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I am aware of the more standard proof that for $Z \subseteq \mathbb{A}^n(k)$, $\overline Z = V\left(I(Z)\right)$, its closure in the Zariski topology. I was wondering if this result could also be had using the definition of closure in general topology as the intersection of all closed sets containing $Z$.

The proof I imagined was along these lines \begin{align}\overline{Z} &= \displaystyle\bigcap_{C\supseteq Z,\\C \subseteq \mathbb{A}^n \text{ closed}} C \\& = \displaystyle\bigcap_{\mathfrak{a} \subseteq k[x_1, \ldots, x_n],\\\mathfrak{a} \text{ vanishes on } Z} V(\mathfrak{a}) \\ &= V\left(\sum_{\mathfrak{a} \text{ vanishes on } Z} \mathfrak{a}\right) \\&\ldots?\\ &= V\left(\bigcap_{\mathfrak{m} \text{ vanishes on } Z,\\\mathfrak{m} \text{ maximal}} \right) \\ &= V\left(\bigcap_{x \in Z} \mathfrak{m}_x\right) \\&= V(I(X)) \end{align}

Have I messed this up anywhere? If not, what are the missing steps?

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  • $\begingroup$ @Shaun why are you bumping so many old posts lately? $\endgroup$
    – KReiser
    Commented Sep 18, 2022 at 15:09
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    $\begingroup$ I'm just making edits where edits are due, @KReiser. I'm also investing a lot of time into algebraic geometry and linear algebraic groups lately, for my formal studies, so I encounter older post with the more basic stuff in. Does it matter? $\endgroup$
    – Shaun
    Commented Sep 18, 2022 at 15:22

1 Answer 1

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This is correct, and there is nothing to add.

Indeed, if $f$ vanishes on $Z$, then by definition $f \in \mathfrak m_x$ for all $x \in Z$. Conversely, if $f \in \bigcap_{x \in Z} \mathfrak m_x$, then $\mathfrak a := (f)$ by definition vanishes on $Z$.

This shows that $$ \bigcap_{x \in Z} \mathfrak m_x = \sum_{\mathfrak a \text{ vanishes on } Z} \mathfrak a$$

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    $\begingroup$ Sorry if this is a somewhat obtuse question, but given that correspondence, how do you go from an intersection of ideals to a sum of ideals? $\endgroup$ Commented May 21, 2017 at 21:50
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    $\begingroup$ No problem ! Recall that $\Sigma \mathfrak a_i$ is the finite linear combinaisons of elements $f_i \in \mathfrak a_i$. If $f \in \cap \mathfrak m_x$, that means that $\forall x \in Z, f(x) = 0$. In particular, $f$ vanishes on $Z$, so the ideal generated by $f$ vanishes on $f$, in particular $f \in \Sigma_{\mathfrak a \text{ vanishes on Z}} \mathfrak a$. Is this clear ? $\endgroup$
    – user171326
    Commented May 21, 2017 at 22:20
  • $\begingroup$ Yes. Intuitively, can I think of the LHS as forming an ideal by cutting it out of very large ideals, and the RHS as building it directly as a $k$-vector space? $\endgroup$ Commented May 21, 2017 at 23:28
  • $\begingroup$ Yes ! (But the RHS is also an ideal as sums of ideals is still an ideal). $\endgroup$
    – user171326
    Commented May 22, 2017 at 5:11

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