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I am trying to understand the concept of multiplicity, I'll copy the definitions from the book I am reading:


If $f$ is a curve and $l$ is a line with equation $Y=aX+b$, the points of $l\cap f$ an be obtained eliminating $Y$ and solving the equation:

$$f_l(X):=f(X,aX+b)=0$$

We have three possibilities:

  • $f_l(X)$ is zero, and $l$ is a component of $f$.
  • $f_l(X)$ is a constant $\neq 0$, when $f\cap l = \emptyset$
  • $f_l(X)$ is a non-constant polynomial that can be written as $f_l(X)=c\prod_{i=1}^{r}(X-x_i)^{m_i}$ where $c$ is a constant and $x_i$ are pairwise distinct roots of the intersection.

Definition: The multiplicity or index of intersection of $l,f$ at the point $P$ is given by:

$$ \begin{equation*} (l,f)_P= \left\{ \begin{array}{ll} 0 & \quad P\notin l\cap f \\ \infty & \quad P\in l \subset f\\ m_i &\quad P=(x_i,ax_i+b) \text{ as in case 3 above.} \end{array} \right. \end{equation*} $$


Here is my confusion: Suppose I want to compute the multiplicity of $f(X,Y)=2X^4-3X^2Y+Y^2-2Y^3+Y^4$ at $(0,0)$. I think I need to take a straight line passing through $(0,0)$, which is $y=AX$. Suppose, for simplicity $A=1$, then we can write

$$f(X,AX)=X^2 - 5 X^3 + 3 X^4=\frac{ X^2\left(6 X-\sqrt{13}-5\right) \left(6 X+\sqrt{13}-5\right)}{12} $$

Then we have a polynomial $f_l(X)=c\prod_{i=1}^{r}(X-x_i)^{m_i}$ just as is given in the description.

My confusion is the following: We have factors with $m_i=1$ and $m_i=2$. Which one is the correct multiplicity? I noticed that in the definition, $x_i$ are pairwise distinct roots, so what do we do here? Because we have a term $(x-0)^2$ that has two equal roots.

EDIT:

Definition: Let $f$ be a curve and $P$ a point in $f$. There exists an integer $m=m_P(f)\geq 1$ such that for any line $l$ passing through $P$, we have $(l,f)_P \geq m$.

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    $\begingroup$ You're asking about two different concepts here. Intersection multiplicity, which takes as its inputs two curves and a point, and multiplicity of a curve, which takes as its inputs a curve and a point. You need to go back to your source and find the correct definition. (There are relations between these concepts of multiplicity, but you should start with the correct definition.) $\endgroup$
    – KReiser
    Commented Nov 16, 2021 at 20:57
  • $\begingroup$ @KReiser I guess I get it. I added the correct definition. So the multiplicity is the largest integer smallest than any $(l,f)_P$? I am still a bit confused though: What is the intersection multiplicity between the given line and curve at the point $(0,0)$? $\endgroup$
    – Red Banana
    Commented Nov 17, 2021 at 1:12
  • $\begingroup$ Why use the word doubt ? It makes no sense $\endgroup$
    – jimjim
    Commented Nov 17, 2021 at 6:06
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    $\begingroup$ @jimjim english.stackexchange.com/questions/2429/… might shed some light on the matter. $\endgroup$
    – KReiser
    Commented Nov 21, 2021 at 7:45

1 Answer 1

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To resolve your first confusion: you've done everything correct so far, and the intersection multiplicity of your curve and line is two, because that's the exponent corresponding to the $X$ term in your single-variable polynomial. So this says that the multiplicity is at most two.

To prove that the multiplicity is exactly two using your definitions, we'd need to check all the lines through the origin. Writing $f(X,AX)=2X^4-3AX^3+A^2X^2-2A^3X^3+A^4X^4$, this simplifies to $X^2(A^2-2(A+A^3)X+(2+A^4)X^2)$ and we see that the minimum order to which $X$ divides this polynomial is 2. We also need to check $X=0$, which gives $Y^2(1-2Y+Y^2)$, which is also divisible by $Y$ to order two, so the multiplicity is two.

The fact that this corresponded to the lowest-degree term, $Y^2$, in the expansion of your polynomial around $(0,0)$ is no accident: translating so the point where we're considering the intersection is the origin, one may note that the lowest-degree homogeneous part of the equation for your curve splits as a product of homogeneous linear factors, and as long as the line you're intersecting with isn't one of those factors, the intersection multiplicity you get from your calculation will be exactly the degree of the smallest nonzero homogeneous part. If the line you're intersecting with is one of those factors, it will be bigger, proving the claim.

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  • $\begingroup$ I'm still a bit confused: It says there that the multiplicity is the exponent $m_i$ from the polynomial. But I have $f(X,AX)=\frac{ X^2\left(6 X-\sqrt{13}-5\right) \left(6 X+\sqrt{13}-5\right)}{12}$ which has exponents $1$ and $2$. Which one of those is the correct one? $\endgroup$
    – Red Banana
    Commented Nov 21, 2021 at 7:36
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    $\begingroup$ $2$, the exponent corresponding to $X$. Would it help to write $X=(X-0)$? $\endgroup$
    – KReiser
    Commented Nov 21, 2021 at 7:43
  • $\begingroup$ I think I get it! The condition says: $$ m_i \quad P=(x_i,ax_i+b) \text{ as in case 3 above.}$$ If we write the given polynomial, we get three "tests": $$\overbrace{(0,0)=(0,0) }^{true}\qquad \overbrace{(0,0)=(\sqrt{13}-5,\sqrt{13}-5)}^{false} \qquad \overbrace{(0,0)=(-(\sqrt{13}-5),-(\sqrt{13}-5))}^{false}$$ So, it must be the exponent of the factor for which the test yields "true". Is that correct? $\endgroup$
    – Red Banana
    Commented Nov 21, 2021 at 7:52
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    $\begingroup$ Sure. ${}{}{}{}{}$ $\endgroup$
    – KReiser
    Commented Nov 21, 2021 at 7:54

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