Exercise 3.16

Proof.

$i)\Rightarrow \mathit{ii})$: From AM prop 3.16, $𝔭\in \mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(A)$ is the contraction of a prime ideal of $B$ iff ${𝔭}^{\mathit{ec}}=𝔭$. From the assumption, ${𝔞}^{\mathit{ec}}=𝔞$ for all ideals $𝔞$, so $𝔭\in \mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(A)$ is the contraction of a prime ideal of $B$. Hence, $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(B)\to \mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(A)$ is surjective.

$\mathit{ii})\Rightarrow \mathit{iii})$: assume ${𝔪}^e=(1)$. There exists a prime ideal $𝔮$ of $B$ such that ${𝔮}^c=𝔪$. Then we see that $(1)={𝔪}^e={𝔮}^{\mathit{ce}}\supseteq 𝔮$, which contradicts to the assumption that $𝔮$ is a prime ideal.

$\mathit{iii})\Rightarrow \mathit{iv})$: Let $0\ne x\in M$, and let $M^{\prime }=\mathit{Ax}$. We see that $0\to M\to M^{\prime }$ is exact. Since $B$ is flat over $A$, $0\to M\otimes B\to M^{\prime }\otimes B$ is exact. Hence, if $M_B^{\prime }=M^{\prime }\otimes B$ is nonzero, $M_B$ is also nonzero. $M^{\prime }\simeq A∕𝔞$ for some ideal $𝔞\ne (1)$ of A, so $M_B^{\prime }\simeq B\otimes A∕𝔞=B∕{𝔞}^e\otimes A\simeq B∕{𝔞}^e$. Then, There is a maximal ideal $𝔪$ that contains $𝔞$ (Corollary 1.4), and ${𝔞}^e\subseteq {𝔪}^e⊊(1)$ from the assumption. Hence, $M_B^{\prime }\ne 0$.

$\mathit{iv})\Rightarrow v)$: Let $M^{\prime }$ be the kernel of $M\to M_B$. Then $0\to M^{\prime }\to M\to M_B$ is exact. Because $B$ if flat, $0\to M_B^{\prime }\to M_B\to {(M_B)}_B$ is exact. Using AM Exercise 2.13, $M_B\to {(M_B)}_B$ is injective, so $M_B^{\prime }=0\Rightarrow M^{\prime }=0$. Hence, the mapping $x\mapsto 1\otimes x$ of $M$ into $M_B$ is injective.

$v)\Rightarrow i)$ Choose an ideal $𝔞$ of $A$. Let $M=A∕𝔞$. Then we see that the mapping $x\mapsto 1\otimes x$ of $M=A∕𝔞$ into $M_B=A∕𝔞\otimes B=A\otimes B∕{𝔞}^e\simeq B∕{𝔞}^e$ is injective. Hence, ${𝔞}^{\mathit{ec}}\subseteq 𝔞$. ${𝔞}^{\mathit{ec}}\supseteq 𝔞$ from AM proposition 1.17, so ${𝔞}^{\mathit{ec}}=𝔞$. □