⎈ captn's log
31.8.2020  captn' on vacation
This website will not be updated anymore. I need to make update posts with huge commutative diagrams which are hard to write in mathjax. I'm keeping this here as an archive.
☸ ⚓ ⛴ 🦜
12.08.2020  a little about the sufficient values of $z$
Let $F$ and $G$ be curves defined by
$$ \begin{align} F &= x^a  y^A, \\ G &= x^b  y^B, \end{align} $$where $x=0$ is their only common tangent at the origin. (therefore $a \lt A$ and $b \lt B$).
Then possible (not smallest possible) value for $z$ is
$$ \begin{align} z &= (B+Aba) + (a + B  1), \text{ if } I_O(F,G) = aB, \\ z &= (B+Aba) + (b + A  1), \text{ if } I_O(F,G) = bA. \end{align} $$Current proof for this is ugly (= nonelegant). I hope I'll find a prettier one.
Some easier cases of $z$:
From now on, let $z_0$ be the lowest possible value of $z$.
 If $a=b$ or $A=B$, then $(F,G,I^k) \sim (x^a,y^A,I^k)$, and $z_0 = a+A1$.
 If $b>a$ or $A>B$, then $(F,G,I^k) \sim (x^a,y^B,I^k)$, and $z_0 = a+B1$.
 I don't have the rest caluclated yet.
12.08.2020  bigger update
It is summed up in separate document (letky.pdf) because of all the commutative diagrams
05.08.2020  curves with only one tangent in common
For any two curves $F$ and $G$ with only one tangent (at origin) in common, we know that $\dim(\ker(\psi))=1$. That means
$\require{AMScd}$ \begin{CD} \dim(\mathcal{O}/(F,G)) @= \dim(\ker(\pi)) @. + \dim(\mathcal{O}/(F,G,I^{m+n}) @.\\ @. @. @ \\ @. \dim(k[x,y]/I^{m+n}) @.  @. \dim(\ker(\varphi)) \\ @. @. @. @ \\ @. @. @. \dim (k[x,y]/I^n \times k[x,y]/I^m)  \dim(\ker(\psi)) \\ \end{CD}is actually
$\require{AMScd}$ \begin{CD} \dim(\mathcal{O}/(F,G)) @= \dim(\ker(\pi)) @. + (mn+1) @.\\ @. @. @ \\ @. \frac{1}{2}(m+n)(m+n+1) @.  @. \left(\frac{1}{2}(m(m+1) + n(n+1))  1\right) \\ @. @. @. @ \\ @. @. @. \frac{1}{2}(m(m+1) + n(n+1))  1. \\ \end{CD}All the interesting stuff happens in $\ker(\pi)$.
05.08.2020  curves with exactly two tangents in common
For the curves with two tangents in common, $\dim(\ker(\psi))$ is either $3$, or $2$. Analogously as in the case with only one tangent in common, this implies that
$\require{AMScd}$ \begin{CD} \dim(\mathcal{O}/(F,G)) @= \dim(\ker(\pi)) @. + (mn+2) @.\\ @. @. @ \\ @. \frac{1}{2}(m+n)(m+n+1) @.  @. \left(\frac{1}{2}(m(m+1) + n(n+1))  2\right) \\ @. @. @. @ \\ @. @. @. \frac{1}{2}(m(m+1) + n(n+1))  2. \\ \end{CD}Or $\dim(\mathcal{O}/(F,G)) = \dim(\ker(\pi)) + (mn+3)$.
30.07.2020  nice version of the supersimple case
How about we use this instead:
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^{m+n}) + \dim(\ker(\pi))  \dim(\ker(\varphi)) $$where
$$ \begin{align} \pi &\colon \mathcal{O}/(F,G) \longrightarrow \mathcal{O}/(I^{m+n},F,G), \\ \varphi &\colon k[x,y]/I^{m+n} \longrightarrow k[x,y]/(I^{m+n},F,G). \end{align} $$Or perhaps
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^{m+n},F,G) + \dim(\ker(\pi)). $$EXAMPLE I:
Let $F = x  y^i$, $G = x  y^j$, where $1 \lt i \lt j$. Then
$$ \mathcal{O}/(F,G) = \mathcal{O}/(x  y^i,x  y^j) \cong \mathcal{O}/(x  y^i,y^i(y^{ji}1)) \cong \mathcal{O}/(x, y^i), $$(obviously, now we already know their intersection multiplicity, because it is $\dim(\mathcal{O}/(F,G)) = \dim(\mathcal{O}/(x, y^i))=i$, but that's not the point)
$$ \mathcal{O}/(I^2,F,G) = \mathcal{O}/(x^2,xy,y^2,x  y^i,x  y^j)\cong \mathcal{O}/(x,y^2), $$ $$ k[x,y]/(I^2,F,G) = k[x,y]/(x^2,xy,y^2,x  y^i,x  y^j) \cong k[x,y]/(x,y^2), $$therefore
$$ \begin{align} \pi \colon \mathcal{O}/(x, y^i) &\longrightarrow \mathcal{O}/(x,y^2), \\ \ker(\pi) &= (y^{i1}, \cdots, y^2) \end{align} $$ $$ \begin{align} \varphi \colon k[x,y]/(x^2,xy,y^2) &\longrightarrow k[x,y]/(x,y^2) \\ \ker(\varphi) &= (x) \end{align} $$and $\dim(\ker(\pi))=i2$, $\dim(\ker(\varphi))=1$. Now
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^2) + \dim(\ker(\pi))  \dim(\ker(\varphi)) = 3 + (i2)  1 = i. $$Or even easier version is
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^2,F,G) + \dim(\ker(\pi)) = 2 + (i2) = i. $$EXAMPLE II:
Let $F = x  y^2 + x^3$, $G = x  y^2 + x^4$
Then
