Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $k$ be an algebraically closed field with characteristic distinct from $2$.

I want to compute the multiplicity of the intersection of the points which lie in $V_{1} \cap V_{2}$ where:

$V_{1}=V(x^{2}+y^{2}-z^{2}) \subseteq \mathbb{P}^{2}$ and $V_{2}=V(x^{2}+y^{2}-2z^{2}) \subseteq \mathbb{P}^{2}$.

Well doing the algebra shows they intersect at two points $[1:i:0]$ and $[1:-i:0]$. By Bezout's theorem we know that the sum of the multiplicities is equal to $4$ right?

Now, as I understand, to compute the multiplicity at $[1:i:0]$ we need to take a chart containing this point yes? so we can take say $x=1$ then we need to compute the dimension of the following vector space:

$k[y,z]/(1+y^{2}-z^{2},1+y^{2}-2z^{2})$

Macaulay says that the dimension is equal to $4$ but isn't this impossible? it would force that that the intersection multiplicity of the other point is zero but this is impossible because the point lies in the intersection.

What am I doing wrong?

EDIT: is the mistake that we are taking a chart which also contains the point $[1:-i:0]$?, that is, do we need to take a chart that contains $[1:i:0]$ but not $[1:-i:0]$? can we take then $x=1$ and $z=1$?

share|improve this question
1  
Am I being naive or isn't the dimension 4 because you've modded out by the ideal generated by \emph{both} of your polynomials? So this is your coordinate ring on the intersection, not on just one point of the intersection. –  Derek Allums May 23 '12 at 13:39
1  
As far as I know $k[y,z]/(1+y^{2}-z^{2},1+y^{2}-2z^{2})$ is giving you the sum of all intersection numbers by definition so 4 is indeed the expected result, by Bezout's theorem. You have to work with the local ring of $k[y,z]/(1+y^{2}-z^{2},1+y^{2}-2z^{2})$ at $[1:i:0]$ to get the intersection number. –  Nils Matthes May 23 '12 at 13:41
    
@Nils Matthes: I'm trying to use the idea suggested by Michael Joyce here math.stackexchange.com/questions/147965/… –  user31509 May 23 '12 at 13:42
1  
If you want to learn more about intersection theory for projective plane curves I can very much recommend to you the book "Algebraic Curves" by William Fulton, which is a wonderful introduction to algebraic geometry and has a nice section on intersection theory. It can be found online here math.lsa.umich.edu/~wfulton/CurveBook.pdf –  Nils Matthes May 23 '12 at 14:00
1  
@user31509: My answer to the previous question was somewhat misleading in that I avoided invoking the local ring in order to simplify the calculation at hand. In general, you have to work in the local ring at the point to compute the intersection multiplicity. In practice, all it means is that any function which does not vanish at the point becomes invertible. –  Michael Joyce May 24 '12 at 1:29
show 3 more comments

3 Answers

up vote 4 down vote accepted

Call $P_1=[1:i:0]$ and $P_2=[1:-i:0]$ the two intersection points of your circles.

All of Euclid's (or Descartes') good old circles $C\subset \mathbb A^2_{\mathbb C}$ with equation $(x-a)^2+(y-b)^2=r^2$ have their closure $\overline C=C\cup \lbrace P_1,P_2\rbrace\subset \mathbb P^2_{\mathbb C}$ obtained by adding the same two points $P_1,P_2$.
For that reason these points used to be called the cyclic points in the projective geometry of the 19th and early 20th century.
In other words the circles are the three-dimensional linear system obtained from the five-dimensional system of all conics by requiring that the conic go through the cyclic points.

Back to your problem after this nostalgic reminder.

In the chart $x=1$ of $\mathbb P^2_{\mathbb C}$ your circles have indeed the equations
$1+y^2-z^2=0$ and $1+y^2-2z^2=0$.
Their intersection correspond to an algebra of dimesion $4$ as your electronic servant says, but you missed the point that this takes into account both $P_1$ and $P_2$ which have coordinates $y=\pm i,z=0$.
If you want to see what is going on at $P_1$ it is convenient to consider a new coordinate system $(\eta=y-i,z)$ for $\mathbb P^2_{\mathbb C}$ in the neighbourhood of $P_1$ (which has coordinates $\eta=z=0$).
In that system the required multiplicity of intersection at $P_1$of the circles $\overline C_1, \;\overline C_2$ is given by

$$dim_\mathbb C (\mathcal O/\langle \eta (\eta+2i)-z^2,\eta (\eta+2i)-2z^2 \rangle )=dim_\mathbb C (\mathcal O/\langle \eta (\eta+2i),z^2\rangle )=2$$

where $\mathcal O=\mathbb C[\eta, z]_{\langle \eta ,z\rangle }$ [Note that $\eta(\eta+2i)=y^2+1$ and that $\eta + 2i$ is invertible in $\mathcal O$]

The two circles $\overline C_1$ and $\overline C_2$ are thus tangent at $P_1$ and of course similarly at $P_2$ .

share|improve this answer
    
just added a question: math.stackexchange.com/questions/148971/… , can you please have a look? –  user31509 May 23 '12 at 21:04
add comment

As I have said before in this similar question, theorem 3 of Section 3.3 of Fulton's Algebraic Curves gives an algorithm for computing the intersection number at a point P. The interestion number is defined as the dimension of the vector space that you are looking for.

For the projective case you need to dehomogenize with respect to the "proper" line to reduce it to the affine case.

There are some examples here.

share|improve this answer
add comment

For variety here's another approach for this specific problem.

In characteristic 3 mod 4, your equations are defined over $\mathbb{F}_p$, and both $i$ and $-i$ are in the same Galois orbit; therefore, so are the two points, and so they must have the same multiplicity.

The same argument works in characteristic zero. In fact (but I haven't thoroughly checked the details), if you're comfortable with schemes, we can apply it to $\mathbb{P}^2_{\mathbb{Z}}$ to see that the intersection of those two equations is the union of:

  • The degree-2 horizontal curve corresponding to the points $(1 : \pm i : 0)$ in $\mathbb{P}^2_{\mathbb{Q}}$, with multiplicity 2
  • $\mathbb{P}^2_{\mathbb{F}_2}$ (with multiplicity 1?)

So, for any characteristic $> 2$, we can obtain the result by taking the intersection with $\mathbb{P}^2_{\mathbb{F}_p} \subseteq \mathbb{P}^2_{\mathbb{Z}}$ must

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.