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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 $mathbb C [x,y,z]$?

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. $a'=0$, $aneq0$. 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.

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.)

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).

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.

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

Let, $i$ be a square root of $-1$ in $K$.Then we have, $x^2+y^2+z^2=(x+iy)(x-iy)+z^2$
Since $x+iy$ is irreducible in $K[x,y]$ and $x+iy$ does not divide $x-iy;$ irreducibly follows from the Eienstein's Citeria in $K[x,y,z]=K[x,y][z]$