Let $\mathfrak{a}$ be an ideal in $\mathbb{k}[x_1, \ldots, x_n]$ and a Gröbner basis of the ideal be $\{g_1, \ldots, g_t\}$. For each $i = 1, \ldots,n$, there exists $j \in \{1, \ldots, t\}$ such that $\mathrm{lp}(g_j) = {x_i}^\nu$ for some $\nu \in \mathbb{N}$. How can one show that an ascending chain of prime ideals in the affine $\Bbbk$-algebra, $ \Bbbk[x_1, \ldots, x_n]/\mathfrak{a}$ is zero?
P.S: I know there is the same way which involves showing that its variety is finite and dimension is therefore is zero but is there a way to show that the chain of prime ideals is itself zero?
First notice that $\dim_{\Bbbk}\Bbbk[x_1,\dots,x_n]/\mathfrak{a}<\infty$.
Then let $\mathfrak p/\mathfrak a$ be a prime ideal in $\Bbbk[x_1,\dots,x_n]/\mathfrak{a}$. The quotient $R=\Bbbk[x_1,\dots,x_n]/\mathfrak p$ is also finitely dimensional over $\Bbbk$ (why?).
Now let's prove that $R$ is a field: if $r\in R$, $r\ne0$, then $1,r,r^2,\dots$ are linearly dependent over $\Bbbk$, so there are $a_i\in\Bbbk$ not all zero such that $a_0+a_1r+\cdots+a_mr^m=0$. We may assume $a_0\ne0$ and then $1=r(-a_0)^{-1}(a_1+\cdots+a_mr^{m-1})$ hence $r$ is invertible.
It follows that $\mathfrak p$ is maximal and we are done.