For ring $R$, the opposite ring $R^{op}$ is the abelian group $(R,+)$ with multiplicative operation $\circ$ defined by $x\circ y=yx$. I am trying to prove the following propositions.
- Show that $R^{op}$ is a ring and that $(R^{op})^{op}\cong R$.
- Show that $R\cong S$ if and only if $R^{op}\cong S^{op}$.
- Show that $I$ is a left ideal of $R$ if and only if $R$ is a right ideal of $R^{op}$.
What I have so far is:
1) To prove that $R^{op}$ is a ring, we must show that $(R, +)$ is an abelian group, multiplication is associative, and multiplication is distributive w.r.t. addition. By definition of $R^{op}$, we trivially have that $(R,+)$ is an abelian group. Now, for associativity observe that our operation is $x\circ y = yx$. Take elements $a,b,c \in R$. Then, $ (a\circ b)\circ c = ba \circ c = cba = (cb)a = a \circ (cb) = a\circ (b\circ c) $ Finally, we can show distribution. $ a \circ (b+c) = (b+c)a = ba + ca = a\circ b + a\circ c $, and $ (a+b) \circ c = c(a+b) = ca + cb = a\circ c + b\circ c $. So $R^{op}$ is a ring.
However, I always struggle with showing isomorphism. I am not sure how to proceed for either (1) or (2) in this regard. Also, for (3) does this seem like a typo? If it said $I$ is a left ideal of $R$ $\iff$ $I$ is a left ideal of $R^{op}$ it is easy, but as stated I am confused.
1) $(R^{op})^{op}$ actually equals $R$: it is the same set with the same operations.
2) Hint: Any isomorphism between $R$ and $S$ is automatically an isomorphism between the opposite rings, and vice versa.
3). It should read “$I$ is a left ideal of $R$ if and only if $I$ is a right ideal of $R^{op}$.”