How can I prove $y^2-(x-z^2)^3$ is irreducible?

Question

Suppose $F$ is an algebraically closed field.

I'm trying to find the set of singular points of the algebraic set $X=V(y^2-(x-z^2)^3)$. So I may need to prove the polynomial $y^2-(x-z^2)^3 \in F[x,y,z]$ is irreducible. If I can do that, then I will get $I(X)=\langle y^2-(x-z^2)^3 \rangle$. I have tried Eisenstein criterion but that didn't work.

Is there any other way to prove this polynomial is irreducible? Or can I find the singular points without doing this step?

Answer 1

Here's a trick to make Eisenstein work:

Let $L/K$ be a finite field extension and let $f \in K[x]$ be irreducible such that $\deg(f)$ and $[L:K]$ are coprime. Then $f$ is irreducible over $L$.

Proof:Let $E/L$ be splitting field of $f$ and $\alpha$ be a root of $f$. We have $[L(\alpha):K]=[L(\alpha):K(\alpha)][K(\alpha):K]=[L(\alpha):K(\alpha)]\deg(f)$ and $[L(\alpha):K]=[L(\alpha):L][L:K]$, so $[L:K]$ and $\deg(f)$ both divide $[L(\alpha):K]$, thus also their product $\deg(f)[L:K]$ does, as they are coprime. On the other hand $[L(\alpha):L] \leq \deg(f)$, so $[L(\alpha):K]\leq [L:K]\deg(f)$. Putting these together, we obtain $[L(\alpha):K]=[L:K]\deg(f)$, which implies $[L(\alpha):L]=\deg(f)$ which means that $f$ is irreducible over $L$.

The $k$-algebra morphism $x \mapsto x+z^2,y \mapsto y, z \mapsto z$ is an automorphism of $k[x,y,z]$, so we can consider $f(x+z^2,y,z)=x^3-y^2$ instead This as a degree $3$ polynomial in $F(y,z)[x]$. $F(y^2,z) / F(y,z)$ is a degree $2$ extension and over $F(y^2,z)$, $f$ is Eisenstein with $p=y^2$. Applying the lemma (and Gauss lemma to go from $F(y,z)[x]$ to $F[x,y,z]$, we obtain that $f$ is irreducible in $F[x,y,z]$.

Answer 2

Since your polynomial (call it $p(x,y,z)$) has degree $2$ in $y$, if it's not irreducible it has a factor of degree $0$ or a factor of degree $1$ in $y$. It's easy to see there is no nontrivial factor of degree $0$ in $y$ (that would mean for some values of $x$ and $z$, $p(x,y,z) = 0$ for all $y$, and that's obviously not true. A factor of degree $1$ would mean there's a rational function $c(x,z)$ such that $p(x,c(x,z),z) = 0$, i.e. $c(x,z)^2 = (x - z^2)^3$. But the right side has degree $3$ in $x$, so that's impossible.