# Finding nilpotent elements in a quotient ring.

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.

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

• 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? – aleio1 Jun 26 '16 at 15:43
• Yes, the intersection of the latter is $(1)\times(6)$. It is more difficult to show that $(x)\cap (x^3-3)=(x^2)$.. – Joel92 Jun 26 '16 at 15:49
• But is not $(x)\cap(x^3-3)=((x)(x^3-3))$? – aleio1 Jun 26 '16 at 16:06