# Submodules of finitely generated modules

I was reading Hungerford's Algebra and I read the next proposition (they present it as a corollary):

Let $R$ be a PID. If $A$ is a finitely generated $R$-Module generated by $n$ elements, then every submodule of $A$ may be generated by $m$ elements with $m \leq n.$

I was trying to prove it, but I don't know how to proceed. I know that since $R$ is PID, it is also noetherian, so any submodule of $A$ must be finitely generated, the only thing I'm missing is to prove that it is generated by $m\leq n$ elements. I've seen that some people use the primary desomposition, but I don't know anything about it. My first instinct was to try induction over the number of generators of $A$, but I don't seem to get anywhere.

Let $$M$$ be an $$R$$-module generated by $$n$$elements, and let $$N$$ be a submodule of $$M.$$
Assume that you know the desired result is true when $$M$$ is free (I will handle this case later). Now, for a general $$M$$, you have an isomorphism $$M\simeq A^n/P$$, where $$P$$ is a submodule of $$A^n.$$ Then submodules of $$M$$ correspond via this isomorphism to submodules of $$A^n/P$$. These submodules have the form $$N'/P$$ where $$N'$$ is a submodule of $$A^n$$ containing $$P.$$ By the case of free modules, $$N'$$ is generated by $$m\leq n$$ elements, and thus so is $$N'/P$$. Hence, it is also true for the submodules of $$M.$$
It remains to handle the case where $$M$$ is free.We will proceed by induction on the rank $$n$$ of $$M$$. If $$n=0$$, then $$M=0$$ and there is nothing to do. Now assume the result true for any free module of rank $$n$$, and let $$M$$ be a free module of rank $$n+1$$. Fix a basis $$(e_1,\ldots,e_{n+1})$$ of $$M$$.
Let $$N$$ be a submodule of $$M$$ and let $$(x_j)_{i\in J}$$ be a family of generators of $$N$$. We have $$x_{j}=\sum_{i=1}^{n+1} a_{ij}\cdot e_i \mbox{ for all }j\in J.$$ Let $$\mathfrak{a}$$ be the ideal generated by the $$a_{n+1 \ j},j\in J$$.Since $$R$$ is a PID, we have $$\mathfrak{a}=(a)$$. We may write $$a=\sum_{j\in J} \lambda_j a_{n+1 \ j},$$ where the $$\lambda_j's$$ are all zero, except for a finite number of them. Set $$y_0=\sum_{j\in J} \lambda_j\cdot x_j\in N.$$ Then, the $$(n+1)$$-th coordinate of $$y_0$$ is $$a$$. One may also write $$a_{n+1 \ j}=\mu_j a$$, and thus the submodule $$N'$$ generated by the elementss $$x_j-\mu_j y_0,j\in J$$ is a submodule of $$M'=Re_1\oplus\cdots\oplus Re_n$$, which is free of rank $$n$$. By induction hypothesis, $$N'$$ is generated by $$y_1,\ldots,y_m\in N'\subset N, m\leq n.$$ One deduce easily that $$x_j$$ is a linear combination of $$y_0,y_1,\ldots,y_{m}$$. Hence, $$N$$ is generated by the $$m+1\leq n+1$$ elements $$y_0,\ldots,y_{m}$$. This finishes the induction step, and the proof.