$$ \mathcal{O}/(F,G) = \mathcal{O}/(x  y^i,x  y^j) \cong \mathcal{O}/(x  y^2 + x^3,x  y^2 + x^4) \cong \mathcal{O}/(xy^2,x^3) \cong \mathcal{O}/(x,y^6), $$ $$ \mathcal{O}/(I^2,F,G) = \mathcal{O}/(x^2,xy,y^2,x  y^i,x  y^j) \cong \mathcal{O}/(x,y^2), $$ $$ k[x,y]/(I^2,F,G) = k[x,y]/(x^2,xy,y^2,x  y^i,x  y^j) \cong k[x,y]/(x,y^2), $$therefore
$$ \begin{align} \pi \colon \mathcal{O}/(x, y^6) &\longrightarrow \mathcal{O}/(x,y^2), \\ \ker(\pi) &= (y^{5}, \cdots, y^2) \end{align} $$ $$ \begin{align} \varphi \colon k[x,y]/(x^2,xy,y^2) &\longrightarrow k[x,y]/(x,y^2) \\ \ker(\varphi) &= (x) \end{align} $$and $\dim(\ker(\pi))=4$, $\dim(\ker(\varphi))=1$. Now
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^2) + \dim(\ker(\pi))  \dim(\ker(\varphi)) = 3 + (4)  1 = 6. $$Or even easier version is
$$ I_O(F,G) = \dim(\mathcal{O}/(F,G)) = \dim(k[x,y]/(I^2,F,G) + \dim(\ker(\pi)) = 2 + 4 = 6. $$RANDOM THOUGHTS
 In this supersimple case (both curves with only one simple tangent $x$), this all depends only on $\pi$, because we always have $$ \begin{align} \dim(\mathcal{O}/(I^2,F,G)) &= \dim(\mathcal{O}/(x,y^2)) = 2, \\ \dim(\mathcal{O}/(I^2)) &= 3, \\ \dim(\ker(\varphi)) &= 1, \\ \dim(k[x,y]/(I) \times k[x,y]/(I) &= 2 \\ \dim(\ker(\psi)) &= 1. \end{align} $$
 $\require{AMScd}$ \begin{CD} \dim(\mathcal{O}/(F,G)) @= \dim(\ker(\pi)) @. + \dim(\mathcal{O}/(F,G,I^{m+n}) @.\\ @. @. @ \\ @. @. \dim(k[x,y]/I^{m+n}) @.  \dim(\ker(\varphi)) \\ @. @. @. @ \\ @. @. @. \dim (k[x,y]/I^n \times k[x,y]/I^m)  \dim(\ker(\psi)) \\ \end{CD}
 $\require{AMScd}$ \begin{CD} \dim(\mathcal{O}/(F,G)) @= \dim(\ker(\pi)) @. + 2 @.\\ @. @. @ \\ @. @. 3 @.  1 \\ @. @. @. @ \\ @. @. @. 2  1 \\ \end{CD}
 Why do all my examples (including those which are not here) have the property $\mathcal{O}/(F,G) = \mathcal{O}/(x,y^i)$?
28.07.2020  hmm...
... to mi nevychadza.
Aha, uz je to ok.
24.07.2020  anyway
So far, we have been concentrating on $\ker(\psi)$, but there is also another option. The function $\varphi$, defined as
$$ \begin{align} \varphi \colon k[x,y]/I^{m+n} &\longrightarrow k[x,y]/(I^{m+n},F,G), \\ z &\longrightarrow z, \end{align} $$has the property
$$ \dim(\ker(\varphi)) = \dim (k[x,y]/I^n \times k[x,y]/I^m)  \dim(\ker(\psi)) = \frac{1}{2}(n(n+1)+m(m+1))  \dim(\ker(\psi)). $$So we can have a look on $\ker(\varphi)$ instead.
EXAMPLE:
Let
$$ \begin{align} F &= x^4 \\ G &= x^2y^5 + (y+x)^{11}. \end{align} $$Then their intersection multiplicity $I_O(F,G) = \mathcal{O}/(F,G)$ is
$$ \begin{align} \mathcal{O}/(F,G) &= \dim(\ker(\pi)) + [\dim(\text{Im}(\pi))] = \\ &= \dim(\ker(\pi)) + [\dim(\mathcal{O}/(I^{m+n},F,G))] = \\ &= \dim(\ker(\pi)) + [\dim(k[x,y]/(I^{m+n},F,G))] = \\ &= \dim(\ker(\pi)) + [\dim(\text{Im}(\varphi))] = \\ &= \dim(\ker(\pi)) + [\dim(k[x,y]/I^{m+n})  \{\dim(\ker(\varphi))\}] = \\ &= \dim(\ker(\pi)) + [\dim(k[x,y]/I^{m+n})  \{\dim(\text{Im}(\psi))\}] = \\ &= \dim(\ker(\pi)) + [\dim(k[x,y]/I^{m+n})  \{ \dim(k[x,y]/I^n \times k[x,y]/I^m)  \dim(\ker(\psi)) \}] \end{align} $$In this case, it is
$$ \begin{align} 44 &= 13 + [31] = \\ &= 13 + [31] = \\ &= 13 + [31] = \\ &= 13 + [31] = \\ &= 13 + [66  \{35\}] = \\ &= 13 + [66  \{35\}] = \\ &= 13 + [55  \{ 38  3 \}] \end{align} $$08.07.2020  the supersimple case
⯈ pdf version ⯇
I think writing down this very simple case could be useful for me right now. Therefore, let $F$ and $G$ be curves, each with only one tangent at $0$, the line $x=0$. That means
$$ \begin{align} F &= x + F_2 + F_3 + \cdots, \\ G &= x + G_2 + G_3 + \cdots. \end{align} $$Their intersection multiplicity is equal to $I_O(F,G) = m\cdot n + t + l = 2 + l$. So far, we know that
$$ \begin{align} mn &= 1.1 = 1 \\ \ker(\psi) &= 1 \text{ (  this is their one common tangent)} \\ \dim(\mathcal{O}/(I^{2},F,G)) &= 1+1 = 2 \\ \ker(\pi) &= ? \\ I_O(F,G) = \dim(\mathcal{O}/(F,G)) &= 2 + ? = ?? \end{align} $$This is actually easy to calculate whole. What happens here, is the following:
We need to search for the lowest index $i$ and $j$, such that $F_i$ and $G_j$ are not divisible by $x$. Then we have two possibilities. Either $ i \lt j $ (without loss of generality), or $i=j$. In the first case, $I_O(F,G) = i$, in the second case $I_O(F,G) \geq i$.
