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Given an irreducible curve $C$ of degree $d$ in $\mathbb{P}^r$ and a general hyperplane $H\subset\mathbb{P}^r$, the uniform position theorem states that any $r$ points on the hyperplane section $H\cap C$ will be linearly independent. (Suppose we are working over the complex numbers) One reference I have found so far is Arbarello-Cornalba-Griffiths-Harris, Geometry of Algebraic Curves I.

Now I wonder, what general really means in this statement:

My first hope is that any transversal intersection will do. Is this true? My feeling is that this might not be good enough.

Should it fail, suppose I am given a subspace of linear forms on $\mathbb{P}^r$ that induces a base point free linear system on the given curve $C$. Can I find a general hyperplane section in this linear system such that the uniform position property holds?

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  • $\begingroup$ If I understand your question, this will fail. Suppose $h^0(O_C(D)) = 2$ and $D$ is bpf, then the entire pencil will fail uniform position for the canonical embedding. $\endgroup$ – meh Jun 29 '14 at 15:42
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The answer to both questions is negative, as pointed out by aginesky in the comments. To complete his answer, let me give a simple concrete counterexample.

Consider a canonical curve $C$ of genus 4 with only one $g^1_3$, so a complete intersection of an irreducible singular quadric $Q$ and a cubic $G$ in ${\mathbb P}^3$.

Pick a line $l$ through the node of the quadric which is NOT contained in $Q$, and consider the pencil of planes containing $l$. Since $l$ intersects $Q$ just in the node, which is not in $C$, the pencil has no base points on $C$, as you require.

On the other hand, the intersection of each of those planes with $Q$ splits as two lines, and therefore gives a divisor of degree 6 on $C$ which fails uniform position: more precisely the 6 points split as two sets of three collinear points.

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