Flatness of $\mathbb{Z}$-modules

I have to prove that :

1) For every positive integer $n$, $\mathbb{Z}_{n}$ is not a flat $\mathbb{Z}$-module

2) Every torsion free abelian group is a flat $\mathbb{Z}$-module

What I have done:

1) We can consider the multiplication by $n$ as a map $\phi_{n} : \mathbb{Z} \to \mathbb{Z}$, this is injective, and then tensor by $\mathbb{Z}_{n}$, we obtain $$\phi_{n} \otimes id : \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{n} \to \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{n}$$ but $\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{n} \cong \mathbb{Z}_{n}$ , $\phi_{n} \otimes id$ is the multiplication by $n$ and so is not injective.

2) If the group is finitely generated, then we apply the structure theorem abou finitely generated abelian groups and we recall that tensor product commutes with direct sum.

So the problem is when the group isn't finitely generated. Any hint ?

(1) is good.

A finitely generated torsionfree (abelian) group is free, hence flat. Now, suppose $P$ is a torsionfree module and let $f\colon M\to N$ be an injective morphism. You want to prove that $$f\otimes P\colon M\otimes P\to N\otimes P$$ is injective. If $t=\sum_{i=1}^n x_i\otimes y_i\in\ker(f\otimes P)$ then you can consider the subgroup $L=\langle x_1,\dots,x_n\rangle\subseteq M$ and the restriction $g\colon L\to N$ of $f$ to $L$. Then you have the commutative diagram $$\require{AMScd}\begin{CD} L\otimes P @>g\otimes P>> N\otimes P \\ @Vj\otimes PVV @VV\mathit{id}V \\ M\otimes P @>f\otimes P>> N\otimes P \end{CD}$$ Since $L$ is finitely generated, $g\otimes P$ is injective. Set $s=\sum_{i=1}^n x_i\otimes y_i$ as an element of $L\otimes P$; then, $j\colon L\to M$ being the inclusion, $$(g\otimes P)(s)=(f\otimes P)\circ(j\otimes P)(s)=(f\otimes P)(t)=0$$ so $s=0$ and therefore $t=0$.

• Why does the statement "Since $L$ is finitely generated, $g \otimes P$ is injective" hold? – probably123 Dec 29 '19 at 0:56

$\DeclareMathOperator{\col}{colim}$ (1) Sounds good.

(2) Note that every module is the directed colimit of its finitely generated submodules, and that the tensor product commutes with any colimit (direct sums are just special cases).

So if $A$ is torsion free and $f : N \to M$ is mono, we have $A \otimes f = A \otimes N \stackrel{\cong}{\longrightarrow} (\col_i A_i) \otimes N \stackrel{\cong}{\longrightarrow} \col_i (A_i \otimes N) \stackrel{A_i \otimes f}{\longrightarrow} \col_i (A_i \otimes M) \stackrel{\cong}{\longrightarrow} A \otimes M$, where $(A_i)_i$ runs over all finitely generated submodules of $A$. By applying the structure theorem for those $A_i$ we see that $A \otimes f$ is a composition of isomorphisms and monomorphisms (note that a directed colimit of monomorphisms is mono), and hence is mono as well.

From the same proof you even see that any directed colimit of flat modules must also be flat.