# Is a height one ideal in a UFD principal?

One of the defining features of a UFD is that any height one prime ideal is principal (see Wikipedia).

Is it also true that any height one (i.e. every prime minimal among those containing it has height $1$) ideal is principal? Is this true even in the case of polynomial rings?

If not, please provide a counterexample.

• You mean "every minimal prime ideal containing it". – MooS Mar 12 '15 at 6:03
• This is a naïve and elementary question and thus a great one! As far as I know it is not explicitly addressed in books on commutative algebra and this gap has now be filled thanks to Dorebell's question. – Georges Elencwajg Mar 12 '15 at 10:08
• Still it holds that in a noetherian UFD, an ideal has height $\le 1$ iff it's contained in a proper principal ideal (this characterization follows from standard facts; anyway it does not make reference to prime ideals). – YCor May 14 '17 at 13:24

You can take any UFD of dimension $\geq 2$, an irreducible element $f$ and a maximal ideal $\mathfrak m$ containing $f$. Then $I=(f) \cdot \mathfrak m$ has height $1$ (the only minimal prime over $I$ is $(f)$), but is not principal.

In particular the polynomial ring over a field admits a counterexample, for instance $(X^2+Y^2-1) \cdot (X-1,Y)$.

The counterexample given in the other answer is obtained by the choice $f=X, \mathfrak m=(2,X)$.

$\mathbb{Z}[X]$ is a counterexample. The ideal $(2X,X^2)$ has height $1$, but is not principal.

Note that any such counterexample must have dimension $2$ or greater, as a UFD with dimension $1$ is a PID. I don't believe that a polynomial ring over a field can be a counterexample, but I can't think of a proof offhand.

Note: As MooS and Georges have pointed out, polynomial rings, as well as any UFDs of dimension $\geq 2$, are counterexamples.

• Dear Slade: Nice post! I have given an example with a polynomial ring in my answer. – Georges Elencwajg Mar 12 '15 at 10:09

A geometric example would be the notorious ideal $I=(XY,X^2)\subset A=\mathbb C[X,Y]$.
It has height one and is not principal: the closed subscheme $V(I)\subset\operatorname {Spec}(A)=\mathbb A^2_\mathbb C$ is the $X$-axis with some "fuzz" added to the origin, rendering $I$ not radical: $I\subsetneq \sqrt I$.
This explains the phenomenon:

$\bullet$ The height of the ideal $I$ corresponds to the codimension of the subscheme $V(I)$, which is $1$.
That codimension is a rough, purely topological invariant of the subscheme, which doesn't see nor care about the fine points of the schematic structure of $V(I)$, i.e. about the nilpotents in $A/I$.
$\bullet \bullet$ Being principal is a more sensitive invariant of an ideal, which does care about these nilpotents, as demonstrated here.