Every subring of a field which contains 1 is an integral domain

Prove that any subring of a field which contains the identity is an integral domain.

Let R \subseteq F be a subring of a field. (We need not yet assume that 1 \in R). Suppose x,y \in R with xy = 0. since x,y \in F and the zero element in R is the same as that in F, either x = 0 or y = 0. Thus R has no zero divisors. If R also contains 1, then R is an integral domain.

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  • Kate  On November 5, 2010 at 5:23 pm

    I am wondering if you need to show that it is not necessarily a field? like somehow show that the multiplicative inverse does not need to be in the subring R?

  • nbloomf  On November 5, 2010 at 8:32 pm

    We can show an example of such a subring which isn’t a field- consider the integers as a subring of the rationals.

    In general though there are fields which have only the trivial subrings, and thus any subring containing 1 *is* a field. (For instance \mathbb{Z}/(p).)

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