Hartshorne's Exercise I.4.9: What is the desired projection explicity?

Here is a proof that we can always find a point and a hyperplane as desired in the statement of the problem. This should also make the importance of the $r+2leq n$ assumption clearer. We assume that the statement of Takumi Muramaya's theorem 4.6A$^star$ has been proven - this is a very important gap to be filled and I thank him for his proof.

Theorem 4.6A$^star$. Let $L$ be a finite separable extension field of a field $K$, and suppose that $K$ contains an infinite subset $S$. Then, there is an element $alpha in L$ which generates $L$ as an extension field of $K$. Furthermore, if $beta_1,beta_2,ldots,beta_n$ is any set of generators of $L$ over $K$, then $alpha$ can be taken to be a linnear combination $$alpha = c_1beta_1 + c_2beta_2 + cdots + c_nbeta_n$$ of the $beta_i$ with coefficients $c_i in S$.

Taking $T_i$ as coordinates on $Bbb P^n$, up to a permutation of coordinates we may assume that $Xcap D(T_0)neqemptyset$. This implies that $k(X)$ is generated by the images of $t_i=frac{T_i}{T_0}$ under $k[D(T_0)]to k[Xcap D(T_0)] to k(X)$, so by theorem I.4.8A, the extension $ksubset k(X)$ is separably generated. By an application of theorem I.4.7A, $t_1,cdots,t_n$ must contain a separating transcendence base, and up to a permutation of coordinates we may assume that this is exactly $t_1,cdots,t_r$ so that $k(t_1,cdots,t_r)subset k(X)$ is a finite separable extension. As $ksubset k(t_1,cdots,t_r)$ is an infinite subset, we may apply our upgraded theorem of the primitive element to the extension $k(t_1,cdots,t_r)subset k(X)$ to find a primitive element $alpha = sum_{r+1}^n a_it_i$ with $a_iin k$ so that $k(X)=k(t_1,cdots,t_r,alpha)$. Up to a linear automorphism of $Bbb P^n$ fixing all the coordinates $T_0$ through $T_r$, we can assume that $alpha=t_{r+1}$. Now I claim that we can find a $Pin V(T_0,T_{r+1})cap D(T_{r+2})$ which isn't in $X$, and the projection $pi$ from $P$ to $V(T_{r+2})$ induces an isomorphism of function fields $k(pi(X))to k(X)$. The reason we can find such a $Pin V(T_0,T_{r+1})$ is the combination of our assumptions that $Xcap D(T_0)neq emptyset$ and $dim X < n-1$: the first assumption means that $dim Xcap V(T_0)< n-2$, so $Xcap V(T_0,T_{r+1})$ is a proper closed subset of $V(T_0,T_{r+1})$ and thus for $Pin V(T_0,T_{r+1})$ the conditions $Pnotin X$ and $Pin D(T_{r+2})$ are both satisfied on dense open sets.

To verify that $k(pi(X))to k(X)$ is an isomorphism, we start by computing the image of a point under the projection. Given a point $[x_0:cdots:x_n]in Bbb P^n$, its projection on to $V(T_{r+2})$ from $P=[0:p_1:cdots:p_r:0:1:p_{r+3}:cdots:p_n]$ is given by $[x_0:x_1-p_1x_{r+2}:cdots:x_r-p_rx_{r+2}:x_{r+1}:0:x_{r+3}-p_{r+3}x_{r+2}:cdots]$. This means that for any $ineq r+2$ the pullback of the function $t_iin k[pi(X)]$ is given by $t_i-p_it_{r+2}$ (in particular, $t_{r+1}$ pulls back to $t_{r+1}$), so the image of the map on function fields contains $k(t_1-p_1t_{r+2},t_2-p_2t_{r+2},cdots,t_r-p_rt_{r+2},t_{r+1})$. As $k(X)=k(t_1,cdots,t_r)(t_{r+1})$ by assumption, this means that $p_it_{r+2}$ can be written as a polynomial in $t_{r+1}$ for all $i$, and therefore $k(t_1-p_1t_{r+2},t_2-p_2t_{r+2},cdots,t_r-p_rt_{r+2},t_{r+1})=k(t_1,cdots,t_r)(t_{r+1})$, which is exactly $k(X)$. This means that $pi(X)$ is birationally equivalent to $X$ by corollary I.4.5.