Skip to main content
MathOverflow

Return to Question

Corrected title of book.
Source Link
edit approved Jan 15, 2018 at 0:43
jeq
edit approved Jan 15, 2018 at 0:43
jeq
  • 1.2k
  • 5
  • 16
  • 21

I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature.

In Silverman's Advanced Topics ofin the Arithmetic of Elliptic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism clasSclass?

In particular the proof of theorem II.2.3, he states "We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$.

Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$.

I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it?

I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature.

In Silverman's Advanced Topics of Arithmetic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism clasS?

In particular the proof of theorem II.2.3, he states "We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$.

Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$.

I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it?

I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature.

In Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism class?

In particular the proof of theorem II.2.3, he states "We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$.

Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$.

I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it?

Source Link
Rdrr
Rdrr
  • 901
  • 6
  • 15

Fields of Definition of Elliptic Curves

I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature.

In Silverman's Advanced Topics of Arithmetic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism clasS?

In particular the proof of theorem II.2.3, he states "We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$.

Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$.

I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it?