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The set of nilpotent elements in a commutative ring is an ideal


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.29

Solution: Let $x,y \in \mathfrak{N}(R)$. Then for some nonnegative natural numbers $n$ and $m$, we have $x^n = y^m = 0$.

Consider $(x+y)^{n+m}$. By the Binomial Theorem, we have $$(x+y)^{n+m} = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k y^{n+m-k}.$$ Note that if $k \geq n$, $x^k = 0$, and if $k < n$, then $n+m-k>m$ abd hence $y^{n+m-k} = 0$. Thus $(x+y)^{n+m} = 0$, and we have $x+y \in \mathfrak{N}(R)$. Moreover, we have $$(-x)^n = (-1)^nx^n = 0,$$ so that $-x \in \mathfrak{N}(R)$. Since $0 = 0^1 \in \mathfrak{N}(R)$, $\mathfrak{N}(R)$ is an additive subgroup of $R$.

Finally, since $R$ is commutative, if $r \in R$ then $$(rx)^n = r^nx^n = 0$$ and likewise $(xr)^n = 0$. Thus $\mathfrak{N}(R)$ absorbs $R$ on the left and the right, and hence is an ideal.


Linearity

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