(This question originates from Pinter's Abstract Algebra, Chapter 19, Exercise C4.)
Let $\phi$ be the function from $\mathscr{F}(\mathbb{R})$ to $\mathscr{F}(\mathbb{Q},\mathbb{R})$ defined as follows: \begin{align*} \phi(f) &= f_{\mathbb{Q}} = \text{the restriction of }f\text{ to }\mathbb{Q} \end{align*} (Note the domain of $f_{\mathbb{Q}}$ is $\mathbb{Q}$ and on this domain $f_{\mathbb{Q}}$ is the same function as $f$.)
Let $J$ be the subset of $\mathscr{F}(\mathbb{R})$ consisting of all $f$ such that $f(x)=0$ for every rational $x$.
Explain why $J$ is an ideal of $\mathscr{F}(\mathbb{R})$ and $\mathscr{F}(\mathbb{R})/J\cong\mathscr{F}(\mathbb{Q})$.
Question: Is there a typo in the question in that the isomorphism should be $\mathscr{F}(\mathbb{R})/J\cong\mathscr{F}(\mathbb{Q},\mathbb{R})$ rather than $\mathscr{F}(\mathbb{R})/J\cong\mathscr{F}(\mathbb{Q})$ ?
The reason is that $\mathscr{F}(\mathbb{R})/J\cong\mathscr{F}(\mathbb{Q},\mathbb{R})$, but $\mathscr{F}(\mathbb{Q},\mathbb{R})$ is not isomorphic to $\mathscr{F}(\mathbb{Q})$. Let $\psi$ be the function from $\mathscr{F}(\mathbb{Q},\mathbb{R})$ to $\mathscr{F}(\mathbb{Q})$ such that $\psi(f)$ is the restriction of the range of $f$ to $\mathbb{Q}$. But $\psi$ is not well defined. For example, $\psi(f)$ would be undefined for the function $f(x) = \pi$ for all $x\in \mathbb{Q}$.
Am I right?
Yes, it is a typo, and should be $\mathscr F(\Bbb R)/J\cong \mathscr F(\Bbb Q, \Bbb R)$.