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Which are nilpotent elements of $\mathbb{Q}[x]/(x^5-3x^2)\times\mathbb{Z}/(12)$?

I tried to decompose in this way: $$\mathbb{Q}[x]/(x^5-3x^2)\times\mathbb{Z}/(12)\cong\mathbb{Q}[x]/(x^2)\times\mathbb{Q}[x]/(x^3-3)\times\mathbb{Z}/(3)\times\mathbb{Z}/(4)$$ so i thought that nilpotent elements are only: $$(0,0,0,2), (x,0,0,2) \ \ \mbox{and} \ \ (x,0,0,0).$$

I don't know if I am right, because i tried another approach considering the intersetion of all prime ideals of that ring and i don't know to understand if the result is the same.

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  • $\begingroup$ In place of $x$ you can also use $ax$ with any $a\in\Bbb{Q}$ as the first component. $\endgroup$ – Jyrki Lahtonen Jun 26 '16 at 15:45
  • $\begingroup$ Thank you, you are right. Are there any other nilpotent elements? $\endgroup$ – aleio1 Jun 26 '16 at 15:47
  • $\begingroup$ No. Joel's answer gives the argument. $\endgroup$ – Jyrki Lahtonen Jun 26 '16 at 15:49
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The nilpotent elements of the product are obtained as the tuples of the nilpotent elements of single factors.

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  • $\begingroup$ But if i consider prime ideals of that ring I obtain $((x)\times(1)), \ ((x^3-3)\times(1)), \ ((1)\times(2)), \ ((1)\times(3))$. Is their intersection equal to those 3 elements I listed in my question? $\endgroup$ – aleio1 Jun 26 '16 at 15:43
  • $\begingroup$ Yes, the intersection of the latter is $(1)\times(6)$. It is more difficult to show that $(x)\cap (x^3-3)=(x^2)$.. $\endgroup$ – Joel92 Jun 26 '16 at 15:49
  • $\begingroup$ But is not $(x)\cap(x^3-3)=((x)(x^3-3))$? $\endgroup$ – aleio1 Jun 26 '16 at 16:06
  • $\begingroup$ Sorry, I'm clearly wrong. The one you said it's the intersection by Chinese Remainder Theorem. $\endgroup$ – Joel92 Jun 26 '16 at 17:11

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