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?
 A: I think this seldom happens. Let $B$ be a finite $A$-algebra and $b\in B$. If $x_1,\dotsc,x_n\in B$ generate $B$ as an $A$-module and $B[1/b]$ is a finitely generated $A$-module, then there is a $d\ge0$, such that $x_1,\dotsc,x_n,\frac{x_1}{b},\dotsc,\frac{x_n}{b},\dotsc,\frac{x_1}{b^d},\dotsc,\frac{x_n}{b^d}\in B[1/b]$ generate $B[1/b]$ as an $A$-module. That means there are $a_{ij}\in A$ for $i=1,\dotsc,n,\,j=0,\dotsc,d$, such that $$\frac{1}{b^{d+1}}=\sum_{j=0}^d\sum_{i=1}^na_{ij}\frac{x_i}{b^j}\qquad\text{ in $B[1/b]$}.$$
Setting $c=\sum_{j=0}^d\sum_{i=0}^na_{ij}x_ib^{d-j}\in B$, this implies that the image of $c$ under the canonical map $B\rightarrow B[1/b]$ is $1/b$. It follows that the canonical map $B\rightarrow B[1/b]$ is surjective. Conversely, if it surjective, $B[1/b]$ is obviously a finite $A$-algebra since $B$ is (the map is automatically $A$-linear, since it is a ring homomorphism). Note in particular that $B[1/b]$ is a finite $A$-algebra if and only if it is a finite $B$-algebra by these considerations, so there's no harm in assuming $A=B$ forthcoming. Let's analyze some special cases to see when this can happen.
If $B$ is an integral domain and $b\neq0$, the canonical map $B\rightarrow B[1/b]$ is always injective, so it is surjective if and only if it an isomorphism, which is the case if and only if $b\in B^{\times}$. So inverting something that is neither zero nor a unit always takes you out of the finite world.
If $B$ is a zero-dimensional local ring, every element of $B$ is either nilpotent or a unit. In the first case, $B[1/b]=0$, so $B\rightarrow B[1/b]=0$ is surjective. In the second case, $B[1/b]=B$ and $B\rightarrow B[1/b]=B$ is the identity, so surjective as well.
Now, if $B\cong B_1\times\dotsc\times B_n$ decomposes as a finite product of rings and $b\in B$ corresponds to $(b_1,\dotsc,b_n)\in B_1\times\dotsc\times B_n$, then we obtain a commutative diagram
$$\require{AMScd}
\begin{CD}
B @>{\sim}>> B_1\times\dotsc\times B_n\\
@VVV @VVV\\
B[1/b] @>{\sim}>> B_1[1/b_1]\times\dotsc\times B_n[1/b_n]
\end{CD},$$
where the vertical maps come from the canonical maps. It follows that the canonical map $B\rightarrow B[1/b]$ is surjective if and only if each of the canonical maps $B_i\rightarrow B_i[1/b_i]$ is surjective for $i=1,\dotsc,n$. Let's use this to generalize the previous two cases.
If $B$ is isomorphic to a finite direct product of integral domains $B_1\times\dotsc\times B_n$ and $b\in B$ corresponds to $(b_1,\dotsc,b_n)\in B_1\times\dotsc\times B_n$, the map $B\rightarrow B[1/b]$ is surjective if and only if $b_i=0$ or $b_i\in B_i^{\times}$ for each $i=1,\dotsc,n$. As an exercise, you can see that this is equivalent to the existence of a non-zero-divisor $c\in B$, such that $bc$ is idempotent.
If $B$ is zero-dimensional and has finite spectrum, then $B$ is canonically isomorphic to the product of its localizations, i.e. it is isomorphic to a finite product of zero-dimensional local rings. The previous observations combined imply that the canonical map $B\rightarrow B[1/b]$ is always surjective in this case. This applies in particular to Artinian rings such as non-trivial quotients of Dedekind domains. Specifically, it also implies to $B=K[X]/I$ for $K$ a field and $0\neq I\subseteq K[X]$ a non-trivial ideal, since $K[X]$ is a PID.
A: 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  $\dim_K S^{-1} B\le \dim_K B$.
