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Define different embedding of a number field

asked 9 years ago

Dianbin Bao gravatar image

updated 9 years ago

kcrisman gravatar image

Hi,

Let f(x) be an irreducible polynomial over ZZ[x]. We can define a number field

K.<a>= NumberField(f(x))

My question is how does one define all the embedding of K into the real or complex in Sage?

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answered 9 years ago

tmonteil gravatar image

You can have a lool at the embeddings method:

sage: K.embeddings?

Then you can gel all embeddings into the (complex) algebraic field QQbar of the real algrbraic field AA.

sage: K.<a>= NumberField(x^2-2)
sage: K
Number Field in a with defining polynomial x^2 - 2
sage: K.embeddings(QQbar)
[
Ring morphism:
  From: Number Field in a with defining polynomial x^2 - 2
  To:   Algebraic Field
  Defn: a |--> -1.414213562373095?,
Ring morphism:
  From: Number Field in a with defining polynomial x^2 - 2
  To:   Algebraic Field
  Defn: a |--> 1.414213562373095?
]
sage: K.embeddings(AA)
[
Ring morphism:
  From: Number Field in a with defining polynomial x^2 - 2
  To:   Algebraic Real Field
  Defn: a |--> -1.414213562373095?,
Ring morphism:
  From: Number Field in a with defining polynomial x^2 - 2
  To:   Algebraic Real Field
  Defn: a |--> 1.414213562373095?
]

Or,

sage: K.<a>= NumberField(x^2+1)
sage: K.embeddings(QQbar)
[
Ring morphism:
  From: Number Field in a with defining polynomial x^2 + 1
  To:   Algebraic Field
  Defn: a |--> -1*I,
Ring morphism:
  From: Number Field in a with defining polynomial x^2 + 1
  To:   Algebraic Field
  Defn: a |--> 1*I
]
sage: K.embeddings(AA)
[]
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Comments

This is exactly what I want. Thank you very much, Tmonteil!

Dianbin Bao gravatar imageDianbin Bao ( 9 years ago )

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Asked: 9 years ago

Seen: 480 times

Last updated: Feb 10 '16