0

The question is : Show that the nilpotent elements of a commutative ring form a subring.

Here is my unsuccessful take on it:

Let $R$ be a commutative ring and let $S = \{a \in R | a^n = 0 \}$ be the set of nilpotent elements from $R$. Now, $0 \in S$, since $0^1 =0$. Hence $S$ is nonempty.

Let $x,y \in S$, then for some positive integers $n$ and $m$ we have $x^n=0$ and $y^m=0$. So far so good. Now we try to show that $(x + y)^k = 0$, for some positive integer k. This is where I am stuck.

Eingalf
  • 88
  • 1
    Hint: try k=n+m, expand with binomial theorem and argue that for each term, one of the exponents is big enough. – Mathemagician Mar 02 '14 at 06:24
  • but big enough for what? – Eingalf Mar 02 '14 at 06:31
  • Wait, I think I have got it. Will try to solve and post solution. – Eingalf Mar 02 '14 at 06:33
  • The proposed duplicate resolves the sum of nilpotent elements being nilpotent (in a manner similar to Mathemagician's Hint). However it doesn't address the multiplicative requirements to be a subring. Although the assumption of commutativity quickly resolves closure under multiplication, it fails to provide a multiplicative identity. Indeed the multiplicative identity is not a nilpotent element of a commutative ring, so apart from the dubious case 0=1, the proposition is always false. It would be more acceptable to say the nilpotent elements form an ideal rather than a subring. – hardmath Mar 02 '14 at 21:49

0 Answers0