Flat Module

Proof. For $f:N'\to N$ an injective homomorphism, $(\bigoplus M_{\lambda})\otimes f= \bigoplus(M_{\lambda}\otimes f)$. Here $\bigoplus(M_{\lambda}\otimes f): M\otimes N'\to M\otimes N$ is injective. Thus $M\otimes -$ preserves injectives if and only if $M_{\lambda}$ preserves injectives.

For faithfully flat, $M\otimes N=\bigoplus ((M_{\lambda}\otimes N))$. Thus if $M\otimes N=0$, then $M_{\lambda}\otimes N=0$ for all $\lambda$. If at least one $M_{\lambda}$ is faithfully flat, then immediately $N=0$. $M$ is faithfully flat. ◻

Proof. For free module $F\cong \bigoplus_{\Lambda} R$. $R$ is obviously faithfully flat, so $F$ is faithfully flat. Since projective modules are direct summands of free modules, they are flat. ◻

Proof.

1. For any $N$, take the free presentation $0\to K\to R^{\Lambda}\to N\to 0$, tensor it with the exact sequence $0\to M'\to M\to M''\to 0$, we have the following commutative diagram.

Since $R^{\Lambda}$ and $M''$ are flat, $\alpha$ and $\beta$ are injective. So each rows and columns are exact. Then just apply the snake lemma. $0\to M'\otimes N\xrightarrow{\gamma} M\otimes N$ is exact. $\gamma$ is injective so $0\to M'\otimes N\to M\otimes N\to M''\otimes N\to 0$ is exact.

2. Assume $f:N'\to N$ be an injective homomorphism, consider the diagram:

each rows are exact.

Assume $M$ is flat, then $\alpha$ is injective, $M'$ is flat.

Assume $M'$ is flat, then $\alpha'$ is injective. By five lemma, $\alpha$ is injective, $M$ is flat. ◻

Proof. For injective $f:A\to B$, we only need to show $id_{M}\otimes f:M\otimes A\to M\otimes B$ is injective. For any $u\in\mathrm{Ker}id_{M}\otimes f$, we construct finite generated submodule $N$ of $M$ and $v\in N$ s.t. $i\otimes id_{A}(v)=u$, $i$ is the inclusion map. Assume $u=\sum\limits_{i=1}^{n}m_{i}\otimes a_{i}$, $(id_{M}\otimes f)(u)=\sum_{i=1}^{n}m_{i}\otimes f(a_{i})$. Let $(id_{M}\otimes f)(u)$ be finite linear combination of elements in $M\times B$, and take all the elements of the $M$-coordinate to form the set $S$. Let $N$ be the submodule of $M$ generated by $S\cup\{m_{1},\dots,m_{n}\}$, $v=\sum\limits_{i=1}^{n}m_{i}\otimes a_{i}\in N\otimes A$, then the conditions are satisfied. ◻

Proof. Omitted. ◻

Proof. Omiited. ◻

Proof. Only need to show for any given $\alpha:R^{m}\to M$ and $k\in\mathrm{Ker}\alpha$, there is $\varphi:R^{m}\to R^{n}$ s.t. $\alpha:R^{m}\xrightarrow{\varphi}R^{n}\to M$ with $\varphi(k)=0$. Then $M$ is filtered direct colimit of free modules with finite rank, by Lazard theorem, $M$ is flat.

