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Here: https://en.wikipedia.org/wiki/Imaginary_hyperelliptic_curve#The_divisor_and_the_Jacobian it says:

"Actually, Bézout's theorem states that a straight line and a hyperelliptic curve of genus $2$ intersect in $5$ points. So, a straight line through two point lying on $C$ does not have a unique third intersection point, it has three other intersection points."

However in the same page, Figure 2 has , let's say a "curved line" that intersects the elliptic curve on 6 points. How can this happen? I am not seeing why the "curved line" does not fulfill the above paragraph....

P.S.: I know there are other questions related on HECC in the current site, but I am asking something very specific, not mentioned before, the way I put it.

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    $\begingroup$ Your quote is about a straight line and a hyperelliptic curve. Your question is about a "curved line" and an elliptic curve. Can you elaborate on why you think the one has anything to do with the other? $\endgroup$ Commented Feb 22 at 21:57
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    $\begingroup$ First, you typed elliptic for hyperelliptic. Second, remember these are real pictures that are not to be taken literally. Bezout’s Theorem applies over an algebraically closed field. Third, the curvy line is a cubic, not a line. By Bezout, a cubic intersects a curve of degree $5$ in $15$ points (counting multiplicities). $\endgroup$ Commented Feb 22 at 22:34

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The short answer: The "curved line" is not a straight line.

The longer answer: It may be useful to have an explicit situation, where we add reduced divisors. Let us consider the Example from the linked wiki page. $$ (C)\ :\qquad y^2 =x(x+1)(x+3)(x-3)(x-5)\ . $$ Consider the points $P,Q,R,S$, and the reduced divisors $D_1,D_2$: $$ \begin{aligned} P &= (1, 8)\ , \\ Q &= (3, 0)\ ,\\ R &= (5, 0)\ ,\\ S &= (0, 0)\ ,\\[3mm] D_1 &= P+Q-2O\ ,&&\text{Mumford representation:} & ((x-1)(x-3),&\ -4(x-3)\ ,\\ D_2 &= R+S-2O\ ,&&\text{Mumford representation:} & (x(x-5),&\ 0)\ . \end{aligned} $$ (There is no $[P]$ notation to pass from geometry, a point $P$, to a linearized object, $[P]$, a divisor. It should be clear from the context.)

It turns out that the sum $D=D_1+D_2$ is the divisor with the Mumford representation $$ (x^2 - 8x+3, \ -12x)\ . $$ So if we want to represent it as $T+U-2O$, we have to take the roots of $x^2-8x+3$, which are $4\pm s$ with $s=\sqrt {13}$ for an easy typing, and so we get the points $T,U$ equal to $(4\pm s,\ 12(4\pm s))$. Details follow.

The cubic that passes through $P, Q, R, S$, is $$ a(x) := x(x-3)(x-5)\ . $$ Consider now the cubic $(C')$ with equation: $$ (C')\ :\qquad y = a(x) = x(x-3)(x-5)\ . $$ We intersect now the cubic $(C')$, $y=a(x)$, with the given hyperelliptic curve of genus two, $y^2 =f(x)$. Intersecting means solving the system with two equations, those for $(C')$ and $(C)$, with two unknowns ($x,y$, the ones that parametrize our affine piece of geometry): $$ \left\{ \begin{aligned} y &= a(x)\ ,\\ y^2 &= f(x)\ . \end{aligned} \right. $$

How many solutions do we have?

This is more or less your question. (I found it easier to understand the situation in a sample case like the chosen one.) Well, we take $y=a(x)$ from the one equation, substitute it in the second one, and obtain one equation in $x$, which is $a(x)^2 =f(x)$, in our case: $$ x^2(x-3)^2(x-5)^2 = x(x+1)(x+3)(x-3)(x-5)\ . $$ We move everything on the L.H.S. - factors are transparent, and we know that
$P,Q,R,S$ (i.e. their $x$-component) should be among the roots: $$ \begin{aligned} a(x)^2 - f(x) &= x^2(x-3)^2(x-5)^2 - x(x+1)(x+3)(x-3)(x-5) \\ &= x(x-3)(x-5)\Big( (x^3 - 8x^2 + 15x)- (x^2 + 4x+3)\Big) \\ &=x(x-3)(x-5)\cdot(x^3 -9x^2 +11x-3)\qquad\text{(and $(x-1)$ from $P$ is a factor)} \\ &=x(x-3)(x-5)\cdot(x-1)(x^2-8x+3)\ . \end{aligned} $$


A plot of $(C)$ and $(C')$, with expected shapes, and their intersection points follows.

