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.
Best Answer
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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