HW9

1. If $R$ is a local ring, then $R$ has no idempotent other than the identity.

\begin{proof} For any idempotent $e\in R$, one have $e^2=e$ and $e(e-1)=0$. If $e\ne 1$, then $e-1\ne 0$ and so $e,e-1$ are not units. Since $R$ is a local ring, $J(R)$ is the set of non-units and so $e,e-1\in J(R)$. It deduces that $1=1-e+e\in J(R)$, which is a contradiction. \end{proof}

2. Let $R$ be an integral domain. If $I$ is a finitely generated projective fractional ideal, prove that

$${\mathrm{Hom}}_R(I,I)=R.$$

\begin{proof} Define $F=\mathrm{Frac}(R)$, then $I$ is a $R$-submodule of $F$ and there exists $\lambda \in R$ such that $\lambda I\subseteq R$. For any nonzero $x,y\in I$ and $\phi \in {\mathrm{Hom}}_R(I,I)$, we have

$$\phi (\lambda xy)=\lambda \phi (xy)=\lambda x\phi (y)=\lambda y\phi (x).$$

Since $R$ is an integral domain, one have $x^{-1}\phi (x)=y^{-1}\phi (y)$. Define $c_{\phi }=x^{-1}\phi (x)$ for any nonzero $x\in I$, then the map

$$\Psi :{\mathrm{Hom}}_R(I,I)\to F,\phi \mapsto c_{\phi }$$

is a injective map. Define $S=\{c\in F:c=c_{\phi }\text{ for some }\phi \in {\mathrm{Hom}}_R(I,I)\}$ and define $T=\{c\in F:cI\subseteq I\}$, then $S\subseteq T$. Conversely, for any $c\in T$, the map $f(x)=cx$ is contained in ${\mathrm{Hom}}_R(I,I)$. Therefore, ${\mathrm{Hom}}_R(I,I)$ can be identified as $T$.

Recall that $I$ is a finitely generated projective $R$-module iff $I^{\ast }I=R$ with $I^{\ast }=\{x\in F:xI\subseteq R\}$. Then there exist finitely many $x_k\in I^{\ast }$ and $y_k\in I$ such that ${\sum }_{k=1}^n{x}_k{y}_k=1$. For any $t\in T$, we have $t={\sum }_{k=1}^n{x}_{k}t{y}_k\in {\sum }_{k=1}^n{x}_{k}I\subseteq R$ and so $T\subseteq R$. As $T\supseteq R$ is trivial, we prove that $T=R$ and so ${\mathrm{Hom}}_R(I,I)=R$. \end{proof}

3. Let $R$ be an integral domain. Let $S$ be a nonempty multiplicatively closed subset of $R\backslash \{0\}$.

• If $R$ is a principal ideal domain, so is $S^{-1}R$.
• If $R$ is a Dedekind domain, so is $S^{-1}R$.

\begin{proof} Define $j:R\hookrightarrow S^{-1}R,r\mapsto r/1$, which is an embedding. For any ideal $J\subseteq S^{-1}R$, we have that

$$I=j^{-1}(J)=\{r\in R:j(r)\in J\}$$

is an ideal of $R$. For any $r/s\in J$, there is $r/1\in J$ and so $r\in I$. Thus $S^{-1}I\supseteq J$. On the other hand, for any $x\in S^{-1}I$, $x$ can be written as $x=r^{\prime }/s^{\prime }$ with $r^{\prime }\in I,s^{\prime }\in S$. Then $r^{\prime }/s^{\prime }\in J$ by $r^{\prime }\in I$ and $r^{\prime }/s^{\prime }=(r^{\prime }/1)(1/s^{\prime })$. Now we proved $S^{-1}I=J$, that is, each ideal $J$ of $S^{-1}R$ can be written as $J=S^{-1}I$ for some ideal $I\subseteq R$.

i) For any ideal $J\subseteq S^{-1}R$, $J=S^{-1}I$ for some ideal $I$ of $R$. Since $R$ is a principal ideal domain, we know $I$ is free. Note that $xy-yx=0$ for any $x,y\in I$, so $\mathrm{rank}I⩽1$. If $I=(0)$, then $J=S^{-1}I=(0)$ is also free. If $I\ne (0)$, then $I=(a)$ for some $a\in I$. In this case, $S^{-1}J=(a/1)$ is also free. By the arbitrary of $J$, $S^{-1}R$ is a principal ideal domain.

