[proofplan] We use the injectivity criterion for flatness: an $R$-module $F$ is flat if, for every injective $R$-[linear map](/page/Linear%20Map) $u: M \to N$, the induced map $u \otimes_R \operatorname{id}_F: M \otimes_R F \to N \otimes_R F$ is injective. Since $F$ is free, we identify it with a [direct sum](/page/Direct%20Sum) of copies of $R$. Tensoring with that direct sum identifies $M \otimes_R F$ with a direct sum of copies of $M$, and under this identification $u \otimes_R \operatorname{id}_F$ becomes the componentwise direct sum of copies of $u$. A componentwise direct sum of injective maps is injective, so tensoring with $F$ preserves injections. [/proofplan] [step:Identify the free module with a direct sum of copies of $R$] Let $F$ be a free $R$-module. Choose a basis $(e_i)_{i \in I}$ of $F$, where $I$ is an index set. By [citetheorem:10013], the map \begin{align*} \Phi: \bigoplus_{i \in I} R \to F \end{align*} defined by \begin{align*} \Phi((r_i)_{i \in I}) = \sum_{i \in I} r_i e_i \end{align*} is an $R$-module isomorphism, where the sum is finite because $(r_i)_{i \in I}$ has finite support. Thus it is enough to prove that $\bigoplus_{i \in I} R$ is flat, since flatness is preserved under $R$-module isomorphism. Indeed, if $\theta: F \to F'$ is an $R$-module isomorphism, then for every $R$-module $M$ the induced map \begin{align*} \operatorname{id}_M \otimes_R \theta: M \otimes_R F \to M \otimes_R F' \end{align*} is an $R$-module isomorphism, so injectivity of tensor maps is unchanged after replacing $F$ by an isomorphic module. [/step] [step:Construct the tensor-direct-sum identification] Let $M$ be an $R$-module. Define \begin{align*} \alpha_M: M \otimes_R \left(\bigoplus_{i \in I} R\right) \to \bigoplus_{i \in I} M \end{align*} to be the $R$-linear map determined on pure tensors by \begin{align*} \alpha_M\left(m \otimes_R (r_i)_{i \in I}\right) = (r_i m)_{i \in I}. \end{align*} This is well-defined because $(r_i)_{i \in I}$ has finite support, so $(r_i m)_{i \in I}$ is an element of $\bigoplus_{i \in I} M$. For each $j \in I$, let $\varepsilon_j \in \bigoplus_{i \in I} R$ denote the element whose $j$-th coordinate is $1_R$ and whose other coordinates are $0_R$. Define \begin{align*} \beta_M: \bigoplus_{i \in I} M \to M \otimes_R \left(\bigoplus_{i \in I} R\right) \end{align*} by \begin{align*} \beta_M((m_i)_{i \in I}) = \sum_{i \in I} m_i \otimes_R \varepsilon_i, \end{align*} where the sum is finite because $(m_i)_{i \in I}$ has finite support. For $(m_i)_{i \in I} \in \bigoplus_{i \in I} M$, we have \begin{align*} (\alpha_M \circ \beta_M)((m_i)_{i \in I}) = (m_i)_{i \in I}. \end{align*} For a pure tensor $m \otimes_R (r_i)_{i \in I}$, we compute \begin{align*} (\beta_M \circ \alpha_M)\left(m \otimes_R (r_i)_{i \in I}\right) = \sum_{i \in I} r_i m \otimes_R \varepsilon_i. \end{align*} By the balancing relation in the [tensor product](/page/Tensor%20Product), \begin{align*} r_i m \otimes_R \varepsilon_i = m \otimes_R r_i \varepsilon_i. \end{align*} Since $(r_i)_{i \in I} = \sum_{i \in I} r_i \varepsilon_i$ in $\bigoplus_{i \in I} R$, additivity in the second tensor factor gives \begin{align*} \sum_{i \in I} r_i m \otimes_R \varepsilon_i = m \otimes_R (r_i)_{i \in I}. \end{align*} Thus $\beta_M \circ \alpha_M$ is the identity on pure tensors, hence on all of $M \otimes_R \bigl(\bigoplus_{i \in I} R\bigr)$. Therefore $\alpha_M$ is an $R$-module isomorphism with inverse $\beta_M$. [/step] [step:Show tensoring with the free module sends injections to injections] Let \begin{align*} u: M \to N \end{align*} be an injective $R$-[module homomorphism](/page/Module%20Homomorphism). Under the isomorphisms $\alpha_M$ and $\alpha_N$ constructed above, the tensor map \begin{align*} u \otimes_R \operatorname{id}_{\bigoplus_{i \in I} R}: M \otimes_R \left(\bigoplus_{i \in I} R\right) \to N \otimes_R \left(\bigoplus_{i \in I} R\right) \end{align*} corresponds to the componentwise direct sum map \begin{align*} \bigoplus_{i \in I} u: \bigoplus_{i \in I} M \to \bigoplus_{i \in I} N \end{align*} defined by \begin{align*} \left(\bigoplus_{i \in I} u\right)((m_i)_{i \in I}) = (u(m_i))_{i \in I}. \end{align*} Indeed, for every pure tensor $m \otimes_R (r_i)_{i \in I}$, \begin{align*} \alpha_N\left((u \otimes_R \operatorname{id})(m \otimes_R (r_i)_{i \in I})\right) = (r_i u(m))_{i \in I}, \end{align*} while \begin{align*} \left(\bigoplus_{i \in I} u\right)\left(\alpha_M(m \otimes_R (r_i)_{i \in I})\right) = (u(r_i m))_{i \in I}. \end{align*} Since $u$ is $R$-linear, $u(r_i m) = r_i u(m)$ for every $i \in I$, so the two expressions agree. Now suppose $(m_i)_{i \in I} \in \bigoplus_{i \in I} M$ lies in the kernel of $\bigoplus_{i \in I} u$. Then \begin{align*} u(m_i) = 0 \end{align*} for every $i \in I$. Since $u$ is injective, $m_i = 0$ for every $i \in I$. Hence $(m_i)_{i \in I} = 0$, so $\bigoplus_{i \in I} u$ is injective. Because $\alpha_M$ and $\alpha_N$ are isomorphisms, $u \otimes_R \operatorname{id}_{\bigoplus_{i \in I} R}$ is injective. [guided] We must prove that tensoring with the [free module](/page/Free%20Module) preserves injections. Let \begin{align*} u: M \to N \end{align*} be an injective $R$-module homomorphism. The previous step gives natural $R$-module isomorphisms \begin{align*} \alpha_M: M \otimes_R \left(\bigoplus_{i \in I} R\right) \to \bigoplus_{i \in I} M \end{align*} and \begin{align*} \alpha_N: N \otimes_R \left(\bigoplus_{i \in I} R\right) \to \bigoplus_{i \in I} N. \end{align*} The point of these isomorphisms is that they turn the tensor map into a map whose injectivity can be checked coordinate by coordinate. The tensor map is \begin{align*} u \otimes_R \operatorname{id}_{\bigoplus_{i \in I} R}: M \otimes_R \left(\bigoplus_{i \in I} R\right) \to N \otimes_R \left(\bigoplus_{i \in I} R\right). \end{align*} Under $\alpha_M$ and $\alpha_N$, this map corresponds to \begin{align*} \bigoplus_{i \in I} u: \bigoplus_{i \in I} M \to \bigoplus_{i \in I} N, \end{align*} where \begin{align*} \left(\bigoplus_{i \in I} u\right)((m_i)_{i \in I}) = (u(m_i))_{i \in I}. \end{align*} We verify this correspondence on pure tensors, which is enough because pure tensors generate the tensor product as an $R$-module. For $m \in M$ and $(r_i)_{i \in I} \in \bigoplus_{i \in I} R$, the route through the tensor map gives \begin{align*} \alpha_N\left((u \otimes_R \operatorname{id})(m \otimes_R (r_i)_{i \in I})\right) = \alpha_N(u(m) \otimes_R (r_i)_{i \in I}) = (r_i u(m))_{i \in I}. \end{align*} The route through the direct sum map gives \begin{align*} \left(\bigoplus_{i \in I} u\right)\left(\alpha_M(m \otimes_R (r_i)_{i \in I})\right) = \left(\bigoplus_{i \in I} u\right)((r_i m)_{i \in I}) = (u(r_i m))_{i \in I}. \end{align*} Since $u$ is $R$-linear, $u(r_i m) = r_i u(m)$ for every $i \in I$, so the two routes agree. It remains to prove that $\bigoplus_{i \in I} u$ is injective. Let $(m_i)_{i \in I} \in \bigoplus_{i \in I} M$ satisfy \begin{align*} \left(\bigoplus_{i \in I} u\right)((m_i)_{i \in I}) = 0. \end{align*} This means exactly that \begin{align*} u(m_i) = 0 \end{align*} for every $i \in I$. Because $u$ is injective, each equality $u(m_i)=0$ implies $m_i=0$. Hence every coordinate of $(m_i)_{i \in I}$ is zero, so $(m_i)_{i \in I}=0$. Therefore $\bigoplus_{i \in I} u$ is injective. Finally, an isomorphic conjugate of an injective map is injective. Since $\alpha_M$ and $\alpha_N$ are isomorphisms and $\bigoplus_{i \in I} u$ is injective, the tensor map \begin{align*} u \otimes_R \operatorname{id}_{\bigoplus_{i \in I} R} \end{align*} is injective. [/guided] [/step] [step:Conclude flatness of the original free module] We have shown that, for every injective $R$-module homomorphism $u: M \to N$, the induced map \begin{align*} u \otimes_R \operatorname{id}_{\bigoplus_{i \in I} R}: M \otimes_R \left(\bigoplus_{i \in I} R\right) \to N \otimes_R \left(\bigoplus_{i \in I} R\right) \end{align*} is injective. Hence $\bigoplus_{i \in I} R$ is flat by the injectivity criterion for flatness. Since $F$ is isomorphic to $\bigoplus_{i \in I} R$, and flatness is invariant under $R$-module isomorphism, $F$ is flat. Therefore every free $R$-module is flat. [/step]