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Let $a$ be an algebraic number, whose minimal polynomial has integral coefficients. Let $K = \Bbb Q(a)$ be an algebraic number field. Let $\mathcal O_K$ be the ring of integers in this algebraic number field. Fix an basis for $\mathcal O_K$ and $\Bbb Z[a]$. Let $A$ be the matrix whose column vectors express the basis elements of $\Bbb Z[a]$ as a $Z$-linear combination of the basis for $\mathcal O_K$. Then

$\det(A) = [\mathcal O_K : \Bbb Z[a]]$

I have no idea why this is the case. Can someone explain this to me?

user153012
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Rioghasarig
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  • Stacked bases theorem? – Jyrki Lahtonen Sep 06 '14 at 20:31
  • Don't know what that is. This is just a statement made in a book on Computational Algebraic Number Theory in the chapter about factoring primes – Rioghasarig Sep 08 '14 at 13:00
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    It is from structure theory f.g. modules over PIDs. If $N\subset M$ are free modules of rank $n$ over a PID $R$, then there is such a basis ${b_1,\ldots,b_n}$ and constants $r_1,r_2,\ldots,r_n\in R$ such that ${r_1b_1,r_2b_2,\ldots,r_nb_m}$ is a basis of $N$. If $R=\Bbb{Z}$ it follows immediately that $[M:N]=|r_1r_2\cdots r_n|=|\det(A)|$. You may have seen this in connection with Smith Normal Forms of matrices over a Euclidean domain (works over a PID, too). By ordering carefully you can further arrange things such that $r_{i+1}\mid r_i$ for all $i$. Invariant factors. – Jyrki Lahtonen Sep 08 '14 at 13:06

3 Answers3

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Expanding the comments to an answer.

As Rene pointed out this fact is a special case of a general result. Assume that $N\subseteq M$ are free abelian groups of the same rank $n$. Let $\mathcal{B}=\{x_1,x_2,\ldots,x_n\}$ be a basis of $M$, and $\mathcal{C}=\{y_1,y_2,\ldots,y_n\}$ be a basis of $N$. As $N\subset M$ we have that $$ y_i=\sum_{j=1}^na_{ij}x_j $$ for some integer matrix $A=(a_{ij})$. Everything follows from finding the Smith normal form of $A$. What that says is that there exists integer matrices $P$ and $Q$ such that their determinants are $\pm1$ and $$ PAQ=\left(\begin{array}{ccccc}d_1&0&0&\cdots&0\\ 0&d_2&0&\cdots&0\\ 0&0&d_3&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&\cdots&0&0&d_n \end{array}\right) $$ for some positive integers $d_i, i=1,2,\ldots,n$. Furthermore, here $d_{i+1}\mid d_i$ for all $i$.

What this means is that if we replace $\mathcal{B}$ with $\mathcal{B}'=Q^{-1}\mathcal{B}$, and $\mathcal{C}$ with $\mathcal{C}'=P\mathcal{C}$, the resulting bases $\mathcal{B}'=\{x_1',x_2,\ldots,x_n'\}$ and $\mathcal{C}'=\{y_1',y_2',\ldots,y_n'\}$ are stacked, IOW $y_i'=d_ix_i'$ for all $i$.

This implies immediately that $$ M/N\cong \Bbb{Z}_{d_1}\oplus\Bbb{Z}_{d_2}\oplus\cdots\oplus\Bbb{Z}_{d_n}. $$ Note that as $\det P=\pm1$ and $\det Q=\pm1$, it follows that $$ |\det(A)|=d_1d_2\cdots d_n. $$ Your claim follows from this.

