projective module, The module not projective

I found a proof " The $$\mathbb{Z}$$ - module $$M=\mathbb{Z} \times\mathbb{Z} \times\dots$$ is not projective" in T.Y.Lam - Exercises in Modules and Rings.

Proof

Suppose that $$M$$ is projective, so $$M$$ is isomorphic to a direct summand of a free module then $$M \subseteq F$$ for a suitable free abelian group $$F$$ with basis $$\{e_i| i \in I\}$$. We have $$P=\mathbb{Z}\oplus \mathbb{Z}\oplus \dots\oplus \mathbb{Z}\subseteq M$$ be countable, so we can decompose $$I$$ into a disjoint union $$I_1\cup I_2$$ such that $$I_1$$ is countable and $$P$$ is contained in the span $$F_1$$ of $$\{e_i|i\in I_1\}$$. Let $$F_2$$ is the span of $$\{e_i|i\in I_2\}$$.\ The group $$M/(M\cap F_1)$$ has an embedding into the free abelian group $$F/F_1\cong F_2$$. We will get a contradiction if we can show that $$M/(M\cap F_1)$$ contains a nonzero element $$\alpha$$ that is divisible by $$2^n$$ for any $$n\ge 1$$. Consider the set $$S=\{(2^1\epsilon_1,2^2\epsilon_2,2^3\epsilon_3,\dots):\epsilon_i=\pm1\}\subseteq M$$ We see that $$F_1$$ is countable but $$S$$ is uncountable, so $$S$$ contains $$a=(2^1\epsilon_1,2^2\epsilon_2,2^3\epsilon_3,\dots) \not \in F_1$$ Hence, we have $$\begin{array}{*{20}{l}} \alpha &=& (2{\epsilon_1},{2^2}{\epsilon_2}, \ldots ,{2^{n - 1}}{\epsilon_{n - 1}},0,0, \ldots ) + (0,0, \ldots ,{2^n}{\epsilon_n},{2^{n + 1}}{\epsilon_{n + 1}}, \ldots ) + M \cap {F_1}\\ &=& {2^n}\left( {0,0, \ldots ,{\epsilon_n},2{\epsilon_{n + 1}}, \ldots } \right) + M \cap {F_1} \end{array}$$ Then $$\alpha=a+(M\cap F_1)\in M/(M\cap F_1)\setminus \{0\}$$ is divisible by $$2^n$$ for any $$n$$.

I have a question: Why we get a contradiction when we show that $$M/(M\cap F_1)$$ contains a nonzero element $$\alpha$$ that is divisible by $$2^n$$ for any $$n\ge 1$$. Thankyou.

You say $$M/(M\cap F_1)$$ has an embedding into some free Abelian group $$F_2$$. Let $$\alpha\in F_2$$. Let $$(e_i)_{i\in I}$$ be a basis of $$F_2$$. Then $$\alpha =\sum_{i\in I_0} a_i e_i$$ where $$a_i\in\Bbb Z$$ and $$I_0$$ is a finite subset of $$I$$. If $$\alpha\ne0$$ then some $$a_i\ne 0$$, and in $$F_2$$, the only positive integers that $$\alpha$$ could possibly be divided by are the factors of $$a_i$$, of which there only finitely many. So in $$F_2$$ no nonzero element is divisible by $$2^n$$ for all $$n$$.