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.)

  • 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

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.


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