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Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other questionyour other question and (yet more closely) to Lior Bary-Soroker's answer to it.

Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

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Pete L. Clark
Pete L. Clark
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Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof immediatelyquickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof immediately reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

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Pete L. Clark
Pete L. Clark
  • 65.4k
  • 12
  • 241
  • 381

Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof immediately reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.