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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.12

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

Solution: 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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