Finding the volume of O Let K be a number field, r 1 denotes the number of real embeddings and 2 r 2 denotes so that r 1 + 2 r 2 = n = [ K : Q ]. Define the ring homomorphism, canonical imbedding of K, σ : K → R r 1 × C r 2 as σ ( x ) = ( σ 1 ( x ) , … , σ r 1 ( x ) , … , σ r 1 + r 2 ( x ) )We only consider the 'first' r 2 complex embeddings since the rest are conjugate of the previous ones.So, there is this lemma that I do not know how to start, what to think:Lemma.If M is a free Z -module of K of rank n, and if ( x i ) 1 ≤ i ≤ n is a Z −-base of M, then σ ( M ) is a lattice in R n , whose volume is v ( σ ( M ) ) = 2 − r 2 | det 1 ≤ i , j ≤ n ( σ i ( x j ) ) | .The determinant on the right hand side is Δ K as far as I know. However, I appreciate any help to understand the proof of this lemma.Edit: The lemma does not say about anything about O , but it is a free Z -module of rank n. So the lemma is more general, O is a specific case but I wrote the title as this way. You can either take a Z -base for O and continue or work on any free Z -module of rank n.

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Finding the volume of       O  Let K be a number field,       r    1   denotes the number of real embeddings and   2      r    2   denotes so that       r    1    +  2      r    2    =  n  =  [  K  :      Q    ]. Define the ring homomorphism, canonical imbedding of K,   σ  :  K  →            R                      r        1              ×            C                      r        2             as  σ  (  x  )  =  (      σ    1    (  x  )  ,  …  ,      σ                  r        1              (  x  )  ,  …  ,      σ                  r        1            +              r        2              (  x  )  )We only consider the &#039;first&#039;       r    2   complex embeddings since the rest are conjugate of the previous ones.So, there is this lemma that I do not know how to start, what to think:Lemma.If M is a free       Z  -module of K of rank n, and if   (      x    i        )          1      ≤      i      ≤      n       is a       Z    −-base of M, then   σ  (  M  ) is a lattice in             R        n  , whose volume is   v  (  σ  (  M  )  )  =      2          −              r        2                  |        det          1      ≤      i      ,      j      ≤      n        (      σ    i    (      x    j    )  )      |  .The determinant on the right hand side is             Δ      K       as far as I know. However, I appreciate any help to understand the proof of this lemma.Edit: The lemma does not say about anything about       O  , but it is a free       Z  -module of rank n. So the lemma is more general,       O   is a specific case but I wrote the title as this way. You can either take a       Z  -base for       O   and continue or work on any free       Z  -module of rank n.

volksgeesr9 · Accepted Answer

Step 1I think the issue here is not about rings of algebraic integers, but just about lattices   Λ in             R        n  , meaning discrete subgroups   Λ  ⊂            R        n    such that the quotient             R        n        /    Λ is compact.We have a fundamental (measure/integral-theory) theorem, sometimes attributed to A. Weil (from his book on integration in topological groups and applications): for example, with abelian topological group A and discrete subgroup   Λ  ⊂  A, with Haar measure da on A, there is a unique invariant measure   d            a      ˙       on   A      /    Λ such that, for all   f  ∈      C    c    o    (  A  ),      ∫    A    f  (  a  )    d  a    =        ∫          A              /            Λ            ∑          λ      ∈      Λ        f  (            a      ˙        +  λ  )    d            a      ˙      With respect to this measure, the total measure of   A      /    Λ  is the &quot;co-volume&quot; of   Λ. Yes, it depends on the Haar measure on A. For   A  =            R        n  , we have a standard Haar measure.

Brylee Shepard · Accepted Answer

Step 1Let       α    1    ,  …  ,      α    n   be a       Z  -basis of M. Then   σ  (  M  ) is spanned by   σ  (      α    1    )  ,  …  ,  σ  (      α    n    ) over       Z  . Let   A  =  (  σ  (      α    1    )  ,  …  ,  σ  (      α    n    )  ) be the matrix with i-th column given by   σ  (      α    i    ), where you split up the complex number       σ    j    (      α    i    ) into its real and imaginary part for   j  =      r    1    +  1  ,  …  ,      r    1    +      r    2  .Step 2If you do some row operations, you will see that the determinant of A is equal to   −      2          −              r        2              det  ⁡  (      σ    i    (      α    j    )  ). Since the trace form is non-degenerate, this will be non-zero, so A defines an isomorphism             R        n    ≅            R        n   taking   [  0  ,  1      ]    n   to the fundamental parallelepiped of   σ  (      α    1    )  ,  …  ,  σ  (      α    n    ). This gives you the desired formula for the volume of the lattice   σ  (  M  ).

If M is a free mathbb{Z}-module of K of rank n, and if (x_{i})_{1 leq i leq n} is a mathbb{Z}-base of M, then sigma(M) is a lattice in mathbb{R}^n, whose volume is v(sigma(M))=2^{-r_{2}}|det_{1 leq i, j leq n}(sigma_{i}(x_{j}))|

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