How can we show that $\dim R/p=0\Leftrightarrow p=(x_{1},\ldots,x_{n})\Leftrightarrow R/p\simeq\mathbb{K}$, where $R=\mathbb{K}[x_{1},\ldots,x_{n}]$ is considered graded with standard grading (i.e. $\deg(x_i)=1$) and $\mathbb{K}$ is an arbitrary field of characteristic zero. Also, $p$ is a graded prime ideal of $R$ and $\dim$ is the Krull dimension.

The Krull dimension of a ring $S$, written $\dim S$, is the supremum of the lengths of chains of prime ideals in $R$ (for example, the chain $p_0\subsetneq p_1\subsetneq \cdots\subsetneq p_n$ has length $n$)

I don't know if the following is of any help on this:

If $\dim R/p=0$, then $p$ is the unique graded maximal ideal of $R$. But how can we proceed to show that $p=(x_1,\ldots,x_n)$?

If anyone can help unstump me on this, I'd be grateful.


$R/p$ is an integral domain. If $\dim R/p=0$, then it is a field, so $p$ is maximal. Since $p$ is graded it is contained in $(x_1,\dots,x_n)$, hence equality. The rest is easy.

  • $\begingroup$ You say that since $p$ is graded it is contained in ($x_1,\ldots,x_n$). Why is that? Can you please elaborate? $\endgroup$ – Vagerman Dec 5 '13 at 16:47
  • 2
    $\begingroup$ @Vagerman A graded ideal is generated by homogeneous elements. (If contains a homogeneous element of degree $0$, then the ideal is the whole ring which is not the case for prime ideals.) Furthermore, in $K[x_1,\dots,x_n]$ the ideal $(x_1,\dots,x_n)$ contains all homogeneous elements of degree greater than $0$. $\endgroup$ – user89712 Dec 5 '13 at 17:56
  • $\begingroup$ Thank you so very much! Indeed, your explanation made perfect sense and helped a lot. Thank you again for a very enlightening answer. $\endgroup$ – Vagerman Dec 7 '13 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.