Proving the Generalized Bezout Theorem

Question

I am trying to prove the higher-dimensional analogue of Bezout theorem. It states the following:

Suppose $F_1,...,F_n\in k[X_1,...,X_{n+1}]$ are homogenous polynomials of degrees $d_1,...,d_n$, and let $C_1,...,C_n\subseteq \mathbb{P}_n$ be their zero loci. Suppose in addition that they have no factor in common. Then their intersection number is given by: \begin{align*} \dim \Gamma(C_1\cap C_2\cap ...\cap C_n)=d_1...d_n\end{align*}

For $n=2$ (which is the Bezout theorem we know and love), a fairly well known way to prove it is as follows.

Suppose $C_1\cap C_2\subseteq\mathbb{A}_2$, so none of their intersection happens in $l_{\infty}=Z(X_3)$. This can be accomplished through coordinate transformations. Letting, $f_1,f_2$ be the dehomogenized versions of $F_1$ and $F_2$ it suffices to prove that: \begin{align*}\dim k[x_1,x_2]/(f_1,f_2)=d_1d_2\end{align*} For $t\geqslant d_1+d_2$, let $R_t\subseteq k[x_1,x_2]$ be the set of polynomials of degree at most $t$. If we prove that $\dim R_t/(f_1,f_2)=d_1d_2$ for all $t\geqslant d_1+d_2$, then it follows immediately that $\dim k[x_1,x_2]/(f_1,f_2)=d_1d_2$ as well. This on its turn can be accomplished by proving $\dim R_t/(f_1,f_2)$ only depends on the degrees of $f_1$ and $f_2$, and compute the dimension using easy examples (like $f_1=x_1^{d_1}$ and $f_2=x_2^{d_2}$). In order to do so, we prove that there exist an exact sequence: \begin{align*} 0\xrightarrow{}R_{t-d_1-d_2}\xrightarrow{\alpha_1}R_{t-d_1}\oplus R_{t-d_2}\xrightarrow{\alpha_2}R_t\xrightarrow{\pi} R_t/(f_1,f_2)\xrightarrow{}0\end{align*} Define the maps by $\alpha_1(g)=(f_2g,f_1g)$ and $\alpha_2(g_1,g_2)=f_1g_1-f_2g_2$, and let $\pi$ be the projection. It remains to prove we obtain an exact sequence. It is immediate that $\alpha_1$ is injective, $\pi$ is surjective, and $\alpha_2\alpha_1=0$ and $\pi\alpha_2=0$.

To prove $\ker(\alpha_2)\subseteq\text{im}(\alpha_1)$, let $(g_1,g_2)\in\ker (\alpha_2)$. Then $f_1g_1-f_2g_2=0$, so $f_1g_1=f_2g_2$. Since $f_1$ and $f_2$ have no prime factors in common, for each prime factor $\pi\mid f_1$, we must have $\pi\mid g_2$. As such, $f_1\mid g_2$. Likewise, $f_2\mid g_1$. Obviously, $g_2/f_1=g_1/f_2$. As such, there exist a $h$ such that $(g_1,g_2)=(hf_2,hf_1)$. We also have $\deg(h)\leqslant t-d_1-d_2$, otherwise $\deg (g_1)>t-d_1$. So $(g_1,g_2)\in\text{im}(\alpha_1)$.

To prove $\ker(\pi)\subseteq \text{im}(\alpha_2)$, let $g\in \ker(\pi)$. Then there exist $u_1,u_2\in k[x_1,x_2]$ such that $h_1f_1+h_2f_2=g$. Assume that $\deg(h_1), \deg(h_2)$ are minimal. It remains to prove $\deg(h_1)\leqslant t-d_1$ and $\deg(h_2)\leqslant t-d_2$.

Suppose $\deg(h_1)>t-d_1$. Then the lead terms of $h_1f_1$ and $-h_2f_2$ are the same. Letting $H_1, H_2$ be the homogenized versions of $h_1,h_2$, and for any $p\in k[X_1,X_2,X_3]$, let $p^*=p[X_1,X_2,0]$. Then we have: $F_1^*H_1^*+F_2^*H_2^*=0$. As $F_1,F_2$ don't intersect on $l_{\infty}$, $F_1^*$ and $F_2^*$ are coprime, so $F_1^*\mid H_2^*$ and $F_2^*\mid H_1^*$. Letting $Q=H_1^*/F_2^*$, we have $-Q=H_2^*/F_1$. Write $H_1=QF_2+R_1$ and $H_2=-QF_1+R_2$. Then $R_1$ and $R_2$ are $0$ on $l_{\infty}$. As such, $X_3\mid R_1,R_2$. But that means that for their restrictions $r_1,r_2$ on $\textbf{A}_2$, we have: \begin{align*} \deg(r_1)<\deg(h_1); \deg(r_2)<\deg(h_2)\end{align*} In addition, we have $R_1F_1+R_2F_2=H_1F_1+H_2F_2$. That means that $g=r_1f_1+r_2f_2$. This is a contradiction to the minimality of the degrees of $h_1,h_2$.

