Let $R \simeq S$ be isomorphic commutative rings with unity. Prove the following:
a). If $R$ is an integral domain then $S$ is an integral domain.
For this, I said that If I let $f: R \to S$ be an isomorphism and $(R, + , \circ), (S, \oplus, \ast)$, then let $x, y\in R$, such that $x \circ y = 0_R$, because $f$ is an isomorphism, $f(x \circ y) = f(x) \ast f(y) = f(0_R)$. If I let $R$ be an integral domain then, as per the definition $R$ is commutative, has unity, and no zero divisors.
So $f(x \circ y) = f(x) \ast f(y) = f(0_R)$ where we have either $f(x \circ y) = f(0_R) \ast f(y) = f(0_R)$ or $f(x \circ y) = f(x) \ast f(0_R) = f(0_R)$.
Since $f$ is an isomomorphism, let $s = f(x), t = f(y)| s,t \in S$. Then $s \ast t = f(0_R)$. By definition, $S$ also has unity so $0_S$ exists. If we map $0_R \to 0_S$. Then either $s = 0_S$ or $t = 0_S$.
So then, since $S$ has unity, is commutative, and has a no zero divisors as we showed above, it is also an Integral domain.
I'm not sure I have this correctly.
b). If $R$ is a field then $S$ is a field.
I don't really understand how to do this one. Since $S$ is an integral domain, it has a $0_S$. So I need to show that $S$ distributes $\ast$ over $\oplus$.
Any help with these is appreciated
Best Answer
For part $b)$ you just have to prove $S$ has inverses.
Take an element of $S$ other than zero, then it is of the form $f(r)$, with $r$ different from zero, now you just need to show $f(r^{-1})$ is an inverse for $f(r)$ (notice $r^{-1}$ exists because $R$ is a field).
Since the ring is commutative all we have to show is $f(r)f(r^{-1})=1_S$.
Notice $f(r)f(r^{-1})=f(rr^{-1})=f(1_R)=1_S$