Generators of a finite intersection of maximal ideals In the polynomial $k$-algebra $A$, in $d$ variables over a field $k=\bar k$, is it true that any finite intersection of maximal ideals is generated by $d$ elements?
If $I=m_1\cap\dots\cap m_n$, with every $m_i$ maximal ideal, of course it is generated by at least $d$ elements, since the minimal primes over $I$ are the $m_i$. My problems are in the other direction; first I thought of Nakayama's lemma, but here doesn't seem useful, and then to use that $A/I$ is a finite $k$-module (by Chinese remainder theorem), that wasn't useful either. I also know that every $m_i$ must be a minimal prime over an ideal generated by $d$ elements, but also this fact doesn't say much to me, because $I$ is not the only ideal over which (any) $m_i$ is minimal. Can you give a hint? Thanks
 A: We can prove this by induction on $d$. The case $d=1$ is clear, because a polynomial ring in 1 variable is a Principal Ideal Domain.
The $n$ maximal ideals correspond to $n$ points $P_1,P_2,\dots,P_n$ in $k^d$. Let $Q_i$ be the projection of $P_i$ onto the first $d-1$ coordinates, and let $r_i$ be the $d$-th coordinate of $P_i$ so that $P_i=(Q_i,r_i)$.
After a linear change of coordinates (which corresponds to a ring automorphism of the polynomial ring), we may assume that $Q_1,\dots,Q_n$ are distinct. By the induction hypothesis, there exist $d-1$ polynomials $f_1,f_2,\dots,f_{d-1}$ in the first $d-1$ variables that generate the vanishing ideals of the projections of the $n-1$ points. Using polynomial interpolation, we can construct a polynomial $g=g(x_1,x_2,\dots,x_{d-1})$ such that $g(Q_i)=r_i$ for all $i$. Now take $f_{d}=x_{d}-g(x_1,x_2,\dots,x_{d-1})$.
The zero set of $f_1,f_2,\dots,f_d$ is the set $\{P_1,P_2,\dots,P_{d-1}\}$. By induction,
$(f_1,f_2,\dots,f_{d-1})$ is a reduced ideals,
and $f_d$ intersects this transversally, so that $(f_1,f_2,\dots,f_d)$ is reduced as well.
So $(f_1,f_2,\dots,f_d)$ is intersection of the maximal ideals.
