Let $V = \{(\alpha,\alpha^2,\alpha^3):\alpha\in\mathbb R\}$. Show that if $\Phi :\mathbb{R}\rightarrow\mathbb{R}^3$ is the polynomial function $\Phi(\alpha)=(\alpha,\alpha^2,\alpha^3)$ (which gives the morphism $\phi :\mathbb{R}\rightarrow V$ ), then $$F|_{\phi} :\mathbb{R}[x,y,z]/I\rightarrow\mathbb{R}[t], f(\overline{x},\overline{y},\overline{z})\mapsto f(t,t^2,t^3)$$ is a ring isomorphism, where $I = \langle y-x^2,z-x^3\rangle$.
From previous steps of this problem, I have shown that $I=\mathscr{I}(V)$ (the vanishing ideal of $V$). I know that I have to show that the kernel of this map is $I$ and the Image is $\mathbb{R}[t]$ so I can use the first isomorphism theorem but I am so confused that I'm unable to proceed.
Let $k$ be a field, and let $I$ be the ideal generated by $y-x^2$ and $z-x^3$ in $A:=k[x,y,z]$. There is a map of algebras $\phi:k[x,y,z]\to k[t]$ such that $\phi(x)=t$, $\phi(y)=t^2$, $\phi(z)=t^3$. This map vanishes on the generators of the ideal $I$, so it induces a map $\bar\phi:A/I\to k[t]$ such that $\bar\phi(x+I)=t$, $\bar\phi(y+I)=t^2$ and $\bar\phi(z+I)=t^3$.
On the other hand, there is obviously a map of algebras $\psi:k[t]\to A/I$ such that $\psi(t)=x+I$.
I claim that $\bar\phi$ and $\psi$ are mutually inverse morphisms abnd, in particular, isomorphisms.
To check that $\bar\phi\circ\psi$ is the identity of $k[t]$ it is enough to check that its value on $t$ is $t$, and that is easy: $$\bar\phi(\psi(t)))=\bar\phi(x+I)=t.$$
Similarly, to check that $\psi\circ\bar\phi$ is the identity of $A/I$ it is enough to check that it maps each of the three classes $x+I$, $y+I$ and $z+I$ to themselves: \begin{align} \psi(\bar\phi(x+I)) &= \psi(t) = x+I, \\ \psi(\bar\phi(y+I)) &= \psi(t^2+I) = x^2+I = y+I, \\ \psi(\bar\phi(z+I)) &= \psi(t^3+I) = x^3+I = z+I, \end{align} since $y-x^2$ and $z-x^3$ are in $I$.
This proves what we want.