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

**hatsoff** · March 31st 2010, 05:56 AM

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!

**tonio** · March 31st 2010, 07:28 AM

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

**Opalg** · March 31st 2010, 07:53 AM

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 $a$ .

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