how can we see that from the blowup?
When using the blowup, the line $x=0$ remains to be the tangent, until we "reach" the first homogeneous part of $F$ (or $G$) which is not divisible by $x$.
example:
Let
$$ \begin{align} F &= x + xy + y^3 \\ G &= x + x^2 + y^4 \\ \end{align} $$Then the blowup of their union at the origin looks like this:
If this occurs at the same step for the curves $F$ and $G$ (that means $i=j$ when using the notation from before), the new curves might have the same tangent. This happens if the coeffitient at $y^i$ in $F$ and $G$ is the same. In this case, we need to continue with the blowups.
example II
Let us try the more complicated case. Let
$$ \begin{align} F &= x + xy  y^4 + xy^5 \\ G &= x + x^2y  y^4 + xy^4 \\ \end{align} $$The first three blowups look like this:
Now we can use the change of coordinates, such that their common tangent $x=y$ is transformed into the tangent $x=0$ and then use the same strategy as before. Let us define the new coodinates $x'$ and $y'$, where $x = y' + x'$, and $y = y'  x'$.
Now the curves are transformed into $F' = xyx+{y}^2+y\left(yx\right)^{5}$, $G' = x+{y}^{2}\left(yx\right)^{4}+y\left(yx\right)^{4}$, (where $F'(x,y) = F(x',y')$ and $G'(x,y) = G(x',y')$).
Now we can use blowups the same way as before. We already know how is this goindg to end up. The homogeneous part of the degree $2$ of the curve $F'$ ($F'_2 = y^2xy$) is not divisible by $x$, so that is where our journey ends. But I'm going to do the blowups anyway.
How can we predict this before the transformation? Each homogeneous part $F_k = n_0x^k + n_1x^{k1}y + \cdots + n_{k}y^k$ gets transformed into $F'_k = n_0(x+y)^k + n_1(x+y)^{k1}(yx) + \cdots + n_{k}(yx)^k$, and the coefficient at $y^k$ in $F'$ is be $n_0 + \cdots + n_k$. Therefore we know that $F'_k$ is not divisible by $x$ if and only if $n_0 + \cdots + n_k \neq 0$. So we just need to check this condition for each homogeneous part of both polynomials $F$ and $G$.
This, of course, works only for our tangent $y=x$. If we had an different tangent, say $f_0x=f_1y$, if would change to $\frac{n_0}{f_0^k} + \frac{n_1}{f_0^{k1}f_1} + \cdots + \frac{n_k}{f_1^k} \neq 0$.
how to put it together?
Altogether, let $F$ and $G$ be curves defined as
$$ \begin{align} F &= x + F_2 + F_3 + \cdots, \\ G &= x + G_2 + G_3 + \cdots. \end{align} $$Let $i$ and $j$ be the lowest integers such that $F_i$ and $G_j$ is not divisible by $x$. In other words the coefficient at $f_i y^i$ (in $F_i$), resp $g_j y^j$ (in $G_j$) is nonzero. Then
 If $i \lt j$, then $I_O(F,G) = i$.
 If $j \lt i$, then $I_O(F,G) = j$.
 If $j=i$, then we have two possibilities:
 If $f_i \neq g_j$, then $I_O(F,G) = i=j$.
 If $f_i = g_j$, then $I_O(F,G) \gt i=j$. To get an exact number, we need to continue with the blowups.
random thoughts:
 Here, the $\ker(\psi)$ is always equal to $1$. All these problems are hidden in the $\ker(\pi)$, which is just the difference of dimensions of $\dim(\mathcal{O}/(I^{2},F,G))$ and $\dim(\mathcal{O}/(F,G))$
 Even for this very simple case of two curves with only one simple tangent at the intersection point, we can create a situation with whatever intersection multiplicity we want. Do we want it to be $1000$? No problem, just define $F=x  y^{1000}$ and $G = xy^{1001}$.
also puiseux is nice here
doplnim neskor
25.4.2020  okay, here we go (again)
I'll write down even things that has been already shown before, i would like to use this post as a summary.
What I'm trying to do right now, is to get rid of the change of basis, where the tangents where either $X=0$ or $Y=0$. This was a good strategy up to two common tangents, but cannot be used for the case of three (or more) different tangents. We still assume that $m \leq n$.
