Prove that the rings and are not isomorphic.

We begin with some lemmas.

Lemma 1: Let and be rings with 1. If is a surjective ring homomorphism, then . Proof: Let . Now for some . Note that , and likewise . Since the 1 in is unique, we have .

Lemma 2: Let and be rings with 1. If is a ring homomorphism such that , then . Proof: If is a unit, we have for some . Then , so that is a unit.

Lemma 3: Let and be rings with 1. If is a surjective ring homomorphism, then . Proof: The direction follows from the previous lemma. Now let . Since is surjective, for some . Note that for some . Then , and thus .

Lemma 4: Let and be rings with 1. If is a ring isomorphism, then is a multiplicative group isomorphism. Proof: By the previous lemma, indeed maps surjectively to . Moreover, as the restriction of an injective function, is injective. Finally, is a multiplicative homomorphism since is.

Note that and are rings with 1. If is a ring isomorphism, then the restriction of to the units is a group isomorphism . However, note that by Proposition 4, has 2 units while has infinitely many units; thus we have a contradiction.

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## Comments

I don`t know if I`m just confused or not. In the second lemma, you proved that a unit element goes to a unit element under ring homomorphism but you stated the statement of the lemme and the image of R is contained in s , is that wrong? I mean I don`t see any connection between the statement of the lemma and what you proved!! Am I missing something?

By I mean the set of units in . So the line just means that units map to units under .

Another proof: In , has a solution for all , namely . But in , has no solution.

Nice!