If $R$ is a commutative ring, then prove that $(s \cdot t) x =(s \cdot x)(t \cdot x); x \in R ~,~ s,t \in Z$
Attempt: I do not understand how $(s \cdot t) x =(s \cdot x)(t \cdot x) = (s \cdot t)x^2$
Is this statement even true? Because I read it on a web page elsewhere?
Thank you.
The statement is in general false in the absence of special restrictions on $x$. As a counterexample, simply take $R = Z$, $0, 1 \ne x \in Z$, and $0 \ne s, t \in Z$; then if
$(st)x = (sx)(tx) =(st)x^2, \tag{1}$
we have, since $st \ne 0$, that
$x = x^2; \tag{2}$
but in $Z$, (2) implies that $x =0$ or $x = 1$, contradicting our assumption that $x \ne 0, 1$. The same conclusion binds if $Z$ is replaced by any integral domain.
It strikes me that Jyrki Lahtonen is "morally correct" when he indicates in his comment that what is intended here might not be what is written. My take on this is that what is wanted is a demonstration of either $(s(tx)) = (st)x$ or $(sx)(tx) = (st)x^2$, and that some careless reader/writer (present company of course excluded! ;-)) got the two equations mixed up. I also surmise that the underlying point of this exercise was to show that when the integers act most obviously on the elements of arbitrary rings, viz. via
$nr = (r + r + \ldots + r), \; n \; \text{ times}, \tag{3}$
where $r \in R$, $n \in \Bbb N$, the natural numbers, and for $n < 0$ we take $nr = (-n)(-r)$ (and of course $0r = 0$), then multiplication of integers behaves in the "obvious" manner as well ; such assertions are easy to prove, by e.g. simple inuctive arguments. But if such was the original intention of this question, it was morphed and mutated into something else altogether. But such is the nature of evolution, I guess.
In closing, I would like to adress our OP VHP's second question:
Be careful about believing what you read on the 'Net!
To One and All: May this 4th of July find your variables independent!
Hope this helps! Cheers,
and as always,
Fiat Lux!!!