[Paper Review] Injectivity on one line
This paper proves that a polynomial mapping $ H: k^2 \to k^2 $ over an algebraically closed field of characteristic zero is a polynomial automorphism if its Jacobian is a non-zero constant and it is injective on some line in $ k^2 $. The proof uses the Abhyankar–Moh theorem and properties of Newton polygons to reduce the problem to a known automorphism criterion, establishing a sufficient condition for the Jacobian conjecture with minimal geometric input.
Let $k$ be an algebraically closed field of characteristic zero. Let $H:k^2 o k^2$ be a polynomial mapping such that the Jacobian $ ext{Jac}\,H$ is a non-zero constant. In this note we prove, that if there is a line $l \subset k^2$ such that $H|_l:l o k^2$ is an injection, then $H$ is a polynomial automorphism.
Motivation & Objective
- To establish a sufficient condition for a polynomial mapping with constant non-zero Jacobian to be a polynomial automorphism.
- To investigate whether injectivity on a single line implies global invertibility in polynomial mappings over algebraically closed fields of characteristic zero.
- To provide a geometric criterion—injectivity on one line—that implies the full automorphism property, strengthening prior results requiring injectivity on multiple lines.
- To contribute to the long-standing Jacobian conjecture by proving a special case using algebraic geometry and Newton polygon theory.
Proposed method
- Apply the Abhyankar–Moh theorem to reparametrize the image of the line under $ H $, ensuring it can be realized as the image of a line under an automorphism.
- Use an affine change of coordinates to assume without loss of generality that the line $ l $ is the x-axis ($ y = 0 $).
- Define $ \gamma(x) = H(x, 0) $, which is injective and has non-vanishing derivative, hence an embedding.
- Construct an automorphism $ H_1 \in \operatorname{Aut}k^2 $ such that $ H_1(x, 0) = \gamma(x) $, so $ H_1 $ maps the line to the same image as $ H $.
- Define $ G = H_1^{-1} \circ H $, so $ G(x, 0) = (x, 0) $, and use Lemma 2.3 to conclude $ G \in \operatorname{Aut}k^2 $.
- Conclude $ H = H_1 \circ G $ is a composition of automorphisms, hence itself a polynomial automorphism.
Experimental results
Research questions
- RQ1Does injectivity of a polynomial mapping $ H: k^2 \to k^2 $ on a single line imply that $ H $ is a polynomial automorphism when the Jacobian is a non-zero constant?
- RQ2Can the Abhyankar–Moh theorem be used to reduce the injectivity condition on a line to a standard automorphism form?
- RQ3What role do Newton polygons and their similarity properties play in analyzing the structure of polynomial mappings with constant Jacobian?
- RQ4Is it possible to prove the Jacobian conjecture under a weaker geometric condition than full injectivity or finite fibers?
- RQ5How does the degree structure of the components of $ H $ constrain the existence of non-injective fibers when injectivity holds on one line?
Key findings
- If $ H: k^2 \to k^2 $ has a constant non-zero Jacobian and is injective on a single line $ l \subset k^2 $, then $ H $ is a polynomial automorphism.
- The injectivity of $ H|_l $ implies that the image of $ l $ under $ H $ is isomorphic to the affine line, allowing application of the Abhyankar–Moh theorem.
- By composing $ H $ with an automorphism $ H_1 $, the problem reduces to showing that a mapping fixing the x-axis pointwise is an automorphism, which is established via Lemma 2.3.
- The proof relies on the fact that if $ H(x,0) = (x,0) $, then $ H $ must be a polynomial automorphism, even without assuming bounded degrees.
- The result strengthens previous work requiring injectivity on three lines by showing that injectivity on just one line suffices when the Jacobian is constant.
- The key insight is that injectivity on one line, combined with the constant Jacobian, forces the mapping to preserve the structure required for global invertibility via automorphism composition.
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This review was created by AI and reviewed by human editors.