Is $y^{7}+x^{3}z^{4}-x^{2}yz^{4}$ irreducible in $\mathbb{C}[x,y,z]$?

Question

During an exam I had this morning, there was this polynomial $$y^{7}+x^{3}z^{4}-x^{2}yz^{4}$$ and at some point I asked myself if it was irreducible (in $\mathbb{C}[x,y,z]$).

I have to say that the focus of the exam was the curve associated with the polynomial so the approach I had may seem strange, but I welcome any kind of demonstration that it is irreducible.

This is broadly how I demonstrated the irreducibility: I found the singular points of the curve which were $(1,0,0)$ and $(0,0,1)$; I then found the places of the curve centered in the singular points and there was one for the first and three for the second; since the curve has only two singular points there are no multiple components so if it is reducible there must be at least a component of degree $5$; finally there can be no components that produce the places i found in the second singular point so the polynomial must be irreducible.

I don't know if there could have been another demonstration or if this one is correct, any help is welcome.

Answer 1

You could check the irreducibility of the curve just in the affine patch $z=1$ of the projective plane, thus leaving away the point $(1:0:0)$. There the curve is given by the equation $y^7+x^3-x^2y=0$. Consider the polynomial $y^7+x^3-x^2y$ as an element of $\mathbb{C}[y][x]$. To prove its irreducibility it suffices to prove its irreducibility in $\mathbb{C}(y)[x]$, where reducibility is equivalent to the existence of a root in the field $\mathbb{C}(y)$. Assume $\frac{z}{n}$ is such a root, where one can assume that $z,n\in\mathbb{C}[y]$ possess no common divisor. Then

$ z^3-z^2ny+y^7n^3=0 $

which implies $n\in\mathbb{C}$ and therefore $z$ divides $y^7$, which means that $z$ is some power of $y$. This is not possible.

Answer 2

COMMENT.-If $f(x,y,z)=y^7+x^3z^4-x^2yz^4$ were reducible then the product $f_1(x,y,z)f_2(x,y,z)=f(x,y,z)$ for integer values $(x,y,z)$ cannot be prime many times (the maximum of times is determined by the degree of $f$).

On the other hand we have, taking $f(x,1,1)=x^3-x^2+1$ this polynomial is irreducible, its only real zero is $x_0=\dfrac13\left(1-\sqrt[3]{\dfrac{2}{25-3\sqrt{69}}}-\sqrt[3]{\dfrac{25-3\sqrt{69}}{2}}\right)$.

It is known the conjecture of Bunyakovsky so $f(x,1,1)$ has infinitely many prime values for $x$ integer. In particular we have for $x=2,3,5,6,8$ the primes $5,19,101,181,449$ respectively.

We can conclude that $f(x,y,z)$ is irreducible.