# Viewing as a Lattice in a Real Vector Space

Let  be a number field of degree . By the primitive element theorem, for some , so we can write , where is the minimal polynomial of . Because is algebraically closed and  is irreducible, it has exactly complex roots. Each of these roots induces a homomorphism given by , whose kernel is the ideal . Thus we obtain  embeddings of into  :

Example 6.1.1   We compute the embeddings listed above for .
sage: K = QQ[2^(1/3)]; K
Number Field in a with defining polynomial x^3 - 2
sage: K.complex_embeddings()
[Ring morphism: ...
Defn: a |--> -0.629960524947 - 1.09112363597*I,
Ring morphism: ...
Defn: a |--> -0.629960524947 + 1.09112363597*I,
Ring morphism: ...
Defn: a |--> 1.25992104989]


Let be the map , and let be the -span of the image of  inside .

Lemma 6.1.2   Suppose is a subgroup of the vector space  . Then the induced topology on  is discrete if and only if for every the set

is finite.

Proof. If  is not discrete, then there is a point such that for every there is such that . By choosing smaller and smaller  , we find infinitely many elements all of whose coordinates are smaller than . The set is thus not finite. Thus if the sets are all finite,  must be discrete.

Next assume that  is discrete and let be any positive number. Then for every there is an open ball that contains  but no other element of . Since is closed and bounded, it is compact, so the open covering of has a finite subcover, which implies that is finite, as claimed.

Lemma 6.1.3   If  if a free abelian group that is discrete in a finite-dimensional real vector space  and , then the rank of  equals the dimension of .

Proof. Let be an -vector space basis for , and consider the -submodule of . If the quotient is infinite, then there are infinitely many distinct elements of  that all lie in a fundamental domain for , so Lemma 6.1.2 implies that is not discrete. This is a contradiction, so is finite, and the rank of  is , as claimed.

Proposition 6.1.4   The -vector space  spanned by the image of has dimension .

Proof. We prove this by showing that the image is discrete. If were not discrete it would contain elements all of whose coordinates are simultaneously arbitrarily small. The norm of an element is the product of the entries of , so the norms of nonzero elements of would go to 0. This is a contradiction, since the norms of nonzero elements of are nonzero integers.

Since is discrete in , Lemma 6.1.3 implies that equals the rank of . Since  is injective, is the rank of , which equals  by Proposition 2.4.5.

Subsections
William Stein 2012-09-24