Is localization of finite algebra also finite? 
Let $A \to B$ a finite algebra, i.e., $B$ is finitely generated as $A$-module. If $S \subseteq B$ is a multiplicative subset, then $S^{-1}B$ is also finite $A$-algebra?

If not, how about $A=K$ a field and $S=\{1,b,b^2,\cdots\}$ for some $ 0 \neq b \in B$?
(In this special case I guess it is true and ask for making sure.)
In fact, I wanted to show the following:
Let $K$ be a field, $I \subseteq K[x]$ a nonzero ideal(so that $K[x]/I$ is finite dimensional), and $0 \neq b\in K[x]/I$ be a nonzero element. Then $(K[x]/I)[b^{-1}]$ is also finite dimensional.
Is general case true?
Can anyone comment?

If $B$ is a finite dimensional $K$-algebra and $b\in B$ then let $g\in K[x]$ be its minimal polynomial, write $g  = x^r f\in K[x]$ with $f(0)\ne 0$, show that $f(b)\in \ker(B\to B[b^{-1}])$ and that $b\in B/(f(b))^\times$, whence $B[b^{-1}]=B/(f(b))$ and  $\dim_K S^{-1} B\le \dim_K B$.
