A reduced ring and zero divisors
Advanced Ring Theory question for CSIR-NET · GATE. Choose an option, check your answer, then read the detailed solution.
Question RT-010
Let $R$ be a ring (not assumed commutative) with the property that for every $a\in R$, $a^2=0$ implies $a=0$. Let $a,b\in R$ be such that $ab=0$. Which of the following must necessarily be true?
Detailed Solution
Using only associativity, $(ba)^2=(ba)(ba)=b(ab)a=b(0)a=0$. Since $R$ has the property that squaring to zero forces the element to be zero, $(ba)^2=0$ immediately gives $ba=0$ — and this argument never used commutativity, only that $ab=0$ was given, proving (A) and ruling out (B) and (C). Option (D) is false: the hypothesis (no nonzero square-zero elements) does not rule out zero divisors in general — for instance $R=\mathbb Z\times\mathbb Z$ has this property (no nonzero nilpotent squares to $0$) yet $(1,0)(0,1)=(0,0)$ is a genuine zero-divisor pair with neither factor zero. Thus the correct option is $\boxed{\mathrm{A}}$.Concepts tested: Reduced ringsZero divisorsAssociativity
💬 Discuss this problem
Have a doubt or a different approach? Continue the mathematics discussion on Telegram.
Join Telegram Discussion
Have a doubt or a different approach? Continue the mathematics discussion on Telegram.
Join Telegram Discussion
Keep Practising
More Ring Theory Questions
Zero divisors in a ring of functions
Which statement about the ring of real-valued functions is false?
Attempt Question →Complement of an Idempotent Element
If $e$ is idempotent in a ring with unity, what can be said about $1-e$?
Attempt Question →Nontrivial Idempotents in $\mathbb Z_3\times\mathbb Z_3$
Does the product ring $\mathbb Z_3\times\mathbb Z_3$ contain nontrivial idempotents?
Attempt Question →