Let $k=\sum_{i}x_{i}e_{i}$ $\{e_{i}\}$ is the basis of $R^{m}$. Define $\varphi:R^{m}\to R^{n}$ from the matrix $(x_{ij})$, then $\varphi(k)=\sum_{i}\varphi(e_{i})=\sum_{i}x_{i}\sum_{j}x_{ij}e_{j}'$, $\{e_{j}'\}$ is the basis of $R^{n}$. Then $\varphi(k)=0$ by assumption. Define $\beta:R^{n}\to M$ by $\beta(e_{j}')=y_{j}'$. Then $\beta(\varphi(e_{j}))=\beta(\sum_{i} x_{ij}e_{j})=\sum_{i}x_{ij}y_{j}'=y_{j}$. Thus $\beta\varphi=\alpha$. ◻

Proof. $M\otimes N$ is generated by the elements of the form $m\otimes n$. If $n=\sum x_{\lambda}n_{\lambda}$ with $x_{\lambda}\in R$, then $m\otimes n=\sum(x_{\lambda}m)\otimes n_{\lambda}$. Thus any element of $M\otimes N$ has the form $\sum m_{\lambda}\otimes n_{\lambda}$. Assume $m_{\sigma}$ and the $x_{\lambda \sigma}$ exist, then

$$\sum m_{\lambda}\otimes n_{\lambda}=\sum_{\lambda}(\sum_{\sigma}x_{\lambda \sigma}m_{\sigma})\otimes n_{\lambda}=\sum_{\sigma}(m_{\sigma}\otimes \sum_{\lambda}x_{\lambda \sigma}n_{\lambda})=0$$

Conversely, take a free presentation $R^{\Sigma}\xrightarrow{\beta}R^{\Lambda}\xrightarrow{\alpha}N\to 0$ with $\alpha(e_{\lambda})=n_{\lambda}$ for all $\lambda$, where $\{e_{\lambda}\}$ is the basis of $R^{\Lambda}$. Then the sequence $M\otimes R^{\Sigma}\to M\otimes R^{\Lambda} \to M\otimes N\to 0$ is exact. $id_{M}\otimes \alpha(\sum m_{\lambda} \otimes e_{\lambda})=0$, so the exactness implies there’s $s\in M\otimes R^{\Sigma}$ s.t. $(1\otimes \beta)(s)=\sum m_{\lambda}\otimes e_{\lambda}$. Let $\{e_{\sigma}\}$ be the basis of $R^{\Sigma}$ and $s=\sum m_{\sigma}\otimes e_{\sigma}$ with $m_{\sigma}\in M$. Write $\beta(e_{\sigma})=\sum_{\lambda}x_{\lambda \sigma}e_{\lambda}$. Then $\sum_{\lambda}x_{\lambda \sigma}n_{\lambda}=\alpha(\beta(e_{\sigma}))=0$ and

$$0=\sum_{\lambda}m_{\lambda}\otimes e_{\lambda}-\sum_{\sigma}m_{\sigma}\otimes(\sum_{\lambda}x_{\lambda\sigma}e_{\lambda})=\sum_{\lambda}(m_{\lambda}-\sum_{\sigma}x_{\lambda\sigma}m_{\sigma})\otimes e_{\lambda}$$

Since $e_{\lambda}$ are independent, $m_{\lambda}=\sum_{\sigma}x_{\lambda \sigma}m_{\sigma}$. ◻

Proof. Assume $N$ is flat. We know $R\otimes N\to N$ is an isomorphism, $N$ is flat so injective homomorphism $\mathfrak{a}\to R$ induces $\mathfrak{a}\otimes N\to R\otimes N=N$. So $\mathfrak{a}\otimes N\to \mathfrak{a}N$ is injective thus an isomorphism.

Conversely, given any $\sum\limits_{i=1}^{n}x_{i}y_{i}=0$ with $x_{i}\in R$, $y_{i}\in N$. Take $\mathfrak{a}=(x_{1},\dots, x_{n})$. If $\mathfrak{a}\otimes N\cong \mathfrak{a}N$, then $\sum_{i}x_{i}\otimes y_{i}=0$, by lemma above, we have $\sum_{\sigma}x_{\lambda \sigma}m_{\sigma}=m_{\lambda}$ for all $\lambda$ and $\sum_{\lambda}x_{\lambda \sigma}n_{\lambda}=0$ for all $\sigma$. Thus $N$ is flat. ◻

Proof. Omitted. ◻