# Thread: [SOLVED] Seeking a better, more accurate and more elegant proof

1. ## [SOLVED] Seeking a better, more accurate and more elegant proof

Let $N$ be the set of all nilpotent elements of a commutative ring $R$, and show that $N\subseteq P$ for each prime ideal $P$ of $R$.
Proof: Let $p\in P$ and $a\in N$. Then there is a positive integer $m\geq 2$ with $a^m=0$. Assume towards a contradiction that $a\notin P$. Since $a^{m-1}a=a^m=0=0p\in P$ and $a\notin P$, then by definition of prime ideals we have $a^{m-1}\in P$. However, the same can be said for any integer $n\geq 2$, that is, if $a^n\in P$ then $a^{n-1}a\in P$ and therefore $a^{n-1}\in P$. By induction, since $a^m\in P$, then $a\in P$, which contradicts our assumption that $a\notin P$. So $a\in P$ whenever $a\in N$, that is, $N\subseteq P$. $\blacksquare$

The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

Thanks!

2. Originally Posted by hatsoff
Proof: Let $p\in P$ and $a\in N$. Then there is a positive integer $m\geq 2$ with $a^m=0$. Assume towards a contradiction that $a\notin P$. Since $a^{m-1}a=a^m=0=0p\in P$ and $a\notin P$, then by definition of prime ideals we have $a^{m-1}\in P$. However, the same can be said for any integer $n\geq 2$, that is, if $a^n\in P$ then $a^{n-1}a\in P$ and therefore $a^{n-1}\in P$. By induction, since $a^m\in P$, then $a\in P$, which contradicts our assumption that $a\notin P$. So $a\in P$ whenever $a\in N$, that is, $N\subseteq P$. $\blacksquare$
The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

Thanks!

Why did you have to use contradiction at all? Let $a$ be any nilpotent, then $a^m=0\in P\Longrightarrow a\in P$ (as you showed this is fine)...and voila! No contradiction and short.

Tonio

3. If you don't like that "finite induction" argument, you can derive the result about nilpotent elements as a corollary of a more general result:
If a product $a_1a_2\cdots a_n$ lies in a prime ideal P, then at least one of the elements $a_i$ belongs to P.
That result is easily proved, for all $n\geqslant2$, by a (standard) induction argument, the base case n=2 being just the definition of primality. The result about nilpotents follows by taking all the $a_i$ to be equal to $a$.