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Suppose $(R,m)$ is a regular, local ring. Let $x_1,x_2,...,x_n$ be a regular system of parameters. Let $I$ be an ideal generated by squarefree monomials in the $x_i$'s. Is $I$ a radical ideal? The motivation for this is the polynomial ring in finitely many indeterminates over a field (which although not local is regular) and the indeterminates generate the homogeneous maximal ideal. Then every ideal generated by squarefree monomials in the indeterminates is radical.

I was trying to see whether the proof from monomial case in polynomial rings carries over. There we express an ideal generated by squarefree monomials as an intersection of ideals generated by subsets of the indeterminates. Each such ideal is prime. In the case of a regular system of parameters in a regular local ring, any subset of a rsop's also generates a prime ideal. So the only obstruction in the proof is whether the modularity law holds for such ideals.

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  • $\begingroup$ Dear Koose, I am not sure I see where the problem is. What do you mean by modularity law? $\endgroup$
    – Hailong Dao
    Commented Feb 14, 2011 at 5:50
  • $\begingroup$ @Hailong: By modularity law, I mean for any if $J,K,L$ are ideals, then $J\cap (K+L)=J\cap K + J\cap L$. This is the terminology used in Atiyah-Macdonald (and some set theory books). If this holds for ideals generated by subsets of a regular system of parameters, then, I think, we are done. By "I am not sure I see where the problem is", do you mean you know/think this is true? $\endgroup$
    – Koose Muniswamy
    Commented Feb 14, 2011 at 12:28
  • $\begingroup$ In the above comment I mean, "if the modularity law holds for ideals generated by squarefree monomials on a regular system of parameters". $\endgroup$
    – Koose Muniswamy
    Commented Feb 14, 2011 at 12:33
  • $\begingroup$ @Koose, I see. Sorry, I am not very familiar with Atiyah-Macdonald. Hope my answer helps. $\endgroup$
    – Hailong Dao
    Commented Feb 14, 2011 at 16:27
  • $\begingroup$ @Koose: I am a bit confused about your second paragraph. If you already know that $I$ is a intersection of primes, then it is radical, right? $\endgroup$
    – Hailong Dao
    Commented Feb 14, 2011 at 17:20

1 Answer 1

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ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$.

Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that:

$$ I = (I,x) \cap (I,F_1)$$

If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$.

Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies $$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$

$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired.

REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question.

So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p).

It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show:

$$I \cap J = P + I'J' $$

Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that this is even true for $I,J$ generated by parts of a fixed regular sequence).

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  • $\begingroup$ @Hailong: Thanks for your answer and the link. I eventually figured out modularity for parameter ideals. However, (as I posted in my follow up comment above), this is perhaps not sufficient for the proof to go through. Its certainly sufficient if modularity holds for ideals generated by monomials on a regular system of parameters. $\endgroup$
    – Koose Muniswamy
    Commented Feb 14, 2011 at 17:04
  • $\begingroup$ @Koose: I see, I was answering your last sentence of the question. Would your proof go through if we assume $J$ is parameter (since perhaps we keep intersecting with such)? Then I think my outline above still works, and I can fix it. But it will perhaps be helpful if you specify exactly what proof you are thinking about? $\endgroup$
    – Hailong Dao
    Commented Feb 14, 2011 at 17:17
  • $\begingroup$ @Hailong: Sorry, I was under the impression that the proof was standard in the polynomial case. Essentially the monomial ideal is expressed as an intersection of its minimal primes which are ideals generated by subsets of the indeterminates. The main ingredient of the proof are (1) that intersection distributes over sum when all ideals are monomial (modularity) and (2) the intersection of principal ideals is a principal ideal generated by the lcm of the generators in a UFD. $\endgroup$
    – Koose Muniswamy
    Commented Feb 14, 2011 at 19:11
  • $\begingroup$ contd: A regular local ring is a UFD, ideals generated by subsets of a regular system of parameters are prime, so I thought the same proof would go through if we can push modularity in some form. I think it would be sufficient if modularity holds when $J$ is a parameter ideal. $\endgroup$
    – Koose Muniswamy
    Commented Feb 14, 2011 at 19:11
  • $\begingroup$ OK, so I added a proof which I think is easy enough. It did not use modularity, which seems to make things unnecessarily complicated. $\endgroup$
    – Hailong Dao
    Commented Feb 15, 2011 at 21:30

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