0
$\begingroup$

In Mumford's red book the statement of the Going-Down Theorem (Chapter II Section 8) is as follows. Let $f: X \to Y$ be a finite morphism. Assume that $Y$ is an irreducible normal scheme. Assume that for all $x \in X$, no non-zero element of $O_{Y, f(x)}$ is a $0$-divisor in $O_{X,x}^{*}$. Then for every pair of points $x_1 \in X$, $y_0 \in Y$ such that $f(x_1) \in \overline{\{ y_0 \}}$, there is a point $x_0 \in f^{-1}(y_0)$ such that $x_1 \in \overline{ \{ x_0 \} }$.

By "no non-zero element of $O_{Y, f(x)}$ is a $0$-divisor in $O_{X,x}^{*}$", does this mean that it is ok if a non-zero element of $O_{Y, f(x)}$ gets mapped to $0$ in $O_{X,x}$? or does it mean every non-zero element of $O_{Y, f(x)}$ gets mapped to a non-zero element which is not a $0$-divisor in $O_{X,x}^*$? Thank you.

PS Instead of editing this question, I decided to ask separately On an application of the going-down theorem of Cohen-Seidenberg in Mumford, which was the reason why I was trying to understand the statement in the first place.

Improve this question
$\endgroup$
12
  • $\begingroup$ To me, a "zero divisor" in a ring $R$ is an element $a$ such that multiplication by $a$ in $R$ is not injective. This includes $0$ if $R$ is not the zero ring. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 24, 2020 at 19:28
  • $\begingroup$ @LaurentMoret-Bailly That is what I first thought as well, but then he does put the $*$, as in a $0$-divisor in $O_{X,x}^{*}$... Do you think that changes the meaning? $\endgroup$
    – Johnny T.
    Commented Mar 24, 2020 at 19:40
  • $\begingroup$ $R^*$ usually denotes non zero elements in a ring $R$.. $\endgroup$
    – Praphulla Koushik
    Commented Mar 24, 2020 at 20:20
  • $\begingroup$ @PraphullaKoushik Right, so I was wondering because he writes a $0$-divisor in $R^{*}$, is he excluding the zero element? $\endgroup$
    – Johnny T.
    Commented Mar 24, 2020 at 21:33
  • 1
    $\begingroup$ That mark * is not a star on the ring but an asterisk referring the reader to a footnote at the bottom of the page. This the one instance where the Springer edition is clearer than the original. $\endgroup$
    – roy smith
    Commented Mar 26, 2020 at 1:02

0

Reset to default

You must log in to answer this question.