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In a ring, $a\neq0$ and $b\neq0$. $aba=0$.

Prove $ab=0$ or $ba=0$.

This is one question in my abstract algebra homework-- it seems pretty easy at first glance, yet I have spent hours thinking about possible solutions...still I haven't find a way out. Could you give me some hint? Thank you~~

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Isn't $\mathbb Z/8\mathbb Z$ with $a=b=\overline{2}$ a counterexample? – marlu Sep 22 '12 at 11:35
Daring person: things that are impossible seem relatively easy to him/her :) – rschwieb Sep 22 '12 at 12:49
Or $\mathbb Z / 4\mathbb Z$ with $a=2,b=1$. – sdcvvc Sep 22 '12 at 12:49
It's a good habit, especially if you are hours into a problem, to switch to thinking about counterexamples for a bit. Sometimes this can cut down on time spent spinning wheels. – rschwieb Sep 22 '12 at 12:54
up vote 15 down vote accepted

Hint $\ $ For $\rm\:b=1\:$ it is $\rm\: a^2 = 0\:\Rightarrow\: a= 0,\:$ which is not true in every ring, e.g. $\rm\ a = n\:$ in $\rm\: \Bbb Z/n^2.$

This is essentially the only obstruction, i.e. the statement holds true iff the ring is reduced, i.e. iff the ring has no nonzero nilpotent elements, i.e. $\rm\: x^n = 0\:\Rightarrow\: x = 0.\:$

Theorem $\ $ The following are equivalent in any ring.

$\rm(1)\quad xyx = 0\ \Rightarrow\ xy=0\ \ or\ \ yx=0$

$\rm(2)\quad x^2 = 0\ \Rightarrow\ x=0$

$\rm(3)\quad x^n = 0\ \Rightarrow\ x=0$

Proof $\ \ \ \ (1\Rightarrow 2)\ \ \ $ Put $\rm\:y = 1.\:$
$(2\Rightarrow 3)\ \ \, $ The least $\rm\:n\:$ such that $\rm\:x^n = 0\:$ is $\rm\:n = 1\:$ since any larger value can be reduced, viz.

$$\begin{eqnarray}\rm x^{2k} = 0\ &\Rightarrow&\rm\ (x^k)^2\! &=& 0\ &\Rightarrow&\rm\ x^k &=& 0\\ \rm x^{2k+1}\!=0\ &\Rightarrow&\rm\ (x^{k+1})^2\! &=& 0\ &\Rightarrow&\rm\ x^{k+1} &=& 0\end{eqnarray}$$

$\rm(3\Rightarrow 1)\ \ \ $ If $\rm\:xyx = 0\:$ then $\rm\: (xy)^2 = (xyx)y = 0,\:$ hence $\rm\: xy = 0\:$ by $(3)$.

Remark $\ $ The absence of nilpotent elements facilitates simpler structure theory, e.g. Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents is isomorphic as a ring to a direct sum of copies of $\rm\:\mathbb R\:$ and $\rm\:\mathbb C\:.\:$Wedderburn and Artin proved a generalization that every finite-dimensional associative algebra without nilpotent elements over a field $\rm\:F\:$ is a finite direct sum of fields.

Such structure-theoretic results greatly simplify classifying such rings when they arise in the wild. For example, I applied a special case of these results here to prove that a finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2\:.\:$ For another example, a sci.math reader once proposed an extension of the real numbers with multiple "signs". This turns out to be a very simple case of the above results. See my answer in Is there a third dimension of numbers? for much further discussion, including references.

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This is false. Consider the ring $\mathbb{Z}_{12}$. Let $a =2, b=3$. Then $aba =2\cdot 3 \cdot 2 =0$, however $2 \cdot 3 = 6 \neq 0$ and $3 \cdot 2 = 6 \neq 0$.

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