|
|
本帖最后由 Czhang271828 于 2023-3-6 22:37 编辑 Proposition 1 Let $R$ be any commutative ring. $I$ and $J$ are ideals of $R$. If $R/I\simeq R/J$ as $R$-modules with the standard $R$-module structure, then $I=J$. In other words, isomorphic modules have the same annihilators.
Proof of Prop 1 For isomorphic $R$-modules $\varphi: R/I\simeq R/J$ and arbitrary $x\in J$, we have that
$$
\varphi (x+I)=x\varphi(1+I)=x(1+J)=0+J.
$$
Thus $x+I\in \mathrm{ker}(\varphi)$. Since $\varphi $ is isomorphism, we have that $x\in I$. It yields that $J\subseteq I$. Similarly, $I\subseteq J$, thus we have $I=J$.
Proposition 2 Let $R$ be any commutative ring. $I$ and $J$ are ideals of $R$. The ring isomorphism $\varphi :R/I\simeq R/J$ no longer implies $I\simeq J$. Here $\varphi$ preserves $\{+,\cdot ,0,1\}$. In other words, isomorphic quotient rings have the isomorphic annihilators.
Proof of Prop 2 A counterexample: (1) set $R:=\mathcal C^0(\mathbb R,\mathbb R)$, $R/(x)\simeq R/(x-1)$ by translation. (2) For any commutative ring $A$, $(A\times A)/(0\times A)\simeq (A\times A)/(A\times 0)$.
Lemma 0 This lemma may help with your initial question. Set $R:=\mathcal C^0(\mathbb R,\mathbb R)$. If we have principal ideals $(f)=(g)$, then there exists invertible $h\in R$ such that $fh=g$.
Proof of Lemma 0 There exsits $h,h'\in R$ such that $fh=g$ and $gh'=f$. Since $fhh'=f$, $gh'h=g$, we have that $hh'=1$ on $\overline{\mathrm{supp}(f)}=\overline{\mathrm{supp}(g)}$, which is the disjoint union of closed intervals.
Set $\overline{\mathrm{supp}(f)}=\mathbb R\setminus (0,1)$ without the loss of generality. Since $h$ and $h'$ are uniformly continuous on $[-1,0]$ and $[1,2]$, they are bounded. Let $h_-$ (resp. $h_+$) be the restriction of $h$ on $(-\infty,0]$ (resp. $[1,+\infty)$), so is $h'_-$ (resp. $h'_+$). We proved that
$$
S=\{h_-(0),h_-'(0),h_+(1),h_+'(1)\}
$$
is bounded. Since $h_-(0)h'_-(0)=h_+(1)h'_+(1)=1$, $0\notin S$. Define
$$
\tilde h(x)=\left\{\begin{align*}
&|h(x)|&&x\in \overline{\mathrm{supp}(f)},\\
&|h(0)|(1-x)+|h(1)|x&&x\in(0,1).
\end{align*}\right.
$$
Set $\tilde h'(x)=1/\tilde h(x)$. Here $\tilde h(x)$ and $\tilde h'(x)$ are non-vanishing and thus invertible. Then we have either
$$
f\tilde h=g\quad g\tilde h'=f,
$$
or
$$
(f-2f_+)\tilde h=(g-2g_+)\quad (g-2g_+)\tilde h'=(f-2f_+).
$$
Here we have the same ideals $(f)=(f-2f_+)$ and $(g)=(g-2g_+)$.
The hidden post may contain ambiguous expression, factual errors, or dogmatic opinions. **Question.** Set $R:=$ the ring of continuous functions from $\mathbb R$ to $\mathbb R$.
We say $f,g\in R$ are equivalent, whenever there exists invertible $h_1,h_2\in R$ such that $h_1 fh_2=g$.
Let $(\tau)$ be the principal ideal generated by $\tau\in R$.
We shall prove that $R/(f)$ and $R/(g)$ are isomorphic $R$-modules whenever $f$ and $g$ are almost equivalent.
**Proof.** When $f$ and $g$ are equivalent, there exists invertible $r\in R$ such that $f=gr$. Then we have the following isomorphism
$$
\varphi :R/(f)\to R/(g)\quad \overline{h}\mapsto \overline{hr}.
$$
To see this,
1. $\varphi(\overline 0)=\varphi(\overline {f})=\overline{fr}=\overline g=\overline 0$, thus $\varphi$ preserves $0$.
2. $\varphi$ preserves the structure of additive group.
3. $\varphi (h\cdot \overline p)=\overline {hpr}=h\overline {pr}=h\cdot \varphi(\overline p)$.
Clearly, $\varphi$ is an homomorphism with inverse
$$
\varphi^{-1} :R/(g)\to R/(f)\quad \overline{h}\mapsto \overline{hr^{-1}}.
$$
Thus is an isomorphism.
We say $f$ and $g$ are almost equivalent, the sets of zeros differ on finite points. Here we only need to prove $R/(x)\simeq R/(1)$ and use induction. Define
$$
\varphi:R/(1)\to R/(x),\quad \overline{f(x)-f(\epsilon)}\mapsto \overline{f(x)-x\epsilon^{-1}f(\epsilon)}.
$$
Here we fix $\epsilon$ such that $f(\epsilon )\neq 0\neq g(\epsilon)$. We see $\varphi$ is a bijection, since both $f(x)-f(\epsilon)$ and $f(x)-x\epsilon^{-1}f(\epsilon)$ are unique representative element in the corresponding equivalence classes.
The converse part is similar.
Here a new question is raised, do we have $\overline{\mathrm{supp}}(f)=\overline{\mathrm{supp}}(g)$ whenever $R/(f)=R/(g)$? I believe the answer is Yes. |
|
Czhang271828
发表于 2023-3-4 13:47
