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Let $J$ be an ideal in the ring $\mathbb{C}[u] := \mathbb{C}[u_1,...,u_n]$ of regular functions on $\mathbb{C}^n$.

  • Given a subspace $\mathbb{C}^k \subset \mathbb{C}^n$ and $I$ its ideal in $\mathbb{C}[u]$, when can we say that the $n-k$ generators of $I$ form a regular sequence in $\mathbb{C}[u] / J$ ? Can we express this condition in terms of the zero sets of $I$ and $J$ ?
  • Suppose that the generators of $I$ form indeed a regular sequence in $\mathbb{C}[u] / J$. Is it still true in the ring $\mathbb{C}[u,u^{-1}] / J\mathbb{C}[u,u^{-1}]$ (considered as a $\mathbb{C}[u]$-module) ?
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    $\begingroup$ Your claim in the second sentence is only valid if $J$ is a radical ideal. Regularity of a sequence is equivalent to the vanishing of the higher homology modules of the Koszul complex associated to the sequence. This vanishing is preserved by every flat base change (and implied by vanishing after any faithfully flat base change). The ring homomorphism $\mathbb{C}[u]/J \to \mathbb{C}[u,u^{-1}]/J\mathbb{C}[u,u^{-1}]$ is flat (usually not faithfully flat). Thus, the image in $\mathbb{C}[u,u^{-1}]/J\mathbb{C}[u,u^{-1}]$ of a regular sequence in $\mathbb{C}[u]/J$ is still a regular sequence. $\endgroup$
    – Jason Starr
    Commented Sep 25, 2018 at 16:37
  • $\begingroup$ Thanks @Jason Starr. I don’t understand your statement about $J$ being radical. Seems that the flatness of the ring homomorphism is enough (does this depend on radicality of $J$ ?). $\endgroup$
    – BrianT
    Commented Sep 25, 2018 at 19:38
  • $\begingroup$ I also don’t understand your statement “and implied by vanishing after any faithlully flat base change”. $\endgroup$
    – BrianT
    Commented Sep 26, 2018 at 5:59
  • $\begingroup$ For $\mathbb{C}[u]/J$ to be isomorphic to the ring of regular functions on the zero zet $V(J)$ you need $J$ to be radical. For a counterexample consider $n = 1$. Then $V(u_1) = V(u_1^2)$, but obviously $\mathbb{C}[u_1]/(u_1) \ncong \mathbb{C}[u_1]/(u_1^2)$. $\endgroup$
    – red_trumpet
    Commented Sep 26, 2018 at 8:12
  • $\begingroup$ Oh ok, thank you. I erased this sentence, which is not necessary anyway. Suppose then that $J$ is not necessarily radical. Is there a way to relate the fact that $I$ is generated by a regular sequence in $\mathbb{C}[u] / J$ and the zero sets $\mathbb{C}^k$ and $V(J)$ ? $\endgroup$
    – BrianT
    Commented Sep 26, 2018 at 8:20

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In answer to the first question:

The zero sets of $I$ and $J$ are not enough to know whether the generators of $I$ remain a regular sequence in $\mathbb{C}[u]/J$. In particular, the zero set of $J$ is not sensitive to changes in $J$ that do not affect its radical, but these changes can be relevant to whether the generators of $I$ remain a regular sequence.

This is shown by the example Jason Starr gave in answer to your previous question. For simplicity, consider the case $m=2$, $k=1$, i.e. $\mathbb{C}[u] = \mathbb{C}[x,y]$, and let $I=(y)$, $J=(x^2,xy)$ (per Jason's example). Then the zero set of $J$ is $\{x=0\}$, and the zero set of $I$ is $\{y=0\}$. In this situation, the generator $y$ of $I$ does not remain regular in $\mathbb{C}[x,y]/J$, since it becomes a zerodivisor.

On the other hand, if $J=(x)$ (note this is the radical of the previous choice of $J$, thus it has the same zero set), then $y$ remains regular in $\mathbb{C}[x,y]/J$.

