Suppose $A = K[x_1, ..., x_n]$ ($n \geq 2$) is the ring of polynomials in $n$ variables over the field $K$. Let $I$ be a proper ideal of $A$. Show that if $A/I$ is Artinian and $I = (f_1, ..., f_n)$ with $f_i \in A$, then $n = \mu(I)$. ($\mu(I)$ denotes the minimum number of $I$ generators).
attempt: the inequality $\mu(I) \leq n$ is obvious. For the other inequality, I tried the following: $\mu(I) \geq ht(I) = ht(P)$, for some minimal prime $P$ of $I$. (Here $ht$ denotes the height of the ideal). By the one-to-one correspondence of ideals of $A/I$ and ideals of $A$ that contain $I$, cheho that $ht(I) \geq 1$, which didn't help much.
We know that $ht(I) = \min \{ht(P)|\ P\ is\ a\ minimal\ prime\ of\ I\}$. Consider then $P$ a minimal prime of $I$ with $ht(I)=ht(P)$ and in this case, $I \subset P$. From the one-to-one correspondence between the ideals of $A$ containing $I$ and the ideals of $A/I$, it follows that $P/I$ is prime and therefore maximal in $A/I$, since $A /I$ is Artinian. Again by one-to-one correspondence $P$ is maximal in $A$, where $A/P$ is field. Hence, how
$$n = \dim(A/P) + ht(P) $$
we see that $ht(I) = ht(P) = n$. Therefore, since $A$ is Noetherian, from Krull's general theorem, $n = ht(I) \leq \mu(I).$