jupyter notebook showing calculation of the valuation

script/research/ec/valuation.ipynb

@@ -0,0 +1,564 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "6a512e0d",
+   "metadata": {},
+   "source": [
+    "# Computing Valuation Manually\n",
+    "\n",
+    "We wish to compute $\\textrm{ord}_P(f)$ where $P = (2, 4)$ and $f = y - 2x$.\n",
+    "\n",
+    "$$E(\\mathbb{F}_11) : y^2 = x^3 + 4x$$\n",
+    "\n",
+    "Let $K[V] = K[x, y] / \\langle y^2 - x^3 - 4x \\rangle$ by the coordinate ring. $K(V)$ is the field of fractions for $K[V]$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "5d554e0c",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "Fraction Field of Quotient of Multivariate Polynomial Ring in x, y over Finite Field of size 11 by the ideal (-x^3 + y^2 - 4*x)"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Define our base polynomial ring over Z_11\n",
+    "K.<x, y> = GF(11)[]\n",
+    "# This is K(V)\n",
+    "S = K.quotient(y^2 - x^3 - 4*x).fraction_field()\n",
+    "S"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "69feb612",
+   "metadata": {},
+   "source": [
+    "Because $P$ lies on the curve $E$, we can take the nontangent lines $x - 2, y - 4$ as a basis for the local curve. The intuition is that these lines describe the coordinate grid around $P$, and we can multiply them by any polynomial to get cosets of $K(V)$.\n",
+    "\n",
+    "More formally we can see this by noting that:\n",
+    "\n",
+    "$$(y - 4)(y + 4) = (x - 2)^3 - 5*(x - 2)^2 - 6(x - 2)$$\n",
+    "\n",
+    "So therefore any function on $E$ can be expressed in terms of $(x - 2)$ and $(y - 4)$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "id": "c430ae30",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "True"
+      ]
+     },
+     "execution_count": 11,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "X, Y = S(x), S(y)\n",
+    "(Y - 4)*(Y + 4) == (X - 2)^3 - 5*(X - 2)^2 - 6*(X - 2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2515224a",
+   "metadata": {},
+   "source": [
+    "Begin by expressing $f = y - 2x$ in terms of this basis.\n",
+    "\\begin{align*}\n",
+    "y &= y - 2x \\\\\n",
+    "  &= -2(x - 2) + 1(y - 4) \\\\\n",
+    "\\mathbf{a} &= (-2, 1, 0) \\\\\n",
+    "\\mathbf{b} &= (x - 2, y - 4, 1)\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "id": "bd415ef9",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-2*x + y"
+      ]
+     },
+     "execution_count": 10,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Components for f\n",
+    "a0, a1, a2 = -2, 1, 0\n",
+    "# Our basis\n",
+    "b0, b1, b2 = x - 2, y - 4, 1\n",
+    "a0*b0 + a1*b1 + a2*b2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "e3087005",
+   "metadata": {},
+   "source": [
+    "Using the identity above for $(y - 4)(y + 4)$, we can see that\n",
+    "\\begin{align*}\n",
+    "(y - 4) &= \\frac{(x - 2)^3 - 5*(x - 2)^2 - 6(x - 2)}{(y + 4)} \\\\\n",
+    "        &= (x - 2) \\frac{(x - 2)^2 - 5*(x - 2) - 6}{(y + 4)}\n",
+    "\\end{align*}\n",
+    "So we can know that $b_1 = (E_f / E_g) b_0$, and can make this substitution in $f$.\n",
+    "\\begin{align*}\n",
+    "f &= -2 (x - 2) + 1 \\cdot \\frac{(x - 2)^3 - 5(x - 2)^2 - 6(x - 2)}{y + 4} \\\\\n",
+    "  &= \\frac{1}{(y + 4)}[(x - 2)^3 - 5(x - 2)^2 - 6(x - 2) - 2(x - 2)(y + 4)]\n",
+    "\\end{align*}"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "id": "9e2be279",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "True"
+      ]
+     },
+     "execution_count": 13,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# and lets double check this\n",
+    "B0 = X - 2\n",
+    "Eg = Y + 4\n",
+    "(B0^3 - 5*B0^2 - 6*B0 - 2*B0*Eg)/Eg == Y - 2*X"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9dfe225e",
+   "metadata": {},
+   "source": [
+    "Now lets quotient out $b_0 = (x - 2)$ from $f$ to get the first power of $k = 1$ for the uniformizer.\n",
+    "\\begin{align*}\n",
+    "f &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 6 - 2(y + 4)]\n",
+    "\\end{align*}\n",
+    "Notice now the constant term is $- 6 - 2(y + 4)$ which should be a multiple of $y - 4$ if we can continue extracting $b_0$ from the expression."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "id": "249b0ed0",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-2*y - 3"
+      ]
+     },
+     "execution_count": 14,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "-6 - 2*(y + 4)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "id": "5894ffc4",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-2*y - 3"
+      ]
+     },
+     "execution_count": 15,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "-2*(y - 4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "dc5c6b41",
+   "metadata": {},
+   "source": [
+    "Great so we continue. Instead of writing $-6 - 2(y + 4)$, lets instead write $-2(y - 4)$\n",
+    "\\begin{align*}\n",
+    "f &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 2(y - 4)] \\\\\n",
+    "  &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 2\\frac{(x - 2)^3 - 5(x - 2)^2 - 6(x - 2)}{y + 4}] \\\\\n",
+    "  &= \\frac{(x + 2)^1}{(y + 4)^2}[(x - 2)^2(y + 4) - 5(x - 2)(y + 4) - 2((x - 2)^3 - 5(x - 2)^2 - 6(x - 2))] \\\\\n",
+    "  &= \\frac{(x + 2)^2}{(y + 4)^2}[(x - 2)(y + 4) - 5(y + 4) - 2((x - 2)^2 - 5(x - 2) - 6)] \\\\\n",
+    "\\end{align*}\n",
