Quick Question:
Let $B$ be some finitely generated $A$-algebra (say A is a domain). Suppose I localize:
If I have some elements in $S^{-1}B$ which are algebraically independent over $S^{-1}A$, can I pull them back to algebraically independent elements (over A) in $B$?
For example, say I have $\frac{b_1}{s_1},...,\frac{b_n}{s_n}$ which are algebraically independent in $S^{-1}B$. Then any nonzero $f(x_1,...,x_n)\in S^{-1}A[x_1,...,x_n]$ has that $f(\frac{b_1}{s_1},...,\frac{b_n}{s_n})\neq 0$. Any polynomial in $A[x_1,...,x_n]$ is a polynomial in $S^{-1}A[x_1,...,x_n]$, but is it necessarily true that $f(b_1,...,b_n)\neq 0$?
I'm starting to think it might not be true, but if not, is there a way I get algebraically independent elements of B over A?
Edit: What if $S^{-1}A=K$ is the field of fractions of $A$? I don't know if that makes a difference.
I think I have it:
Suppose $\frac{b_1}{s_1},...,\frac{b_n}{s_n}\in S^{-1}B$ are algebraically independent over $S^{-1}A$.
Now as a matter of notation, let $I=(i_1,...,i_n)$, $|I|=\sum_{j=1}^n i_j$ and denote $x_1^{i_1}...x_n^{i_n}$ by $x^I$. Take any polynomial $f(x_1,...,x_n) = \sum_{|I|\leq m}a_Ix_I\in S^{-1}A[x_1,...,x_n]$.
If $f(b_1,...,b_n)=0$, then by replacing $b_i$ with $s_i\frac{b_i}{s_i}$, we obtain an algebraic relation $\tilde{f}(\frac{b_1}{s_1},...,\frac{b_n}{s_n})=0$ where the coefficients of $\tilde{f}$ are $s^Ia_I$. Then by the algebraic independence of the $\frac{b_i}{s_i}$ over $S^{-1}A$, we have all the coefficients $s^Ia_I$ of $\tilde{f}$ are zero.
Now it turned out to be important that $A$ is an integral domain and $S$ contains nonzero elements of $A$ so that $s^I\neq 0$ and $s^Ia_I=0\implies a_I=0\implies f=0$ in $S^{-1}A[x_1,...,x_n]$. Therefore, $b_1,...,b_n$ are algebraically independent over $S^{-1}A$, and hence over $A$.