# Flatness via factoring homomorphisms

Theorem (4.32) of "Lectures on modules and rings" by T.Y. Lam says that a module $P_R$ is flat iff any $R$-homomorphism $λ:M→P$ where $M$ is any finitely presented $R$-module can be factord through a finitely generated free module: there exist $ν:M→R^m , μ:R^m→P$ (for some finite $m$) with $λ=μoν$. My question leans on the 'if" part if one wants to use Theorem (4.24)(3). I could not realize how to reach a finitely presented module $M$ and a $λ$ from the hypothesis of Theorem (4.24)(3). Thanks for any help!

Assume you have relations $\sum_j a_j r_{j,l} = 0$ with $a_j \in P$, $r_{j,l} \in R$, $1 \le j \le n$, and $1 \le l \le p$, as in Theorem 4.24(3).
Let $e_j$ denote the $j$-th standard unit vector of $R^n$. Now let $K$ be the submodule of $R^n$ generated by $\{ \sum_{j} e_j r_{j,l} \mid 1 \le l \le p\}$. Set $M=R^n/K$ and let $\lambda \colon M \to P$ be the homomorphism induced from the homomorphism $R^n \to P$, $e_j \mapsto a_j$. ($K$ is indeed in the kernel of this homomorphism due to the given relations.)
• Thanks for answering! I went through this way , but I could only find the necessary relations $a_j= \sum b_is_{ij}$. How the other equality $\sum s_{ij}r_{jl}=0$ is obtained? Thanks again! – karparvar Sep 2 '15 at 17:20
• Defined the $(s_{i,j})$-matrix such that it describes the homomorphism $R^n \to R^n/K \to R^m$ with respect to the standard basis vectors. Then you see that the relation holds because $K$ is in the kernel of this homomorphism. – moonlight Sep 2 '15 at 18:05
• And in the proof of "only if" part, I really do not understand "well-definedness" of definition of $ν$. If $m=0\in M$ we should prove that $ν(m)=0$. Now, $m$ is a linear combination of $x_1,...x_n$ like $\sum x_jr_j=0$. But, what we see in the proof in the book is "$p$" such linear relations not just one linear relation with coefficients in $R$. How could we get the result? – karparvar Sep 5 '15 at 7:15
• He's essentially applying the homomorphism theorem with the chosen representation of $M$, but this is not stated very explicitly. Maybe the following helps: Let $y_1$, $\ldots$, $y_n$ denote a basis of $R^n$ and $M=R^n/K$ with $K=\langle \sum_{j} y_j r_{j,l}\rangle$. Write $x_i$ for the image of $y_i$ in $M$. Define $\nu_0\colon R^n \to R^m$ by $\nu_0(y_i)=\sum_i e_i s_{i,j}$. Now we check $K \subset \ker(v_0)$ (this is his calculation). By the homomorphism theorem, there exist $\nu\colon M \to R^m$ with $\nu(x_i)=\nu_0(y_i)$. – moonlight Sep 5 '15 at 7:38