Show that the zero ideal is a product of maximal ideals (not necessarily distinct) in the ring $k[x,y,z]/(x(x-1),y^2,z^3)$.

I tried using Nullstellensatz and then the 4th Isomorphism Theorem, to get that the maximal ideals in this ring are of the form $(x-a_1,y-a_2,z-a_3)$ where $a_i\in k$. Then essentially I should get a product of these ideals to be equal $(x(x-1)\cdot p_1,y^2\cdot p_2,z^3\cdot p_3)$ where $p_i\in k[x,y,z]$, but I always end up with unwanted elements, e.g., $$(x(x-1),xy,xz,y(x-1),y^2,yz,z(x-1),yz,z^2)\not=(0).$$

  • 3
    $\begingroup$ Maybe you just need to add more ideals. I think you need two different ideals, and a product of eight ideals. $\endgroup$ – Tobias Kildetoft Nov 8 '18 at 18:52
  • $\begingroup$ Did you forget to mention that $k$ is algebraically closed? $\endgroup$ – user26857 Nov 8 '18 at 19:10
  • 1
    $\begingroup$ Anyway, the maximal ideals of your ring are $m_1=(x,y,z)$ and $m_2=(x-1,y,z)$ no matter if $k$ is algebraically closed or not. Now let's see what is $m_1m_2$ and what you need in order to get $0$. $\endgroup$ – user26857 Nov 8 '18 at 19:11
  • 1
    $\begingroup$ @user26857 Ahh, true, the appeal to Nullstellensatz does use it. $\endgroup$ – Tobias Kildetoft Nov 8 '18 at 19:14
  • 2
    $\begingroup$ Multiplying again is not actually that bad, since more and more things will disappear. But as I said, I think you need a product with eight factors. $\endgroup$ – Tobias Kildetoft Nov 8 '18 at 19:36

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.