A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $\mathfrak p$ is defined to be Krull dimension of the ring $R/\mathfrak p$) and every maximal ideal has the height same as dimension of the ring.

I want to have an example of a Noetherian domain which is not equidimensional.

Note that $R$ cannot be a finite type $k$ algebra, neither it can be local.

Thank you.

  • 1
    $\begingroup$ "An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1." $\endgroup$ – user26857 Nov 18 '18 at 22:22
  • $\begingroup$ Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property. $\endgroup$ – user26857 Nov 18 '18 at 22:24
  • $\begingroup$ @user26857 Ok but this is the definition given in Eisenbud. $\endgroup$ – Rtk427 Nov 19 '18 at 14:07

Your Answer

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

Browse other questions tagged or ask your own question.