HW7

16. Let $B$ be a flat $A$-algebra. Then the following conditions are equivalent:

• i) $a^{ec}=a$ for all ideals $a$ of $A$.
• ii) $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is surjective.
• iii) For every maximal ideal $m$ of $A$ we have $m^e\ne$ (1).
• iv) If $M$ is any non-zero $A$-module, then $M_B\ne 0$.
• v) For every $A$-module $M$, the mapping $x\mapsto 1\otimes x$ of $M$ into $M_B$ is injective. (Hint: see here.)

$B$ is said to be faithfully flat over $A$.

\begin{proof} i)→ii) By Proposition 3.16, for any given prime ideal $a$ of $\mathrm{Spec}(A)$, $a^{ec}=a$ yields $p$ is a contraction of a prime ideal $q\subseteq B$. So $a\in \mathrm{im}(\mathrm{Spec}(B)\to \mathrm{Spec}(A))$ for all $a\in \mathrm{Spec}(A)$, that is, $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ is surjective.

ii)→iii) Assume that there exists $m\in \mathrm{Spec}(A)$ such that $m^e=(1)$, then $m^{ec}=(1)\in \mathrm{Spec}(B)$ and so $m^{ec}\ne m$. By Proposition 3.16, $m$ is not a contraction of any prime $m^{\prime }\subseteq B$ and so $m\notin \mathrm{im}(\mathrm{Spec}(B)\to \mathrm{Spec}(A))$, contradicting with $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ surjective.

iii)→iv) Let $x$ be a non-zero element of $M$ and let $M^{\prime }=Ax$. Then there is an injective map $M^{\prime }\hookrightarrow M$. Since $B$ is flat over $A$, the map $M_B^{\prime }\to M_B$ is also injective. Thus, it is enough to show that $M_B^{\prime }\ne 0$.

Note that there is a natural map

$$A\to Ax=M^{\prime },a\mapsto ax,$$

then $M^{\prime }\simeq A/a$ for some ideal $a$. By Text-Ex. 2.2, we have $M_B^{\prime }=B{\otimes }_{A}A/a=(B{\otimes }_{A}A)/(aB)=B/a^e$. If $M_B^{\prime }=0$, then $a^e=(1)$. By iii), it deduces $a=A$ and then $M^{\prime }=0$, which is impossible. Therefore, $M_B^{\prime }\ne 0$ and so $M_B\ne 0$.

iv)→v) Define $\phi :M\to M_B,x\mapsto 1\otimes x$. Let $M_0$ be the kernel of $\phi$. Then

$$0\to M_0\to M\to M_B$$

is exact. Since $B$ is flat, one have

$$0\to (M_0{)}_B\to M_B\to (M_B{)}_B\text{\hspace{1em}(*)}$$

is exact. Take $N=M_B$ and apply Text-Ex. 2.13, then $M_B\to (M_B{)}_B$ is injective. Then by exact $(\ast )$, we know $(M_0{)}_B=0$ and so $M_0=0$ by iv). Therefore, $\phi$ is injective.

v)→i) For any ideal $a\subseteq a$, take $M=A/a$ and then by v) $f:M=A/a\to B/a^e=M_B$ is injective. If $a^{ec}\ne a$, then $a^{ec}⊋a$ and $a^{ec}/a\subseteq A/a$ is non-empty. However, $f(a^{ec}/a)=(a^e{)}^{ce}/a^e=a^e/a^e=0$ by Proposition 1.17, which contradicts with $f$ injective. \end{proof}

22. Let $A$ be a ring and $p$ a prime ideal of $A$. Then the canonical image of $\mathrm{Spec}\left(A_p\right)$ in $\mathrm{Spec}(A)$ is equal to the intersection of all the open neighborhoods of $p$ in $\mathrm{Spec}(A)$.

\begin{proof} By Proposition 3.11, there is a $1$-$1$ correspondence between prime ideals of $A_p$ and prime ideals of $A$ which do not meet $A-p$. Thus the canonical image of $\mathrm{Spec}\left(A_p\right)$ in $\mathrm{Spec}(A)$ is the set

$$\{q:q\subseteq A\text{ is a prime ideal, and }q\cap (A-p)=\varnothing \}=\{q:q\subseteq A\text{ is a prime ideal, and }q\subseteq p\}.$$

Recall that $X_f$ is a set of topological basis of $\mathrm{Spec}(A)$. If $X_f=\mathrm{Spec}(A)\setminus V(f)$ is an open neighborhood of $p$, then $X_f\supseteq p$ and $V(f)\cap p=\varnothing$ and so $f\notin p$. Therefore, the intersection of all the open neighborhoods of $p$ in $\mathrm{Spec}(A)$ is

$${\cap }_{f\notin p}{X}_f={\cap }_{f\notin p}(\mathrm{Spec}(A)\setminus V(f))=\mathrm{Spec}(A)\setminus {\cup }_{f\notin p}V(f).$$

For any prime ideal $q\in {\cup }_{f\notin p}V(f)$, there exists $f\notin p$ and $f\in q$. It deduces that $q⊈p$. Conversely, if $q\notin {\cup }_{f\notin p}V(f)$, then for any $f\notin p$, $f\notin q$, which yields $q\subseteq p$. So

$${\cap }_{f\notin p}{X}_f=\mathrm{Spec}(A)\setminus {\cup }_{f\notin p}V(f)=\{q:q\subseteq A\text{ is a prime ideal, and }q\subseteq p\}.$$

Now we finish the proof. \end{proof}