# Codimension and Krull's principal ideal theorem

I know that in an UFD, each minimal prime will be principal. So, let $k[x_1,...,x_n]$ be a polynomial ring over a field. Further, set $S =k[x_1,..,x_n]/P$, and suppose that this is an UFD. A subvariety $V(Q)$ of $V(P)$ of codimension one will give that $Q$ is a minimal prime in $S$, and thus principal there. What does this imply for the generators of the subvariety? Must it be defined by one equation, and so why? My thinking so far is that: Say that $Q$ is minimal in $S$. This corresponds to a prime ideal $R$ of $k[x_1,..,x_n]$ properly containing $P$ and so that there are no principal ideals "in between" them. So, if the zero-set was generated by more, I suppose we should find a contradiction but I'm not sure how to find one. Any ideas, and in general, what's the connection between codimension of something, and how many equations that generate it?

Edit: Here's my current thinking. If we have $V(Q) < V(P)$, where $V(Q)$ has codimension 1, we have that $I(V(Q))$ will be prime, and we have that for every element $f$ in $I(V(Q))$ that $V(Q) < V(f)$, where $V(Q)$ is a component of $V(f)$. But then I'm kinda stuck.

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To be honest, I don't really know what you're asking. Here are a couple takes.

For the General Set Up: Let $R$ be a polynomial ring (over an algebraically closed field) and $P\subset Q$ be primes of $R.$ Set $\pi$ to be the projection map from $R$ to $R/P.$

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Question 1 If $\pi(Q)$ has height 1 and $R/P$ is a UFD is $Q$ principal?

Answer No. For a counterexample, consider $R = \mathbb{C}[X,Y],$ $P = (X),$ and $Q = (X,Y).$

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Question 2 If $\pi(Q)$ has height 1 and $R/P$ is a UFD, $V(Q)$ is given by the vanishing of a single polynomial $p\in R$? [Note that as $ht(\pi(Q)) = 1$ and $R/P$ is a UFD, $\pi(Q)$ is principal.]

Answer No. For this to be true, $Rad(p) = Rad(Q) = Q.$ As no nonzero element of $R = \mathbb{C}[X,Y]$ is both a power of $X$ and a power of $Y,$ considering the same example as given in the answer above yields a counterexample.

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Question 3 If $\pi(Q)$ has height 1 is $\pi(Q)$ principal?

Answer This is only guaranteed if $R/P$ is a UFD.

A noetherian ring is factorial iff every prime over a minimal over a principal ideal is itself principal. Hence, any quotient of $R$ which is an integral domain but not a UFD will contain a height $1$ prime which is not principal.

For example consider $R = \mathbb{C}[X,Y]$ and $Q = (X,Y)$ and $P=(Y^2−X^3)$. Then $\pi(Q)$ is not principal.

Geometrically the problem of $\pi(Q)$ not having the "right" number of generators arises from the fact that $(0,0)$ is a singular point on the curve $V(Q).$

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Oh, sorry, my question was probably put incorrectly. I was more curious as to whether the zero-set of the the minimal prime will always be generated by one equation. And more in general, if some prime ideal of R/P has height n, will it then be generated by n equations? – Dedalus Jun 10 '11 at 14:57
Sorry for a very badly worded question. My question is about the zero-set of Q, under the conditions that $\pi(Q)$ is principal in $R/P$, an UFD. Will one equation generate the zero-set then? And if so , why? And more in general, if $\pi(Q)$ would have height n in $R/P$, UFD, would the zero-set be generated by n equations? – Dedalus Jun 10 '11 at 16:06
Okay see answer 2. – jspecter Jun 10 '11 at 16:18
Okay, now I see you answer, great answer! Thank you so much. Okay, so it must not be generated by one equation. It is probably me who've misunderstood something. The question I have been working on says : "A subvariety of V(P) is the zero-set of a prime ideal Q that contains P, and so, if $\pi(Q)$ is a minimal prime in $R=k[x_1,...,x_n]/P$, which is an UFD, show that the subvariety is given by one equation." I get that it is given by one equation in R, since it's an UFD, and that's probably what they meant? – Dedalus Jun 10 '11 at 17:22
yes I think so. – jspecter Jun 10 '11 at 17:38