Wiki says that the coordinate ring $\mathbb{C}[x,y,z]/(x^2+y^2+z^2−1)$ of the complex sphere is not a unique factorization domain. I want to know why it is not a UFD.
We denote $X,Y,Z$ the residue class of $x,y,z$. Obviously, we have $(X+iY)(X-iY)=(1+Z)(1-Z)$ in $\mathbb{C}[x,y,z]/(x^2+y^2+z^2−1)$. But how to prove the irreducibility? Maybe there are some deep techniques of commutative algebra or algebraic geometry should be used in this question.
Best Answer
Techniques from algebraic geometry may be used for this problem. Everything in this answer is available in Hartshorne chapter II section 6 and probably also the section of Vakil dealing with divisors.
It's clear that with $A=\Bbb C[x,y,z]/(x^2+y^2+z^2-1)$, $X$ is normal (it's even smooth), so we can compute $\operatorname{Cl} X$ to see whether $A$ is a UFD or not.
Let $X'=V(-x_0^2+x_1^2+x_2^2+x_3^2)\subset\Bbb P^3$. Then $A$ is the coordinate algebra of $D(x_0)\cap X'$, and the class groups of $X$ and $X'$ are related by the exact sequence $$\Bbb Z\to \operatorname{Cl} X'\to\operatorname{Cl} X\to 0,$$ where the first map sends $1$ to the class of $V(x_0)\cap X$. As the middle group is $\Bbb Z^2$ from identifying $X'$ with $\Bbb P^1\times\Bbb P^1$, $\operatorname{Cl} X\neq 0$ and thus $A$ is not a UFD. (NB: if you're reading this and wondering how this all breaks for the real case mentioned at wikipedia, the class group of $X'$ depends on the base field!)
One can probably also use a norm map to prove that the elements in your factorization are indeed irreducible. See here for an example.