In general, we are working with the curves
$$F = F_{m} + F_{m+1} + F_{m+2} + \cdots , $$ $$G = G_{n} + G_{n+1} + G_{n+2} + \cdots . $$The possible elements of the kernel
$$(A,B) = (A_{n1} + \cdots + A_0, B_{m1} + \cdots + B_0).$$Since we are finding the solutions of $AF  BG = 0$, we are actually solving the set of equations
$$ \begin{align} F_m A_{nm} & = G_n B_0 \\ F_m A_{nm+1} + F_{m+1}A_{nm} & = G_n B_1 + G_{n+1}B_0 \\ & \cdots \\ F_m A_{n1} + \cdots + F_{2m1}A_{nm} & = G_n B_{m1} + \cdots + G_{n+m1}B_{0} \end{align} $$We already know that
 If $(A,B) = (A_{n1} + \cdots + A_{ni} + 0 + \cdots + 0, B_{m1} + \cdots + B_{mj} + 0 + \cdots + 0)$, then $i=j$.
 If $F$ and $G$ have exactly $k$ tangents in common, then $j=i \leq k$.
Let $F$ and $G$ have exactly $1$ common tangent.
This is a very simple case.
That means that $F_m = (aX + bY)\mathcal{F}_{m1}$, and $G_n = (aX + bY)\mathcal{G}_{n1}$.
Then we are looking for elements of the form $(A,B) = (A_{n1},B_{m1})$. The only equation to solve here is $F_m A_{n1} = G_n B_{m1}$, which is easy, because it is equal to $(aX + bY)\mathcal{F}_{m1} A_{n1} = (aX + bY)\mathcal{G}_{n1} B_{m1}$, and $\gcd(\mathcal{F}_{m1},\mathcal{G}_{n1})=1$.
The elements of the $ker(\psi)$ are the elements $(A,B) = (b_0(\mathcal{G}_{n1}),b_0(\mathcal{F}_{m1}))$, for any $b_0 \in k$. Therefore
$$\dim(\ker(\psi)) = 1.$$Let $F$ and $G$ have exactly $2$ common tangents.
That means that $F_m = (\alpha X + \beta Y)(\gamma X + \delta Y)\mathcal{F}_{m2}$, and $G_n = (\alpha X + \beta Y)(\gamma X + \delta Y)\mathcal{G}_{n2}$.
Then we are looking for elements of the form $(A,B) = (A_{n1} + A_{n2},B_{m1} + B_{m2})$.
What we need to solve is the system
$$ \begin{align} F_m A_{n2} & = G_n B_{m2} \\ F_m A_{n1} + F_{m+1}A_{n2} & = G_n B_{m1} + G_{n+1}B_{m2} \end{align} $$(UPDATE 19.5.2020)
The first equation gives us the conditions
$$ \begin{align} B_{m2} &= c_0 \mathcal{F}_{m2}, \\ A_{n2} &= c_0 \mathcal{G}_{n2}, \end{align} $$for some $b_0 \in k$. Then the second equation of the system can be written as
$$ \begin{align} (\alpha X + \beta Y)(\gamma X + \delta Y)\mathcal{F}_{m2} A_{n1} + F_{m+1}c_0 \mathcal{G}_{n2} &= (\alpha X + \beta Y)(\gamma X + \delta Y)\mathcal{G}_{n2} B_{m1} + G_{n+1}c_0 \mathcal{F}_{m2} \\ (\alpha X + \beta Y)(\gamma X + \delta Y)[\mathcal{F}_{m2} A_{n1}  \mathcal{G}_{n2} B_{m1}] &= c_0[G_{n+1} \mathcal{F}_{m2}  F_{m+1} \mathcal{G}_{n2}] \end{align} $$Now we have two possibilities. Either $[G_{n+1} \mathcal{F}_{m2}  F_{m+1} \mathcal{G}_{n2}]$ is divisible by $(\alpha X + \beta Y)(\gamma X + \delta Y)$ (but what does it mean for the curve?), or $c_0=0.$

First, if it is not divisibile, then $b_0=0$, and $\mathcal{F}_{m2} A_{n1} = \mathcal{G}_{n2} B_{m1}$. Then (since $\gcd(\mathcal{F}_{m2},\mathcal{G}_{n2})=1$, we get the solution
$$ \begin{align} B_{m1} &= (b_1X + b_2Y) \mathcal{F}_{m2}, \\ A_{n1} &= (b_1X + b_2Y) \mathcal{G}_{n2}, \end{align} $$so all the solutions of the system are $(A,B) = ((b_1X + b_2Y) \mathcal{G}_{n2},(b_1X + b_2Y) \mathcal{F}_{m2})$ and the dimenstion of kernel is $2$.
