By a domain I mean a non trivial ring without any zero-divisors (not necessarily commutative).
Let $R$ be a ring and $M$ be a left $R$-module. We say an element $m\in M$ is a torsion element iff there exist some regular element $r\in R$ ($r$ is not a zero-divisor) such that $rm=0$
Now, it is easy to check that when $R$ is a commutative domain, the set of torsion elements of $M$, written $tM$, is a submodule of $M$. However, an exercise from the book I'm reading (Basic Algebra by P. M. Cohn) asks me to show that the same conclusion holds if we start with a (left) Noetherian domain $R$!
After some googling I found out that this property of a ring is called 'Ore condition', and that Noetherian rings are Ore rings, but the reference there was too advanced for me to understand. Can this be explained using only ACC conditions for the ring $R$?
Best Answer
Here's an argument why from basics, essentially spelling out a blog post on why Noetherian rings satisfy the Ore condition.
Let $R$ be a Noetherian domain and $M$ a left $R$-module. We will show that for all nonzero $r_1, r_2 \neq 0$, $R r_1 \cap R r_2 \neq (0)$. This exactly says that for any $r_1, r_2 \in R$, there are $s, s' \in R$ so that $s r_1 = s' r_2 \neq 0$.
Once we have this, showing that $tM$ is a submodule is straightforward. If $m_1, m_2 \in tM$, let $r_1$ and $r_2$ satisfy $r_1 m_1 = r_2 m_2 = 0$. Then taking $s, s'$ as above, $$ sr_1 (m_1 + m_2) = sr_1 + s' r_2 = 0, $$ so $r_1 + r_2 \in tM$. Similarly, if $m \in tM$, $r_1 m = 0$ and $r_2 \in R - \{0\}$, with $s, s'$ as above, $s'r_2 m = sr_1 m = s 0 = 0$. So $r_2 m \in tM$ and so $tM$ is a submodule.
So we just need to show our claim above. Suppose for contradiction that $R r_1 \cap R r_2 = (0)$ for $r_1, r_2$ nonzero. We show that $$ Rr_1 \subset Rr_1 + R r_1 r_2 \subset Rr_1 + Rr_1 r_2 + R r_1 r_2^2 \subset \cdots $$ is an infinite ascending chain of left ideals, contradicting Noetherian-ness.
Notice first that for all $n \in \mathbb{Z}^+$, $r_1 r_2^n$ is not $0$ as $R$ is a domain. So in order to show that this chain strictly increases it remains for us to show that $R r_1 r_2^n$ has trivial intersection with $\Sigma_{i=0}^k R r_1 r_2^i$ for all $k < n$. Assume otherwise, and let $n$ be the smallest number for which there is a nontrivial intersection. (Clearly $n > 0$).
Accordingly, write $- a_n r_1 r_2^n = \sum_{i=0}^{n-1} a_i r_1 r_2^i$ for some $a_n \neq 0$ so that after reorganizing, $$a_0 r_1 = \left(\sum_{i=1}^{n-1} a_i r_1 r_2^{i-1} \right)r_2.$$ But this is in $Rr_1 \cap Rr_2$, hence must be $0$ by hypothesis! So we have forced $\sum_{i=0}^{n-1} a_{i+1} r_1 r_2^{i} = 0$ and hence $- a_n r_1r_2^{n-1} = \sum_{i=0}^{n-2} a_{i+1} r_1 r_2^{i}$, and hence $R r_1 r_2^{n-1} \cap \Sigma_{i=0}^{n-2} R r_1 r_2^i \neq (0),$ contradicting minimality of $n$ and proving the result.