On the nil but not nilpotent ideal counterexample.

Question

In this post, there is a counterexample where a nil ideal (ie every element is nilpotent) is not nilpotent.

More precisely, let's consider the polynomial ring $$R:=\Bbb C[X_1,X_2,X_3,\dots]/\langle X_1^2,X_2^3,X_3^4,\dots\rangle,$$ and set $I:=\langle X_1^2,X_2^3,X_3^4,\dots\rangle$. Then we can consider the ideal $$\tilde I:=\langle X_1+I,X_2+I,X_3+I,\dots\rangle \trianglelefteq R.$$ But I can not see why this is a nil ideal. Normally, we should take an arbitrary element $\bar x$ of $\tilde I$ and show that there exists an $m\in \Bbb Z^+$ such that $\bar x^m=0_R=I$.

Could you please help?

Thanks.

Answer 1

It seems that your definition of a nil ideal is that every element is nilpotent.

It is an easy exercise that the sum of two nilpotent elements is still nilpotent.

Solution:

suppose $x^m = y^n = 0$, then $(x + y)^{m + n - 1} = 0$ by binomial expansion.

Using induction, one can deduce that any finite sum of nilpotent elements is nilpotent.

Equally clear is that any multiple of a nilpotent element is also nilpotent.

Therefore, any ideal generated by (possibly infinitely many) nilpotent elements is nilpotent, because every element in the ideal is a finite sum of multiples of the generators.

This in particular applies to your $\tilde I$, which is generated by images of the $X_i$'s.

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It seems that your definition of a nilpotent ideal is that there is an $n$ such that $I^n = 0$.

Then it's clear that $\tilde I$ is not nilpotent, since for every $n$, the ideal $\tilde I^n$ contains the image of $X_n^n$, which is nonzero in $R$.