All answers I have seen regarding this question (showing that the set of nilpotent elements of $R$ form an ideal) make use of the Binomial theorem in order to prove that $(N,+)$ is an additive subgroup of $(R,+)$. My book (Abstract Algebra A First Course by Dan Saracino) makes no reference to the Binomial theorem, and so I am unsure as to whether I can use it (perhaps that is a silly question).
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6If you can prove it, you can use it. You don't even need to know the coefficients. Induct on $n$ to show $$(x+y)^n = \sum_{i=0}^n c_{i,n}\cdot x^iy^{n-i}$$ where $c_{i,n} \in \mathbb{N}$. – Brian Moehring Jan 30 '21 at 02:53
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Presumably "using the Binomial Theorem" refers to proofs like these, correct? – Bill Dubuque Jan 30 '21 at 19:12
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Suppose $a,b$ are nilpotent elements of $R$, say $a^n=0=b^m$. Consider the ring $\overline{R}=R/Ra$ and use $\overline{\phantom{b}}$ to denote cosets in $\overline{R}$. Then $\overline{b}\,^m=0$. Since $\overline{a+b}=\overline{b}$, we have $\overline{a+b}\,^m=\overline{0}$. Thus $(a+b)^m=ra$ for some $r\in R$, and so $(a+b)^{mn}=r^na^n=0$.
This proof can be modified slightly to prove that any finite sum of nilpotent ideals in any ring (including noncommutative rings) is nilpotent and any (possibly infinite) sum of nil ideals is nil.
Allen Bell
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Given $\ \color{#c00}{b^k} = 0 = \color{#0a0}{a^n},\,$ Factor Theorem $\Rightarrow (a\!+\!b)^n-a^n$ has factor $\,(a\!+\!b)-a = b,\,$ therefore $\, bc = (a\!+\!b)^n - \color{#0a0}{a^n}= (a\!+\!b)^n\,$ $ \overset{\large (\,\ )^{\Large k}}\Longrightarrow\, (a\!+\!b)^{nk} = (bc)^k = \color{#c00}{b^k}c^k = 0$.
Note $ $ The proof only needs $\,(a+b)^n = a^n + bc\,$ for some $\,c.\,$ There are many elementary ways to prove that besides the Factor Theorem (or Binomial Theorem), including an obvious simple inductive proof. These allow us to replace the Factor Theorem by various closely related well-known facts, e.g. the Congruence Power Rule $\Rightarrow \bmod b\!:\ a\!+\!b\equiv a\,\Rightarrow\, (a\!+\!b)^n\equiv a^n\, $ (or equivalent arithmetic in the quotient ring $R/b\,).\, $ Alternatively we can employ the Polynomial Remainder Theorem, or Taylor's Theorem (to first order), etc.
Bill Dubuque
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Yes, you could show that the set of nilpotent elements is equal to the intersection of all the prime ideals! The intersection of a family of ideals is an ideal.
But typically showing that everything in that intersection is nilpotent uses the concept of localization, which is more advanced than just the binomial theorem.
If I were you and squeamish about assuming the binomial Theorem, I’d just prove it.
rschwieb
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