Again I'm here to ask for clarification on some already existed posts on the Hartshorne Exercise I.4.9, which goes like this:

Exercise: Let $X$ be a projective variety of dimension $r$ in $\mathbf{P}^n$ with $n\geq r+2$. Show that for suitable choice of $P\notin X$, and a linear $\mathbf{P}^{n-1}\subseteq \mathbf{P}^n$, the projection from $P$ to $\mathbf{P}^{n-1}$ induces a birational morphism of $X$ onto its image $X'\subseteq \mathbf{P}^{n-1}$.

In the post Exercise 4.9, Chapter I, in Hartshorne, @Takumi Murayama provided a great answer. But by looking up in the algebraic geometry note by J. Milne (especially Chapter 6, Section 6.27) and Shafarevich's book (especially Volume I, Chapter 1, Example 1.27), it seems that the $\pi$ constructed is a projection with center "the $(n-r-1)$-plane $Z(x_1/x_0,\ldots,x_r/x_0,\alpha)$" and project to the plane $Z(x_{r+2}/x_0,\ldots,x_{n}/x_0)$. However, this seems not the projection that the exercise demands, which should be the projection from a point $P \not\in X$ to a linear plane $\mathbf{P}^{n-1}$.

EDIT on Aug 1st, 2020 @Takumi Murayama has updated the answers there, which is of great help.

In the post On an exercise from Hartshone's Algebraic Geometry, Ch I sect 4, there is a proof which seems been adapted from this proof:

We can assume that $X$ is affine and is contained in $\mathbb{A}^n$, the set of points in $\mathbb{P}^n$ with first $x_0 = 0$. The field of fractions $K(X)$ is generated by $x_1, \ldots, x_n$, so we can assume that $x_1, \ldots, x_r$ is a separating transcendence basis for $K(X)/k$ by 4.7A and 4.8A, and K(X) is generated by $a_{r+1} x_{r+1} + \cdots + a_n x_n$ for some $a_i$'s in $k$, by 4.6A. As $r \leq n - 2$, we can find a form $b_{r+1} x_{r+1} + \cdots + b_n x_n$ not proportional to $a_{r+1} x_{r+1} + \cdots + a_n x_n$. Choose any point at infinity not in this plane (denoted by $\Sigma$) or in $\bar{X}$. Then the projection from this point to the plane maps $K(\Sigma)$ onto $K(X)$, so it is an isomorphism from the function field of the image of $X$ to $K(X)$, and therefore a birational isomorphism.

But just as the comments there, I think the coefficients $a_i$'s should be in $k(x_1, \ldots, x_r)$ instead of $k$. However, I cannot figure out what the projection is if this holds. So could someone explain how this projection is constructed?

In a word, my question is:

  1. How to construct the demanding projection? and I hope to get an explicit expression on the projection, so as to clarify that $X$ and the image $X^\prime$ are birational.

  2. After the construction, how to show the birationality.

Thank you all in advance! :)


1 Answer 1


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 $r+2\leq n$ assumption clearer. We assume that the statement of Takumi Murayama'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 $$\alpha = c_1\beta_1 + c_2\beta_2 + \cdots + c_n\beta_n$$ 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 $X\cap D(T_0)\neq\emptyset$. This implies that $k(X)$ is generated by the images of $t_i=\frac{T_i}{T_0}$ under $k[D(T_0)]\to k[X\cap D(T_0)] \to k(X)$, so by theorem I.4.8A, the extension $k\subset k(X)$ 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 $k\subset k(t_1,\cdots,t_r)$ 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 $a_i\in k$ 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 $P\in V(T_0,T_{r+1})\cap D(T_{r+2})$ 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 $P\in V(T_0,T_{r+1})$ is the combination of our assumptions that $X\cap D(T_0)\neq \emptyset$ and $\dim X < n-1$: the first assumption means that $\dim X\cap V(T_0)< n-2$, so $X\cap V(T_0,T_{r+1})$ is a proper closed subset of $V(T_0,T_{r+1})$ and thus for $P\in V(T_0,T_{r+1})$ the conditions $P\notin X$ and $P\in D(T_{r+2})$ 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 $i\neq r+2$ the pullback of the function $t_i\in k[\pi(X)]$ 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.


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