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Number field basis containing 1

In Sage, the default basis for a maximal order $O_K$ often does not contain the element $1$:

sage: QuadraticField(-3).ring_of_integers().basis()
[1/2*a + 1/2, a]


However, $1$ is always a primitive lattice vector in $O_K$. Is there an elegant way to produce a basis containing $1$?

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Found the answer! Basically the method is to reverse the coordinates on $O_K$ and re-echelonize the basis.

def basis_with_1(L):
assert L.is_field();
n = L.degree();
R = matrix([[1 if i+j==n+1 else 0 for j in [1..n]] for i in [1..n]]);
A = L.maximal_order().module().matrix();
A = A*R;
M = (ZZ^n).span_of_basis([vector(v) for v in A])
A = matrix(M.echelonized_basis());
A = R*A*R;
return [L(v) for v in A];

sage: L = QuadraticField(-3)
sage: basis_with_1(L)
[1, 1/2*a + 1/2]

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( 2018-04-25 17:56:48 +0200 )edit

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Asked: 2018-04-24 21:21:01 +0200

Seen: 89 times

Last updated: Apr 25 '18