0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $f(rr')=f(r)f(r')=ss' \neq 0_{S}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.