On the openness of the map $\mathbb A^{n+1}_k\smallsetminus 0\rightarrow \mathbb P^n_k$.

Question

In this question, it was asked whether the map $\pi:\mathbb A^{n+1}_k\smallsetminus 0\rightarrow \mathbb P^n_k$ is open, and I tried to draft to an answer but ran into a snag.

First, we reduce to the case of affine opens, by noting that it suffices to check that $\pi$ is open on $U_i=D(x_i)\subset \mathbb A^{n+1}_k\smallsetminus 0$. In this setting, $\pi$ is given by the inclusion: $$\phi_i:(k[x_0,\dots, x_n]_{x_i})_0\longrightarrow k[x_0,\dots, x_n]_{x_i}$$ It suffices to check this on the basis $D(g/1)\subset U_i=\operatorname{Spec} k[x_0,\dots, x_n]_{x_i}$, where $g$ ia polynomial in $k[x_0,\dots, x_n]_{x_i}$. If $g$ is homogenous then this is easy, and we have that $\pi(D(g/1))=D(g/x_i^{\deg g})\subset \operatorname{Spec}(k[x_0,\dots, x_n]_{x_i})_0$.

If $g$ is not homogenous, then I defined a new polynomial by: $$h=g(x_0/x_i,\dots, x_n/x_i)=g_0+\frac{g_1}{x_i}+\cdots +\frac{g_d}{x_i^d}$$ where $g_j$ is the degree $j$ part of $g$, and $g$ is assumed to have top degree $g$.

I then conjectured that: $$\pi(D(g/1))=D(h)\subset \operatorname{Spec}(k[x_0,\dots, x_n]_{x_i})_0$$ as $h$ is a polynomial of degree $0$, and I would hope things would work out this way.

One inclusion seems to be clear, indeed, if $\mathfrak p\in D(h)$, then there is a homogenous prime ideal $\mathfrak q\in \operatorname{Spec}k[x_0,\dots, x_n]_{x_i}$ such that $\phi_i^{-1}(\mathfrak q)=\mathfrak p$. It follows that if $g\in \mathfrak q$, then $g_j\in \mathfrak q$ for all $j$, hence $g_j/x_i^j\in \mathfrak p$ for all $j$ as well, implying that $h\in \mathfrak p$, a contradiction; hence $\mathfrak p\in \pi(D(g/1))$.

The other inclusion is trickier, and I haven't gotten very far. If $\mathfrak p=\phi_i^{-1}(\mathfrak q)$, then we can assume that $\mathfrak q$ is homogenous by taking the ideal generated by homogenous elements of $\mathfrak q$, as this will have the correct degree zero part (and still be prime). It remains to show that $g/1\notin \mathfrak q$ then then $h\notin \mathfrak p$.

However, I think this may not be true? Indeed consider $\mathfrak q=\langle xy+zw\rangle \subset k[x,y,z,w]_y$, and the polynomial $g/1=x+zwy^4$, then $h=x/y+zw/y^2$, which lies in the degree zero part of $\mathfrak q$, whilst $g\notin \mathfrak q$.

So, my questions are: 1) is my counter example correct as I am concerned I have oversimplified and there is something I am not catching that exempts something like this from happening. 2) if so how can I see that the image of $\pi(D(g/1))$ is open/if not how can I show the reverse inclusion?

Any help would be appreciated.

Answer 1

Your idea to reduce to the situation over a standard affine open and prove this for the case of $D(g)$ is good. Let's continue from there. We observe that a point $p$ in $\Bbb A^n$ is not in the image of the projection of $D(g)\subset\Bbb A^n\times(\Bbb A^1\setminus 0)$ iff $V(g)$ contains the $\Bbb A^1\setminus 0$ over $p$. What's the condition for that? Writing $g=\sum_{j=-n}^n g_jx_i^j$, the condition is that $g_j(p)=0$ for all $j$, which is a closed condition on the base cut out by $V(g_{-n},\cdots,g_n)$. So the image is open.

Your attempt is flawed, as FShrike pointed out in the comments, because it's attempting to show that the image of a basic open is again a basic open. For instance, if $n=2$ and $i=2$, then one can take $g=x_0+x_1x_2$ and see that the image of $D(g)$ is $\Bbb A^2\setminus 0$, well-known to not be affine and hence also not a basic open in $\Bbb A^2$.