ii) For any ideal $J\subseteq S^{-1}R$, $J=S^{-1}I$ for some ideal $I$ of $R$. Since $R$ is a Dedekind ring, $I$ is finitely generated projective, that is, $I$ is a direct summand of $R^n$ for some $n$. Assume that $R^n=I\oplus I^{\prime }$. Note that $J=S^{-1}R\otimes I$ and $S^{-1}R$ is flat, so we have

$$(S^{-1}R{)}^n=S^{-1}{R}^n=S^{-1}R\otimes R^n=(S^{-1}R\otimes I)\oplus (S^{-1}R\otimes I^{\prime })=J\oplus J^{\prime }$$

and $J$ is a direct summand of $(S^{-1}R{)}^n$. So $J$ is also a finitely generated projective ideal. Now we finish the proof. \end{proof}

4. Let $R$ be a Dedekind domain and $I$ be a fractional ideal. There exist integral ideals $J$ and $K$ such that $I=J{K}^{-1}$.

\begin{proof} Since $I$ is a fractional ideal, there exists $\lambda \in R$ such that $\lambda I\subseteq R$. Let $J=\lambda I$ and $K=\lambda R$. Then $K^{-1}=(1/\lambda )R$ and $J{K}^{-1}=\lambda I\cdot (1/\lambda )R=I$. Now we finish the proof. \end{proof}

5. Let $R$ be a Dedekind domain. We say that two fractional ideals $I,J$ are coprime or relatively prime if $I+J=R$. More generally, given a family $I_1,\dots ,I_r$ of fractional ideals, we say that $I_1,\dots ,I_r$ are relatively prime if $I_1+\cdots +I_r=R$.

• (a) If $I$ and $J$ are coprime, they are both integral.
• (b) Let $I_1,\dots ,I_r$ be a family of integral ideals. The following assertions are equivalent:
  • (i) $I_1,\dots ,I_r$ are relatively prime,
  • (ii) for each prime ideal $\mathfrak{p}$ of $R$, $\min \left\{v_{\mathfrak{p}}{\left(I_i\right)}_{1\le i\le r}\right\}=0$,
  • (iii) no prime ideal of $R$ divides (i.e., contains) all the $I_i$ ‘s.

\begin{proof} a) If $I$ and $J$ are coprime, then $I+J=R$ and so $I,J\subseteq R$. Thus they are both integral.

b) i)→ii) If $I_1,\cdots ,I_r$ are relatively prime, then $I_1+\cdots +I_r=R$ and so $I_j$ are all integral. It deduces that $I_j={\prod }_p{p}^{{\nu }_p(I_j)}$ with ${\nu }_p(I_j)⩾0$. If there exists $\mathfrak{p}$ such that $\min \left\{v_{\mathfrak{p}}{\left(I_i\right)}_{1\le i\le r}\right\}>0$, then $I_j\subseteq \mathfrak{p}$ for all $j$ and so $I_1+\cdots +I_r\subseteq \mathfrak{p}$, which is a contradiction. Therefore, for each prime ideal $\mathfrak{p}$ of $R$, $\min \left\{v_{\mathfrak{p}}{\left(I_i\right)}_{1\le i\le r}\right\}=0$.

ii)→iii) If there exists a prime ideal $\mathfrak{p}$ such that $\mathfrak{p}$ divides all $I_i$, then ${\nu }_{\mathfrak{p}}(I_i)>0$ for all $i$, leading to a contradiction.

iii)→i) If $I_1+\cdots +I_r⊊R$, then it can be contained in some maximal ideal $m$. As $m$ is prime and $I_j\subseteq m$, we have $m$ divides $I_j$, which is impossible. \end{proof}

6. The following holds for a valuation $v$ of $K$ :

• (i) $v(\pm 1)=0$
• (ii) $v(a)=v(-a)$
• (iii) $v\left(a^{-1}\right)=-v(a)$
• (iv) $v(b)<v(a)\Rightarrow v(a+b)=v(b)$.

\begin{proof} i) Since $\nu (ab)=\nu (a)+\nu (b)$ and $K$ is a field, we have $\nu (1)=\nu (1\times 1)=\nu (1)+\nu (1)$ and so $\nu (1)=0$. Then $\nu (1)=\nu ((-1){)}^2=\nu (-1)+\nu (-1)$ yields $\nu (-1)=0$.

ii) Note that $\nu (-a)=\nu (-1)+\nu (a)=\nu (a)$.

iii) Note that $0=\nu (1)=\nu (a\cdot a^{-1})=\nu (a)+\nu (a^{-1})$ and so $v\left(a^{-1}\right)=-v(a)$.

iv) Since $\nu (a+b)⩾\min \{\nu (a),\nu (b)\}=\nu (b)$, it suffices to show $\nu (a+b)>\nu (b)$ is impossible. Otherwise assume $\nu (a+b)>\nu (b)$ holds, then $\nu (b)=\nu (a+b+(-a))⩾\min \{\nu (a+b),\nu (a)\}>\nu (b)$, which is impossible. Hence $\nu (a+b)=\nu (b)$. \end{proof}