$$\dim(\ker(\psi)) = 2.$$ 
Now let $[G_{n+1} \mathcal{F}_{m2}  F_{m+1} \mathcal{G}_{n2}]$ be divisible by $(\alpha X + \beta Y)(\gamma X + \delta Y)$, that means $G_{n+1} \mathcal{F}_{m2}  F_{m+1} \mathcal{G}_{n2} = (\alpha X + \beta Y)(\gamma X + \delta Y) \Delta_{m+n3}$. Now, we are trying to solve the equation
$$ \begin{align} \mathcal{F}_{m2} A_{n1}  \mathcal{G}_{n2} B_{m1} & = c_0\Delta_{m+n3} \\ (f_0X^{m2} + \cdots + f_{m2}Y^{m2}) (a_0X^{n1} + \cdots + a_{n1}Y^{n1})  (g_0X^{n2} + \cdots + g_{n2}Y^{n2}) (c_0X^{m1} + \cdots + b_{m1}Y^{m1}) & = c_0\Delta_{m+n3} \end{align} $$When writing down the equation for each power of $X^iY^j$ seperately, we get the linear system of $m+n2$ equations vith $m+n$ variables $a_0, \cdots a_{n1}, b_0, \cdots , b_{m1}$
$$ \begin{pmatrix} f_0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & g_0 & 0 & 0 & 0 & 0 \\ f_1 & f_0 & 0 & 0 & 0 & 0 & 0 & 0 & g_1 & g_0 & 0 & 0 & 0 \\ \vdots & & \ddots & 0 & 0 & 0 & 0 & 0 & \vdots & & \ddots & 0 & 0 \\ f_{m2} & f_{m3} & \cdots & f_0 & 0 & 0 & 0 & 0 & g_{m2} & g_{m3} & \cdots & g_0 & 0 \\ 0 & f_{m2} & \cdots & f_1 & f_0 & 0 & 0 & 0 & g_{m1} & g_{m2} & \cdots & g_1 & g_0 \\ 0 & 0 & \ddots & & & \ddots & 0 & 0 & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & f_{m2} & \cdots & f_1 & f_0 & 0 & g_{n2} & g_{n3} & \cdots & g_{nm} & g_{nm1}\\ 0 & 0 & 0 & 0 & f_{m2} & \cdots & f_1 & f_0 & 0 & g_{n2} & \cdots & g_{nm+1} & g_{nm} \\ 0 & 0 & 0 & 0 & 0 & \ddots & & \vdots & 0 & 0 & \ddots & & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & f_{m2} & f_{m1} & 0 & 0 & 0 & g_{n2} & g_{n3} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & f_{m2} & 0 & 0 & 0 & 0 & g_{n2} \\ \end{pmatrix} \begin{pmatrix} a_0 \\ a_1 \\ \vdots \\ a_{n1} \\b_0 \\b_1 \\ \vdots \\b_{m1} \end{pmatrix} = \begin{pmatrix} c_0 \delta_{0} \\ c_0 \delta_{1} \\ \vdots \\ \vdots \\c_0 \delta_{m+n4} \\c_0 \delta_{m+n3} \end{pmatrix} $$What is the rank of this matrix? It looks almost like the Sylvester matrix of $\mathcal{F}_{m2}(X,1)$ and $\mathcal{G}_{n2}(X,1)$, but it is not square. But either way, it is somewhere between $n1$ and $m+n2$.
That means that with the condition $G_{n+1} \mathcal{F}_{m2}  F_{m+1} \mathcal{G}_{n2} = (\alpha X + \beta Y)(\gamma X + \delta Y) \Delta_{m+n3}$, the dimension of the kernel is at least $3$. Possibly more, with some more conditions? I'm not sure. But at least we have
$$ m+2 \geq \dim(\ker(\psi)) \geq 3.$$UPDATE 3.7.2020:
Okay, looks like the rank of this matrix is always full. I'm almost certain. (almost? wtf??) Therefore
$$ \dim(\ker(\psi)) = 3.$$
24.4.2020  it's' all about number of common tangents
Looks like the best approach to make this work is to use induction by number of common tangents after all. I should write the results down before I forget everything and my notes stop making sense. Evetually, they turn into a pile of abstract art. It's the circle of life.
17.4.2020  the whole universe is linear algebra
The task of finding the dimensions of the kernel for different cases is turning into simple system of linear equations. But the conditions of the cases are getting more and more ugly. But at least they are still in some sort of relationship with the tangents of the curves.
UPDATE: I've made some good examples, but I'm too angry to write it down right now. Men can be very frustrating somethimes. Grr >:(
UPDATE 2: Maybe I shouldn't write these when I'm angry with my boyfriend.
10.4.2020  Let $m=3$.(updated 17.4.)
I don't have this solved completely yet. This is just a list of partial solutions of simple cases.
Let $m=3$.
Then $F = F_3 + F_4 + F_5 + \cdots$, $G = G_n + G_{n+1} + G_{n+2} + \cdots$. (the terms in $\cdots$ have no impact on $\ker(\psi)$)There are six (or more) cases:
 $\deg(\gcd(G_n,F_3))=0$ (they share no tangents) Then the dimension of the kernel is $0$.
 $\deg(\gcd(G_n,F_3))=1$ (they share exactly one tanget). Then the dimension of the kernel is $1$.
 $\deg(\gcd(G_n,F_3))=2$ (they share exactly two tangets). Then there are two possibilities.
 First, if we satisfy the following condition for each common tangent $X$, then the dimension of the kernel is $3$. The condition is different for simple and double tangents.
 If $X$ is a simple tangent (of multiplicity $1$), then if we write $F$ and $G$ in the form: $$F = \Big[X \cdot (X\cdot(\cdots) + AY^2\Big] + \Big[X\cdot(\cdots) + BY^4 \Big] + \cdots $$ $$G = \Big[X(X\cdot(\cdots) + CY^{n1})\Big] + \Big[ X\cdot(\cdots) + DY^{n+1} \Big] + \cdots ,$$ then the condition is $AD = BC$.
 Ket $X$ be a double tangent. Then we write $F$ and $G$ as: $$F = \Big[X^2 \cdot (aX + AY)\Big] + \Big[X^2\cdot(\cdots) + bXY^3 + BY^4 \Big] + \cdots $$ $$G = \Big[X^2(X^2\cdot(\cdots) + cXY^{n3} + CY^{n2})\Big] + \Big[ X^2\cdot(\cdots) + dXY^{n} + DY^{n+1} \Big] + \cdots ,$$ then the conditions are $AD=BC$, and $aD + Ad = bC + Bc$.
