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 \begin{equation} I = (\sum_{j \ne i} a_j x_j + x_i) \end{equation}

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?

  • 3
    $\begingroup$ 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 $x_j-a_jx_i$. $\endgroup$
    – Mohan
    Commented Feb 6, 2016 at 16:35

3 Answers 3


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 $\mathbb P^1_\mathbb R$.

  • $\begingroup$ Can I ask why the classical point corresponds to that ideal? How to see that? $\endgroup$
    – user557
    Commented Sep 22, 2021 at 1:13
  • $\begingroup$ @user6344267 The system of equations $a_ix_j-a_jx_i=0$ has as unique projective solution the point $[x_0:\cdots:x_n]=[a_0:\cdots:a_n]$. $\endgroup$ 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:





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 $a_1,a_2 \neq 0$.


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