Examine with justification whether the following pairs of rings are isomorphic:
(a) $\mathbb{R}[x]$ and $\mathbb{C}[x]$
(b) $\mathbb{Q}[x]/(x^2-x)$ and $\mathbb{Q}\times\mathbb{Q}$
My attempt for (a): We can prove by the First Isomorphism Theorem that $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}[x]$. If we assume that $\mathbb{R}[x] \cong \mathbb{C}[x]$, then by transitivity $\mathbb{R}[x]/(x^2+1) \cong \mathbb{R}[x]$ which implies $(x^2+1)=(0)$. Contradiction. Hence $\mathbb{R}[x] \ncong \mathbb{C}[x]$
My attempt for (b): $(x^2-x) \subseteq (x)$ implies $(x^2-x)$ is not maximal, therefore $\mathbb{Q}[x]/(x^2-x)$ is not a field, but $\mathbb{Q}\times\mathbb{Q}$ is. Therefore not isomorphic.
This is a question that appeared in a past question paper of an exam I'm preparing for. Is my reasoning correct?
You cannot argue like this in a). If $I \neq 0$, then $R$ and $R/I$ are definitely not isomorphic as $R$-modules, but there are examples, where they are isomorphic as rings. For instance take an infinite direct product of one given ring and let $I$ be one factor.
Here is a much simpler argument: The two rings are not isomorphic, because their unit groups $\mathbb R^*$ and $\mathbb C^*$ are not isomorphic. One sees this by looking at elements of finite order (roots of unity).
For b): The Chinese remainder theorem shows they are isomorphic. $\mathbb Q \times \mathbb Q$ is not a field.