Mfrhtyj

It is about exercise 4.9:

Let $X$ be a projective variety of dimension $r$ in $mathbb{P}^n$ with $ngeq r+2$. Show that for suitable choice of $P notin X$ and a linear $mathbb{P}^{n-1}subseteq mathbb{P}^n$, the projection from $P$ to $mathbb{P}^{n-1}$ induces birational morphism of $X$ onto its image $X' subseteq mathbb{P}^{n-1}$. You will need (4.8A), (4.7A) and (4.6A).

Here is my thinking:

WLOG we can suppose that $X$ is an affine variety. The idea is that after a suitible change of coordinates, we can choose the hyperplane $H$ defined by $lbrace x_n=0 rbrace$ and take $P=(0,dots,0,1)$ so that the projection is defined by $(x_1,dots,x_n) mapsto (x_1,dots,x_{n-1},0)$. We want to prove that the $k$-algebra homomorphism

begin{align}
frac{k[x_1,dots,x_{n-1}]}{mathcal{I}(X)cap k[x_1,dots,x_{n-1}]} & hookrightarrow
frac{k[x_1,dots,x_n]}{mathcal{I}(X)} \
x_i & mapsto x_i
end{align}

induces an isomorphism of extensions of $k$

begin{equation}
phi:text{Frac} left( frac{k[x_1,dots,x_n]}{mathcal{I}(X)cap k[x_1,dots,x_{n-1}]} right) rightarrow
text{Frac} left( frac{k[x_1,dots,x_n]}{mathcal{I}(X)} right)
end{equation}

Now let $K$ be the field of rational fuctions of X. Reasoning as in Proposition 4.9, it is possible to find a trascendence base such that, after changing coordinates, it is formed by rational functions $x_1,dots,x_r in K$ so that $K$ is a finite separable extension of $k(x_1,dots,x_n)$. Consider the following extensions:

begin{equation}
k subseteq k(x_1,dots,x_r,x_{r+1},dots,x_{n-2}) subseteq k(x_1,dots,x_r,x_{r+1},dots,x_{n-2})[x_{n-1},x_n]=K
end{equation}

the second one is a finite separable extension, so by (4.6A) there is a rational fuction $alpha$ which generates K as an extension of $k(x_1,dots,x_r,x_{r+1},dots,x_{n-2})$. Furthermore, there exist $f_1,f_2,g_1,g_2 in k[x_1,dots,x_n]$ such that
begin{equation}
alpha = frac{f_1(x_1,dots,x_{n-2})}{g_1(x_1,dots,x_{n-2})}x_{n-1} + frac{f_2(x_1,dots,x_{n-2})}{g_2(x_1,dots,x_{n-2})}x_n
end{equation}

At this point, I would like to ask if there is some continuation in order to prove that $phi$ is surjective.

Thank you very much for your answers.

We assume $r = n-2$ and that we have $x_1,ldots,x_{n-2}$ algebraically independent over $k$ (selected from $X_1/X_0,ldots, X_n/X_0$):

$$K(X) = K = k(x_1,ldots,x_{n-2})[x_{n-1},x_n] = F[x_{n-1},x_n]$$

with $x_{n-1}, x_{n}$ algebraic and separable over $F$.
We assume, that $x_{n-1}$ and $x_n$ are linear independent over $k$ otherwise everything would be even easier.

Consider the linear expression
$(1-t) x_{n-1} + t x_n = x(t)$ with $t in k$. Let $sigma_1,ldots, sigma_d:K to bar{F}$ be the different embeddings over $F$ of $K$ into the algebraic closure of $F$. Call $W_{ij}$ the $F$-vector space of $x in K$ with $sigma_i(x) = sigma_j(x)$.

Then $x(t)$ can not lie in a single $W_{ij}$ for all $t$, because otherwise $sigma_i$ would be equal to $sigma_j$ on $K$. So the intersection of $W_{ij}$ with the $x(t)$ is in an affine subspace of $k$, that is in a single $t_{ij}$. Assuming that $k$ is an infinite field (true because $k$ is assumed algebraically closed), we conclude, there is a $t' in k$, such that

$$x(t') = (1-t') x_{n-1} + t' x_n = alpha$$

is in no $W_{ij}$ and therefore generates $K$ algebraically over $F$.
This gives the coefficients of a linear projection:

$$(x_1,ldots,x_n) mapsto (x_1,ldots,x_{n-2}, x(t'), 0)$$