Ring homomorphisms map an identity element to an identity or a zero divisor

Let R and S be nonzero rings with identies 1_R and 1_S, respectively. Let \varphi : R \rightarrow S be a nonzero ring homomorphism.

  1. Prove that if \varphi(1_R) \neq 1_S, then \varphi(1_R) is a zero divisor in S. Deduce that if S is an integral domain then every ring homomorphism R \rightarrow S sends the identity of R to the identity of S.
  2. Prove that if \varphi(1_R) = 1_S then if u \in R is a unit, then \varphi(u) \in S is a unit and \varphi(u)^{-1} = \varphi(u^{-1}).

  1. Suppose \varphi(1_R) = r, with r \neq 1. First, if r = 0, then \varphi(x) = \varphi(x \cdot 1_R) = \varphi(x) \varphi(1_R) = \varphi(x) \cdot 0 = 0, so that \varphi = 0, a contradiction. Thus r \neq 0. Now \varphi(1_R) = \varphi(1_R \cdot 1_R) = \varphi(1_R) \cdot \varphi(1_R) = r^2, so that r^2 = r, and we have r(r-1) = 0. Thus r is a left zero divisor. Similarly, (1-r)r = 0, and r is a right zero divisor. Hence r is a zero divisor in S.

    If S is an integral domain, this yields a contradiction. Thus if \varphi : R \rightarrow S is a nonzero ring homomorphism and R and S have identities, then \varphi(1) = 1. (I.e. \varphi is a unital ring homomorphism.)

  2. Suppose u \in R is a unit. Now \varphi(u) \varphi(u^{-1}) = \varphi(uu^{-1}) = \varphi(1) = 1. Similarly, \varphi(u^{-1}) \varphi(u) = 1. Thus \varphi(u) is a unit, and by the uniqueness of inverses, \varphi(u)^{-1} = \varphi(u^{-1}).
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