Krull dimension of a $\Bbbk$-algebra

Question

Given an ideal, $\mathfrak{a}\subseteq \Bbbk[x_1,\ldots,x_n]$, where $\Bbbk$ is a field. Let the maximal set of indeterminates independent modulo the ideal $\mathfrak{a}$ be of cardinality $k$. Therefore the dimension of $\Bbbk[x_1,\ldots, x_n]/\mathfrak{a}$ is $k$. How can we construct the chain of prime ideals in $\Bbbk[x_1,\ldots,x_n]/\mathfrak{a}$ of length $k$?

Answer 1

Without loss of generality, assume that $S=\{x_1,\ldots, x_k\}$. Then for any $x_m, m>k$, we have $f_m(x_1,\ldots, x_k, x_m)=0$ in $R=\mathbb{k}[x_1,\ldots, x_n]/\mathfrak{a}$, for some non-zero polynomial $f$ in the polynomial ring. Let $a_m=a_m(x_1,\ldots, x_k)$ be the leading coefficient of $f_m$ as a polynomial in $x_m$. Let $s=\prod_{m>k} a_m$. Then, you have $\mathbb{k}[x_1,\ldots x_k]_s\subset R_s$ an integral extension and so by going up theorem, you can find a prime chain of length $k$ in $R_s$, which when pulled back to $R$ gives you such a chain in $R$.