Calculation of the principal divisor

Question

I have quite a problem in solving this excercise:

Given $P=(0,0)$ and the curve $C: x+y+x^4+y^3=0.$

I want to calculate the principal divisor of $f=y(x^2-y)\in k(C).$

So I want to calculate $$(f)=\sum_{Q\in C}v_Q(f)\cdot Q.$$ Here we have defined $$v_Q(g)=\max_{k\in\mathbb{Z}}(g\in m_Q^k)$$ for any function $g$ defined on $C$. $m_Q$ is the ideal of every function $g$ defined on $C$ with $g(Q)=0$.

I don't know where to start.

Here I will write steps of the solution found by your help.

We have $div(f)=div(y)+div(x^2-y).$ Let us start with $div(y)$. We need to find the points $P\in C$ with $v_P(y)\ne 0$. The points with $v_P(y)\ge 1$ satisfy $y=0$ and $x+y+x^4+y^3=0$, so $x(1+x^3)=0$. That are $(a,0)$ with $a\in\{0,-1,(-1)^{1/3},-(-1)^{2/3}\}$. But what we can say about the points with $v_P(y)\le -1$? There are no such points because $y$ has no poles.

For example: Calculation of $v_{(-1,0)}(y)$: Clearly $y\in m_{(-1,0)}$. But we can't find two functions $g,h\in m_{(0,-1)}$ with $y=gh$ and $g(0,-1)=h(0,-1)=0$. So $y\notin m_{(0,-1)}^2$. So $v_{(0,-1)}(y)=1$. The same argument shows in general $v_{(a,0)}(y)=1$ for all $a$ like above.

That means that we have $$div(y)=(0,0)+(-1,0)+((-1)^{1/3},0)+(-(-1)^{2/3},0)+z\cdot\infty.$$ Since a principle divisor has degree $0$ we can conclude $z=-4$. So we have found the divisor $div(y)$.

Now: Calculate $div(x^2-y)$. There are no poles so we are only searching points with $v_P(x^2-y)\ge 1$. That means $x^2-y=0=x+y+x^4+y^3$. So $x^2=y$ and $x+x^2+x^4+x^6=0.$ And now I am in trouble.

Answer 1

$\DeclareMathOperator{\div}{\operatorname{div}}$ First, a recap of what you've done so far. Let $V$ be the projective closure of the affine curve $C$ specified in the problem, and assume $k$ is algebraically closed and has characteristic zero. Note that $V$ is given by the homogeneous equation $V: X Z^3 + Y Z^3 + Y^3 Z + X^4 = 0$, which has a single point $[0:1:0]$ at infinity.

Note that $\div(y(x^2 - y)) = \div(y) + \div(x^2 - y)$. We begin by computing $\div(y)$. Setting $y = 0$ in the equation defining $C$, we find $0 = x^4 + x = x(x+1)(x - e^{\pi i/3})(x - e^{5 \pi i/3})$. Recall that a principal divisor on a projective curve has degree $0$. Since $y$ has no poles in the affine plane, then it must have a pole of order $4$ at infinity. Thus $$ \div(y) = (0,0) + (-1,0) + (e^{\pi i/3}, 0) + (e^{5\pi i/3}, 0) - 4 \infty \, . $$

Now we compute $\div(x^2 - y)$. Setting $x^2 - y = 0$, i.e. $y = x^2$, we obtain $$ 0 = x^6 + x^4 + x^2 + x = x(x^5 + x^3 + x + 1) \, . $$ Let $g = x^5 + x^3 + x + 1$. This quintic is not solvable since it has Galois group $S_5$. But since $k$ is algebraically closed, we know $g$ has $5$ roots (counted with multiplicity), even if we can't express them in terms of radicals. Using the Euclidean algorithm, we find that $\gcd(g,g') = 1$, so $g$ has only simple roots. Denoting these roots by $\alpha_1, \alpha_2, \ldots, \alpha_5$, then the divisor of zeroes is $$ \div_0(x^2 - y) = (0,0) + (\alpha_1, \alpha_1^2) + (\alpha_2, \alpha_2^2) + \cdots + (\alpha_5, \alpha_5^2) \, . $$ Note that $x^2 - y$ is generically $6$-to-$1$, hence has degree $6$. Since it has no poles in the affine plane, it must have a pole of order $6$ at infinity. Thus $$ \div(x^2 - y) = (0,0) + (\alpha_1, \alpha_1^2) + (\alpha_2, \alpha_2^2) + \cdots + (\alpha_5, \alpha_5^2) - 6 \infty \, . $$

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We've been sneaky in using the fact that a principal divisor has degree $0$ to (not) compute the order of poles at infinity. As Jürgen Böhm points out in the comments, another strategy would be to work with affine charts. When we write $C: x+y+x^4+y^3=0$, we are really working in the affine open $Z \neq 0$ where $x = X/Z$ and $y = Y/Z$. To compute the order of pole at infinity, we could instead work in the chart $Y \neq 0$ (since $\infty = [0:1:0]$), with $u = X/Y$ and $v = Z/Y$. For an example of this more computational approach, see here.

Answer 2

If you can live with $k=\mathbb{Q}$ it is easy to calculate with Macaulay2 (although perhaps a manual solution is intended, I give here the mechanical calculation, first to illustrate the mechanism and second to give the OP the possibility to calculate further examples by hand and check with Macaulay2).

Call $g = x + y + x^4 + y^3$ and $f = y (x^2-y)$. Then calculate the primary decomposition of the ideal $(f,g) \subseteq k[x,y]$:

$$(f,g) = \mathfrak{q}_1 \cap \mathfrak{q}_2 \cap \mathfrak{q}_3 \cap \mathfrak{q}_4$$

and assure yourself that $\dim k[x,y]/\mathfrak{q}_i = 0$, where $\dim$ is the Krull-dimension.

Then $(f) = \sum_{i=1}^4 v_{P_i}(f) P_i$ with $P_i = V(\sqrt{\mathfrak{q}_i})$ and $v_{P_i}(f) = \dim_k (k[x,y]/\mathfrak{q}_i)$.

The following session of Macaulay 2 does this:

+ M2 --no-readline --print-width 134
Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R=QQ[x,y]

o1 = R

o1 : PolynomialRing

i2 : g = x + y + x^4 +y^3

      4    3
o2 = x  + y  + x + y

o2 : R

i3 : f = y * (x^2 - y)

      2     2
o3 = x y - y

o3 : R

i4 : idI1 = ideal(f,g)

             2     2   4    3
o4 = ideal (x y - y , x  + y  + x + y)

o4 : Ideal of R

i5 : pdecomp=primaryDecomposition idI1

                     2                                2                   2       3    2             2
o5 = {ideal (x + y, y ), ideal (y, x + 1), ideal (y, x  - x + 1), ideal (x  - y, y  + y  + x + y, x*y  + x*y + x + 1)}

o5 : List

i6 : apply(pdecomp,(xx)->dim(R/xx))

o6 = {0, 0, 0, 0}

o6 : List

i7 : apply(pdecomp,(xx)->degree(R/xx))

o7 = {2, 1, 2, 5}

o7 : List