A)$. A ring is zero-dimensional if and only if every prime ideal in the ring is maximal. If you take a prime $\mathfrak{p}$ of $A$, $A/\mathfrak{p}$ is a finite $k$-algebra which is moreover a domain. What can you say about such a ring? What can you conclude about $\mathfrak{p}$?