So I want to repeat this technique for $n>2$. Take the assumption that $C_1\cap ...\cap C_n\subseteq \textbf{A}_n$, so it suffices to prove that $\dim k[x_1,...,x_n]/(f_1,...,f_n)=d_1...d_n$. Letting $R_t=k[x_1,...,x_n]_{\deg\leqslant t}$, we want to prove $\dim R_t/(f_1,...,f_n)=d_1...d_n$ for $t\geqslant d_1+...+d_n$. In order to do so, for any $S\subseteq \{1,...,n\}$, let $R_{t,S}=R_{t_S}$ with $t_S=t-\sum_{i\in \{1,...,n\}\setminus S}d_i$. We obtain the following sequence: \begin{align*} 0\xrightarrow{} \prod_{|S|=0} R_{t,S}\xrightarrow{\alpha_1}\prod_{|S|=1}R_{t,S}\xrightarrow{\alpha_2}...\xrightarrow{\alpha_{n}}\prod_{|S|=n}R_{t,S}=R_t\xrightarrow{\pi}R_t/(f_1,...,f_n)\xrightarrow{}0\end{align*} Again, $\pi$ is just the projection. For each $i$, $\alpha_i$ is the sum of maps in the form $\beta_{S,k}$ with $|S|=i-1$ and $k\notin S$. Such $\beta_{S,k}$ is given by:

\begin{align*} \beta_{S,k}: R_{S,t}\to R_{S\cup\{k\},t}; g\mapsto\begin{cases} -gf_k&\text{ if }|S_{<k}|\text{ is odd}\\ gf_k&\text{ if }|S_{<k}|\text{ is even}\end{cases}\end{align*} The question is: how do I prove that this sequence is exact? It is again obvious that $\alpha_1$ is injective, $\pi$ is surjective and $\pi\alpha_n=0$. It is also fairly easy to see $\alpha_{i+1}\alpha_{i}=0$ since $\beta_{S\cup\{k\},l}\beta_{S,k}+\beta_{S\cup\{l\},k}\beta_{S,l}=0$.

Answer 1

Ok, I have figured it out myself. Thanks to Mohan for pointing out that $F_1,...,F_n$ has to be a regular sequence.

The dehomogenisation of $F_i$, and the restriction to a certain degree, has to wait untill later. Instead, let each $R_S=k[X_1,...,X_{n+1}]$ for each $S\subseteq \{1,...,n\}$, and for each pair $(S,t)$ with $t\notin S$, define:

\begin{align*} \beta_{S,t}: R_S\to R_{S\cup\{t\}}, G\mapsto (-1)^{|S_{<t}|}GF_t\end{align*}

Now let $\alpha_i: \prod_{|S|=i-1}R_S\to \prod_{|S|=i}R_S$ be the sum of all $\beta_{S,t}$ with $|S|=i$, and let $R=R_{1,...,n}$. This gives us the following chain:

\begin{align*} 0\to \prod_{|S|=0}R_S \xrightarrow{\alpha_1}\prod_{|S|=1}R_S\xrightarrow{\alpha_2}... \xrightarrow{\alpha_{n}}\prod_{|S|=n}R_S=R\to R/(F_1,...,F_n)\to 0 \text{ }(*)\end{align*} We will prove this is exact. In order to do so, we will rely on the following result.

For each $S\subseteq\{1,...,n\}$, let $M_S$ be an $R$-module ($R$ is any ring). Let $(\beta_{S,t}\in\text{Hom}(M_S,M_{S\cup\{t\}}))_{t\notin S}$ be a series of morphisms satisfying the following conditions:

(1) The morphism $\beta_{\emptyset, t}$ is injective for any $t\in \{1,...,n\}$

(2) For each $S$ with $|S|\geqslant 2$, the following induced sequence is exact: \begin{align*} \prod_{T\subseteq S: |T|=2}M_{S\setminus T}\xrightarrow{\phi_S}\prod_{t\in S}M_{S\setminus \{t\}}\xrightarrow{\psi_S} M_S\end{align*}

Then the following induced sequence is exact: \begin{align*} 0\to \prod_{|S|=0}M_S \xrightarrow{\alpha_1}\prod_{|S|=1}M_S\xrightarrow{\alpha_2} .... \xrightarrow{\alpha_n}\prod_{|S|=n}M_S \end{align*} Proof: Induction to $n$, and some tasty diagram chasing.

It is immediately clear that in fact all $\beta_{S,t}$ are injective, so condition (1) certainly holds. So it remains to prove (2). As $\beta_{S\cup\{t\},u}\circ\beta_{S,t}=-\beta_{S\cup\{u\},t}\circ\beta_{S,u}$ for all $S,u,t$, it is immediate that $\ker{\psi_S}\subseteq\text{im}\phi_S$ for all $S$. So it remains to prove that $\ker{\psi_S}\supseteq\text{im}\phi_S$.

