Proof Checking: Submodules of a Free Module of Finite Rank over a PID is Also Free of Finite Rank

[caffeinemachine]

I have been trying to prove the following:

Let $R$ be a P.I.D. and $M$ be a free module of finite rank over $R$.
Then every submodule $N$ of $M$ is a free module with
$$\text{rank} N\leq \text{rank} M$$

and here is my "proof":

Using the fact that $R$ is a P.I.D., the theorem is trivial when $\text{rank} M=1$.
Inductively assume that the theorem is true whenever the rank of $M$ is less than $n$, where $n>1$.
Let $\text{rank} M$ be $n$ and $(e_1,\ldots,e_n)$ be a basis of $M$.
By the second isomorphism theorem, we have
$$\frac{N+Re_n}{Re_n}\ \cong \ \frac{N}{N\cap Re_n}$$
Now $(N+Re_n)/Re_n$ is a submodule of $M/Re_n$.
Since $\text{rank}(M/Re_n)=n-1$, by induction we infer that $(N+Re_n)/Re_n$ is a free module of rank no more than $n-1$.
Let $\bar v_1,\ldots,\bar v_k$ be a basis of $(N+Re_n)/Re_n$, where $k\leq n-1$.
Also, $N\cap Re_n$ is a submodule of $Re_n\cong R$, and hence $N\cap Re_n$ is a free module of rank no more than $1$.
Assume first than $\text{rank}(N\cap Re_n)=1$ and $v$ be a basis of the same.
We claim that $(v,v_1,\ldots,v_k)$ forms a basis for $N$.
Let $n\in N$.
Then $\bar n=a_1\bar v_1+\cdots+a_k\bar v_k$ for some $a_i$'s in $R$.
Therefore, $n-(a_1v_1+\cdots+a_kv_k)\in Re_n$, which means, there exists $a\in R$ such that $n=av+a_1v_1+\cdots+a_kv_k$.
So we see that $(v,v_1,\ldots,v_k)$ spans $N$.
Now we show linearly independence.
Say $av+a_1v_1+\cdots a_kv_k=0$ for some $a,a_1,\ldots,a_k\in R$.
Then $a_1\bar v_1+\cdots +a_k\bar v_k=\bar 0$, giving $a_1=\cdots=a_k=0$.
This in turn leads to $a=0$ and we are done.
Now if $\text{rank}(N\cap Re_n)=0$ then $N\cap Re_n=\{0\}$.
From here we can show that $(v_1,\ldots,v_k)$ is a basis for $N$ and we are done.What bugs me about this proof is that I don't seem to use the fact that $R$ is a P.I.D. anywhere after the base case for induction.

Can anybody see if I have made a mistake?

Thanks.

 

[jvicens]

After reviewing your proof, I can confirm that you have indeed made a mistake. While your proof is correct in showing that $N$ is a free module, it does not prove that $\text{rank} N \leq \text{rank} M$. In order to prove this, you need to use the fact that $R$ is a P.I.D. in the following way:

In the case where $\text{rank}(N \cap Re_n) = 1$, you correctly show that $N$ is a free module with basis $(v,v_1,\ldots,v_k)$. However, you need to use the fact that $R$ is a P.I.D. to show that the ideal generated by $v$ is contained in $Re_n$. This is because in a P.I.D., any ideal generated by a single element is equal to the ideal generated by the GCD of all elements in the ideal. Since $v$ and $e_n$ are both elements of $Re_n$, their GCD must also be an element of $Re_n$. This allows you to conclude that $v \in Re_n$ and therefore $N$ has a basis of $n$ elements or less.

In the case where $\text{rank}(N \cap Re_n) = 0$, you correctly show that $N$ is a free module with basis $(v_1,\ldots,v_k)$. However, you need to use the fact that $R$ is a P.I.D. to show that the ideal generated by $v_1,\ldots,v_k$ is contained in $Re_n$. This is because in a P.I.D., the ideal generated by multiple elements is equal to the ideal generated by their LCM. Since $v_1,\ldots,v_k$ are all elements of $Re_n$, their LCM must also be an element of $Re_n$. This allows you to conclude that $N$ has a basis of $n$ elements or less.

I hope this helps clarify the use of the P.I.D. property in your proof.