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The following is from David Mumford's book Algebraic geometry I, page 29-30.

Let $X$ be a projective variety of the form $X=V[\mathscr{\beta}]$, with $\mathscr{\beta}\subset \mathbb{C}[X_{1},...,X_{n}]$. Let $f,g\in \mathbb{C}[X_{1},...,X_{n}]$ are homogeneous of the same degree where $g\not \in \mathscr{\beta}$. Let $Y_{0},Y_{1}$ be homogeneous coordinates in $\mathbb{P}_{1}$. Then consider the closed algebraic set $$Z=V(Y_{1}g-Y_{0}f)\subset X\times \mathbb{P}^{1}$$

Mumford claim that if we decompose $Z$ into irreducibles, then we will have $$Z=Z^{*}\cup Y_{1}\cup Y_{2}...\cup Y_{k}$$ where $$Y_{i}\subset \text{proper subvariety of X}\times \mathbb{P}^{1}$$

Further he asserts that $p_{1}:Z^{*}\rightarrow X$ is surjective by Zariski's main theorem. Why is this true? Mumford says "..Thus $Z^{*}$ is the Zariski-closure of the graph of the map $x\rightarrow \frac{f(x)}{g(x)}$ and is a rational map from $X$ to $\mathbb{P}^{1}$". Since $Z^{*}$ is not given explicitly I feel I am a bit lost why this must be true.

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up vote 4 down vote accepted

I happen to have a copy of the book here, and $Z^\ast$ is actually given quite explicitly: It is the largest irreducible component of the closure of $Z\cap(X_0\times\mathbb P_1)$ inside $X\times\mathbb P_1$, which was earlier argued to be irreducible (for everyone without a copy: $X_0$ is the nonvanishing set of $fg$). Since $Z^\ast$ is closed, the main theorem states that its image is closed in $X$. He has earlier argued that the restriction $(X_0\times\mathbb P_1)\to X_0$ is bijective, so the image of $Z^\ast$ under $p_1$ must be the closure of $X_0$, i.e. $X$.

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Thanks for the helpful response! –  Bombyx mori Nov 10 '12 at 18:34
    
Glad to be helpful! Wasn't sure if I was. If I left questions open, don't hesitate to inquire. –  Jesko Hüttenhain Nov 10 '12 at 18:43
    
Mumford is clear, but he did not type the words you have written, and I could not think through this. So your interpretation is very helpful. –  Bombyx mori Nov 10 '12 at 19:08
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