# $Z\subseteq\mathbb{A}^n(k)$ such that $\Gamma(Z)$ is no UFD

Let $k$ be an algebraically closed field. Do you have an example for a Zariski closed irreducible subset $Z\subseteq\mathbb{A}^n(k)$ such that $\Gamma(Z)\cong k[X_1,\dots,X_n]/I(Z)$ is no unique factorisation domain anymore? What is the least $n$ such that such a set exists?

Case $n=1$: Is $k[X]/(p)$ always UFD for irreducible polynomials $p$?

In the case $n=1$, $k[X]$ is a PID, so irreducible elements are prime and primes are either zero are maximal. So $k[X]/(p)$ is either $k[X]$ or a field.

In the case $n=2$, take the polynomial $y^3 - x^2$. You can check that it is irreducible in $k[X,Y]$, and that the quotient is isomorphic as such: $k[X,Y]/(y^3-x^2) \cong k[X^2,X^3]$. But in the second ring, $X^2$ and $X^3$ are irreducible and not associate, but $X^6 = X^2 X^2 X^2 = X^3 X^3$ has two factorizations. This is because in the parlance of lattices, if we look at the look the lattice of the positive integers by divisibility, we have removed the meet of $2$ and $3$.

For the case $n>2$, we can just take the ring $k[X_1,...X_n]/(X_2^3-X_1^2)$, where we have simply adjoined more indeterminates, and the same factorization problem remains. So it is not a UFD.

• Thx. Asking for case $n=1$ was really dumb. I should have seen that by myself. At least I assumed that it wouldn't work for $n=1$ but for $n=2$. – principal-ideal-domain Jun 8 '14 at 21:38

The standard example is the ring of a cusp. Let $n=2$ and consider $I=(x^3-y^2)$. Then $k[x,y]/I \approx k[t^2,t^3]$. In the ring on the right hand side, we have the factorization $t^6 = t^3 \cdot t^3 = t^2 \cdot t^2 \cdot t^2$. Clearly $t^3$ and $t^2$ are irreducible.

Another way to see this is to note that $k[t^2,t^3]$ is not integrally closed, and any UFD is integrally closed. See this answer for details.

• To show that $k[x,y]/I\cong k[t^2,t^3]$ I considered $\phi:k[x,y]\to k[t^2,t^3]$ with $x\mapsto t^2$ and $y\mapsto t^3$. Clearly $I\subseteq \ker(\phi)$. But how do we show that $\ker(\phi)\subseteq I$? – principal-ideal-domain Jun 8 '14 at 21:41
• @principal-ideal-domain One way is to observe that both $I$ and $\ker \phi$ are prime ideals of Krull dimension one. This implies that they must be equal. – Fredrik Meyer Jun 9 '14 at 6:57
• You mean the height of $I$ and $\ker(\phi)$? Okay, we are in $k[x,y]$, a ring of Krull dimension 2. Since neither $I$ nor $\ker(\phi)$ are trivial or maximal we get immediately that $\ht(I)=1=\ht(\ker(\phi))$. But I'd like to rely the proof not on the fact that $k[x_1,\dots,x_n]$ has Krull dimension $n$ since I only know the proof for $n=1$. Do you have another argument? – principal-ideal-domain Jun 9 '14 at 7:58