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

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

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

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 $\displaystyle p\in P$ and $\displaystyle a\in N$. Then there is a positive integer $\displaystyle m\geq 2$ with $\displaystyle a^m=0$. Assume towards a contradiction that $\displaystyle a\notin P$. Since $\displaystyle a^{m-1}a=a^m=0=0p\in P$ and $\displaystyle a\notin P$, then by definition of prime ideals we have $\displaystyle a^{m-1}\in P$. However, the same can be said for any integer $\displaystyle n\geq 2$, that is, if $\displaystyle a^n\in P$ then $\displaystyle a^{n-1}a\in P$ and therefore $\displaystyle a^{n-1}\in P$. By induction, since $\displaystyle a^m\in P$, then $\displaystyle a\in P$, which contradicts our assumption that $\displaystyle a\notin P$. So $\displaystyle a\in P$ whenever $\displaystyle a\in N$, that is, $\displaystyle N\subseteq P$. $\displaystyle \blacksquare$
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 $\displaystyle a$ be any nilpotent, then $\displaystyle a^m=0\in P\Longrightarrow a\in P$ (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 $\displaystyle a_1a_2\cdots a_n$ lies in a prime ideal P, then at least one of the elements $\displaystyle a_i$ belongs to P.
That result is easily proved, for all $\displaystyle n\geqslant2$, 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 $\displaystyle a_i$ to be equal to $\displaystyle a$.