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Let $p \subset R$ be a prime ideal. Prove that for any nilpotent $r \in R$, it follows that $r \in p$.

One of my classmates told me to use induction.

Also, show that for all $s \in R$, $1-sr$ is a unit.

Note: This was the last assignment (due date already passed [April 30] so I'm not directly asking for hw answers) for my introduction to abstract algebra course. I'm posting it here because my professor decided not to post solutions for this last assignment since she'll be busy making the final exam. Still, I feel like these questions are very relevant and may pop up on the final.

user72625
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  • I'm adding the commutative algebra tag, since the first statement doesn't work for general noncommutative rings. – rschwieb Dec 06 '13 at 17:25

2 Answers2

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$r$ nilpotent means there exists a natural number $n>0$ such that $r^n=0$. But $0\in\mathfrak{p}$, hence $r^n=r\cdot r\ldots r\in\mathfrak{p}$. If a prime ideal contains a product, then it contains at least one factor, in this case all factors equal $r$, so that $r$ is in $\mathfrak{p}$.

For second question,you have $$(1-rs)(1+rs+\ldots+r^{n-1}s^{n-1})=1+rs-rs+\ldots-(rs)^n=1-(rs)^n=1-0=1$$ hece $rs$ is an invertible element of $R$.

Federica Maggioni
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Hint: If $r ∈ R$ in nilpotent, then by definition there is an $n ∈ ℕ$ such that $r^n = 0$.

For the first one, write $rr^{n-1} = 0$ and use that $0$ is in the prime ideal $p$. Then use induction.

For the second one, check out $(1- sr)(\sum_{k=0}^{n-1} (sr)^k)$.

k.stm
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