$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$

Question

Is $x^2 +y^2 + z^2$ irreducible in $\mathbb C [x,y,z]$?

As $(x^2+y^2+z^2)= (x+y+z)^2- 2(xy+yz+zx)$,

$$(x^2+y^2+z^2)=\left(x+y+z+\sqrt{2(xy+yz+zx)}\right)\left(x+y+z-\sqrt{2(xy+yz+zx)}\right).$$

But how to show that none of these factors belong to $\mathbb C [x,y,z]$?

Answer 1

I don't quite see why you are interested in this particular square root, $\sqrt{xy+yz+zx}$. In general, adjoining an element to a ring may or may not affect irreducibility of a given element. For example, $13$ is irreducible in $\mathbb{Z}$ factors as $(3+2i)(3-2i)$ in $\mathbb{Z}[i]$, remains irreducible in $\mathbb{Z}[\sqrt{-5}]$, factors as $(4+\sqrt3)(4-\sqrt3)$ in $\mathbb{Z}[\sqrt3]$, remains irreducible in $\mathbb{Z}[\sqrt7]$ et cetera.

Anyway, you can prove irreducibility of $x^2+y^2+z^2$ for example as follows. $\mathbb{C}[x,y,z]=\mathbb{C}[u,v,z]$ with $u=x+iy$, $v=x-iy$. Your polynomial then looks like $x^2+y^2+z^2=uv+z^2$. If this were not irreducible, it would be a product of two linear polynomials. As this polynomial is homogeneous, so are the presumed factors. So we need to rule out the possibility $$ uv+z^2=(au+bv+cz)(a'u+b'v+c'z) $$ for some constants $a,b,c,a',b',c'$. As $aa'=0$ one of those constants is zero, w.l.o.g. $a'=0$, $a\neq0$. Similarly from $bb'=0$ we see that one of those also needs to be zero. Clearly we must assume $b=0, b'\neq0$. This leaves us $$ uv+z^2=(au+cz)(b'v+c'z) $$ with $a,c,b',c'$ all non-zero ($cc'=1$). This forces non-zero coefficients to terms $vz$ and $uz$, so no factorization is possible.

Answer 2

If your polynomial were reducible, it would be the product of two linear polynomials. Just check that it isn't.

Answer 3

First notice that both $\mathbb C[x,y,z]$ and $\mathbb C[x,y]$ are UFDs since $\mathbb C$ is a field.

Hence to see that $x^2+y^2+z^2$ is irreducible in $\mathbb C[x,y,z]$ it suffices to show that the monic polynomial $x^2+y^2+z^2$ in the variable $z$ has no root in the ring $\mathbb C[x,y]$. But this holds since $x^2+y^2=(x+iy)(x-iy)$ is the factorization into irreducible elements of $x^2+y^2$ in $\mathbb C[x,y];$ the powers of the irreducible elements in this factorization are not even.

Answer 4

Also, this is part of a more general pattern, using Gauss' Lemma and Eisenstein criterion: for positive integers $\ell,m,n$, $x^\ell+y^m+z^n$ is irreducible over any field $k$ of characteristic not dividing the exponents. Prove this via Gauss/Eisenstein by first noting that $k[x,y,z]$, $k(z)[x,y]$, and such are UFDs. By Gauss, irreducibility in $k[z,y,z]$ is equivalent to that in $k(z)[x,y]$. By Eisenstein, the polynomial is irreducible in $k(z)[y][x]$ if $y^m+z^n$ has some prime factor in $k(z)[y]$ that does not occur twice. Since the char of the field does not divide the exponents, this is easily so. (For $m=n$ and $k$ alg closed it is easy to write down factors $y+z$, etc.)

Answer 5

There is a more geometrical way to prove that $x^2+y^2+z^2$ is irreducible in $\mathbb{C}[x,y,z]$. Notice that, since the polynomial is homogeneous, the equation $x^2+y^2+z^2 = 0$ describes a curve $\mathcal{C}$ in $\mathbb{P}^2(\mathbb{C})$.

If the polynomial is reducible, $\mathcal{C}$ has two irreducible components (counted with multiplicity, i.e. $\mathcal{C}$ is made of two possibly coincident straight lines). By Bézout's theorem, these components must intersect in at least one point, which must be a singular point of $\mathcal{C}$. Then it suffices to check that $\mathcal{C}$ has no singular points to conclude that the polynomial is irreducible, and this is trivial.

This argument can also be applied for the more general case of $x^n+y^n+z^n$ (which correspond to the so-called Fermat curves).

Answer 6

We can consider $X^2+Y^2+Z^2$ as a polynomial in $R[X]$, where $R=\mathbb C[Y,Z]$. Now apply the Eisenstein criterion with the irreducible $Y+\mathrm iZ$ in $R$:

• It divides $Y^2+Z^2$;
• It does not divide $1$, the coefficient of $X^2$;
• It does not divide $(Y^2+Z^2)^2$.

Irreducibility follows.