1
$\begingroup$

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., $$(x(x-1),xy,xz,y(x-1),y^2,yz,z(x-1),yz,z^2)\not=(0).$$

$\endgroup$
  • 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 agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.