Suppose I have an integral basis for a number field. The exercise I'm looking at says I should "describe all other integral bases". I have an idea how to do this for a quadratic field, but for higher degree, I have questions.
First, my idea. Suppose my integral basis, written as a row vector, is $\begin{bmatrix} \alpha & \beta \end{bmatrix}$. Then every other integral basis is given by:
$$\begin{bmatrix} \alpha & \beta \end{bmatrix}A$$
where $A$ is a $2\times 2$ integer matrix with determinant $\pm 1$. Now, this answer isn't quite satisfying, because of overcounting: each basis will be listed twice, in different orders. I can compensate for this by requiring that $\det A=1$, because for any suitable choice of $A$, swapping the columns reverses the order of the resulting basis, and the sign of the determinant.
My first question: Is that valid? Am I going to obtain every integral basis, exactly once, in this fashion?
My second question: What is the equivalent trick when the dimension of my space is $n>2$? In that case, if I just use $n\times n$ integer matrices with determinant $\pm 1$, I should get every integer basis, not just twice, but $n!$ times. What can I do to "mod out" the repetitions? Do I take the group of matrices and literally mod out the permutation matrices, which presumably form a normal subgroup? What is the resulting group? What does it look like; has it got a name? Is there a usual way of talking about this?
I guess my second question was more than one question. I hope my query is clear. Thanks in advance.
Almost everything you say is correct, but the permutation matrices will never form a normal subgroup (if $n>1$). So when you mod them out, you aren't getting a group; you're just getting a set of cosets of the subgroup $S_n$ of permutation matrices inside the group $GL_n(\mathbb{Z})$ of all invertible integer matrices.
For what it's worth, it is rather rare that one cares about talking about all the integral bases like, or more generally all the bases of some free module over a ring. It is much more common to care about all the ordered bases. In that case, you don't have to mod out by permutations, and instead very nicely have a torsor over the group of invertible matrices.