# Prove that R is commutative if and only if R' is commutative?

R and R' denote arbitrary rings. Suppose R and R' are isomorphic rings. Prove that R is commutative if and only if R' is commutative

Relevance

Let f: R -> R' be an isomorphism. If R is commutative, and a', b' are elements of R', then a' = f(a) and b' = f(b) for some a,b in R. Then we have: a'b' = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = b'a', so R' is commutative. Conversely if R' is commutative, and a,b are elements of R, then f(a) = a' and f(b) = b' for some a',b' in R'. Then we have: f(ab) = f(a)f(b) = a'b' = b'a' = f(b)f(a) = f(ba), so f(ab) = f(ba), and by injectivity of f, ab = a. Hence R is commutative.

A quick note: isomorphism is the "right" notion so that any nontrivial algebraic property carries over isomorphism. If you can say something in algebraic language, like "R is commutative," it's a safe bet that the same holds for isomorphic rings. You don't have to go proving these things all the time.

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• Let R and R' be isomorphic rings. Then there exists a bijection from R to R'. Let f:R -> R' be this bijective.

First, let R be commutative. We need to show that R' is commutative. We need to show that for all a', b' in R' that a' b' = b' a'. Let a' and b' be in R'. Then by surjectivity of f we know there exists a and b in R such that a' = f(a) and b' = f(b). Then a' b' = f(a) f(b) = f(ab) by homorphism property. Then f(ab) = f(ba) since a,b are in R which is commutative. Then f(ba) = f(b) f(a) = b' a'. Thus, a' b' = b' a'. Thus, R' is commutative.

Now, let R' be commutative. We need to show that R is commutative. We need to show that for all a, b in R that ab = ba. Let a, b be in R. Since R' is commutative, a' b' = b' a'. Then f(a) f(b) = f(b) f(a). Then f(ab) = f(ba). Then by injectivity of f we see that ab = ba. Thus, R is commutative.

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