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Let $X\subset \mathbb{A}_k^r$ be an irreducible hypersurface defined by a polynomial $g$, where $k$ is an algebraically closed field. Embed $\mathbb{A}^r\hookrightarrow\mathbb{P}^r$ in the usual way. Consider the stereographic projection of $X$ from a point $p$ in $\overline{X}\setminus X$ to some hyperplane not containing $p$, where $\overline{X}$ is $X$'s projective closure. I want to show that this projection is never a finite map from $X$ to this hyperplane. (This comes from an exercise in Shafarevich, specif. section 1.5 exercise 8 in Basic Algebraic Geometry 1.)

(Aside: Actually the desired result is false! See Addendum below for revised question.)

I have reduced the problem to the following algebraic one, but haven't seen how to complete the proof in these terms. My ideal answer would show me how to fill in the needed steps in this algebraic situation, but I would also be interested to hear more geometric arguments as well.

Algebraic reformulation: Let $g$ be an irreducible polynomial in $k[x_1,\dots,x_r]$. Let $L_1,\dots,L_{r-1}$ be $r-1$ linear functions on $\mathbb{A}^r$, and let each $L_j = \ell_j +\alpha_j$ where each $\ell_j$ is a linear form and each $\alpha_j\in k$. Suppose the $\ell_j$'s are linearly independent. Let $g_m$ be $g$'s maximum-degree homogeneous component, and suppose we are given that $g_m$ vanishes on the common zero set of $\ell_1,\dots,\ell_{r-1}$. Prove that in this case, $k[x_1,\dots,x_r]/(g)$ is not finite over the subring generated by the $L_j$'s.

Explanation of algebraic reformulation: The $L_j$'s are the restrictions to $\mathbb{A}^r$ of linear forms $L_j'$ on $\mathbb{P}^r$ that, together with $x_0=0$, cut out the point $p$ that is the center of the stereographic projection. Then the projection is, for suitable parametrization of the hyperplane being projected into, given by $(x_0:x_1:\dots:x_r)\mapsto (x_0:L_1':\dots:L_{r-1}')$; so its restriction to $\mathbb{A}^r$ is given by $(x_1,\dots,x_r)\mapsto (L_1,\dots,L_{r-1})$. The fact that $p$ is in $X$'s projective closure is expressed by the condition that when $x_0=0$ and all the $L_j'$'s are zero too, then $g$'s homogenization is zero. This translates into the condition that $g_m$ vanishes when all the $\ell_j$'s vanish. The question of whether the map is finite is then the question of whether $X$'s coordinate ring $k[x_1,\dots,x_r]/(g)$ is finite over the pullback image of $k[y_1,\dots,y_{r-1}]$ (the coordinate ring of the hyperplane), which is the subalgebra generated by the $L_j$'s.

Work so far: By the Nullstellensatz, the condition that $g_m$ vanishes when the $\ell_j$'s vanish amounts to saying that $g_m$ is in the radical of the ideal generated by the $\ell_j$'s; call it $L$. It seems to me that this ideal is already radical, since the $\ell_j$'s are homogeneous of degree $1$, though I'm not sure of this. Then $g_m$ is actually in $L$. Also, not every $x_i$ can be in $L$ because since it is a homogeneous ideal, an $x_i$ could only be in $L$ if it is a linear combination of the $\ell_j$'s, but there are only $r-1$ of them and $r$ $x_i$'s, so they cannot span the $k$-vector space containing the $x_i$'s. So, consider an $x_i$ that is not in $L$. If $k[X]$ were finite over the subalgebra generated by the $L$'s, there would be an integral equation for $x_i$ over this subalgebra, occurring in $k[X] = k[x_1,\dots,x_r]/(g)$:

$$x_i^s + \sum_{j=0}^{s-1} F_j(L_1,\dots,L_{r-1}) x_i^j = 0$$

Lifting this to an equation in $k[x_1,\dots,x_r]$, and noting that polynomials in the $L_j$'s are equally well polynomials in the $\ell_j$'s (and changing the $F_j$'s accordingly), yields:

$$x_i^s + \sum_{j=0}^{s-1} F_j(\ell_1,\dots,\ell_{r-1}) x_i^j = Hg$$

for some $H\in k[x_1,\dots,x_r]$.

