Suppose $R$ is a regular local ring. Let $m$ be the maximal ideal. Then, if the dimension of $R$ is $n$, there is a regular sequence of size $n$, say $x\_1,x\_2,...,x\_n$ s.t. $m=(x\_1,x\_2,...,x\_n)R$. Further, the ideals $(x\_{i\_1},...,x\_{i\_j})$ with $i\_1,...,i\_j\in {1,...,n}$, are prime. 


Can we make similar statements about any other kind of prime ideals in a regular local ring $R$? Specifically, do any other prime ideals satisfy the condition: if the ideal is minimally generated by a certain set of generators, then every subset of the generators defines a prime ideal? One example in light of the first paragraph, are the prime ideals generated by a subset of the regular sequence that generates the maximal ideal.


Also, when does a regular sequence define a prime ideal in a regular local ring $R$, and when does a maximal regular sequence define a maximal ideal?



prime ideals in regular local rings
As Sandor pointed out, a necessary condition is that the prime ideal $P$ is a complete intersection. Here is a proof that it is also sufficient. It will suffice to prove the following:


**Claim**: Let $(R,m)$ be a Noetherian local ring and $x\in m$ a regular element on $R$. If $R/(x)$ is a domain, then so is $R$. 


**Proof**: Suppose $ab=0$ in $R$. Then modulo $x$, one of them say $a$, must be $0$. So $a=xa\_1$, thus $x(a\_1b)=0$. As $x$ is regular, $a\_1b=0$, and continuing in this fashion one of $a,b$ must be divisible by arbitrary high power of $x$, so it must be equal to $0$.


As for an example which is not a part of a regular s.o.p, take something like $P=(x^2+y^2+z^2, u^2+v^2+w^2)$ in $\mathbb C[[x,y,z,u,v,w]]$.


