Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

1 Answer

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.