Let $S\subseteq\{1,...,n\}$, and write $S=\{i_1,...,i_k\}$. Let $(G_t)_{t\in S}\in\ker\psi$, so we have: \begin{align*} > G_{i_1}F_{i_1}-G_{i_2}F_{i_2}+...+(-1)^kG_{i_k}F_{i_k}=0\end{align*} We recursively construct a sequence $(H^{(p)})_{p\in\{0,...,k\}}$ such that $(\phi_{S}(H^{(p)}))_{i_q}=G_{i_q}$ for $q\leqslant p$. First, let $H^{(0)}=0$. Now let $p\in [1,k]$ and suppose we have constructed $H^{(p-1)}$. Then we can rewrite: \begin{align*} 0&=\sum_{q=p}^{k}(-1)^q G_{i_q}F_{i_q}+\sum_{q=1}^{p-1}(-1)^qG_{i_q}F_{i_q} + \\&=\sum_{q=p}^{k}(-1)^q G_{i_q}F_{i_q}+\sum_{q=1}^{p-1}\sum_{r\neq q}(-1)^q(-1)^{r+\{r>q\}}H^{(p-1)}_{\{i_q,i_r\}}F_{i_r}F_{i_q}\\ &=\sum_{q=p}^{k}(-1)^q G_{i_q}F_{i_q}+\sum_{q=1}^{p-1}\sum_{r=p}^{k}(-1)^{q+r+1}H^{(p-1)}_{\{i_q,i_r\}}F_{i_r}F_{i_q}\\ &=\sum_{q=p}^k(-1)^q[G_{i_q}+\sum_{r=1}^{p-1}(-1)^{r+1}H_{\{i_q,i_r\}}^{(p-1)}F_{i_r}]F_{i_q}\end{align*} Since all $F_i$ are homogeneous, the sequence $F_{i_k},F_{i_{k-1}},...,F_{i_p}$ is a regular sequence. As a result, $F_{i_p}$ is a nonzero devisor in $R/(F_{i_{p+1}},...,F_{i_k})$. But that implies that: \begin{align*} G_{i_q}\equiv \sum_{r=1}^{p-1}(-1)^{r}H_{\{i_q,i_r\}}^{(p-1)}F_{i_r}\mod (F_{i_{p+1}},...,F_{i_k})\end{align*} As such, there exist $P_{p+1},...,P_{k}\in R$ such that: \begin{align*}G_{i_q}\equiv \sum_{r=1}^{p-1}(-1)^{r}H_{\{i_q,i_r\}}^{(p-1)}F_{i_r}+\sum_{r=p+1}(-1)^{r+ \{p>r\}}P_rF_{i_r}\end{align*} Now define $H^{(p)}$ by putting: \begin{align*}H^{(p)}_T=\begin{cases}P_r&\text{if }T=\{i_p,i_r\}\text{ with }r>p\\H^{(p-1)}_T&\text{else}\end{cases}\end{align*} This satisfies $\phi_S(H^{(p)})_{i_p}=G_{i_p}$, while maintaining the values of $\phi_S(H^{(p)})_{i_r}$ for $r<p$. As desired. Our contstruction of $(H^{(p)})_p$ is done. As $\phi_S(H^{(k)})=G$, this proves that $G\in\text{Im}(\phi_S)$.

So the sequence (*) is exact (exactness at the final section is evident from the definition of $\alpha_{\{1,...,n\}}$). Now for $d\geqslant d_1+...+d_n$, let $R_D=R_{\deg=d}$, and let $R_{S,d}=R_{\deg=d-\sum_{s\notin S}d_s}$. Then by nature of homogenous polynomials, we obtain an exact sequence:

\begin{align*} 0\to \prod_{|S|=0}R_{S,d} \xrightarrow{\alpha_1}\prod_{|S|=1}R_{S,d}\xrightarrow{\alpha_2}... \xrightarrow{\alpha_{n}}\prod_{|S|=n}R_{S,d}=R_d\to R_d/(F_1,...,F_n)\to 0\end{align*}

This proves that $\dim R_d/(F_1,...,F_n)$ doesn't depend on the choices of $F_1,..., F_n$, only on the degrees. By checking out the case $F_i=X_i^{d_i}$, one finds easily that: \begin{align*}\dim R_d/(F_1,...,F_n)=d_1...d_n\text{ for all }d\geqslant d_1+...+d_n\end{align*}

It remains to prove that $\dim k[X_1,...,X_n]/(f_1,...,f_n)=d_1...d_n$ when $\bigcap_i F_i\subseteq \mathbb{A}_n$. By not having any zeros on $\{X_{n+1}=0\}$, it follows that $X_{n+1}$ is a nonzero divisor in $R/(F_1,...,F_n)$. As such, the multiplication map $R/(F_i)_i\to R/(F_i)_i$ is injective. This yields a chain of injections: \begin{align*} R_d/(F_i)_i\to R_{d+1}/(F_i)_i\to R_{d+2}/(F_i)_i\to...\end{align*} Since all of these spaces are of dimension $d_1...d_n$, these injections are in fact bijections. As $k[X_1,...,X_n]/(f_i)_i$ is the colimit of the diagram, it follows that \begin{align*}\dim k[X_1,...,X_n]/(f_i)_i=d_1...d_n\end{align*}