I need to prove that $x^3 + y^3 + z^3$ is irreducible in $k[x, y, z]$, where $k$ is a field of characteristic $\ne 3$.
I think Eisenstein's criterion could be applied somehow here, but I'm not sure how to do it.
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See this question. – Dietrich Burde Jan 19 '20 at 15:52
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Apply Eisenstein using the prime $,y+z,$ as in the linked dupes. – Bill Dubuque Jan 20 '20 at 01:18
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Assume $P=x^3+y^3+z^3$ is reducible, write $P=fg$.
Let $s=y^3+z^3 \in R:=k[y,z]$, $f,g \in R[x]$ have $x^3+s$ as a product.
So either $g$ is a constant in $R$, or $g$ has degree $1$ (up to exchanging $f$ and $g$).
In the first case, $g$ must divide (in $R$) the dominant coefficient of $fg=x^3+s$, so $g|1$, so $g \in R^{\times}$ so $g \in k^{\times}$ and the factorization is trivial.
In the second case, as above, the dominant coefficient of $g$ is in $R^{\times}=k^{\times}$, so we may assume it is $1$. Write then $g=x-r_0$, $r_0 \in R$. Then $r_0(y,z)^3=y^3+z^3$.
It follows that $r_0$ must be a linear polynomial (its degree wrt $y$ or $z$ must be a third of that of $y^3+z^3$ which is $3$, so $\deg{r_0}=1$, and $r_0(0,0)=0$ so $r_0$ is linear. As $k$ hasn’t characteristic $3$, we get a contradiction by expanding the equation.
Aphelli
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Note that Eisenstein-based arguments can’t really work on their own here, as all the terms are of multiplicity $\geq 2$. – Aphelli Jan 19 '20 at 15:52
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Oh, good trick. I didn’t even think of applying Eisenstein to a polynomial not $y$ or $z$. – Aphelli Jan 20 '20 at 08:19