$d(h) = \gcd(a_1, \cdots, a_n)$ does not depend on a chosen basis for a finitely generated abelian group.

Question

Let $H$ be a subgroup of a finitely generated torsion free abelian group $G$. Let $H$ be non-trivial. Let $y_1, \dots, y_n$ be a basis of $G$. For any $h = \sum a_i y_i \in H$ set $d(h) = \gcd(a_1, \dots, a_n)$. This does not depend on the chosen basis.

How do I show that the change of basis matrix $P$ has determinant $\pm 1$? Since that is used in the proof in Cohen H. Number Theory vol 1.

Answer 1

You can avoid matrices altogether by finding a basis-independent definition.

Consider the set $M$ of all homomorphisms $G\to \mathbb Z$. Then define $D(h)$ for $h\neq 0$ as:

$$D(h)=\min_{\{\phi\in M\mid \phi(h)>0\}} \phi(h)$$

Define $D(0)=0$.

Now if $h=a_1y_1+\cdots + a_ny_n$, then we see that $d(h)\mid D(h)$, since any common divisor of the $a_i$ must be a divisor of $\phi(h)=\sum a_i\phi(y_i).$

On the other hand, we can find $u_1,\dots,u_n\in\mathbb Z$ so that $a_1u_1+\cdots+a_nu_n=\gcd(a_1,\dots,a_n)=d(h).$ Then we can define $\phi\in M$ as $\phi(y_i)=u_i.$ So we have $\phi(h)=d(h)$, which then means that $D(h)\leq d(h).$

So we get that $D(h)=d(h)$ when $h\neq 0$ and $D(0)=0=d(0).$

But $D(h)$ is defined independent of the basis, so $d(h)$ is independent of the basis.

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This all requires that a finitely generated torsion-free abelian group is a free abelian group (which is true, but may require proof.) (Free abelian lets you define the $\phi$ based on the $u_i$.)

This theorem and proof is true for any free abelian group, including infinitely-generated free abelian groups.

Answer 2

Assume that $P$ is an integer change of basis matrix. Then $P x = y$, where $x, y$ are two basis vectors. Let $x = Qy$. Then $PQ y = y \implies (PQ - I) y = 0$. But by a previous lemma, there are no nontrivial solutions $(a_i)$ for $\sum_{i=1}^n a_i y_i = 0$, thus $PQ - I = 0$.

Since $\det : M_n(\Bbb{Z}) \to \Bbb{Z}$ is multiplicative we have $\det(P)\det(Q) = 1$, which gives the result.