Question

Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in R o+ S if and only if a or b is a zero-divisor or exactly one of a or b is 0.

Commutative Algebra
ANSWERED
asked 2020-12-14
Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in \(\displaystyle{R}\oplus{S}\) if and only if a or b is a zero-divisor or exactly one of a or b is 0.

Answers (1)

2020-12-15

Let (a,b) is a zero-divisor in \(\displaystyle{R}\oplus{S}\)
Since (a,b) is zero-divisor, there exists (c,d) in \(\displaystyle{R}\oplus{S}\) such that
\((a,b)(c,d)=(0,0)\)
\(\displaystyle\Rightarrow\)\( ac=0, bd=0\)
Since \(ac=0\) \(\displaystyle\Rightarrow\) a is a zero divisor in R or \(a=0\)
Also, \(bd = 0\) \(\displaystyle\Rightarrow\) b is zero divisor in R or \(b=0\)
Conversely:
Let a and b are zero divisors.
Since a is zero divisor,there exist element \(\displaystyle{c}\in{R}\) such that ac=0
Also, b is zero divisor, there exist element \(\displaystyle{d}\in{S}\) such that bd=0
\(\displaystyle\Rightarrow\) \(ac\times bd=0\)
\(\displaystyle\Rightarrow\) \((a,b)\times (c,d)=(0,0)\)
Hence, for \(\displaystyle{\left({a},{b}\right)}\in{R}\oplus{S}\), there exists (c,d) such that \((a,b)\times(c,d)=(0,0)\)
By definition (a,b) is zero diisor in \(\displaystyle{R}\oplus{S}\)
Hence, proved

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