# Abstract Algebra: Rings with zero divisors involving Cartesian Product

• Oct 7th 2010, 04:42 PM
jamestrodden
Abstract Algebra: Rings with zero divisors involving Cartesian Product
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

Can anyone help?
• Oct 7th 2010, 08:38 PM
tonio
Quote:

Originally Posted by jamestrodden
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

Can anyone help?

$(r,0)\cdot(0,s)=(0,0)\,,\,\,0\neq r\in R\,,\,0\neq s\in S$

Tonio