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I am having trouble with another homework problem (from Algebra by Hungerford).

Let $R$ be a ring without identity and with no zero divisors. Let $S$ be the ring whose additive group is $R \times \mathbb{Z}$ with multiplication given by $(r_1,n_1)(r_2,n_2)=(r_1r_2+n_2r_1+n_1r_2,n_1n_2)$. Let $A=\{(r,n) \in S \mid rx+nx=0 \text{ for all } x \in R\}$.

(a) $A$ is an ideal in $S$.

(b) $S/A$ has an identity and contains a subring isomorphic to $R$.

(c) $S/A$ has no zero divisors.

I've done (a) and (b), but haven't been able to crack (c). Here's what I have so far:

(c) Suppose $((r, n) + A)((s, m) + A) = (0, 0) + A$. Then $(r, n)(s, m) \in A$, so $(rs + mr + ns)x + (nm)x = 0$ for all $x \in R$. But $(rs + mr + ns)x + (nm)x = rsx + mrx + nsx + nmx = r(sx + mx) + n(sx + mx)$.

Now here we have an expression involving $r(something) + n(something)$ and $s(something) + m(something)$, and I want to say that if it's zero for all $x$ then $rx + nx = 0$ for all $x$ or $sx + mx = 0$ for all $x$, but all I seem to be able to say is that $rx + nx = 0$ for all $x = sx' + mx'$, which is not (or not obviously) what I want.

I appreciate any help!

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up vote 5 down vote accepted

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.

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I see! $rt + nt = 0 \Leftrightarrow t(rt + nt) = 0 \Leftrightarrow trt + nt^2 = 0 \Leftrightarrow (tr + nt)t = 0 \Leftrightarrow tr + nt = 0$. Then we can use that to say $tr + nt = 0 \Leftrightarrow trx + (nt)x = 0 \Leftrightarrow trx + t(nx) = 0 \Leftrightarrow t(rx + nx) = 0 \Leftrightarrow rx + nx = 0$ and we're done! – smackcrane Nov 26 '12 at 22:29

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