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Let $A$ be a commutative ring, $a, a+b \in A$ are nilpotent. Does this imply that $b$ is nilpotent?

user26857
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Daniel Tum
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  • @Koushik: Do you have an example in mind? – Daniel Tum Sep 11 '14 at 10:49
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    Yes, the nilpotent elements form an ideal of $A$, in particular, closed under addition (though this argument would be "backward"). – Quang Hoang Sep 11 '14 at 10:50
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    @QuangHoang : You should have posted that as answer... +1 from my side... –  Sep 11 '14 at 10:51
  • @Hoang: Thanks for the answer. – Daniel Tum Sep 11 '14 at 10:54
  • Welcome to our site! – kjetil b halvorsen Sep 11 '14 at 11:03
  • Let $x:=a$ and $y:=a+b$. Then you know $x^n=0$ for some natural number $n$ and $y^m=0$ for some natural number $m$. What you have to show is ${(y-x)}^k=0$ for some natural number $k$. Since your ring is commutative, you can use the binomial formula: ${(y-x)}^k=\sum\limits_{s=0}^{k} {\begin{pmatrix} k\ s\end{pmatrix} y^{k-s}\cdot {(-1)}^sx^s}$. If you choose $k:=2\cdot n\cdot m$, then you will see that every summand on the rhs of the binomial formula will be equal to $0$, either because of $x^n=0$ or because of $y^n=0$, depending on the summand. – LSt Sep 11 '14 at 11:15

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Yes b is also a nilpotent element in R. If n is the nilpotence degree of a and m is the nilpotence degree of a+b, considering b=(a+b-a)^nm and using binomial theorem, you can easily obtain b is nilpotent. Here with no loss of generalization, take m>n or m>= n+1 and after expanding the above expression and playing m and n, the result will follow.

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Yes. The set of all nilpotent elements in $A$, denoted by $\operatorname{Nil}(A)$, is an ideal, in particular it's an additive group. Therefore $(a + b) - a = b \in \operatorname{Nil}(A)$.

The important part you need to show that $\operatorname{Nil}(A)$ is an ideal is the following: say $a, b \in \operatorname{Nil}(A)$, then there exist $m, n \in \mathbb N$ such that $a^m = 0$ and $a^n = 0$. Now observe that, by the Binomial Theorem, $$(a + b)^{m + n} = \sum_{k = 0}^{m + n} \binom{m + n}{k} a^{m + n - k}b^k = 0.$$ Why is that? Well, if $k > n$, then $b^k = 0$ and if $k \leq n$, then $n - k \geq 0$ which means $m + n - k \geq m$ which means $a^{m + n - k} = 0$.

Robert Cardona
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