Assume the field $\mathbb{K}$, we're working with is algebraically closed. Let $V\subset\mathbb{A}^n$ be an irreducible affine variety. Suppose the coordinate ring $\mathbb{K}[V]$ is NOT a UFD. Choose $f\in \mathbb{K}[V]$ and suppose the factorization of $f$ into its irreducible factors is $f=f_1^{\alpha_1}\cdots f_r^{\alpha_r}$. Then, $\mathcal{V}(f)=\mathcal{V}(f_1)\cup\dotsb\cup\mathcal{V}(f_r)$. But, is it still true that $\mathcal{V}(f_i)$'s are all the irreducible components of $\mathcal{V}(f)$?
I know this is true if $\mathbb{K}[V]$ is a UFD.
Best Answer
An example: consider the hyperboloid with one sheet of equation $$x^2 + y^2 = z^2 + 1$$
There exist lines lying on this surface, for instance
$$x = \frac{4}{5} z+\frac{3}{5}\\ y = \frac{3}{5} z -\frac{4}{5}$$
Notice that the intersection of the surface with the plane $x - \frac{4}{5} z - \frac{3}{5}$ consists of two lines, the one above and $$x = \frac{4}{5} z+\frac{3}{5}\\ y = -\frac{3}{5} z +\frac{4}{5}$$
It is not hard to see that the element $f = x - \frac{4}{5} z - \frac{3}{5}$ is irreducible in the ring of functions of the surface ( a quadratic extension of a ring of polynomials). Of course, it is not prime, since we have the decomposition $$(x-\frac{4}{5} z - \frac{3}{5}) (x+\frac{4}{5} z + \frac{3}{5}) = -(y +\frac{3}{5} z-\frac{4}{5} )(y-\frac{3}{5} z+\frac{4}{5})$$