dimension of quotient by algebraically independent elements Let $f_1,\dots,f_s$ be algebraically independent polynomials of $A:=k[x_1,\dots,x_n]$, $s \le n$. Recall that algebraically
independent means that there is no non-zero polynomial $g \in k[y_1,\dots,y_s]$
such that $g(f_1,\dots,f_s)=0$.
Question 1: What can we say about the Krull dimension of 
$A/(f_1,\dots,f_s)$?
Question 2: In thinking about it, i was misled to conclude that
that the dimension will drop precisely by $s$. The way i thought
about it, was that since $f_1,\dots,f_s$ are algebraically independent,
then i can extend $f_1,\dots,f_s$ to a transcendence basis 
$f_1,\dots,f_s,g_{s+1},\dots,g_n$ of $A$. Then taking the quotient
would "kill" $f_1,\dots,f_s$ and so a transcendence basis of the quotient
would be $\bar{g}_{s+1},\dots,\bar{g}_n$, from where the Krull dimension
of the quotient would simply be $n-s$. Apparently, this argument is
not correct. Any insights as to why is that?

$x,xy∈k[x,y]$ are algebraically independent, and $k[x,y]/(x,xy)=k[x,y]/(x)≃k[y]$ has dimension one. I'm afraid there is not much we can say about that dimension. 
However, a few trivial remarks: modulo $m=(x,y)$ the Jacobian matrix of the above example has rank one, exactly the dimension. I think this can be extended to a more general setting provided the polynomials are in the maximal irrelevant ideal, or $k$ is algebraically closed (this helping us with the form of the maximal ideals).
