How to show that the ring $S/A$ has no zero divisors? (Hungerford, Algebra, Problem 12, Chapter III, Section 2)

Solution 1:

If $(s,m) \in A$ then you are done, so suppose not.

Then $sx+mx \neq 0$ for some choice of $x \in R$; denote the resulting value of $sx+mx$ by $t$. What we have is that $rt + nt = 0$, while $0 \neq t \in R$. We want to deduce that $(r,n) \in A$.

Note that this had better be true if $S/A$ is to have no zero-divisors, since the image of $t$ (or, more precisely, $(t,0)$) in $S/A$ is non-zero, while the image of $(r,n)$ in $S/A$ multiplies with $t$ to give $0$.

So we can rephrase the problem as follows: given $(r,n) \in S$ and $0 \neq t \in R$ such that $rt + n t = 0,$ prove that $rx + nx = 0$ for all $x \in R$.

For this, you have to use the no-zero-divisor property of $R$ (since it is the one non-trivial way you have to verify that an element of $R$ is zero).

Here are some more precise hints:

• You want to conclude that $r x + n x $ is zero, and so you need to multiply it by something non-zero and get zero. The only non-zero element you have at hand is $t$, so you will have to use it.

• You will find that, annoyingly, in the equation $r t + n t = 0$, the element $t$ is on the wrong side of the element $r$; see if you can move it to the other side, i.e. prove that this is equivalent to $t r + n t = 0,$ which will be more useful in carrying out the first step.