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?


1 Answer 1


$x,xy∈k[x,y]$ 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).


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