Exhibit a basis for an ideal in an algebraic integer ring which is upper triangular in an appropriate sense

Let K be an algebraic extension of \mathbb{Q} of degree n and let \{ \omega_i\} be an integral basis. Let \mathcal{O} be the ring of integers in K. Exhibit a nonzero rational integer k and a basis (over \mathbb{Z}) \{\alpha_i\} for the ideal (k) in \mathcal{O} such that, with \alpha_i = \sum c_{i,j}\omega_j, the matrix [c_{i,j}] is upper triangular. (See Theorem 9.9 in TAN.)


Let K = \mathbb{Q}(i), so that \mathcal{O} = \mathbb{Z}[i]. Recall that \{1,i\} is an integral basis for this K. We saw in this previous exercise that \{2,2i\} is a basis for the ideal (2), and moreover has the desired property.

More generally, it is clear that \{k\omega_i\} is a basis for (k). The associated matrix is then not just triangular but diagonal.

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