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?

  • $\begingroup$ Ideals that have the Grobner basis in that form are zero-dimensional and I am interested in those ideals. I am thinking if that will help in determining the structure of prime ideals. What I mean by " Ascending chain of prime ideals is zero " is how do I show that the krull dimension is zero using the definition of prime ideal chain and not equivalent definitions like transcendence degree. $\endgroup$
    – Zoey
    Jun 17, 2015 at 8:11
  • $\begingroup$ Related: math.stackexchange.com/questions/340527/… and math.stackexchange.com/questions/368448/… $\endgroup$
    – user26857
    Jun 17, 2015 at 8:56

1 Answer 1


Let $\mathfrak{a}$ be an ideal in $\mathbb{k}[x_1,\dots,x_n]$ and $\{g_1,\dots,g_t\}$ a Gröbner basis of $\mathfrak{a}$. For each $i=1,\ldots,n$ there exists $j\in\{1,\dots,t\}$ such that $\mathrm{lm}(g_j)=x_i^{\nu_i}$ for some $\nu_i\in\mathbb{N}$. Show that $\dim \Bbbk[x_1,\dots,x_n]/\mathfrak{a}=0$.

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.

  • $\begingroup$ thank you! what happens if for all $i$ except one $lm(g_j) = x_i^{\nu_i}$. This means the $k$-algebra is of dimension $1$. Then we cant show P is maximal. $\endgroup$
    – Zoey
    Jun 17, 2015 at 8:59
  • $\begingroup$ @Anup Then there is a chain of prime ideals of length one. In this case the Krull dimension of the affine $k$-algebra is one. $\endgroup$
    – user26857
    Jun 17, 2015 at 9:02
  • $\begingroup$ Consider any noetherian ring $A$ and the polynomial ring $A[x]$. Consider the quotient ring, $A[x]/\langle x^2 + 1 \rangle$ . Is the dimension of this quotient ring equal to dimension of $A$? $\endgroup$
    – Zoey
    Jun 17, 2015 at 9:07
  • $\begingroup$ So it cannot be directly extended from fields. What am I missing? If there is a chain of ideals in $A[x]$ given by $P_0[x] \subseteq P_1[x] \cdots, P_r[x] \subseteq P_r + \langle x \rangle$. This chain will still be prime in the quotient ring? $\endgroup$
    – Zoey
    Jun 17, 2015 at 9:14
  • $\begingroup$ Asked : math.stackexchange.com/questions/1328630/… $\endgroup$
    – Zoey
    Jun 17, 2015 at 9:28

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