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Emerton
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This was my own motivation to learnfor learning schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is isomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

This was my own motivation to learn schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is isomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

This was my own motivation for learning schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is isomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

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KConrad
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This was my own motivation to learn schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is equalisomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

This was my own motivation to learn schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is equal to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

This was my own motivation to learn schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is isomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

Source Link
Emerton
  • 57.6k
  • 6
  • 209
  • 259

This was my own motivation to learn schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is equal to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.