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If we consider $f \in \mathbb{C}[x,y]$ an irreducible polynomial, then it is true that the domain $ \mathbb{C}[x,y]/(f)$ is normal iff it is UFD?

I think this is false. I was trying to prove that $\mathbb{C}[x,y]/(x^2-y)$ isn't a UFD using the equality $x^2 = y$, although it is normal, because $x^2 - y$ is regular curve.

user26857
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Hugo Rafael Oliveira Ribeiro
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    An exercise for you: $\mathbb{C}[x,y]/(y-x^2) \cong \mathbb{C}[t]$ via the map $p(x,y) + (y-x^2) \mapsto p(t,t^2).$ So that ring is actually a UFD. – Ragib Zaman Jun 03 '14 at 04:42
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    This is untrue. UFD implies normal but not conversely. normal + Cl(R) = 0 is equivalent to UFD. The reference I know is Hartshorne's algebraic geometry (the section on divisors), which probably actually references Matsumura. I don't have an example offhand though, sorry – user148177 Jun 03 '14 at 05:35
  • Actually, here's an example: k[x,y,z]/xy - z^2. This is normal but its not a UFD since xy = z^2 is not uniquely factorizable – user148177 Jun 03 '14 at 05:36

2 Answers2

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This is untrue. Being a UFD implies normal but not conversely. It is true that normal and Cl(R) = 0 is equivalent to UFD.

For a counterexample, let's look for a normal variety whose class ring of Weil divisors is nonzero. I know one: $$k[x,y,z]/(xy - z^2)$$ I know it is normal because it is Cohen Macaulay (complete intersection) and it is regular in codimension one (i.e. its singularities are in codimension $\geq$ 2). It's class group of Weil divisors is $\mathbb{Z}/2$ but it's easier to just see directly that it's not a UFD: $z^2 = xy$.

Some of the above is probably too much theory for this question, but I find it easier to think about these things "geometrically."

For a reference, see Hartshorne's Algebraic Geometry, the section on divisors in chapter 2. This will do a better job than me on the finer details.

Edit: Apologies: in your situation normal is equivalent to UFD.

A normal dimension one variety is regular. You can probably find this result in Hartshorne, but it's also on Wikipedia: http://en.wikipedia.org/wiki/Normal_scheme#The_normalization Regular local rings are UFDs by Auslander-Buchsbaum, and I think a ring that is locally a UFD is a UFD, which you can check by using one of these equivalent definitions (http://en.wikipedia.org/wiki/Unique_factorization_domain#Equivalent_conditions_for_a_ring_to_be_a_UFD).

Edit 2: Normal dimension one varieties are regular using Serre's characterization of normal varieties as (i) regular in codimension one and (ii) some condition which is satisfied by Cohen-Macaulay rings, in particular for complete intersections. A reference is http://stacks.math.columbia.edu/tag/033P and the conditions are R1 and S2 in their notation. An integral (reduced) dimension one variety regular in codimension one is regular.

Pedro
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user148177
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    Yes, I know this example, but my question is in 2 variables. – Hugo Rafael Oliveira Ribeiro Jun 03 '14 at 12:16
  • Okay I updated my answer – user148177 Jun 03 '14 at 13:23
  • It is not quite true that a ring that is locally a UFD is globally a UFD - e.g. a Dedekind domain is locally a DVR, but is a UFD iff it is a PID. Thus Nagata's criterion really requires that the multiplicative set is generated by prime elements. Your last sentence is also confusing: an integral scheme is necessarily reduced, and even if it is of dimension one, is not necessarily regular in codimension one. – zcn Jun 04 '14 at 21:07
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This is indeed not true: an explicit counterexample is $R = \mathbb{C}[x,y]/(y^2 - x^3 + x)$. This is a Dedekind domain that is not a UFD, as $(x,y)$ is a height $1$ prime that is not principal (although it is so locally). To see that $y^2 - x^3 + x$ is irreducible, one can use Eisenstein's criterion at the prime element $x$.

Geometrically, rings of the form $\mathbb{C}[x,y]/(f)$, $f$ irreducible, correspond to affine plane (integral) curves in $\mathbb{A}^2_{\mathbb{C}}$. Requiring $\mathbb{C}[x,y]/(f)$ to be normal is equivalent to requiring the curve to be nonsingular, since as pointed out in the other answer, $\mathbb{C}[x,y]/(f)$ is always Cohen-Macaulay. However, it turns out that the coordinate ring of a nonsingular affine plane curve is a UFD iff the curve is rational, i.e. has genus $0$ (see e.g. Corollary 3.23 here). The example above is in some sense the simplest counterexample, being the standard example of an affine elliptic curve, which has genus $1$.

zcn
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