Faithfully Flat

Proof. If $M$ is faithfully flat. Then $0\to N'\to N\to N''\to 0$ exact implies $\Leftrightarrow 0\to N'\otimes M\to N\otimes M\to N''\otimes M\to 0$ exact.

Consider $0\to N'\otimes M\xrightarrow{\bar{f}} N\otimes M\xrightarrow{\bar{g}} N''\otimes M\to 0$, there are $f:N'\to N$ and $g:N\to N''$ s.t. $f\otimes id_{M}=\bar{f}$, $g\otimes id_{M}=\bar{g}$. Since functor $-\otimes M$ commutes with composition, $f$ injective and $g$ surjective, $g\circ f=0$. So $0\to N'\xrightarrow{f}N\xrightarrow{g}N''\to 0$ is exact.

Conversely, we can immediately have $M$ flat. We only need to show $0_{\mathrm{Hom}(N'\otimes M, N\otimes M)}=f\otimes M$ implies $f=0_{\mathrm{Hom}(N',N)}$.

Since

$$0\to \mathrm{Ker}f\to N'\to N\to 0$$

exact, by flatness,

$$0\to \mathrm{Ker}f\otimes M\to N'\otimes M\to N\otimes M\to 0$$

exact. Thus $\mathrm{Ker}f\otimes M\cong N'\otimes M$. Let $i:\mathrm{Ker}f\to N'$ be the inclusion map, then $0\to \mathrm{Ker}f\otimes M\xrightarrow{i\otimes M} \mathrm{Coker}i\otimes M\to 0$ is exact. Since tensor preserves cokernels, $\mathrm{Coker}i\otimes M=\mathrm{Coker}(i\otimes M)=0\Rightarrow \mathrm{Coker}i=0$. So $f=0_{\mathrm{Hom}(N',N)}$. ◻

Proof. 1 $\Rightarrow$ 2: Suppose $N\otimes M=0$, since $0\to N\otimes M\to 0$ exact, $0\to N\to 0$ exact. So $N=0$.

2 $\Rightarrow$ 3: $M /\mathfrak{m}M\cong (R /\mathfrak{m})\otimes M$. Since $R /\mathfrak{m}\neq 0$, $M /\mathfrak{m}M\neq 0$.

3 $\Rightarrow$ 2: Take $x\in N$, $x\neq 0$. Then $Rx\cong R /Ann(x)$. Let $\mathfrak{m}$ be the maximal ideal containing $Ann(x)$, then $Ann(x)M\subset \mathfrak{m}M\subset M$. Thus $(R /Ann(x))\otimes M\cong M /Ann(x) M\neq 0$. Since $M$ is flat, $0\to (R /Ann(x))\otimes M\to N\otimes M\to 0$ is exact. Thus $N\otimes M\neq 0$.

2 $\Rightarrow$ 1: For any $f\otimes M=0_{\mathrm{Hom}(N'\otimes M, N\otimes M)}$, we need to show $f=0_{\mathrm{Hom}(N',N)}$. Take $i:\mathrm{Ker}f\to N'$ and apply similar arguments as lemma. ◻

Proof. Let $\mathfrak{m}, \mathfrak{n}$ be the maximal ideals of $R$ and $R'$ respectively. Then $\varphi(\mathfrak{m})\subset\mathfrak{n}$ by the definition of local homomorphism. Then $\mathfrak{m}M \subset\mathfrak{n}M$, by Nakayama lemma, $\mathfrak{n}M\neq M$. Thus $\mathfrak{m}M\neq M$. Then apply the previous theorem. ◻

Proof. For any exact $R$-module sequence $0\to N'\to N\to N''\to 0$. $M$ is faithfully flat $R$-module, so the exactness above is equivalent to the exactness of $0\to N'\otimes_{R}M\to N\otimes_{R}M\to N''\otimes_{R}M\to 0$. Thus its equivalent to the exactness of $0\to (N'\otimes_{R}R')\otimes_{R'} M\to (N\otimes_{R} R')\otimes_{R'}M\to (N''\otimes_{R}R')\otimes_{R'}M\to 0$. Since $M$ is faithfully flat $R'$ module, the exactness is equivalent to the exactness of $0\to N'\otimes_{R}R'\to N\otimes_{R}R'\to N''\otimes_{R}R'\to 0$. Thus $R'$ is faithfully flat over $R$. ◻

Proof. 1 $\Rightarrow$ 2: Consider the exact sequence

$$0\to \mathrm{Ker}\alpha\otimes_{R}R'\to M\otimes_{R}R'\xrightarrow{\alpha\otimes_{R}id_{R'}}M\otimes_{R}R'\otimes_{R}R'$$

$\alpha\otimes id_{R'}:m\otimes x\otimes y\mapsto m\otimes xy$ is injective, so $\mathrm{Ker}(\alpha\otimes_{R} id_{R'})=\mathrm{Ker}\alpha\otimes_{R}R'=0$. Thus $\mathrm{Ker}\alpha=0$, $\alpha$ is injective.

2 $\Rightarrow$ 3: By assumption, $R /\mathfrak{a}\to R /\mathfrak{a}\otimes R'=R' /\mathfrak{a}R'$ is injective. $\mathfrak{a}R=\mathfrak{a}^{e}$. So $\forall x\in \mathfrak{a}^{e}$, there is $y\in \mathfrak{a}$ s.t. $f(y)=x$ where $f:R\to R'$. So $f^{-1}(x)\in\mathfrak{a}$, $\mathfrak{a}^{ec}\subset\mathfrak{a}$,
$\mathfrak{a}^{ec}=\mathfrak{a}$.

3 $\Rightarrow$ 4: Take $S=f(R-\mathfrak{p})$, then $S$ is multiplicative closed. $\mathfrak{p}^{e}\cap S=\varnothing$ since $\mathfrak{p}^{ec}=\mathfrak{p}$. Take $\Sigma=\{\mathfrak{b}|\mathfrak{b}\supset \mathfrak{p}^{e}, \mathfrak{b}\cap S=\varnothing\}$. By Zorn’s lemma, $\Sigma$ has a maximal element $\mathfrak{q}$. It’s easy to show $\mathfrak{q}$ is prime. $f^{-1}(\mathfrak{q})\supset \mathfrak{p}$, $f^{-1}(\mathfrak{q})\cap (R-\mathfrak{p})=\varnothing$, then $\mathfrak{q}^{c}=p$.

4 $\Rightarrow$ 5: $\mathfrak{m}=\mathfrak{q}^{c}$ for some prime $\mathfrak{q}$ in $R'$, then $\mathfrak{m}R'\subset\mathfrak{q}\neq R'$.

5 $\Rightarrow$ 6: Take $m\in M, m\neq 0$. Let $M'=Rm$, thus $i:M'\to M$ will give injective $i\otimes id_{R'}:M'\otimes R'\to M\otimes R'$. If $M\otimes R'=0$, we have $M'\otimes R'=0$. Since $M'\cong R /Ann(m)$, we have $R' /Ann(m)R'=0$. Take $\mathfrak{m}$ be the a maximal ideal containing $Ann(m)$, then $Ann(m)R'\subset \mathfrak{m}R'\subsetneq R'$. Contradiction!

6 $\Rightarrow$ 1: Already done. ◻