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I've tried to prove that $x^2+y^2-1$ is irreducible in $C[x,y,z]$.

Attempt: Let $f[y](x) = x^2 + (y^2-1) \in C[y][x]$, then $f[y](x+1) = x^2 + 2x + y^2$. Let $g[x](y) = y^2 + (x^2+2x)$, then it's irreducible by Eisenstein's Criterion (x is irreducible). Hence $x^2+y^2-1$ is irreducible.

Now I know that $x^2+y^2-1$ is actually reducible, so which step of my proof goes wrong?

leo
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  • You have $x^2-z^2-1$ and $x^2-Z^2-1$. I suppose then $Z=z$. – Dietrich Burde Mar 10 '24 at 16:00
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    Why not ask a new question instead of completely rewriting this one? Anyways, $x^2+y^2-1$ is irreducible over the complex numbers, so how do you know it's reducible? – KReiser Mar 10 '24 at 17:29
  • What does z have to do with anything? Your polynomial only involves $x$ and $y$. Writing it as $x^2 + (y+1)(y-1)$, we see it is Eisenstein at $y+1$ and at $y-1$, so it is irreducible as a two-variable polynomial over the complex numbers. – KCd Mar 10 '24 at 17:30
  • See here in the linked dupe. – Bill Dubuque Mar 10 '24 at 18:27

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