# What is wrong with this proof on ring of integers being finitely generated

I have seen some proofs about the theorem that the ring of integers is a finitely-generated $\mathbb{Z}$-module, but I thought I came up with a more straightforward proof. However, I believe there is some loophole in the argument because I don't see this argument being presented anywhere.

Let $K$ by a number field (i.e. finite extension of $\mathbb{Q}$), with $[K:\mathbb{Q} ] = n$. The ring of integers $O_K$ is the integral closure of $\mathbb{Z}$ in $K$. We can define a map $O_K \otimes_\mathbb{Z} \mathbb{Q} \to K$ where we define $\alpha \otimes q \mapsto \alpha q$ (and extend linearly). In my class, I saw that this is an isomorphism of $\mathbb{Q}$-vector spaces, so I don't think there is anything wrong with that claim.

Anyway my class left this point to prove the theorem with some use of trace etc, but I thought the following reasoning should work. In search of contradiction, suppose $O_K$ is not finitely generated over $\mathbb{Z}$. Then I may find $\alpha_1,...,\alpha_{n+1} \in O_K$ such that they are $\mathbb{Z}$-independent. Let $M = \mathbb{Z} \alpha_1 + ... + \mathbb{Z}\alpha_{n+1}$. Then $M \subset O_K$ and so $M \otimes_\mathbb{Z} \mathbb{Q} \hookrightarrow K$. But $M \simeq \mathbb{Z}^{n+1}$ so $M \otimes_\mathbb{Z} \mathbb{Q} \simeq \mathbb{Q} ^ {n+1}$. Thus the injectivity is a contradiction.

Is there anything wrong with the above proof?

It is not true that if $A$ is not finitely generated, you can find $\mathbb Z$-independent elements.

For example, $\mathbb Q$ is a non finitely generated $\mathbb Z$-module, but, clearly, you cannot find two $\mathbb Z$-independent rationals $\frac ab$, $\frac cd$, for $ac = bc\times \frac ab = ad\frac cd$.

When we prove that $\mathcal O_K$ is finitely generated, we are not afraid that $\mathcal O_K$ might turn out to be a free module with an infinite basis (essentially because of the arguments you gave); we are afraid it might turn out to be something like $\mathbb Q$ or $\mathbb Z \left[\frac 12\right]$...

Non f.g. $\mathbb Z$-modules are quite suprising beasts. You might want to read the beginning of Kaplansky's Infinite Abelian Groups to know them a bit better.

• Very nice, I like your specific example of the fractions. My class notes also wrote "If $O_K$ is finitely generated, then $O_K$ will be of the form $\mathbb{Z}^m$" which was why I thought $O_K$ is already free. So what exactly is happening in my confusion? Is the freeness clear only after we know that it is finitely-generated? Feb 6, 2015 at 16:11
• The classification of f.g. Abelian groups (or modules over a PID, to pick a wider context) tells you that a torsionfree f.g. Abelian group is automatically free. Since your module sits in a $\mathbb Q$-vector space, it has no torsion. But as the example of the $\mathbb Z$-module $\mathbb Q$ shows, for non f.g. Abelian groups, not having torsion is a much weaker property than freeness. Feb 6, 2015 at 16:23
• @suncup224 fundamental theorem of finitely generated abelian groups. Feb 6, 2015 at 16:23
• I get it now. Thanks! Feb 6, 2015 at 16:28

The problem with this argument is that a torsion free $\mathbb Z$-module that is not finitely generated need not be free and need not have free submodules of arbitrarily large rank. $\mathbb Q$ itself is an example.

The flaw is your assumption that $\mathscr{O}_K$ not being finitely generated implies that it has $n+1$ $\mathbf{Z}$-linearly independent elements. Without knowing that $\mathscr{O}_K$ is a finite $\mathbf{Z}$-module, one can still prove that $\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{Q}\to K$ is an isomorphism of $\mathscr{O}_K$-algebras. This implies that the $\mathbf{Z}$-rank of $\mathscr{O}_K$ is exactly $n$, i.e., any set of $\mathbf{Z}$-linearly independent elements in $\mathscr{O}_K$ has at most $n$ elements. But for example there are torsion (i.e. rank-zero) $\mathbf{Z}$-modules which are not finitely generated (an example being $\mathbf{Q}/\mathbf{Z}$).