No field is a nontrivial direct product

Prove that if R and S are nonzero rings then R \times S is not a field.


Let a \in R and b \in S be nonzero. Then (a,0),(0,b) \in R \times S are nonzero, but (a,0)(0,b) = 0. Thus R \times S contains zero divisors and cannot be a field.

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Comments

  • Kate  On November 18, 2010 at 1:26 am

    How can (a,0) exist in R x S if 0 does not exist in S? wouldn’t it have to be of the form (c, d) such that c in R and d in S. In the case of (a, 0) a is in R but 0 is not in S.

  • Kate  On November 18, 2010 at 1:26 am

    Just kidding on that last comment, I’m tired.

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