Jyrki Lahtonen
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  • When you construct $\mathcal{B}', \mathcal{C}'$ I suppsoe we need to see $\mathcal{B}'$ as the new basis for $M$ and $\mathcal{C}'$ the new basis for $N$. But how does one argue that this is true? Or is there a hidden argument that implies the wanted isomorphism $M/N\cong \mathbb{Z}{d_1}\oplus \cdots \oplus \mathbb{Z}{d_n}$? – slowpoke Jan 05 '22 at 14:32
  • @slowpoke I'm not sure I understand what is unclear to you? $P$ and $Q$ act, in a sense, as the change of bases matrices. Because they have determinant $\pm1$ they give you different bases. – Jyrki Lahtonen Jan 05 '22 at 16:10
  • I'm just wondering if the new bases $\mathcal{B}', \mathcal{C}'$ generates $M,N$ respectively, up to isomorphism. (As I guess the basis of the groups need not to be unique?) Or if they generate other abelian groups (or I mean, isomorphically different abelian groups.) – slowpoke Jan 05 '22 at 16:29
  • Or, maybe it is cleat that $P,Q$ are bijective as the determinant is $\pm 1$. I'm just trying to convince myself that $M/N\cong \mathbb{Z}{d_1}\oplus \cdots \oplus \mathbb{Z}{d_n}$ is in fact true, as I suppose you did the calculation $M/N\cong \left( \mathbb{Z}{x_1'}\oplus \cdots \oplus \mathbb{Z}{x_n'} \right)/\left( \mathbb{Z}{d_1x_1'}\oplus \cdots \oplus \mathbb{Z}{d_nx_n'} \right) \cong \left( \mathbb{Z}{x_1'}/\mathbb{Z}{d_1x_1'} \right) \oplus \cdots \oplus \left( \mathbb{Z}{x_n'}/\mathbb{Z}{d_nx_n'} \right) \cong \mathbb{Z}{d_1}\oplus \cdots \oplus \mathbb{Z}{d_n}$. – slowpoke Jan 05 '22 at 16:47
  • Hence I'm trying to convince myself that $M\cong\mathbb{Z}{x_1'} \oplus \cdots \oplus \mathbb{Z}{x_n'}$ and $N\cong \mathbb{Z}{d_1x_1'} \oplus \cdots \oplus \mathbb{Z}{d_nx_n'}$ – slowpoke Jan 05 '22 at 16:54
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    The new bases still generate the same groups (or modules) $M$ and $N$. They are free abelian groups so your two last comments are a little bit off. After finding the stacked bases the situation looks a lot like the following. $M=\Bbb{Z}^n$ with the usual natural basis, and $N$ is the submodule spanned by $(d_1,0,\ldots,0)$, $(0,d_2,0,\ldots,0)$, $\ldots$, $(0,0,\ldots,0,d_n)$. – Jyrki Lahtonen Jan 05 '22 at 20:16
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    Both $M$ and $N$ are isomorphic to the free abelian group $\Bbb{Z}^n$, but here the focus is on their relation to each other (and in the quotient group $M/N$). – Jyrki Lahtonen Jan 05 '22 at 20:18
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This is a standard argument from group theory. Let $G$ be a free abelian group on $n$ generators $x_1, \ldots ,x_n$, and $H$ a subgroup generated by $y_i=\sum a_{ij}x_j$ then the index of $H$ is the determinant of $(a_{ij})$. This can be proved by chosing a basis for $H$ which is upper diagonal. First look at all $a_1$ such that for some $a_i$, $a_1x_1+a_2x_2+\cdots \in H$ amongst all such ther is a least positive and chose that element of $H$ as the first element of a basis.

To elaborate on this last sentence.

let $$I=\{ a_1 | \exists a_2, \ldots ,a_n \in \mathbb{Z} \ \ \text{such that } \ a_1x_1+a_2x_2+\cdots +a_nx_n \in H\}$$

$I$ is an ideal in $\mathbb{Z}$ and since every ideal in $\mathbb{Z}$ is principal $I=(b_1)$ for some $b_1$, which we chose without loss to be positive.

Then we see that $b_1$ divides the first or $a_1$ coefficient of all elements of $H$.

By induction we get a basis whose $k$ th element is of the form $b_kx_k +\cdots$ where $x_1, \ldots x_{k-1}$ do not occur. The discriminant of this basis is $b_1b_2 \cdots b_n$ and now one can prove that the elements of the form

$$c_1x_1+c_2x_2 +\cdots +c_nx_n$$ where

$$0\leq c_k < b_k$$ form a complete residue system.

Rene Schipperus
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  • I'm sorry, but I'm not following this explanation at all.

    "First look at all $a_1$ such that for some $a_i$, $a_1x_1+a_2x_2+\cdots \in H$ amongst all such ther is a least positive and chose that element of $H$ as the first element of a basis"

    What are you trying to say there? There is a least positive what?

    And why is it clear that the number of residue classes is equal to the determinant?

     – Rioghasarig Sep 06 '14 at 11:54
  • @Rioghasarig I have edited to answer your conscerns – Rene Schipperus Sep 06 '14 at 16:29
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This is a heuristic argument, but it can be made rigorous without a lot of fuss.

We can think of both $\mathbb{Z}[a]$ and $\mathcal{O}_K$ as lattices in $\mathbb{Q}^n$. The index $c=[\mathcal{O}_K : \mathbb{Z}[a]]$ measures the relative density of the lattices, that is, if we pick an element of $\mathcal{O}_K$ at random, it has a $1/c$ chance of belonging to $\mathbb{Z}[a]$.

On the other hand, $d=|\det (A)|$ (note the absolute values) measures the volume scaling of a change of basis. That is, if we have a figure of volume $V$, and apply the change of basis matrix $A$, the new figure will have volume $d\cdot V$.

But if $A$ scales volume by a factor of $d$, it must also stretch a given lattice to one of relative density $1/d$. In other words, $1/c=1/d$, so $c=d$.

Andrew Dudzik
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