6
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

Yes, I think that the standard example is to take $f(x,y)=x^3-y^2$. Then when you look at it this is the same as the set of polynomials in $K[t]$ with no degree-one term. That is, things that look like $c_0+\sum_{i>1}c_it^i$, the sum being finite, of course. You prove this by mapping $K[x,y]$ to $K[t]$ by sending $x$ to $t^2$ and $y$ to $t^3$. I’ll leave it to you to check that the kernel is the ideal generated by $x^3-y^2$, that the image is what I said, and that this is not UFD.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.