# On an exercise in section 4 of Chapter I from Hartshorne's Algebraic Geometry

It is about exercise 4.9:

Let $$X$$ be a projective variety of dimension $$r$$ in $$\mathbb{P}^n$$ with $$n\geq r+2$$. 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 $$k$$

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

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:

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

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 $$$$\alpha = \frac{f_1(x_1,\dots,x_{n-2})}{g_1(x_1,\dots,x_{n-2})}x_{n-1} + \frac{f_2(x_1,\dots,x_{n-2})}{g_2(x_1,\dots,x_{n-2})}x_n$$$$

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:

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