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The truncated multiplication map is injective exactly when the tangent cones are coprime

Statement

Assume the Axiom of Choice, inherited from the cited local-length, smoothness or Bezout suppliers.

Let $O$ be the local ring of $A^{2}$ at the origin over an algebraically closed field $k$ , let $f, g \in m$ have orders $m, n \geq 1$ , and define

$\overset{ˉ}{ψ} : O / m^{n} \times O / m^{m} ⟶ O / m^{m + n}, \overset{ˉ}{ψ} (A, B) = A f + B g,$

on representatives (the truncation of the $f$ -coefficient is by the order of $g$ , and conversely, so that changes of representatives change both products by elements of $m^{m + n}$ ). Then $\overset{ˉ}{ψ}$ is well defined and $k$ -linear, and it is injective if and only if the lowest-degree forms $f^{*}$ and $g^{*}$ have no common factor in $k [x, y]$ . If $f^{*}, g^{*}$ share a factor, the kernel is nonzero. In particular

$\dim_{k} im \overset{ˉ}{ψ} = \dim_{k} (O / m^{n}) + \dim_{k} (O / m^{m})$

exactly when the tangent cones are coprime.

Facts & Assumptions

Given: AC The Axiom of Choice, an algebraically closed field $k$ , the local ring $O$ of $A^{2}$ at the origin, $m = (x, y)$ , elements $f, g \in m$ of orders $m, n \geq 1$ , and their lowest-degree forms $f^{*}, g^{*}$ .

[F1]

$O$ is a two-dimensional regular local ring and ${gr}_{m} O ≅ k [X, Y]$ is a domain, so initial forms multiply and the order of a product is the sum of the orders associated graded ring of a regular local ring, embedding dimension and regular local ring, A local ring is a nonzero commutative ring with a unique maximal ideal, Multiplicity of a plane curve at a point.

[F2]

$k [x, y]$ is a unique factorisation domain; the classes of the monomials of degree $< n$ form a $k$ -basis of $k [x, y] / m^{n}$ , so every class has a unique representative of degree $\leq n - 1$ , and $\dim_{k} (O / m^{n}) = (\binom{n + 1}{2})$ Lengths of truncated plane local rings, Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis. Over the algebraically closed field $k$ a nonzero binary form of positive degree is a product of linear forms, so any common factor of positive degree of two such forms has a linear factor common to both Tangent cone and tangent lines at a point (Remarks, binary-form factorisation).

[F3]

The product of nonzero homogeneous forms of degrees $a, b$ is nonzero in the polynomial domain and homogeneous of degree $a + b$ homogeneous polynomial and homogeneous ideal, Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes. The kernel and image of a $k$ -linear map between finite-dimensional $k$ -vector spaces satisfy $\dim \ker + \dim im = \dim (domain)$ Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Module homomorphism and isomorphism, kernel, image and cokernel, Vector space over a field, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis.

Proof

1.1F1algebra

The map is well defined: if $A^{'} \equiv A$ modulo $m^{n}$ , then $(A^{'} - A) f \in m^{n} m^{m} \subseteq m^{m + n}$ , and if $B^{'} \equiv B$ modulo $m^{m}$ , then $(B^{'} - B) g \in m^{m + n}$ ; so the class of $A f + B g$ in $O / m^{m + n}$ depends only on the classes of $A$ and $B$ . Additivity and $k$ -linearity are immediate from the ring operations.

1.2F1F3algebra

Suppose $f^{*}$ and $g^{*}$ have no common factor, and let $(A, B)$ with $A \in O / m^{n}$ , $B \in O / m^{m}$ satisfy $A f + B g \in m^{m + n}$ . Choose representatives with $A = 0$ or $r = ord (A) \leq n - 1$ , and $B = 0$ or $s = ord (B) \leq m - 1$ . If $A = 0$ then $B g \in m^{m + n}$ forces $s + n \geq m + n$ , so $s \geq m$ and $B = 0$ ; symmetrically for $B = 0$ . If both are nonzero, the lowest terms of $A f$ and $B g$ have orders $r + m \leq m + n - 1$ and $s + n \leq m + n - 1$ . Since the sum lies in $m^{m + n}$ , these two lowest terms must cancel: $r + m = s + n$ and $A^{*} f^{*} = - B^{*} g^{*}$ . Coprimality forces $g^{*} ∣ A^{*}$ and $f^{*} ∣ B^{*}$ , so $r \geq n$ and $s \geq m$ , contradicting $r \leq n - 1$ , $s \leq m - 1$ . Hence $(A, B) = (0, 0)$ and $\overset{ˉ}{ψ}$ is injective.

1.3F1F2F3algebra

Suppose $f^{*}, g^{*}$ have a common factor. By [F2] they have a common linear factor $L$ , so $f^{*} = L f^{'}$ , $g^{*} = L g^{'}$ with nonzero forms $f^{'}$ of degree $m - 1$ and $g^{'}$ of degree $n - 1$ . Then $g^{'} f - f^{'} g = g^{'} f_{> m} - f^{'} g_{> n}$ , a sum of products each of order at least $m + n$ , so its class in $O / m^{m + n}$ is zero; the pair $(g^{'}, - f^{'})$ is nonzero in $O / m^{n} \times O / m^{m}$ because $\deg g^{'} = n - 1 < n$ and $\deg f^{'} = m - 1 < m$ . Thus the kernel is nonzero and $\overset{ˉ}{ψ}$ is not injective.

2.1step 1.2step 1.3F3∎

By step 1.2 injectivity holds when the initial forms are coprime and fails by step 1.3 when they are not, proving the equivalence and the nonzero-kernel assertion; and by [F3] the displayed dimension formula holds exactly in the injective case, i.e. exactly when the tangent cones are coprime.

Depends on

Used by

Intersection multiplicity dominates the product of multiplicities, with equality for separated tangent cones Theorem

Dependency tree · two levels

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Sources

William Fulton, Algebraic Curves: An Introduction to Algebraic Geometry (2008 electronic edition; Internet Archive copy of the author's PDF) (standard reference, not scraped)