If I could rig the left side to guarantee that the highest-degree homogeneous component is degree $s$, then this would yield a contradiction: reducing mod $L$ would reduce the degree of the right side but not the left, since $x_i^s\notin L$ (if I am right that $L$ is radical). But this is where I am stuck.

Thanks in advance for your thoughts.

Addendum, 3/27: It seems that the result is false! user119882 gives more or less the easiest possible counterexample: $r=2$, $L_1=x$, $g=y+x^2$. Clearly $k[x,y]/(y+x^2)$ is finite over the subring $k[L_1] = k[x]$.

This is also a counterexample to the original geometric statement: we have a plane conic tangent to the line at infinity; the stereographic projection away from this point of tangency is a finite map (actually an isomorphism) to $\mathbb{A}^1$.

Unless I am very very confused, I lifted this statement directly from Shafarevich. The following is on p. 66 of the paperback second edition:

  1. Let $X\subset\mathbb{A}^r$ be a hypersurface of $\mathbb{A}^r$ and $L$ a line of $\mathbb{A}^r$ through the origin. Let $\varphi_L$ be the map projecting $X$ parallel to $L$ to an $(r-1)$-dimensional subspace not containing $L$. Write $S$ for the set of all lines $L$ such that $\varphi_L$ is not finite. Prove that $S$ is an algebraic variety. [Hint: Prove that $S = \overline{X}\cap \mathbb{P}^{r-1}_\infty$.]

So the parenthetical hint must be wrong! $S$ is a subset of $\overline{X}\cap\mathbb{P}^{r-1}_\infty$ but not necessarily the whole thing!

Revised question: What is the condition (either geometric or algebraic) on the point $p$ of $\overline{X}\setminus X$ such that stereographic projection from this point is not a finite map on $X$?

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@user119882 - yup, that's a counterexample! I've edited the question to reflect this. –  Ben Blum-Smith Mar 28 at 3:19
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Why might $\varphi_L$ not be finite? Because points might be "missing" from certain fibres.

If $X$ has degree $d$, then typically, a line through $P$ will meet $\overline{X}$ in $d$ points, one of which will be $P$, and the remainder of which will be $d-1$ other points of $X$. (If a line passes through two distinct points of $\overline{X} \setminus X$, say $P$ and another point at infinity $Q$, then it lies in the plane at infinity, and so can't contain any point of $X$. In other words, a line through $P$ and a point of $X$ can't contain any point at infinity besides $P$.)

But if the line has a multiple intersection at $P$, i.e. lies in the tangent plane to $\overline{X}$ at $P$ (here I am assuming that $\overline{X}$ is smooth, so that it makes sense to talk about the tangent plane), then it meets $X$ in $< d-1$ points, and so finiteness will be violated (the number of points in the fibre, counted with multiplicity, it not being preserved).

But ... what if this line lies entirely at infinity (as in the counterexample)? Then it won't contain any point of $X$, and so the problem with having a fibre that is too small goes away.

Now every tangent line to $P$ is at infinity if and only if the tangent hyperplane to $P$ of $\overline{X}$ is just the hyperplane at infinity. So if $P$ is such that the hyperplane at infinity is tangent to $\overline{X}$ at $P$, we shouldn't get a violation of finiteness. (And this is what is happening in the counterexample.)

[The above is only approximate. I am not describing finiteness in a precise way, but just using the intuition of "having a constant number of points in the fibre", which is more like finite flat than simply finite. Although, if $X$ is smooth, then by miracle flatness, if the projection to the hyperplane is finite, then it is automatically flat. So I am not being too egregious.]

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