# Ring with finitely many maximal ideals whose localizations are Noetherian injects into $\bigoplus_{\mathfrak{m}}R_{\mathfrak{m}}$

Let $R$ be a commutative ring with finitely many maximal ideals. If $R_{\mathfrak{m}}$ is Noetherian for all maximal ideals $\mathfrak{m}\unlhd R$, prove that the product of localization maps $R\rightarrow \bigoplus_{\mathfrak{m}}R_{\mathfrak{m}}$ is an embedding.

The wording is a little unusual (?), but I think the point is to show that the stated map is injective. I'm fairly certain the map in question is $\varphi: R\rightarrow \bigoplus_{i=1}^n R_{\mathfrak{m_i}}$, $\varphi(r)= (\varphi_i(r))$, where $\varphi_i:R \rightarrow R_{\mathfrak{m}_i}$, $r\mapsto r/1$. Here are my rather fruitless attempts and thoughts.

Suppose $\varphi$ is not injective. Then there is some $r\neq 0$ for which $\varphi(r)=0$, hence $r/1=0$ in $R_{\mathfrak{m}_i}$ for all $i$. By definition, we then have $u_i\in R\backslash {\mathfrak{m}_i}$ such that $u_ir=0$ for all $i$. If $r$ is a unit then $r\in R\backslash {\mathfrak{m_i}}$ for all $i$ so $r/1$ is a unit in $R_{\mathfrak{m}_i}$, a contradiction since $r/1=0$. If $r$ is not a unit then let ${\mathfrak{m}_1},\dots,{\mathfrak{m}_k}$ be all the maximal ideals in $R$ which contain $r$, so $r\in \cap_{i=1}^k {\mathfrak{m}_i}$. Then $(r/{u_i})\subsetneq {\mathfrak{m}_i}_{{\mathfrak{m}_i}}$ for all $i=1,\dots,k$. Since $R_{\mathfrak{m}_i}$ is Noetherian, each maximal ideal ${\mathfrak{m}_i}_{{\mathfrak{m}_i}}\unlhd R_{{\mathfrak{m}_i}}$ is finitely generated...

At this point things were just getting unnecessarily messy, with no obvious contradiction, and after messing around with this for some time, I feel like I must be going down the wrong route. I'm wondering if there is a different characterization of Noetherian that I should employ? I can't seem to figure out how to use the ascending chain condition in this case. I also thought about trying to somehow incorporate the fact that $R=\cap_{\mathfrak{m}} R_{\mathfrak{m}}$?

I'll note that I think it's sufficient to prove that $R\rightarrow R_{{\mathfrak{m}_i}}$ is injective for at least one $i$, since in this case if $\varphi(r)=0$ then $\varphi_i(r)=0$ which would imply $r=0$, but I can't seem to accomplish even this much.

Any help would be much appreciated.

UPDATE: It appears that neither assumption (Noetherian, finitely many maximal ideals) is needed for the result. See the comments.

• If $x$ is in the kernel, then $x\in \displaystyle\bigcap_{m\triangleleft R}m$ (because $u_ir =0$, hence $u_i \in m_i$ or $r\in m_i$) – Max Jul 2 '18 at 21:01
• What is the annihilator of $r$? From what you write, it contains all the $u_i$ and hence the ideal generated by them. Can this ideal be a proper ideal? – Mohan Jul 2 '18 at 21:05
• @Mohan, this makes sense! So essentially if $(u_1,\dots,u_n)\subsetneq R$ then $(u_1,\dots,u_n)\subseteq \mathfrak{m}_i$ for some $i$, in which case $u_i\in \mathfrak{m}_i$, a contradiction. So $Ann(r)=R$, so $r$ has to be zero. Very nice. I guess the Noetherian property is not needed for the statement then? I suppose this does make sense - the question formed part of an (oddly worded) past exam question, so I guess the Noetherian property is needed only lated on. Thank you! – Arbutus Jul 2 '18 at 21:20

The result does not need the assumptions of finitely many ideals (if you change direct sum by product) and the localizations to be noetherian. In general it is true that the kernel of $R\rightarrow \prod_{\mathfrak{m}}R_{\mathfrak{m}}$ is always zero. One possible argument uses the Spectrum: if $x$ is in the kernel, then its annihilator is an ideal $I$ such that $V(I)$ has no closed point, hence $I$ is the unit ideal, hence $x=0$.