0
$\begingroup$

I am just reading Fulton's book. I don't understand something in the proof. I understand that the image is subset of the kernel. I don't understand how does the proof get the reverse inclusion?

Proposition 1.8 Let $Y$ be a closed subscheme of a scheme $X$, and let $U = X - Y$. Let $i \colon Y \to X$, $j \colon U \to X$ be the inclusions. Then the sequence $$ A_k Y \xrightarrow{\;i_\bullet\;} A_k X \xrightarrow{\;j^{\,\bullet}\;} A_k U \to 0 $$ is exact for all $k$.

Proof. Since any subvariety $V$ of $U$ extends to a subvariety $\bar{V}$ of $X$, the sequence $$ Z_k Y \xrightarrow{\;i_\bullet\;} Z_k X \xrightarrow{\;j^{\,\bullet}\;} Z_k U \to 0 $$ is exact. If $\alpha \in Z_k X$ and $j^* \alpha \sim 0$, then $$ j^* \alpha = \sum [\operatorname{div}(r_i)] $$ for $r_i \in R(W_i)^*$, $W_i$ subvarieties of $U$. Since $R(W_i) = R(\bar{W}_i)$, $r_i$ corresponds to a rational function $\bar{r}_i$ on $\bar{W}_i$, and $$ j^* ( \alpha - \sum [\operatorname{div}(\bar{r}_i)] ) = 0 $$ in $Z_k U$. Therefore $$ \alpha - \sum [\operatorname{div}(\bar{r}_i)] = i_* \beta $$ for some $\beta \in Z_k Y$, which implies the proposition.

(Original image here.)

$\endgroup$
  • 1
    $\begingroup$ Is your trouble in understanding why $Z_kY\to Z_kX\to Z_kU\to 0$ is exact, or why this implies the result about the exact sequence of $A_k$, or both? $\endgroup$ – KReiser Aug 17 '18 at 1:47
  • $\begingroup$ I don't understand both $\endgroup$ – Adeek Aug 17 '18 at 3:13
2
$\begingroup$

For the sequence of $Z_k$, exactness in the middle follows from the fact that $j^*(S)=S\cap U$ for a closed integral subscheme $S$. Since $S\cap U$ is an open subscheme of $S$ and $S$ is integral, it has the same dimension as $S$. So if $j^*(S)=0$, then $S\cap U$ must be either empty or of dimension less than $k$. Since the second option is impossible, $S\cap U$ must be empty and therefore $S$ was actually in $\operatorname{im} i_*$.

For the sequence of $A_k$, once you (or Fulton) have shown that the maps $i_*,j^*$ descend from $Z_k$ to $A_k$, the only thing to do is to check exactness in the middle, and more specifically, check that $\ker j^*\subset \operatorname{im} i_*$. In fact, it's enough to check that if $\alpha\in Z_kX$ is sent to something rationally equivalent to $0$ by $j^*$, then $\alpha$ is already rationally equivalent to something in the image of $i_*$.

In Fulton's proof, the first equation $j^*(\alpha)=\sum [div(r_i)]$ follows from the definition of rational equivalence. The relation on $r_i=\overline{r_i}$ follows from the fact that since $W_i$ is an open subset of $\overline{W_i}$, it has the same field of rational functions, so $j^*(div(\overline{r_i}))= r_i$. So now we can pull the $[div(r_i)]$ inside $j^*$ to get the second equation, and then we can use our logic from earlier about the exactness of the sequence of $Z_k$. Thus $\alpha$ is rationally equivalent to something in the image of $i_*$, and therefore upon passing from $Z_k$ to $A_k$, we see that $\ker j^*\subset \operatorname{im} i_*$, which is what we wanted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.