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Let $k$ be an algebraically closed field such that $\textrm{char(k)} \neq 2$ and let $n$ be a fixed positive integer greater than $3$

Suppose that $m$ is a positive integer such that $3 \leq m \leq n$.

Is it always true that $f(x_{1},x_{2},\ldots,x_{m})=x_{1}^{2}+x_{2}^{2}+\cdots+x_{m}^{2}$ is irreducible over $k[x_{1},x_{2},\ldots,x_{n}]$?

I think yes. For $m=3$ we need to check that $f(x,y,z)=x^{2}+y^{2}+z^{2}$ is irreducible, yes? can't we use Eisenstein as follows?

Note $y+iz$ divides $y^{2}+z^{2}$ and $y+iz$ is irreducible over $k[y,z]$ and $(y+iz)^{2}$ does not divide $y^{2}+z^{2}$.

Therefore $f(x,y,z)=x^{2}+y^{2}+z^{2}$ is irreducible. Now we induct on $m$. Suppose the result holds for $m$ and let us show it holds for $m+1$.

So we must look at the polynomial $x_{1}^{2}+\cdots+x_{m}^{2}+x_{m+1}^{2}$. Consider the ring $k[x_{m+1}][x_{1},..,x_{m}]$, we have a monic polynomial and by hypothesis $x_{1}^{2}+\cdots+x_{m}^{2}$ is irreducible over $k[x_{1},\ldots,x_{m}]$ and $(x_{1}^{2}+\cdots+x_{m}^{2} )^{2}$ does not divides $x_{1}^{2}+\cdots+x_{m}^{2}$ so Eisenstein applies again and we are done.

Question(s): Is this OK? In case not, can you please provide a proof?

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It might be easier to deal with the case $m=3$, since it implies the rest. – André Nicolas Jun 22 '12 at 23:59

Let $A$ be a UFD. Let $a$ be a non-zero square-free non-unit element of $A$. Then $X^n - a \in A[X]$ is irreducible by Eisenstein.

$Y^2 + Z^2 = (Y + iZ)(Y - iZ)$ is square-free in $k[Y, Z]$. Hence $X^2 + Y^2 + Z^2$ is irreducible in $k[X, Y, Z]$ by the above result.

Let $m \gt 2$. By the induction hypothesis, $X_{1}^{2}+\cdots+X_{m}^{2}$ is irreducible in $k[X_{1},\ldots,X_{m}]$. Hence $X_{1}^{2}+\cdots+X_{m+1}^{2}$ is irreducible in $k[X_{1},\ldots,X_{m+1}] $by the above result.

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I thought your proof was wrong. Perhaps it was just a typo? – Makoto Kato Jun 23 '12 at 2:55
Why is wrong? Where is the flaw? – user31509 Jun 23 '12 at 3:08
Since you corrected your flaw, there's no flaw now. – Makoto Kato Jun 23 '12 at 3:14
Wait but doesnt the last line is flawed? i.e the square part – user31509 Jun 23 '12 at 3:18
Since you corrected it by my suggestion, there's no flaw now. – Makoto Kato Jun 23 '12 at 3:20

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