Subgroups of $\mathbb Z^5$

Question

Is there a nice way to see that any subgroup of $\mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z \times \mathbb Z$ can have at most 5 generators? I know that $\mathbb Z$ is Noetherian, so $\mathbb Z^5$ is also Noetherian, which tells me any submodule is also finitely generated. I am just having trouble showing that this is bounded above by 5.

Answer 1

I believe it is easier to prove by induction that any subgroup of $\mathbb{Z}^n$ has at most $n$-generators. The case $n=1$ follows from the fact that every subgroup of $\mathbb{Z}$ is cyclic.

Now we show the inductive step. As an exercise, you can show that if $H$ is a subgroup of $\mathbb{Z}^{n+1}$, then $\pi(H)=\text{projection of $H$ on the first $n$ coordinates}$ is also a subgroup.

Now, let $\alpha_1,\ldots,\alpha_k$ be generators of $\pi(H)$ with $k\leq n$ (which we know is possible by induction hypothesis). We have two cases:

1. $\pi_{n+1}(H)=\{0\}$ (here $\pi_{n+1}$ is the projection on the $(n+1)$-th coordinate). In this case, the elements $(\alpha_1,0),\ldots,(\alpha_k,0)$ will generate $H$.
2. $\pi_{n+1}(H)\neq \{0\}$. Note that $\pi_{n+1}(H)$ is also a subgroup, so in particular there are positive elements in it. Pick $(\beta,b_{n+1})=(b_1,\ldots,b_n,b_{n+1})\in H$ such that $b_{n+1}=\min\{b\in\pi_{n+1}(H):b>0\}$. We will now show that the elements $\{(\alpha_1,0),\ldots,(\alpha_k,0),(\beta,b_{n+1})\}$ generate $H$.

First, note that for every $(\alpha,a_{n+1})=(a_1,\ldots,a_{n},a_{n+1})\in H$, we have $b_{n+1}|a_{n+1}$. Otherwise, we would have $a_{n+1}=b_{n+1}q+r$ with $0<r<b_{n+1}$, and the element $(\alpha,a_{n+1})-q(\beta,b_{n+1})=(\alpha-q\beta,r)\in H$, contradicting the minimality of $b_{n+1}$.

So, if $(\alpha,a_{n+1})$ is an arbitrary element in $H$ we know that for some $q\in\mathbb{Z}$, $(\alpha,a_{n+1})-q(\beta,b_{n+1})=(\alpha-q\beta,0)\in\pi(H)$, and since $\alpha_1,\ldots,\alpha_k$ generate $\pi(H)$, there are integers $n_1,\ldots,n_k$ such that

$$n_1\alpha_1+\ldots+n_k\alpha_k=\alpha-q\beta,$$ and thus,

$$n_1(\alpha_1,0)+\ldots+n_k(\alpha_k,0)+q(\beta,b_{n+1})=(\alpha,qb_{n+1})=(\alpha,a_{n+1}).$$

We conclude then that $\{(\alpha_1,0),\ldots,(\alpha_k,0),(\beta,b_{n+1})\}$ is a set of $k+1\leq n+1$ generators of the subgroup $H$.

Answer 2

Another way to see this, is by tensoring with $\mathbb{Q}$.

In general, tensor with a module is only right exact. In our case, $\mathbb{Q}$ is flat as a $\mathbb{Z}$-module. Thus applying $-\otimes\mathbb{Q}$ to the exact sequence $$0\to K\to \mathbb{Z}^n\to C\to 0$$ we get $$0\to K\otimes\mathbb{Q}\to \mathbb{Q}^n\to C\otimes\mathbb{Q}\to 0$$ which implies that $K\otimes\mathbb{Q}$ is a vector space over $\mathbb{Q}$ of dimension $\le n$.

As you have mentioned, $K$ is finitely generated. It is also torsion free, and hence isomorphic to $\mathbb{Z}^k$ for some $k$.

Thus we conclude that $k\le n$, as $k$ is the dimension of the vector space $K\otimes \mathbb{Q}$.

(note 1: since it seems you like commutative algebra, note that one way of proving that $\mathbb{Q}$ is a flat abelian group is by noting that any localisation is flat)

(note 2: this proof does not generalize to an arbitrary ring $R$, as we have used the fact that any sub module of a free $R$ module is free)