 Otherwise, the dimension of the kernel is $2$
 First, if we satisfy the following condition for each common tangent $X$, then the dimension of the kernel is $3$. The condition is different for simple and double tangents.
 $\deg(\gcd(G_n,F_3))=3$ (they share all three tangets). I don't have this case solved yet. EDIT (17.4): there are two cases:
 If ugly conditions are satisfied, then the dimension of the kernel is $6$.
 Otherwise, the dimension of the kernel is $5$.
3.4.2020  what do we have solved so far?
Without loss of generality, we assume that $m \leq n$. Enjoy our fresh table of solved cases:
m\n  1  2  3  4  5  6  7  8  ... 
1  🦕  🦕  🦕  🦕  🦕  🦕  🦕  🦕  🦕 
2  🦕  🦕  🦕  🦕  🦕  🦕  🦕  🦕  
3  🦖  🦖  🦖  🦖  🦖  🦖  🦖  
4  
5  
6  
7  
8  
... 
What are the results?
Let $m=1$.
Then $F = F_1 + F_2 + \cdots$, $G = G_n + G_{n+1} + \cdots$. There are two cases:
 $\deg(\gcd(G_n,F_1))=0$ (they don't share the tangent) Then the dimension of the kernel is $0$.
 $\deg(\gcd(G_n,F_1))=1$ (they share the tanget) Then the dimension of the kernel is $1$.
Let $m=2$.
Then $F = F_2 + F_3 + \cdots$, $G = G_n + G_{n+1} + \cdots$. There are four cases:
 $\deg(\gcd(G_n,F_2))=0$ (they share no tangents) Then the dimension of the kernel is $0$.
 $\deg(\gcd(G_n,F_2))=1$ (they share only one tanget). Then the dimension of the kernel is $1$.
 $\deg(\gcd(G_n,F_2))=2$ (they share both tangents). Then there are two possibilities.
 First, if $$F = \Big[XY\Big] + \Big[XY(U_1) + R_3 \Big] + \cdots $$ $$G = XY\Big[XY(V_{n4}) + Q_{n2}\Big] + \Big[ XY(W_{n1}) + R_3 Q_{n2} \Big] + \cdots .$$ Then the dimension of the kernel is $3$
 Otherwise, the dimension of the kernel is $2$
3.4.2020  all the solutions for $m=1$ (with computation)
Ok, I think I know how to give this whole thing into a inductionlike type of order. First, we take the care of the case, where $m=1$ ( and $n \geq 1$ can be arbitrary). This is actually a case, which is solved in my Project of disseration, but I want to include it here anyway. It is better for my current system.
Now, we can write $F$ and $G$ as $$ F = F_1 + F_2 + F_3 + \cdots $$ $$ G = G_n + G_{n+1} + G_{n+2} + \cdots $$
So far, from my Project of dissertation, we know that
 If $(A,B) \in \ker(\psi)$, then $(A,B) = (A_{n1},B_0)$ (lemma 4.2.2).
 The existence of $(A,B) \in \ker(\psi)$ depends only on $F_1$ and $G_n$. (lemma 4.2.4)
With these conditions, $AF  BG=0$ if and only if $$ A_{n1}F_1 = B_0G_n.$$
Let us introduce new coordinates $X$ and $Y$, such that $F_1 = X$. This brings us to two possible cases.
Let $G_n = X\cdot (G'_{n1})$ (for some homogeneous polynomial $G'_{n1}$). Visually, the polynomials look like this: $$ F = X + \cdots , $$ $$ G = X\cdot(G'_{n1}) + \cdots .$$
Then we have a solution $(A,B) = (b_0G'_{n1},b_0)$ for any $b_0 \in k$. In this case, the dimension of the kernel is $1$.
Let $G_n \neq X\cdot (G'_{n1})$ (meant as $G_n$ is not divisible by $X$.)
Then the only solution is $(A,B)=(0,0)$, and the dimension of th kernel is $0$.
3.4.2020  all the solutions for $m=2$ (with computation)
Now by similiar process, we solve the case, where $m=2$ (and $n \geq 2$). By the same lemmas,as in the first case, we know that $(A,B)\in \ker(\psi)$ depends only on $F_2$, $F_3$, $G_n$ and $G_{n+1}$, and (beside of the solution $(A,B)=(0,0)$), either $(A,B) = (A_{n1},B_1)$ (both nonzero), or $(A,B) = (A_{n1} + A_{n2},B_1 + B_0)$ (where $A_{n2}$ and $B_0$ are nonzero).
Now the equation $AFBG=0$ provides us two conditions
 $A_{n2}F_2 = B_0G_n$
 $A_{n1}F_2 + A_{n2}F_3 = B_1G_{n} + B_0G_{n+1}$
Now we can use the same simplification as in the first case. But this time, we have to split our computation into two cases. First, where $F$ has two distinct tangents, and second, where $F$ has one double tangent.
First, let $F$ have two distinct tangents. This means we can pick coordinates $X$ and $Y$, such that $F_2 = XY$. Now we can write the second condition as $$A_{n2}XY = B_0G_n,$$ which means, we have two possibilities. First, where $B_0= A_{n2} = 0$, and second, where both $B_0$ and $A_{n2}$ are nonzero.
We start with the simple case, where $B_0= A_{n2} = 0$. Then the only condition we need to astisfy is $A_{n1}XY = B_1G_n$. The two possible outcomes of this are:
 $G_n = X (G'_{n1})$ for some homogeneous polynomial $G'_{n1}$. Then $B_1 = b_2Y$ (for some $b_2 \in k$), and all the possible solutions are $(A,B) = (b_2G'_{n1},b_2)$. The dimension of the kernel is $1$.
