Question: "Let's denote $g' = g+ J$ for some $g\in R:=k[T_1, \dots T_l]$. Consider multiplicative closed set $D:=\{ 1,g ,g^2 ,\dots \}$ and its image $\bar{D} $ in $B= R/J$. Then we have $D^{-1}R/D^{-1}J \cong \bar{D}^{-1}(B=R/J) = B_{g'}$ ( C.f. Localization commutes with quotient ). Since $D^{-1}R = R_g = R[X]/(gX-1) = k[T_1,\dots T_l , X]/(gX-1)$ is finitely generated over $k$ and the homomorphic image of finitely generated algebra is itself finitely generated, $B_{g'}$ is finitely generated over $k$. And I guess that one of finite generators for $B_{g'}$ is $\{ (T_1 + (gX-1))+ D^{-1}J , \dots (T_l + (gX-1))+ D^{-1}J, (X + (gX-1))+ D^{-1}J \}$
True? Or is there more simpler form of generators?
Can anyone help?"
Answer: Parts of this has been proved elsewhere on this site, here is another elementary and explicit proof of this fact that does not use the universal property of localization:
Lemma: There is an isomorphism of rings $R_g \cong A:=R[t]/(tg-1)$ for any $g\in R$.
Proof define a map $\phi': R[t] \rightarrow R_g$ by $\phi'(t):=1/g$. This map satisfies $\phi'(tg-1)=0$ hence we get an induced map $\phi: R[t]/(tg-1) \rightarrow R_g$. Define a map $\rho': R \rightarrow R[t]/(tg-1)$ by $\rho'(a):=a$. Since $tg=1$ in $A$ it follows the element $g$ is a unit, hence there is (by the universal property of localization) a canonical map $\rho: R_g \rightarrow A$. Let us define this explicitly and prove it is well defined.
Define $\rho(a/g^n):=at^n$. If $a/g^n \cong b/g^m$ there is an integer $k\geq 0$ with
$$ g^{k+m}a = g^{k+n}b $$
in $R$. Using $\rho$ we get a similar equality in $A$. Since $g$ is a unit in $A$ it
follows
$$ t^k g^{k+m}a= t^k g^{k+n}b $$
hence ( in $A$ )
$$ g^m a = g^n b.$$
we get
$$t^m g^m a = t^m g^n b$$
hence $a=g^nt^m b$.
We get
$$t^n a= t^ng^n t^m b=t^mb$$
and it follows $\rho(a/g^n)=t^na=t^mb=\rho(b/g^n)$.
Hence the map is well defined. You may check $\rho$ and $\phi$ are inverses of each other and it follows $\rho$ is an isomorphism of rings.
Hence for any ring $R$ and any element $g\in R$ it follows $R_g$ is a fintely generated $R$-algebra since $R_g \cong R[t]/(tg-1)$ is generated by one element $\overline{t}$.
As mentioned in the comments: If $R:=k[t]$ and if $S:=R-\{(0)\}$ it follows $S^{-1}(R) \cong k(x)$ is not a finitely generated $R$-algebra for the following reason:
Assume there is a finite set of generators $f_i/g_i \in k(x)$ for $i=1,..,n$. Let $g:=\prod g_i, g(i):=g/g_i$. It follows $f_i/g_i=F_i/g$ with $F_i:=f_ig(i)$. Let $h$ be a polynomial that is relatively prime to $g$. A linear combination
$$ \sum_i a_i(f_i/g_i) =(\sum_i a_iF_i)/g $$
can never equal $1/h$ since $g$ and $h$ are relatively prime. Hence $k(x)$ does not have a finite set of generators over $k[x]$.
@Plantation - you should always define maps between commutative rings "constructively". This means if $A:=k[x_1,..,x_n]/I$ is the quotient of a polynomial ring in a finite set of variables $x_i$ by an ideal $I:=(h_1,..,h_l)$, a map of k-algebras $f:A \rightarrow B$ is defined by giving a set of elements $f(x_i):=b_i \in B$, such that for any element $h \in I$ it follows $h(b_i)=0$ for all $h \in I$.
By this I mean the following: A map $f$ induce a map
$$F: k[x_1,..,x_n] \rightarrow B$$
defined by $F(x_i):=b_i \in B$. Since the map $F$ sends the ideal $I$ to zero in $B$ it follows
$$(*) F(h_j(x_1,..,x_n)):=h_j(b_1,..,b_n)=0$$
for the generators $h_1,..,h_l$ of the ideal $I$.
Hence you must define a map $F$ such that the equations $(*)$ are fulfilled.
