Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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~~

share|improve this question
14  
Isn't $\mathbb Z/8\mathbb Z$ with $a=b=\overline{2}$ a counterexample? –  marlu Sep 22 '12 at 11:35
1  
Daring person: things that are impossible seem relatively easy to him/her :) –  rschwieb Sep 22 '12 at 12:49
1  
Or $\mathbb Z / 4\mathbb Z$ with $a=2,b=1$. –  sdcvvc Sep 22 '12 at 12:49
1  
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
add comment

2 Answers

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.

share|improve this answer
add comment

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 $3 \cdot 2 = 6 \neq 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.