# Example of a projective variety that is not projectively normal but normal

I want to prove the following statement:

Let $Y$ be the quartic curve in $\mathbb{P}^3$ given parametrically by $(x,y,z,w)=(t^4,t^3u,tu^3,u^4)$. Then $Y$ is normal but not projectively normal.

To show it is normal, we can consider the affine patches parametrized by $(x,y,z)=(t^4,t^3,t)$ and $(y,z,w)=(u,u^3,u^4)$, each of the patches has affine coordinate ring isomorphic to $k[t]$, which is a UFD and hence normal.

But I couldn't show that $Y$ is not projectively normal. The homogeneous coordinate ring $S(Y)$ of $Y$ is isomorphic to $k[t^4,t^3u,tu^3,u^4]$, and I want to show that $S(Y)$ is not integrally closed.

Suppose $f\in k(t^4,t^3u,tu^3,u^4)$ is integral over $k[t^4,t^3u,tu^3,u^4]$, then $f$ should be integral over $k[t,u]$ and hence $f\in k[t,u]$. So I want prove $k(t^4,t^3u,tu^3,u^4)$ contains something like $t,t^2,t^3,ut,ut^2$ or $u^2t^2$, but I can't find an example.

Any help is appreciated!

• $k(t^4,t^3u,tu^3,u^4)$ contains $t^2u^2 = t^4 \cdot (tu^3)/(t^3u)$, which is integral over $k[t^4,t^3u,ut^3,u^4]$ since it is a root of $X^2 - t^4u^4 = 0$. May 19, 2016 at 3:23
• Oh I see, thank you so much! May 19, 2016 at 3:25

A good way to visualize this is to look the set $Q$ of all possible exponents. In this case, it's the set of all $\Bbb N$-linear combinations of $(4,0), (3,1),(1,3),(0,4)$. For our ring to have even a chance to be normal, $Q$ certainly (prove it!) must satisfy the following criterion (in which case we say $Q$ is saturated):

If $\mathbf a \in \Bbb Z Q$ and there's a positive integer $k$ such that $k \mathbf a \in Q$, then $\mathbf a \in Q$, where $\Bbb Z Q$ is the additive group generated by $Q$. (In our case, $\Bbb Z Q= \Bbb Z^2$)

Taking a look at our $Q$, we run into a glaring obstruction: $(2,2)$ is missing! Specifically, $2(2,2)=(4,0)+(0,4)\in Q$, yet $(2,2)\notin Q$.

Returning to our ring, this tells us that $t^2u^2$ satisfies the monic polynomial $X^2-t^4u^4$ over $S(Y)$, yet $t^2u^2\notin S(Y)$. Thus, $S(Y)$ isn't normal.

The above analysis works because $S(Y)$ is an affine semigroup ring (with affine semigroup Q). It turns out that for an affine semigroup ring, being normal is equivalent to its affine semigroup being saturated.

• Thanks, I see what you mean. But is there any simple method to decide whether $Q$ is saturated or not? May 19, 2016 at 4:32
• @chankifung Not really. Another description is that Q is the set of lattice points in a polyhedral cone. May 19, 2016 at 5:03
• Actually, come to think of it, there's probably an algorithm to compute saturation. I don't know if one off the top of my head, though. May 19, 2016 at 16:42