Given the affine variety $A:=Z(y^{2}-P(x)) \subset \mathbb{C} ^{2} $, where $P(x)$ is a polynomial with $\deg P \geq 2$, I need to show that $A$ is not isomorphic to $ \mathbb{C}$.
I know it has something to do with the number of singularities, but I didn't manage showing two varieties with different number of singularities are not isomorphic, and even that $A$ must have singularities. Detailed answer would be much appreciated.
Let $k$ be a field of characteristic $\neq 2$ and $P \in k[X]$ with $k[X,Y]/(Y^2-P(X)) \cong k[T]$. I claim that $\mathrm{deg}(P) =1$.
By tensoring with an algebraic closure, we may assume that $k$ is algebraically closed. It is clear that $n:=\mathrm{deg}(P) \geq 1$. If $l$ is the leading coefficient of $P$, then we may use the automorphism $Y \mapsto \sqrt{l} \cdot Y$ of $K[Y]$ and reduce to the case $l=1$. Write $P=(X-a_1) \cdot \dotsc \cdot (X-a_n)$ for $a_i \in k$. We may assume that the $a_i$ are pairwise distinct, because otherwise $V(Y^2-P(X))$ is singular. Let $\phi : k[X,Y]/(Y^2-P(X)) \to k[T]$ an isomorphism of $k$-algebras. It is determined by $g=\phi(X)$ and $f=\phi(Y)$, which satisfy $f^2=P(g)=(g-a_1) \cdot \dotsc \cdot (g-a_n)$. Now, $\phi$ induces an isomorphism $$k[Y]/(Y^2) = k[X,Y]/(Y^2-P(X)) / (X-a_i) \cong k[T]/(g-a_i).$$ It follows that $g-a_i=\lambda_i (T-b_i)^2$ for some $\lambda_i \in k^*$ and $b_i \in k$. Observe that $f$ is associated to $(T-b_1) \cdot \dotsc \cdot (T-b_n)$. Since $\phi$ induces an isomorphism $$k[X]/(P(X)) = k[X,Y]/(Y^2-P(X)) / (Y) \cong k[T]/(f),$$ the $b_i$ are pairwise distinct. On the other hand, $g=\lambda_i (T-b_i)^2+a_i = \lambda_i T^2 - 2 \lambda_i b_i T + \lambda_i b_i^2 + a_i$ does not depend on $i$. It follows that $\lambda_i$ and $\lambda_i b_i$ don't depend on $i$, hence also $b_i$ does not depend on $i$. This means $n=1$.