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

  • $\begingroup$ For b), all you have to prove is that, if every non-zero element in $R$ has an inverse, the same is true for $S$. $\endgroup$ – Bernard Apr 18 '16 at 21:14
  • $\begingroup$ Isomorphisms are like mirrors that reflect all algebraic properties. $\endgroup$ – Santiago Apr 19 '16 at 0:18
  • $\begingroup$ The whole point of an isomorphism $f$ from $R$ to $S$ is that if x and y are expressions formed from elements of R by addition and multiplication, x = y iff f(x) = f(y). Therefore any true statements about R that are given in the form of identities are true in S when you apply $f$ to both sides of the identity. $\endgroup$ – Vik78 Apr 19 '16 at 0:36

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 $f(r)f(r^{-1})=f(rr^{-1})=f(1_R)=1_S$


I think (a) is basically right, but it can be stated more simply.

Where $f: R \xrightarrow{\sim} S$ is as above, choose $s, s'$ in $S$ such that $s \neq 0_{S}$, $s' \neq 0_{S}$.

Let $r, r'$ in $R$ be such that $f(r) = s, f(r') = s'$.

$r \neq 0_{R}$, else $f(r)=f(0_{R})=0_{S}=s$, and similarly for $r'$.

Since $R$ is an integral domain, $rr' \neq 0_{R}$, and the same argument applies to show $f(rr')=f(r)f(r')=ss' \neq 0_{S}$.


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