A necessary condition on $V(I)$ and $V(J)$ for the generators of $I$ to remain regular in $\mathbb{C}[u]/J$ (the example above shows it is not sufficient) is that

$$\dim V(J) - \dim V(I)\cap V(J) \geq n-k.$$

Proof: The quotient of the noetherian ring $\mathbb{C}[u]/J$, of Krull dimension $\dim V(J)$, by a regular sequence of length $r$, has Krull dimension $\leq \dim V(J) - r$. Proof by induction on $r$: quotienting by a single regular element $a$ must reduce the dimension. If the chains of primes containing $a$ included any of maximal length, then $a$ would be contained in a minimal prime ideal, which would be an associated prime of zero, and therefore $a$ would be a zerodivisor, contrary to assumption. Thus, if $I$ is generated by a regular sequence of length $r=n-k$, the dimension of $\mathbb{C}[u]/(I+J)$ ($=\dim V(I)\cap V(J)$) must be at least $r=n-k$ lower than the dimension of $V(J)$. This is the stated claim.

Jason answered the second question in comments: a regular sequence in $\mathbb{C}[u]/J$ remains regular when you invert $u$, as it does in all flat extensions. See for example Proposition 1.1.2 in Winfried Bruns and Jurgen Herzog's book Cohen-Macaulay Rings. The argument is basically that an element being regular means multiplication by it is injective, and flat base change preserves injectivity, followed by induction on the length of the sequence.

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  • $\begingroup$ Thank you very much for your answer. I’m wondering though how to translate “regularity of I” in $\mathbb{C}[u] / J$ in terms of intersections of spaces in $\mathbb{C}^n$. In my situation for instance, I know that $J$ is generated by monomials whose degrees are in a certain subspace $\mathbb{Z}_J \subset \mathbb{Z}_{>0}^k \subset \mathbb{C}^k$. Moreover, I have shown the following : if a coordinate subspace $\mathbb{C}^d \subset \mathbb{C}^n$ intersects $\mathbb{C}^k$ non trivially, then it intersects $\mathbb{Z}_J$ non trivially. Does this kind of statements say something about regularity ? $\endgroup$
    – BrianT
    Commented Sep 27, 2018 at 4:48
  • $\begingroup$ It helps that $J$ is a monomial ideal. Is $I$ as well? Or is it at least homogeneous? (Incidentally, I'm having trouble following your inclusions. $\mathbb{Z}_J$ seems to be a subspace of the lattice of exponents. I thought $\mathbb{C}^k$ was a linear subspace of the affine variety $\mathbb{C}^n$. Can you state the claim in more detail?) $\endgroup$
    – benblumsmith
    Commented Sep 27, 2018 at 18:14
  • $\begingroup$ $I$ is the ideal generated by $n-k$ polynomials of degree $1$, since it is the ideal of a linear subspace $\mathbb{C}^k \subset \mathbb{C}^n$. On the other hand, J is a monomial ideal generated by monomials whose exponents are in $\mathbb{Z}_J \subset \mathbb{Z}^k_{>0} \subset \mathbb{Z}^n$. The claim is that the generators of $I$ form a regular sequence in the quotient $\mathbb{C}[u] / J$. $\endgroup$
    – BrianT
    Commented Sep 27, 2018 at 19:35
  • $\begingroup$ The claim I really want to prove is with a slightly different $J$: the $\mathbb{C}[u]$-submodule of $\mathbb{C}[u,u^{-1}]$ defined by monomials with exponents in a “half lattice” $\mathbb{Z}_J \subset \mathbb{Z}^k \subset \mathbb{Z}^n$ (note that $J$ is not an ideal in $\mathbb{C}[u,u^{-1}]$ anymore so localizing might not help, and that $J \cap \mathbb{C}[u]$ plays the role of the ideal $J$ from above). The real claim is: the generators of $I$ form a regular sequence in the quotient $\mathbb{C}[u,u^{-1}] /J$. Proving the statement for $\mathbb{C}[u] / J \cap \mathbb{C}[u]$ might be enough. $\endgroup$
    – BrianT
    Commented Sep 27, 2018 at 20:01
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    $\begingroup$ Without dimension info about $V(J\cap \mathbb{C}[u])$ (at least enough to know that it is at least $n-k$), I don't think you can guarantee that the generators of $I$ remain regular. There is no way for them to be regular if $\mathbb{C}[u]/(J\cap \mathbb{C}[u])$ doesn't have at least $n-k$ dimensions. $\endgroup$
    – benblumsmith
    Commented Sep 29, 2018 at 19:02

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