Isomorphism of Cuspidal cubic to $\mathbb{C}^{1}$ "Lemma: "The affine variety $\mathcal{C}$ is not isomorphic to $\mathbb{C}^{1}$, where $\mathcal{C}$ is given by $y^{2}=x^{3}$. 
I don't understand the proof given in the book. From what I can see:


*

*By definition of isomorphism of algebraic subsets, if there were an isomorphism, then: 
$$ \mathbb{C}[\mathcal{C}] := \mathbb{C}[x,y]/\langle y^{2}-x^{3} \rangle \cong \mathbb{C}[t] $$ 

*So try to construct this and show that it leads to some contradiction. Let $\phi: \mathbb{C}[\mathcal{C}] \rightarrow \mathbb{C}[t]$ define this attempt at an isomorphism, and
$ \alpha(t) = \phi(x),\beta(t)=\phi(y)$. 
$$ \phi(0_{\mathbb{C}[\mathcal{C}]}) =0_{\mathbb{C}[t]}  $$
$$ x^{3}-y^{2} = 0_{\mathbb{C}[\mathcal{C}]} \implies \phi(x^{3}-y^{2})=\alpha^{3}(t)-\beta^{2}(t) = 0_{\mathbb{C}[t]} $$


*

*The claim is then made that because $\mathbb{C}[t]$ is PID: 
$$ \exists h(t) \in \mathbb{C}[t] \; : \alpha(t)=h(t)^{2}\;, \beta(t) = h^{3}(t)$$
But why is that true? $\langle \alpha^{3}(t) -\beta^{2}(t)\rangle$ is a valid ideal in $\mathbb{C}[t]$ but I don't see why the rest follows. 

*It's then claimed that as $\phi$ is surjective, $h(t) \not \in \mathbb{C}$ (and end of proof) But $h(t) \not \in \mathbb{C}$ does not imply $h(t) \not \in \mathbb{C}[t]$? 


 A: [Edited to make argument more similar to the textbook quoted.]
So far, you have shown that $\alpha^3 (t) = \beta^2 (t)$. If you consider how $\alpha (t)$ and $\beta (t)$ factorise into linear factors, you'll see that the only possible factorisation is something of the form
$$ \alpha(t) = k^2 (t - c_1)^{2m_1} \dots (t - c_n)^{2m_n},$$
$$ \beta(t) = k^3 (t - c_1)^{3m_1} \dots (t - c_n)^{3m_n},$$
for some $k \in \mathbb C^\star$, $n\geq 1$, $c_1, \dots c_n \in \mathbb C$, $m_1, \dots m_n \geq 1$.
(What's important here is that $\mathbb C[t]$ is a UFD.)
Thus
$$ \alpha(t) = h(t)^2, \ \ \ \beta(t) = h(t)^3,$$
where
$$ h(t) = k(t - c_1)^{m_1} \dots (t - c_n)^{m_n}.$$
So how can we derive a contradiction from this? We will prove that $h(t)$ is not in the image of $\phi$, which would contradict $\phi$ being surjective.
Indeed, suppose for contradiction that
$$ \phi \left( \sum_{p,q \geq 0} a_{p,q} x^p y^q \right) = \sum_{p,q \geq 0} a_{p,q} \alpha(t)^p \beta(t)^q = h(t)$$
for some $a_{p,q} \in \mathbb C$.
Taking cosets modulo $h(t)$, we have
$$ a_{0,0}\equiv 0 {\rm \ mod \ } h(t),$$
since $\alpha(t)$ and $\beta(t)$ are both divisible by $h(t)$. Hence $a_{0,0} = 0$, which is to say that the constant term vanishes in $\sum_{p,q \geq 0} a_{p,q} x^p y^q $.
Now, taking cosets modulo $h(t)^2$, we have
$$ 0 \equiv h(t) {\rm \ mod \ } h(t)^2$$
in view of the fact that $\alpha(t)$ and $\beta(t)$ are also divisible by $h(t)^2$, and in view of the fact that we have just shown that the constant term in vanishes in $\sum_{p,q \geq 0} a_{p,q} x^p y^q $.
This is saying that $h(t)^2$ divides $h(t)$, which is impossible, because $h(t) \notin \mathbb C$.
