There's a question in Gallian: Suppose that $R$ is a commutative ring with no zero divisors. Characteristic of $R$ is $0$ or prime. I am wondering what could be the role of $R$ given as being commutative.

Attempt: The characteristic of a ring is either $0$ or finite. If it's $0$, we are done.

If it's not zero, then $\exists~ n\in N$ such that $(n \cdot x) = 0 ~~\forall~~x \in R$

If $n$ is composite, then $n=s \cdot t \implies (s \cdot t)~ x =0 \implies (s \cdot x) (t \cdot x) =0 \implies (s \cdot x)=0 $ or $(t \cdot x)=0$

[ .... This is because if $m$ and $n$ are integers and $a$ and $b$ elements from a ring, then :

$(m \cdot a)(n \cdot b)=(mn) \cdot (ab)$ . It doesn't need that R be commutative as :

$(m \cdot a) ( n \cdot b) = (a + a + .... + a) \{m ~~times~~\} (b + b + ....+ b) \{n~~ times~~\} = ab + ab + .... + ab ~~\{mn ~~times~~\} $

$ = (mn) \cdot (ab) $ ..... ]

Since, $n$ as per definition is the least integer satisfying $n \cdot x=0~\forall~x \in R$ . Hence, $n$ must be prime or $0$.

All the above also applies to a non commutative ring, Why was commutative ring specifically mentioned in the question? Thanks .

  • $\begingroup$ $(s \cdot t)~ x =0 \implies (s \cdot x) (t \cdot x) =0$ Error! Error! Error! $\endgroup$ – Giulio Bresciani Jul 4 '14 at 14:51
  • $\begingroup$ $(s.t).x$ cannot be written as $(s.x).(t.x)$ $\endgroup$ – Swapnil Tripathi Jul 4 '14 at 14:52
  • $\begingroup$ $(mn)e=(me)(ne)$ if $e$ is multiplicative identity and $R$ is commutative. $\endgroup$ – Swapnil Tripathi Jul 4 '14 at 14:56
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    $\begingroup$ Well, $(s\cdot t)x = 0$ still implies $(s\cdot x)(t\cdot x) = 0$, since $(s\cdot x)(t\cdot x) = ((s\cdot t)x)x = 0x = 0$. $\endgroup$ – Daniel Fischer Jul 4 '14 at 15:14
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    $\begingroup$ Daniel is right, $(s \cdot t) x =0 \implies (s \cdot x) (t \cdot x) =0$ is true. The reason of my correction is that you seemed to suppose $(s\cdot t)x=(s\cdot x)(t\cdot x)$, and this is wrong. Anyway, the identity is important because you may use $x=e$ and work with a single element. If you show that $ne=0$, then $nx=n(e\cdot x)=(ne)\cdot x=0$ for every $x$. If you don't have the identity, you need to work with all the elements of $R$ at the same time, and you have problems like the one I've shown in the previous comment. $\endgroup$ – Giulio Bresciani Jul 4 '14 at 15:47

Let's see how far we get without assuming commutativity.

First, for every $x\in R\setminus\{0\}$, there is an ideal $A_x \subset \mathbb{Z}$ such that $n\cdot x = 0 \iff n\in A_x$. Let $a_x$ be the non-negative generator of $A_x$.

Then, since $R$ has no zero divisors, for every $x\in R\setminus\{0\}$, $a_x$ is either $0$ or a prime. For suppose $a_x = m\cdot n$ with $m,n > 1$. Then $mx\neq 0$ and $nx\neq 0$, whence $0 \neq (mx)(nx) = ((mn)x)x$, and therefore $0 \neq (mn)x = a_x x$ contradicting the definition of $a_x$.

Next, if there is an $x\in R\setminus\{0\}$ with $a_x \neq 0$, then $a_x y = 0$ for all $y\in R$: If $y = 0$, that is clear, so suppose $y\neq 0$. Then

$$0 = (a_x x)y = x(a_x y),$$

and since $x\neq 0$ and $R$ has no zero divisors, it follows that $a_x y = 0$, and since $a_x$ and $a_y$ are prime, further $a_y = a_x$, and we have $\operatorname{char} R = a_x$.

If on the other hand $a_x = 0$ for all $x\in R\setminus\{0\}$, then $\operatorname{char} R = 0$ is clear.

So we don't need commutativity.

  • $\begingroup$ thank you . I feel at peace now. :-) $\endgroup$ – MathMan Jul 4 '14 at 15:53

It's easy to see this by using the fact that $n:= \text{char}(R)$ is the natural number that generates the kernel of the ring morphism $f \colon \mathbb Z \to R$, $k \mapsto k \cdot 1_R$. The image of this morphism is central in $R$, and heuristically speaking any argument about the characteristic occurs in this image, so commutativity of $R$ is irrelevant.

More specifically, suppose $n \neq 0$ and $n \mathbb Z = \ker f$ and let $p$ be the smallest prime dividing $n$ (which assumes $R$ is not the zero ring), so that $n = ps$ for some some integer $s < n$. We have $0=f(n)=f(ps)=f(p)f(s)$. Since $R$ has no zero-divisors we must have $f(p)=0$ or $f(s)=0$. But $f(s) \neq 0$ since then $s \mathbb Z \subseteq \ker f = n \mathbb Z \subsetneq s \mathbb Z$, a contradiction. So $f(p)=0$, ergo $p \mathbb Z = n \mathbb Z = \ker f$, as desired.

  • $\begingroup$ The ring is, as I understood, not assumed to be unital, so we can't use the natural homomorphism from $\mathbb{Z}$. $\endgroup$ – Daniel Fischer Jul 4 '14 at 18:21

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