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Hailong Dao
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Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.

A natural way to provideI believe the existence of an irreducible element whose ideal is not radical ismight be related to create   non-trivial torsions in the class group (I will assume the ring is normal, one can avoid it by using Chow group of codimension one instead). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that

$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime and $y \notin P$, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal since $z\notin (x)$: if it is, it would have to be generated by assumption and $x\notin (z)$ by irreducibility of$x$ because $x$ is irreducible, but $z\notin (x)$ by assumption.

This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).

So it seems to me the proof you want would rule out certain torsions in the class group but without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a hardharder problem (think about elliptic curves!)

Does that makeI hope this makes some sense?.

Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.

A natural way to provide an irreducible element whose ideal is not radical is to create non-trivial torsions in the class group (I will assume the ring is normal). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that

$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal since $z\notin (x)$ by assumption and $x\notin (z)$ by irreducibility of $x$.

This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).

So it seems to me the proof you want would rule out certain torsions in the class group without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a hard problem (think about elliptic curves!)

Does that make sense?

Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.

I believe the existence of an irreducible element whose ideal is not radical might be related to   non-trivial torsions in the class group (I will assume the ring is normal, one can avoid it by using Chow group of codimension one instead). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that

$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime and $y \notin P$, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal: if it is, it would have to be generated by $x$ because $x$ is irreducible, but $z\notin (x)$ by assumption.

This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).

So it seems to me the proof you want would rule out certain torsions in the class group but without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a harder problem (think about elliptic curves!)

I hope this makes some sense.

Source Link
Hailong Dao
  • 30.5k
  • 5
  • 102
  • 188

Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.

A natural way to provide an irreducible element whose ideal is not radical is to create non-trivial torsions in the class group (I will assume the ring is normal). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that

$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal since $z\notin (x)$ by assumption and $x\notin (z)$ by irreducibility of $x$.

This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).

So it seems to me the proof you want would rule out certain torsions in the class group without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a hard problem (think about elliptic curves!)

Does that make sense?