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Let $a,b,c$ be positive integers. Then $f = x^a + y^b + z^c$ is irreducible in $\mathbb{C}[x,y,z]$.

By Gauss, $f$ is irreducible in $\mathbb{C}[x,y,z]$ iff is so in $\mathbb{C}(z)[x,y]$, and so iff in $\mathbb{C}(y, z)[x]$.

So by Eisenstein, it is sufficient to show that $y^b + z^c$ has a single prime factor.

If $b=c$, this factors through $\Pi(y + \zeta ^i z)$ for some primitive root of unity $\zeta$, so ok.

But if $b\neq c$?

This question is related to this post.

Thank you very much!

user26857
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k.j.
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  • $y^2+z^4=(y+iz^2)(y-iz^2)$, so your guess isn't quite right. You'll need to look at something more number-theoretic than just equality of $b$ and $c$... – KReiser Jun 27 '19 at 20:34
  • If $b\mid c$ you can factor as $\prod (y+\zeta^iz^{c/b})$ and likewise for $c\mid b.$ – Thomas Andrews Jun 27 '19 at 20:34
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    More generally $y^b+z^c$ is reducible if $b$ and $c$ share a common factor; if $d=\gcd(b,c)$ then $$y^b+z^c=(y^{b/d})^d+(z^{c/d})^d,$$ which factors as a product of homogenized cyclotomic polynomials. – Servaes Jun 27 '19 at 20:35
  • So can you prove that it is irreducible if $a,b$ relatively prime? – Thomas Andrews Jun 27 '19 at 20:37
  • Thanks for your comments. So I want to show: $y^b + z^c$ has no multiple prime factors. – k.j. Jun 27 '19 at 20:43
  • @agababibu It suffices to show that there is at least one irreducible prime factor with multiplicity $1$. – Servaes Jun 27 '19 at 20:55
  • @Servaes You're right. It's my mistake. – k.j. Jun 27 '19 at 20:56
  • The phrase "...it is sufficient to show that $y^b+z^c$ has a single prime factor." is not exactly what you mean; instead it is sufficient that it has an irreducible factor with multiplicity $1$. Or equivalently, that there exists a nonconstant $f\in\Bbb{C}[y,z]$ such that $f$ divides $y^b+z^c$ but $f^2$ does not. – Servaes Jun 27 '19 at 21:00
  • @Servaes Polynomials is a generic tag. Multivariate-polynomials is a kind of subtag, not used often btw. – user26857 Jun 28 '19 at 07:17

1 Answers1

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Let $f\in\Bbb{C}[y,z]$ be a factor of $y^b+z^c$ with multiplicity $m\geq1$, and let $g\in\Bbb{C}[y,z]$ be such that $y^b+z^c=f^mg$. Then taking derivatives with respect to $y$ and $z$ shows that $$by^{b-1}=mf^{m-1}f_yg+g_yf^m=f^{m-1}(mgf_y+g_yf),$$ $$cz^{c-1}=mf^{m-1}f_zg+g_zf^m=f^{m-1}(mgf_z+g_zf).$$ In particular we see that $f^{m-1}$ divides both $by^{b-1}$ and $cz^{c-1}$, and hence it is a constant. This shows that $y^b+z^c$ has no repeated irreducible factors. Hence by Eisensteins criterion, the polynomial $$x^a+y^b+z^c\in\Bbb{C}[y,z][x],$$ is irreducible as it is Eisenstein w.r.t. every irreducible factor of $y^b+z^c$.

Servaes
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