 $G_n$ is not divisible by neither $X$, nor $Y$. Then the only solution is $(A,B)=(0,0)$, and the dimension of the kernel is $0$.
This being taken care of, we can move to the second case. From now on, we assume that both $B_0$ and $A_{n2}$ are nonzero. Then from $A_{n2}XY = B_0G_n$, we know that $G_n=XY\cdot(G'_{n2})$ (for some homogeneous polynomial $G'_{n2}$), and $A_{n2} = B_{0}G'_{n2}$. With this information, we write down $F$, $G$, $A$ and $B$ as $$F = \bigg[F_2\bigg] + \bigg[F_3\bigg] + \cdots = \bigg[XY\bigg] + \bigg[f_1X^3 + f_2X^2Y + f_3XY^2 + f_4Y^3\bigg] + \cdots $$ $$G = \bigg[G_n\bigg] + \bigg[G_{n+1}\bigg] + \cdots = \bigg[XY(\mathfrak{g}_1X^{n2} + \mathfrak{g}_2X^{n1}Y + \cdots + \mathfrak{g}_{n+1}Y^{n2})\bigg] + \bigg[ g_1X^{n+1} + g_2X^{n}Y + \cdots + g_{n+2}Y^{n+1} \bigg] + \cdots$$ $$ A = \bigg[A_{n1}\bigg] + \bigg[A_{n2}\bigg] = \bigg[a_1X^{n1} + a_2X^{n2}Y + \cdots + a_nY^{n1}\bigg] + \bigg[b_0(\mathfrak{g}_1X^{n2} + \mathfrak{g}_2X^{n1}Y + \cdots + \mathfrak{g}_{n+1}Y^{n2})\bigg]$$ $$ B = \bigg[B_1\bigg] + \bigg[B_0\bigg] = \bigg[(b_1X + b_2Y)\bigg] + \bigg[b_0\bigg] $$
Now we simply expand $AF  BG$ and compare the parts of the same degree. What we obtain is
 Two conditions for the polynomials $F$ and $G$. These are $b_0\mathfrak{g}_1f_1 = b_0g_1$ and $b_0\mathfrak{g}_{n1}f_4 = b_0g_{n+2}$. Since $b_0$ is nonzero, we can simplify them to $$\mathfrak{g}_1f_1 = g_1,$$ $$\mathfrak{g}_{n1}f_4 = g_{n+2}.$$
 $n$ linear equations of the form $a_i = L_i(b_0,b_1,b_2)$.
If we apply the conditions into $F$ and $G$, we get $$F = \bigg[XY\bigg] + \bigg[f_1X^3 + f_2X^2Y + f_3XY^2 + f_4Y^3\bigg] + \cdots $$ $$G = \bigg[XY(\mathfrak{g}_1X^{n2} + \mathfrak{g}_2X^{n1}Y + \cdots + \mathfrak{g}_{n+1}Y^{n2})\bigg] + \bigg[ (\mathfrak{g}_1f_1)X^{n+1} + g_2X^{n}Y + \cdots + g_{n+1}XY^{n+1} + (\mathfrak{g}_{n1}f_4)Y^{n+1} \bigg] + \cdots ,$$ and the dimension of the kernel is $3$.
We can do analogous computation for the case, where $F$ has a double tangent $X^2$. The results is very similiar. The kernel is again $3$, and the conditions for $F$ and $G$ are $$F = \bigg[X^2\bigg] + \bigg[f_1X^3 + f_2X^2Y + f_3XY^2 + f_4Y^3\bigg] + \cdots $$ $$G = \bigg[X^2(\mathfrak{g}_1X^{n2} + \mathfrak{g}_2X^{n1}Y + \cdots + \mathfrak{g}_{n+1}Y^{n2})\bigg] + \bigg[ g_1X^{n+1} + \cdots + g_nX^{2}Y^{n1} + (\mathfrak{g}_{n1}f_3 + \mathfrak{g}_{n2}f_4)XY^{n} + (\mathfrak{g}_{n1}f_4)Y^{n+1} \bigg] + \cdots ,$$
How to look at this result from a different angle.
The more easytoread version could be this. (I'm using completely unrelated letters for the coeffitients on purpose. It helps me to look at the result from a different angle). This is for the case with different tangents: $$F = \bigg[XY\bigg] + \bigg[XY(U_1) + \alpha X^3 + \beta Y^3\bigg] + \cdots $$ $$G = XY\bigg[XY(V_{n4}) + (aX^{n2} + bY^{n2})\bigg] + \bigg[ XY(W_{n1}) + \alpha a X^{n+1} + \beta b Y^{n1} \bigg] + \cdots .$$
Hm... I think I have an idea. How about we write $F$ and $G$ as $$F = \bigg[XY\bigg] + \bigg[XY(U_1) + R_3\bigg] + \cdots $$ $$G = XY\bigg[XY(V_{n4}) + Q_{n2}\bigg] + \bigg[ XY(W_{n1}) + R_3 Q_{n2} \bigg] + \cdots .$$ where $R_3$ and $Q_{n2}$ are not divisibile by $XY$. I know that $(\alpha X^3 + \beta Y^3)(aX^{n2} + bY^{n2}) \neq \alpha a X^{n+1} + \beta b Y^{n1}$, but their difference gets eaten up by $W_{n1}$ anyway. This works also for the case of double tangents (we simply need to write $X^2$ instead of $XY$).
The interesting part is, that aside from the divisibility condition ($R_3$ and $Q_{n2}$ are not divisibile by $XY$), and $F_2 \neq 0$, $G_n \neq 0$, we have no more conditions. The polynomials $U_1$, $V_{n4}$, $W_{n1}$, $R_3$ and $Q_{n2}$ can be whatever (of the right degree), or even equal to $0$.
