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?
Best Answer
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.