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Let $\mathbb{C}[u]:= \mathbb{C}[u_1,...,u_n]$ the algebra of polynomial functions on $\mathbb{C}^n$. I consider two ideals $I$, and $J$, where $I$ is the ideal generated by the polynomials vanishing on some fixed subspace $\mathbb{C}^k \subset \mathbb{C}^n$ (and $J$ is whatever you want). Note that $I$ is generated by $n-k$ elements in $\mathbb{C}[u]$.

Suppose that we have proved that the zero set of $I + J$ in $\mathbb{C}^n$ is just given by the zero polynomial. By a strong form of the Nullstellensatz, the ideal $I+J$ contains powers of all $u_i$, $i=1,...,n$. In particular, the quotient $$\mathbb{C}[u] / (I+J)$$ is finite dimensional.

My question is the following: is it true that the elements generating $I$ form a regular sequence in the quotient $\mathbb{C}[u] / J$ ?

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    $\begingroup$ That is not true. Let $J$ be the maximal ideal $\langle u_1,\dots,u_n\rangle$. For a more interesting case, let $J$ be $\langle u_iu_j | 1\leq i \leq k, 1\leq j \leq n \rangle$ and let $I$ be $\langle u_r | k<r\leq n \rangle$. The quotient $\mathbb{C}[u]/J$ has Krull dimension $n-k$; it is "dimensionally transverse" with respect to $I$. Yet every generator $u_r$ of $I$ is a zerodivisor in $\mathbb{C}[u]/J$. If $\mathbb{C}[u]/J$ is a Cohen-Macaulay ring of Krull dimension $n-k$, then your claim holds, cf. Matsumura, "Commutative ring theory", p. 135. $\endgroup$
    – Jason Starr
    Commented Sep 21, 2018 at 9:49
  • $\begingroup$ Thanks @Jason Starr, is it obvious that the zero set of $I+J$ in your counter example is reduced to $0$ ? $\endgroup$
    – BrianT
    Commented Sep 21, 2018 at 10:03
  • $\begingroup$ For the zero set of $J$, since $u_i^2 = u_iu_i$ is zero for every $i=1,\dots,k$, the zero set is just a linear space of dimension $n-k$ complementary to the linear space of dimension $k$ that is the zero set of $I$. $\endgroup$
    – Jason Starr
    Commented Sep 21, 2018 at 10:15
  • $\begingroup$ Thanks a lot. Would it be different if we would work in $\mathbb{C}[u,u^{-1}]$ instead of $\mathbb{C}[u]$ with $I$ being now replaced by the ideal $I \mathbb{C}[u,u^{-1}]$ ? $\endgroup$
    – BrianT
    Commented Sep 21, 2018 at 10:37
  • $\begingroup$ "Would it be different ..." I do not understand what you are proposing. If $I$ is the ideal $\langle u_r | k<r \leq n\rangle$ as in my comment, then the ideal $I\mathbb{C}[u_1,\dots,u_n,u_1^{-1},\dots,u_n^{-1}]$ equals the entire ring. $\endgroup$
    – Jason Starr
    Commented Sep 21, 2018 at 10:52

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