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Is $x^3+y^3+z^3-1$ irreducible over a field $k$ of characteristic $\neq 3$?

If I can show that $y^3+z^3-1$ is irreducible, I believe I can use Eisenstein's criterion. But I don't see how to show even this.

The other approach on my mind is to show that $k[x,y,z]/(x^3+y^3+z^3-1)$ is an integral domain, but this doesn't seem particularly promising either.

user26857
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Dominic Wynter
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    That is not promising because it is essentially the same statement. It is the Principle Of Conservation Of The Difficulty™ at work. – Mariano Suárez-Álvarez Jul 05 '16 at 01:04
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    A fancier, but general method is the following. Your polynomial is irreducible if and only if $F=x^3+y^3+z^3-t^3$ is irreducible-- the advantage is that the new polynomial is homogeneous. So, if it is reducible, then it can be written as a product of two homogeneous polynomials, $f,g$. But, then there exists a point $p\in\mathbb{P}^3$ (you may clearly assume $k$ is algebraically closed) where both $f(p)=g(p)=0$ and then $F$ is singular there. But $F_x,F_y, F_y, F_t$ have no common zeroes in 3-space (char $\neq 3$), so $F=0$ is non-singular. – Mohan Jul 05 '16 at 01:10

1 Answers1

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To show that $y^3+z^3-1$ is irreducible, you may try to show that $z^3-1$ has no repeated prime factor. This is where the restriction on the characteristic comes into play. Then, you can use Eisenstein's Criterion.

user26857
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Batominovski
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