# Closed points in projective space correspond to which homogenous prime ideals in $k[x_0,...,x_n]$

I'm trying to think about exercise 4.5.O in Vakil's notes on Algebraic Geometry.

Before we defined the scheme $$\mathbb{P}^n_k := \operatorname{Proj}(k[x_0,...,x_n])$$ and showed that that for $$k$$ algebraically closed the closed points of $$\mathbb{P}^n_k$$ correspond to the points of "classical" projective space.

Now the question is: To which homogeneous prime ideal in $$k[x_0,...,x_n]$$ does the point $$[a_0:\dots :a_n]$$ correspond?

My thinking so far: Since there is an $$a_i \ne 0$$ we can pick the representative with $$a_i = 1$$. Then we get the homogeneous ideal $$$$I = (\sum_{j \ne i} a_j x_j + x_i)$$$$

Is this the one we are looking for? If so, how do you prove all the details, e.g. why is $$I$$ prime, doesn't contain the inessential ideal and really corresponds to this point?

• I find your notation a bit strange. I hope you did not mean a principal ideal when you wrote the summation sign. The ideal in question is the homogeneous ideal generated by (assuming $a_i=1$ as you did) the linear forms $x_j-a_jx_i$. Commented Feb 6, 2016 at 16:35

The classical point $[a_0:a_1:\dots:a_n]$ corresponds to the homogeneous ideal $$\langle a_ix_j-a_jx_i\vert i,j=0\dots n\rangle\subset k[x_0,\dots,x_n]$$ 1) Do not break the beautiful symmetry of this ideal by ugly choices (like $a_i=1$) for the homogeneous coordinates of your point.

2) The formula is still valid for an arbitrary field $k$.
However if $k$ is not algebraically closed $\mathbb P^n_k$ will contain closed points not of the form above.
For example the ideal $\langle x_0^2+x_1^2\rangle\subset \mathbb R[x_0,x_1]$ also corresponds to a closed point of $\mathbb P^1_\mathbb R$.

• Can I ask why the classical point corresponds to that ideal? How to see that? Commented Sep 22, 2021 at 1:13
• @user6344267 The system of equations $a_ix_j-a_jx_i=0$ has as unique projective solution the point $[x_0:\cdots:x_n]=[a_0:\cdots:a_n]$. Commented Sep 22, 2021 at 8:55

To provide an answer along the lines of what you were thinking:

Take the classical point $$[a_0: a_1 : \dots :a_n]$$. For simplicity, let $$a_0 \ne 0$$.

Then $$[a_0: a_1 : \dots :a_n]=[1: \frac{a_1}{a_0}: \dots: \frac{a_n}{a_0}]$$.

This point corresponds to the maximal ideal $$(\frac{x_1}{x_0}-\frac{a_1}{a_0}, \dots, \frac{x_n}{x_0} -\frac{a_n}{a_0}) \in k[\frac{x_1}{x_0}, \dots, \frac{x_n}{x_0}]=(k[x_0, \dots, x_n]_{x_0})_0$$, which then corresponds to the homogeneous prime ideal $$(a_0 x_1-a_1x_0, \dots, a_0 x_n-a_n x_0) \in k[x_0, \dots, x_n]$$ (see exercise 4.5.E.).

The case when some $$a_i \ne 0$$ for $$1\le i\le n$$ is similar.

Question: Try to prove that the classical point correspond to the homogenous ideal $$I:=\langle a_ix_j-a_jx_i\vert i,j=0\dots n\rangle\subset k[x_0,\dots,x_n]$$

Answer: I give the "main idea of the proof" in the case when $$n=2$$. When you have understood this example you may generalize.

Note: The ideal $$I:=(a_ix_j-a_jx_i)$$ for $$i,j=0,..,n$$ contain generators that are not needed.

