Let and be positive integers with dividing . Prove that the natural surjective ring homomorphism is also surjective on the units.
[With help from Ravi Vakil’s notes.]
Given a unital ring homomorphism , it is certainly the case that . If we have equality, we say that is surjective on the units of . If and are unital ring homomorphisms which are surjective on the units and the composition makes sense, then so is . Clearly isomorphisms are surjective on the units.
We begin with some lemmas.
Lemma 1: Let be a nonempty set of rings with . Then is a unit if and only if is a unit for all . Proof: Let be a unit. Then there exists such that , and thus for all . Similarly, . Thus is a unit for each . If is a unit, then there exists such that . Then .
Lemma 2: Let be a prime and let . Then the natural surjective ring homomorphism given by is surjective on the units. Proof: Now let be a unit in . Then , so that , and thus . So is a unit in . Since , is surjective on the units.
Lemma 3: Let and be rings with , and let be a surjective ring homomorphism which is surjective on the units; that is, . Then is surjective on the units. Proof: Follows from Lemma 1.
Now let . Since , we have where for all . Now the are pairwise comaximal, so that by the Chinese remainder Theorem we have isomorphisms and given by and . Note also that given by is a well-defined ring homomorphism. Likewise we have well-defined surjective ring homomorphisms for each given by . It is clear that .
By Lemmas 2 and 3, is surjective on the units. Thus is surjective on the units.