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We are looking for a somewhat direct proof that all $\mathbb{Z}$-submodules of $\mathbb{Z}\oplus\mathbb{Z}$ are free.

We already know that if $A$ is a P.I.D. and we have a free $A$-module $M$, then every sub $A$-module of $M$ is free. However, the proof for this requires quite advanced arguments and we have a strong feeling that a simple proof can be provided for our specific case.

We tried using projections but did not manage to write a coherent proof of the statement.

Thank you very much in advance for your tips and tricks.

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Thanks to @Hwang for the idea on this proof. Here it goes.

Suppose $M$ is a sub-$\mathbb{Z}$-module of $\mathbb{Z}^2$. We take $v_1=(p,0)\in M$ such as $p$ is positive and minimal. If such a $p$ does not exist, we can simply not include this element in our basis.

Now choose $v_2=(q,r)\in M$ for $r$ positive and minimal. Again, if such a point does not exist, it means that all points are on the axis and we can take $v_1$ as our basis.

First of all, we have $\langle\{v_1,v_2\}\rangle\subseteq M$, since $M$ is a module and $v_1,v_2\in M$.

Now, choose $(a,b)\in M$. We show that it can be expressed as a linear combination of $v_1$ and $v_2$. If $b\neq 0$, it means that a choice for $r$ exists and thus $r\neq 0$. We must have that $r|b$. Indeed, if $r$ does not divide $b$, we have that $d=\gcd(r,b)<r$ and by linear combination of $(a,b)$ and $(q,r)$ we could by Bézout's identity find an element for which the 2nd coordinate is $d<r$, a contradiction with our choice of $r$.

Since $r|b$, we have $b=kr$. We can then look at $(a,b)-k(q,r)=(a-kq,0)\in M$. Now, by a similar argument as before, if $a-kq\neq 0$ (and thus $p\neq0$), we can find that $p|(a-kq)$, say $pt=a-kq$. This means we have found that $(a,b)=t v_1+k v_2$.

Now, for linear independance, if we had that no such $p$ or $r$ existed, we excluded respectively $v_1$ or $v_2$ from the basis, so we can assume both are in the basis. Then, if we have a linear combination $k_1(p,0)+k_2(q,r)=(0,0)$, we can deduce $k_2=0$, then $k_1=0$.

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  • $\begingroup$ Wow thank you for the details! That was a very helpful and elementary proof. $\endgroup$ – Alexis Leroux-Lapierre Apr 21 '17 at 3:11
  • $\begingroup$ If you ask me it is extremely convoluted way of proving, universal property makes it simpler. $\endgroup$ – Zelos Malum Apr 21 '17 at 5:57
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A $\mathbb{Z}$-module is just an abelian group. Since any submodule $M\subset \mathbb{Z}^2$ has rank $\dim_{\mathbb{Q}} M\otimes \mathbb{Q} \leq 2$, it is finitely generated. It's clearly torsion-free, so it's free as a $\mathbb{Z}$-module.

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    $\begingroup$ Dear @anomaly, this may be a very stupid question, but what is the definition of the rank of a module that is not necessarily free, and how can we prove that this rank is equal to the dimension of the $\mathbb Q$-vector space obtained by tensoring? $\endgroup$ – Kenny Wong Apr 21 '17 at 0:50
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    $\begingroup$ There is a definition of rank for a module that may not be free, but it's not necessary here. Use the dimension of $M\otimes \mathbb{Q}$ to reduce the size of a generating set for M (using Zorn's lemma to get around the infinite case)-- which, come to think of it, might involve the Euclidean algorithm. $\endgroup$ – anomaly Apr 21 '17 at 1:00
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How about finding a basis explicitly? Let $M$ be a submodule. Pick primitive $v_1 \in M$. We may assume $v_1 = (p,0)$. Pick $v_2 = (q,r) \in M$ with smallest $r>0$ if such element exists. We can check $\{v_1, v_2\}$ is a basis for $M$. Write an element of $M$ as a linear combination of $v_1$ and $v_2$ and show coefficients are integers.

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  • $\begingroup$ Thank you very much! I just completed the details and it works beautifully. $\endgroup$ – Alexis Leroux-Lapierre Apr 21 '17 at 2:48
  • $\begingroup$ @Hwang What do you mean by primitive? $\endgroup$ – user26857 Apr 21 '17 at 14:58
  • $\begingroup$ I meant an element which is not a multiple of an element of $M$. Maybe a bad choice of terminology. $\endgroup$ – Hwang Apr 21 '17 at 21:55
  • $\begingroup$ So far, so good. Then why you may assume that $v_1=(p,0)$? $\endgroup$ – user26857 Apr 22 '17 at 15:03
  • $\begingroup$ We can change the coordinates by applying an element of $GL_2(\mathbb{Z})$. $\endgroup$ – Hwang Apr 23 '17 at 3:05
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I can suggest a slight simplification, coming from the fact that $\mathbb Z$ is a Euclidean domain (not just a principal ideal domain) - though I would be very pleased if somebody could suggest a more elementary answer than this!

Suppose $M'$ is a submodule of $\mathbb Z^{\oplus m}$. (In your example, $m = 2$.) Then $M'$ is finitely generated as a $\mathbb Z$-module. (For example, $\mathbb Z$ is a PID, hence $\mathbb Z$ is noetherian, hence $\mathbb Z^{\oplus m}$ is noetherian, hence all submodules of $\mathbb Z^{\oplus m}$ are finitely generated...)

So let $x_1, \dots, x_n$ be a finite set of generators for $M'$. We can define a homomorphism $f : \mathbb Z^{\oplus n} \to \mathbb Z^{\oplus m}$ by sending $(a_1, \dots, a_n)\mapsto \sum_i a_i x_i$.

Since $\mathbb Z$ is a Euclidean domain, it is possible to choose bases for $\mathbb Z^{\oplus n}$ and $\mathbb Z^{\oplus m}$ in which $f$ is represented by a diagonal matrix. This is a consequence of the Smith normal form algorithm. Therefore, $M'$ is a free $\mathbb Z$-module, whose rank is equal to the number of non-zero diagonal entries in the matrix.

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