Abstract Algebra, rings, zero divisors, and cartesian product

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SUMMARY

The discussion focuses on the proof that the Cartesian product of two nonzero rings, R and S, contains zero divisors. It establishes that for elements (r1, s1) and (r2, s2) in R x S, the multiplication defined as (r1r2, s1s2) can yield (0,0) without either r1, r2, s1, or s2 being zero. This confirms the presence of zero divisors in the Cartesian product of nonzero rings.

PREREQUISITES
  • Understanding of nonzero rings in abstract algebra
  • Familiarity with the Cartesian product of sets
  • Knowledge of the definition and properties of zero divisors
  • Basic operations in ring theory, including addition and multiplication
NEXT STEPS
  • Study the properties of zero divisors in ring theory
  • Explore the implications of the Cartesian product in algebraic structures
  • Learn about nonzero rings and their characteristics
  • Investigate examples of zero divisors in specific rings
USEFUL FOR

Students and researchers in abstract algebra, mathematicians exploring ring theory, and educators teaching concepts related to rings and zero divisors.

jamestrodden
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The problem states:
Let R and S be nonzero rings. Show that R x S contains zero divisors.

I had to look up what a nonzero ring was. This means the ring contains at least one nonzero element.

R x S is the Cartesian Product so if we have two rings R and S
If r1 r2 belong to R and s1 s1 belong to S

(r1, s1) + (r2,s2) = (r1+r2, s1 + s2)
I am using * to denote multiplication here.
(r1, s1)*(r2,s2) = (r1r2,s1s2)

Since we are talking about zero divisors I am going to need the definition of multiplication in the Cartesian Product.

I mean i going to need that (r1r2, s1s2 ) = (0,0)

so r1r2 = 0 and s1s2 = 0 . But neither r1 = 0 = r2 or s1 = 0 = s2
 
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jamestrodden said:
so r1r2 = 0 and s1s2 = 0 . But neither r1 = 0 = r2 or s1 = 0 = s2
Why'd you stop here?
 

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