+    "So $k = 2$, and lets evaluate the constant terms."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "id": "7b343008",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-5*y + 3"
+      ]
+     },
+     "execution_count": 16,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "-5*(y + 4) - 2*(-6)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "id": "595834a3",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-5*y - 2"
+      ]
+     },
+     "execution_count": 17,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "-5*(y - 4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "8928eeda",
+   "metadata": {},
+   "source": [
+    "Now there's a remainder left over so the expression terminates."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "5575757f",
+   "metadata": {},
+   "source": [
+    "# Implementation Details\n",
+    "\n",
+    "Several points of interest:\n",
+    "\n",
+    "* We don't need to keep track of $g$ although it's done for completeness.\n",
+    "* For the uniformizer part, we just need to keep track of $k$.\n",
+    "* The inner expression can be optimized just by looking at the constant term when viewed from $(x - 2)$. Although we are not doing that."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "id": "a0a83e30",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(x - 2, y - 4, 1)"
+      ]
+     },
+     "execution_count": 18,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "b0, b1, b2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "id": "d042e184",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(-2, 1, 0)"
+      ]
+     },
+     "execution_count": 24,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Decompose polynomial function into basis components\n",
+    "def decomp(f):\n",
+    "    a0, r = f.quo_rem(b0)\n",
+    "    a1, r = r.quo_rem(b1)\n",
+    "    a2, r = r.quo_rem(b2)\n",
+    "    return a0, a1, a2\n",
+    "\n",
+    "f = y - 2*x\n",
+    "decomp(f)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "id": "c804233f",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "-x^3 + y^2 - 4*x"
+      ]
+     },
+     "execution_count": 23,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "Px, Py = (2, 4)\n",
+    "EC_A = 4\n",
+    "# Calculate substitution polynomials\n",
+    "Ef = b0^2 + binomial(3,2)*Px*b0^1 + (3*Px^2 + EC_A)\n",
+    "Eg = (y + Py)\n",
+    "# Should be EC equation\n",
+    "Eg*b1 - Ef*b0"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "id": "cbbb63c7",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(x^2 + 2*x - 2*y, y + 4)"
+      ]
+     },
+     "execution_count": 27,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "a0, a1, a2 = decomp(f)\n",
+    "g = 1\n",
+    "\n",
+    "# Perform first reduction\n",
+    "a0 = Eg*a0 + Ef*a1\n",
+    "a1 = 0\n",
+    "g *= Eg\n",
+    "assert a1 == a2 == 0\n",
+    "a0, g"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "9357c553",
+   "metadata": {},
+   "source": [
+    "Set $k = 1$ since we have now factored out $(x - 2)$. Continue for second decomposition followed by reduction."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "id": "d5810612",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(x + 4, -2, 0)"
+      ]
+     },
+     "execution_count": 28,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "a0, a1, a2 = decomp(a0)\n",
+    "a0, a1, a2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2a5b3561",
+   "metadata": {},
+   "source": [
+    "as expected the remainder a2 is zero"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "id": "22671821",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "True"
+      ]
+     },
+     "execution_count": 31,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "U = (X - 2)\n",
+    "F = (X + 4)*(X - 2) - 2*(Y - 4)\n",
+    "G = Y + 4\n",
+    "U^1 * F / G == Y - 2*X"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "id": "411c7216",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(-2*x^2 + x*y + 4*y, y^2 - 3*y + 5)"
+      ]
+     },
+     "execution_count": 32,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Second reduction\n",
+    "a0 = Eg*a0 + Ef*a1\n",
+    "a1 = 0\n",
+    "g *= Eg\n",
+    "assert a1 == a2 == 0\n",
+    "a0, g"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "de5d1ed9",
+   "metadata": {},
+   "source": [
+    "Set $k = 2$. Now decompose and check remainder is zero again before performing reduction."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "id": "eceb52f7",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(-2*x + y - 4, -5, 5)"
+      ]
+     },
+     "execution_count": 33,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "a0, a1, a2 = decomp(a0)\n",
+    "a0, a1, a2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "daab41c5",
+   "metadata": {},
+   "source": [
+    "Now the remainder is 5 so the algorithm stops. Our final valuation is $k = 2$."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 44,
+   "id": "1c654c46",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Convert our function to the function field K(V)\n",
+    "original_f = Y - 2*X\n",
+    "k = 2\n",
+    "\n",
+    "f = a0*b0 + a1*b1 + a2*b2\n",
+    "fprime = b0^k * f/g\n",
+    "assert fprime == S(original_f)\n",
+    "assert g(Px, Py) != 0\n",
+    "assert f(Px, Py) != 0\n",
+    "assert b0(Px, Py) == 0\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "SageMath 9.6",
+   "language": "sage",
+   "name": "sagemath"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}