Example: Let $$n=2$$ and consider $$a_ix_j-a_jx_i$$ for $$i,j=0,1,2$$. You get an ideal with 9 generators:

$$a_0x_0-a_0x_0,a_0x_1-a_1x_0,a_0x_2-a_2x_0$$

$$a_1x_0-a_0x_1,a_1x_1-a_1x_1,a_1x_2-a_2x_1$$

and

$$a_2x_0-a_0x_2,a_2x_1-a_1x_2,a_2x_2-a_2x_2.$$

Pick out the "obvious minimal set" of generators and you get the ideal

$$I:=(a_0x_1-a_1x_0, a_0x_2-a_2x_0,a_1x_2-a_2x_1)\subseteq k[x_0,x_1,x_2]$$

If $$a\neq 0$$ and you localize at $$x_0$$ you get the ideal

$$I_{(x_0)}=(\frac{x_1}{x_0}-\frac{a_1}{a_0}, \frac{x_2}{x_0}-\frac{a_2}{a_0}, a_1\frac{x_2}{x_0}-a_2\frac{x_1}{x_0})\subseteq k[\frac{x_1}{x_0}, \frac{x_2}{x_0}] .$$

Then you notice that

$$a_1\frac{x_2}{x_0}-a_2\frac{x_1}{x_0}= a_1( \frac{x_2}{x_0}-\frac{a_2}{a_0} + \frac{a_2}{a_0}) -a_2(\frac{x_1}{x_0}-\frac{a_1}{a_0}+\frac{a_1}{a_0})=$$

$$a_1( \frac{x_2}{x_0}-\frac{a_2}{a_0}) -a_2(\frac{x_1}{x_0}-\frac{a_1}{a_0})$$

Hence there is an equality of ideals

$$I_{(x_0)}= (\frac{x_1}{x_0}-\frac{a_1}{a_0}, \frac{x_2}{x_0}-\frac{a_2}{a_0}).$$

Hence when $$p:=[a_0:a_1:a_2]$$ is a "point" with $$a_i \in k$$ and $$a_0 \neq 0$$, its "corresponding ideal" in $$D(x_0)$$ is the maximal ideal $$I_0:=I_{(x_0)}$$. And by construction the maximal ideal $$I_0$$ corresponds to the $$k$$-rational point

$$(\frac{a_1}{a_0},\frac{a_2}{a_0}) \in \mathbb{A}^2_k \cong D(x_0).$$

You get a similar calculation if $$a_1,a_2\neq 0$$. Hence the original ideal $$I$$ has 9 generators, but the localized ideal $$I_{(x_0)}$$ has 2. As you have commented: This is not mentioned in the linked thread. The ideal $$I$$ has too many generators and the main point exercise is to find a "minimal set of generators" along the lines I suggest above.

Note: You need all 3 generators of the ideal. If $$a_1\neq 0$$ you get the ideal

$$I_{(x_1)}=(\frac{a_0}{a_1}-\frac{x_0}{x_1}, a_0\frac{x_2}{x_1}-a_2\frac{x_0}{x_1}, \frac{x_2}{x_1}-\frac{a_2}{a_1})\subseteq k[\frac{x_0}{x_1}, \frac{x_2}{x_1}] .$$

And it follows

$$a_0\frac{x_2}{x_1}-a_2\frac{x_0}{x_1}= a_0(\frac{x_2}{x_1}-\frac{a_2}{a_1})-a_2(\frac{x_0}{x_1}-\frac{a_0}{a_1}).$$

It follows

$$I_{(x_1)}=(\frac{x_0}{x_1}-\frac{a_0}{a_1}, \frac{x_2}{x_1}-\frac{a_2}{a_1}).$$

Similarly for $$D(x_2)$$.

The ideal is a prime ideal: If $$a_0 \neq 0$$ it follows there is an equality of ideals

$$I=(x_1-\frac{a_1}{a_0}x_0, x_2-\frac{a_2}{a_0}x_0)$$

hence $$k[x_i]/I \cong k[x_0]$$ which is a domain, hence $$I$$ is a prime ideal. A similar construction holds when $$a_1,a_2 \neq 0$$.