Question

Assume that A and B are isomorphic commutative rings with unity. Prove that if A is a field, so is B.

Answer 1

Answer on Question #57913 - Math - Abstract Algebra

Assume that $A$ and $B$ are isomorphic commutative rings with unity. Prove that if $A$ is a field, so is $B$.

Solution.

By definition field is a

1. commutative ring with identity.

2. $\forall a \in A, a \neq 0, \exists a^{-1}: a a^{-1} = 1$

By condition A and B isomorphic, i.e. exists $f: A \to B$ a

$$f(a)f(b) = f(ab)$$ $$f(a) + f(b) = f(a + b)$$

, where $a, b \in A$

Evidently that $f(0_A) = 0_B$, $f(1_A) = 1_B$ and $\forall b \in B, \exists a \in A: f(a) = b$

So we only need to prove that for each $b \in B, b \neq 0$ exists $b^{-1} \in B: b b^{-1} = 1$.

Lets take for each $b \in B, b \neq 0$ the $b' = f(a^{-1})$, where $f(a) = b$.

It is possible because $b \neq 0$, than $a \neq 0$, and $a^{-1}$ exists.

$$b b' = f(a)f(a^{-1}) = f(a a^{-1}) = f(1) = 1, \text{ i.e. we found } b^{-1}$$

Answer: We proved that for each nonzero element in $B$ exists inverse element. And it means that $B$ is a field too.

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