1
$\begingroup$

Let $R,S$ be rings and let $\theta:R \rightarrow S$ be an isomorphism. Show that R is an integral domain if and only if S is an integral domain.

I'm not sure how to use the isomorphism to help me achieve this.

  • 0
    This is pretty straight forward. First assume $R$ is an integral domain and consider $x_s$, $y_s$ in $S$ such that $x_sy_s=0$. The isomorphism tells you there are $x_r$ and $y_r$ in $R$ such that $\theta(x_r)=x_s$ and $\theta(y_r)=y_s$. Moreover, $x_ry_r=0$ and $R$ is an integral domain. You get the pattern? The other direction is the same. – 2017-02-26

1 Answers 1

0

If $a,b \in R$ are nonzero elements such that $ab = 0_R$, then $\theta(a)\theta(b) = 0_S$ where the images of $a$ and $b$ are also nonzero since $\theta$ is an isomorphism. Therefore if $R$ has zero divisors, then so does $S$, and the reverse direction applies since $\theta^{-1}$ is also an isomorphism. Further, if $R$ is commutative, then for all $p,q \in S$, $p = \theta(a)$ and $q = \theta(b)$ for some $a,b \in R$. Then $$pq = \theta(a)\theta(b) = \theta(ab) = \theta(ba) = \theta(b)\theta(a) = qp$$ so $S$ is also commutative. Again, the reverse direction also applies since $\theta^{-1}$ is also an isomorphism.

So if $R$ and $S$ are isomorphic, and $R$ has no zero divisors, then $S$ has no zero divisors, and if $R$ is commutative, then $S$ is also commutative. Therefore if $R$ is an integral domain, then so is $S$.

  • 0
    By 'the reverse direction', do you mean for the 'if and only if' part of the statement? – 2017-02-26
  • 0
    Yes. The reverse direction of "If $R$ is commutative then $S$ is commutative" would be "If $S$ is commutative then $R$ is commutative". – 2017-02-26