(The points $P,Q,R,S$ we start with are green, the "new" points $T,U$ are painted magenta.)

intersection of a hyperelliptic curve with a cubic to compute addition of tuples of points / of Mumford divisors


Sage code producing the figure:

K.<s> = QuadraticField(13)    # makes s represent sqrt(13)
r.<x> = K[]

f = x*(x + 1)*(x + 3)*(x - 3)*(x - 5)
C = HyperellipticCurve(f)
P = C(1, 8)
Q = C(3, 0)
R = C(5, 0)
S = C(0, 0)
T = C((4 + s, 48 + 12*s))
U = C((4 - s, 48 - 12*s))

var('y');

g = implicit_plot(y^2 - f, (x,-10, 20), (y, -150, 150)
                  , aspect_ratio=1/10
                  , figsize=(10, 10))
g += implicit_plot(y - x*(x - 3)*(x - 5), (x,-10, 20), (y, -150, 150)
                  , aspect_ratio=1/10
                  , figsize=(10, 10)
                  , color='brown')
g += point([(1, 8), (3, 0), (5, 0), (0, 0)]
           , size=30, color='green')
g += point([(4 + s, 48 + 12*s), (4 - s, 48 - 12*s)]
           , size=30, color='magenta')
 

And let now show $g$ with the default viewer... (Type g for instance in the interpreter.)

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    $\begingroup$ $D_1$ and $D_2$ are my choice, any other choices of the shape$$[\text{point}]+[\text{(same or other) point}]-2[O]$$would work as a sample. However, i had to take a good example, so that the final picture fits in a "small area", and such that the intersection points also show up in $\Bbb R$. In general, the remained polynomial of degree two may have no real roots. This happens for instance by taking $(15,\pm 720)$ instead of $(0,0)$ or so, if i remember correctly. Also, note that the linear combination must have degree $0$. So for $2P+3Q-O$ we have $2+3-1\ne 0$. $P,Q,R,S$ are my chosen points.. $\endgroup$
    – dan_fulea
    Commented Feb 23 at 12:16
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    $\begingroup$ .. on the curve, for instance $P=(1,8)$ satisfies the equation $y^2=f(x)$, because plugging in the coordinates $1$ for $x$ and $8$ for $y$ leads to$$8^2=1(1+1)(1+3)(1-3)(1-5)\ .$$A true equality. This also happens for $(15,720)$,$$720^2=15(15+1)(15+3)(15-3)(15-5)$$because factorizing we get$$(2^4\cdot 3^2\cdot 5)^2=(3\cdot5)(2^4)(2\cdot3^2)(2^2\cdot 3)(2\cdot5)\ .$$But finding other "human" (didactic) points with integer coordinates, good to make computations with bare eyes, is not easy. I took the values $P,Q,R,S$ mentioned in my answer for exemplification... $\endgroup$
    – dan_fulea
    Commented Feb 23 at 12:22
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    $\begingroup$ The answer to the initial question was written to show how to add two divisors $D_1=P+Q-2O$ and $D_2=R+S-2O$. The sum was denoted by $D$. Why do we add them? Because there is a geometric construction that reduces the form $(P+Q-2O+(R+S-2O)=(P+Q+R+S)-4O$ to something of the shape $T+U-2O$, and the geometric construction was the subject of the question. We can stop at any point on the road, when we want to, if the initial question was answered. So yes, we do not need to know $D$, but sage delivers $D$ first. Then for the last question, at which point exactly shall we take $a(x)=(x-1)(x-3)(x-5)$? $\endgroup$
    – dan_fulea
    Commented Apr 18 at 11:55
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    $\begingroup$ (Maybe an other question with clear focus would be the good way to proceed, also other members of the community can see it and answer. Then please just comment on the new question and mention me, i will try to answer in short time, if this hard job during the week allows it...) $\endgroup$
    – dan_fulea
    Commented Apr 18 at 11:57
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    $\begingroup$ The point $P$ is $P=(x_P,y_P)=(1,8)$. Let $a$ be the function $a(x)=x(x-3)(x-5)$. Then $$a(x_P)=a(1)=1(1-3)(1-5)=1(-2)(-4)=8=y_P\ ,$$which means that (the graph of) $a$ passes through $P=(x_P,y_P)=(1,8)$. In the same manner, for $Q=(3,0)$,$$a(3)=3(3-3)(3-5)=0\ ,$$, for $R=(5,0)$ then$$a(5)=5(5-3)(5-5)=0\ ,$$and for $S=(0,0)$,$$a(0)=0(0-1)(0-5)=0\ .$$The other suggested choice for $a$, $a_1(x)=(x-1)(x-3)(x-5)$ does not work. For instance for $S$ we have $a_1(0)\ne 0$, for $P$ then $a_1(1)=0$. $\endgroup$
    – dan_fulea
    Commented Apr 19 at 15:13

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