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Jack Huizenga
Jack Huizenga
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There is nothing wrong here. Let $X$ be irreducible of dimension $n$. Any sheaf of dimension less than $n$ is torsion, since it is annihilated by a function vanishing on the support of the sheaf. Sheaves which are pure of dimension $n$ are torsion free, since if they were not torsion free they would have a torsion subsheaf supported on a proper subvariety.

A pure sheaf with irreducible support is torsion-free when view as a sheaf on its support.

There is nothing wrong here. Let $X$ be irreducible of dimension $n$. Any sheaf of dimension less than $n$ is torsion, since it is annihilated by a function vanishing on the support of the sheaf. Sheaves which are pure of dimension $n$ are torsion free, since if they were not torsion free they would have a torsion subsheaf supported on a proper subvariety.

There is nothing wrong here. Let $X$ be irreducible of dimension $n$. Any sheaf of dimension less than $n$ is torsion, since it is annihilated by a function vanishing on the support of the sheaf. Sheaves which are pure of dimension $n$ are torsion free, since if they were not torsion free they would have a torsion subsheaf supported on a proper subvariety.

A pure sheaf with irreducible support is torsion-free when view as a sheaf on its support.

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Jack Huizenga
Jack Huizenga
  • 5.9k
  • 1
  • 28
  • 42

There is nothing wrong here. Let $X$ be irreducible of dimension $n$. Any sheaf of dimension less than $n$ is torsion, since it is annihilated by a function vanishing on the support of the sheaf. Sheaves which are pure of dimension $n$ are torsion free, since if they were not torsion free they would have a torsion subsheaf supported on a proper subvariety.