This means, that from the $\psi$ point of view, the combination $$F = \bigg[XY\bigg] + \bigg[XY(U_1) + R_3 \bigg] + \cdots $$ $$G = XY\bigg[XY(V_{n4}) + Q_{n2}\bigg] + \bigg[ XY(W_{n1}) + R_3 Q_{n2} \bigg] + \cdots $$ is as good as ($R_3=0$) $$F = XY + XY(U_1) + \cdots $$ $$G = XY(P_{n2}) + XY(W_{n1}) + \cdots ,$$ and also as good as $$F = XY + F_3 \cdots + , $$ $$G = A_{n2}(XY + F_3) + \cdots .$$
2.4.2020  I have decided to reverse the order of these posts,
so I don't have to scroll down so much. I'm a genius. 🐦
30.3.2020  what is the upper bound
Funny, I've never thought about the upper bound of the kernel $\ker(\psi)$ until today.
Without loss of generality, let $m\leq n$. Then the general member of the kernel for $F = F_m + F_{m+1} + \cdots$ and $G = G_n + G_{n+1} + \cdots$ is (A,B), where $$ A = A_{n1} + \cdots + A_{nm}, $$ $$ B = B_{m1} + \cdots + B_0. $$ All the other $A_i$'s have to be equal to zero.
If our solution would be any $(A,B)$ as defined above, the dimension of the kernel would be (dimension of the Apart) + (dimension of the Bpart) $$ \sum_{i=1}^m (nm+i) + \sum_{i=1}^m (i) = m(nm) + \frac{(m+1)m}{2} + \frac{(m+1)m}{2} = m(nm) + (m+1)m = m(n+1).$$ But I think this would happen only for the case, where $F=0$ and $G=0$. And that is not very interesting case.
The highest intersection multiplicity (relative to values $m$ and $n$) should be for the cases, where $F=G$ (or $F=cG$, $c \in k$). In this case, $m=n$, and $A=B$. Then the dimension is $\sum_{i=1}^m (i) = (m+1)m/2$.
Is there anything in between? I don't know, I'll think about it later.
27.3.2020  all the solutions for $F$ and $G$ of order $2$.
After the success with finding the conditions for all the solutions of the form $(A,B) = (A_{ni},B_{mi})$, i've been trying to go for the solutions of the form $(A,B) = (A_{ni} + A_{ni1},B_{mi} + B_{mi1})$, but with no luck so far.
Finally yesterday, i've managed to get it done at least for the polynomials of the order $2$, therefore the solutions are of the form $(A,B) = (A_{1} + A_{0},B_{1} + B_{0})$. If $F$ and $G$ are of the order $2$, that means they can be written as $$F = F_2 + F_3 + \cdots,$$ $$G = G_2 + G_3 + \cdots.$$ There are five (EDIT: plus 2 at the end) possible cases of the form of $(A,B)$ (where $A_i$, $B_i$ are homogeneous of degree $i$):
 $(A,B) = (A_0,B_0)$  already solved
 $(A,B) = (A_1,B_1)$  already solved
 $(A,B) = (A_1 + A_0,B_1)$  no such solutions exist
 $(A,B) = (A_1,B_1 + B_0)$  no such solutions exist
 $(A,B) = (A_1 + A_0,B_1 + B_0)$  the solution follows:
Let $L$ and $K$ be the tangents of the curve $G$ at $0$. If we divide $G_3$ by the polynomial $LK$, we get $G_3 = (KL)V_1 + G'_3$, where $V_1$ is a polynomial (homogeneous of the degree $1$, or equal to zero) and $G'_3$ is the remainder (homogeneous of the degree 3, or equal to $0$). Analogously, $F_3 = (KL)W_1 + F'_3$
THE RESULT IS: The solutions of the form $(A,B) = (A_1 + A_0,B_1 + B_0)$ exist if and only if $F$ and $G$ are of the form $$ F = F_2 + F_3 + \cdots = cKL + c((KL)W_1 + G'_3)) + \cdots $$ $$ G = G_2 +G_3 + \cdots = KL + ((KL)V_1 + G'_3) + \cdots, $$ i.e. they have to share the same tangents, and at the homogeneous part of the degree $3$, the part divisible by $(KL)$ can be arbitrary, but the remainder part have to be identical. This is all up to a multiple by some scalar $c \in k$.
Then all the solutions $(A,B) \in \ker(\psi)$ are $$A = a_0 + a_1L + a_2K,$$ $$B = c(a_0 + (a_1  a_0(jf))L + (a_2  a_0(kg)K)),$$ where $a_i \in k$ (the numbers $j,f,k,g \in k$ are defined by some parts of $V_1$ and $W_1$, i'll write it down later). Therefore the dimension of the kernel is $3$.
If the remainder part is equal to zero, we fall into $(A_1,B_1)$ cathegory. If $V_1 = W_1$, we fall into the $(A_0,B_0)$ cathegory.
 EDIT: i just realized, i completely forgot $(A,B) = (A_1 + A_0,B_0)$ and $(A,B) = ( A_0,B_1 + B_0)$, but i think they are part of the solution in the case $5$.
I also have the proof writen down, but in a very ugly form.
These are all the solutions, where $F$ and $G$ are of the order $2$. But these are NOT all the polynomials $F$ and $G$, which have the solutions of the form $(A,B) = (A_1 + A_0,B_1 + B_0)$.
The solution of this simple case could suggest what kind of solutions should we expect in more complicated parts.