TITLE: Quotient $K[x,y]/(f)$, $f$ irreducible, which is not UFD
QUESTION [7 upvotes]: Does anyone know an irreducible polynomial $f \in K[x,y]$ such that the quotient $K[x,y]/(f)$ is not a UFD? Is it known when this quotient is a UFD?
Thanks.

REPLY [3 votes]: The thing is: a UFD $A$ is always normal, i.e. every element in its field of quotients is integral over $A$ if and only if it belongs to $A$. The above example, $f(x,y)=x^3-y^2$, is one where the corresponding curve is not normal, namely:
$$t=\frac{y}{x}\ \mbox{ satisfies }\ t^3=y,$$
but $t$ itself does not belong to the ring.
A word of caution, though: if $f(x,y)=0$ is smooth, then the corresponding ring is locally factorial, i.e. its local rings at all maximals are factorial, though the global ring itself may not be: it will be Dedekind, no doubt, as $A$ Dedekind is equivalent to $A$ being (Noetherian and) a normal domain of Krull dimension $1$.
The question of global factoriality for rings of hypersurfaces is very difficult to answer, though you will find interesting examples in Hartshorne's Ample Subvarieties book.
MORAL: You need to get used to leaving the concept of global UFD in Algebraic Geometry. A variety being nonsingular at a point does imply that its local ring is factorial, but otherwise this condition is difficult to conceive geometrically.