prime ideals in regular local rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:49:18Z http://mathoverflow.net/feeds/question/54650 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54650/prime-ideals-in-regular-local-rings prime ideals in regular local rings Koose Muniswamy 2011-02-07T15:45:03Z 2011-02-08T02:39:42Z <p>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. </p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/54650/prime-ideals-in-regular-local-rings/54673#54673 Answer by Sándor Kovács for prime ideals in regular local rings Sándor Kovács 2011-02-07T18:07:33Z 2011-02-07T22:03:06Z <p><strong>EDIT:</strong> Rephrased the answer in light of Koose's comment.</p> <p><em>Claim.</em> Any ideal satisfying the required property has to be a complete intersection.</p> <p><em>Proof.</em> Let $\mathfrak p\subset R$ be a prime ideal. Assume that the minimal number of generators for $\mathfrak p$ is <code>$r$</code> and let $a_1,\dots,a_r\in\mathfrak p$ be a set of generators. Let $I_t=(a_1,\dots,a_t)$ and one has the sequence of ideals:</p> <p>$$0\subsetneq I_1 \subsetneq \dots \subsetneq I_t\subsetneq I_{t+1}\subsetneq \dots\subsetneq I_r=\mathfrak p$$</p> <p>The containments cannot be equalities, because that would make the corresponding $a_i$ unneeded to generate $\mathfrak p$. If all the $I_t$'s are prime, then $\mathfrak p$ has height at least $r$, but it cannot be more than that, so the claim is proven. $\square$</p> <p><em>Example.</em> Take an irreducible projective variety (say the twisted cubic curve) that is <em>not</em> a complete intersection (in projective space). Then the ideal of the affine cone over this in the affine cone over projective space will be a prime, but if you take away one of the generators, that will not be.</p> http://mathoverflow.net/questions/54650/prime-ideals-in-regular-local-rings/54715#54715 Answer by Hailong Dao for prime ideals in regular local rings Hailong Dao 2011-02-08T01:49:25Z 2011-02-08T02:39:42Z <p>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:</p> <p><strong>Claim</strong>: 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 $R$. </p> <p><strong>Proof</strong>: 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 $0$.</p> <p>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 $\mathbb C[[x,y,z,u,v,w]]$.</p>