Show that $R = \bigcap_mR_m$ whenever $R$ is an integral domain

Question

Show that $R = \bigcap_mR_m$ whenever $R$ is an integral domain, where the intersection is indexed by all maximal ideals of $R$.

$R \subset \bigcap_mR_m$ is clear since $R \subset R_m$ for all $m$ maximal because $R$ is an integral domain which tells us that $R \twoheadrightarrow R_m $ is injective.

Now I am not sure how the other directions works. My idea:

Choose $x \in(\bigcap_mR_m)\setminus R$. Then for all $m$ maximal x must be a unit in $R_m$ since otherwise $x \in m$ which is a contradiction to $x \notin R$. But for $x\notin R$ to be a unit in $R_m $ for all $m$ maximal, $x^{-1} \in R \setminus m$ for all $m$ maximal. But that means $x^{-1}$ is a unit in R, because otherwise there exist a maximal ideal containing m. We conclude that $x$ must already be in $R$ which is a contradiction to the coice of x. Therefore $(\bigcap_mR_m)\setminus R = \emptyset$. Thus the equality $R = \bigcap_mR_m$ holds, whenever $R$ is an integral domain.

Is this correct?

do you know a correct proof? and could you tell me whats wrong with that proof? — CookieMaster, Jul 13 '15 at 13:33
its the localisation of $R$ at the maximal ideal $m$. Thus $R_m=(R \setminus m)^{-1}R$. So $R_m$ is a local ring with maximal ideal $m$. — CookieMaster, Jul 13 '15 at 13:59
Hint - Suppose $x$ is an element of the intersection. Consider the ideal $I = { r \in R \mid r\cdot x \in R}$. Is it contained in a maximal ideal? — peter a g, Jul 13 '15 at 14:24

score 1 · Accepted Answer · answered Jul 13 '15 at 14:27

1

The first claim: "$x$ must be a unit in $R_m$ since otherwise $x\in m$..." it's wrong. What means that $x$ is not a unit in $R_m$? Obviously $x\in mR_m$ which is equivalent to $\exists s\in R-m$ such that $sx\in m$, and this is not exactly what you get.

For the proof you start with $x\in R_m$ for all $m$, so there is $s_m\in R-m$ such that $s_mx\in R$, and then notice that the ideal generated by all the $s_m$ equals $R$.

answered Jul 13 '15 at 14:27

user26857

53,190

notice that the ideal generated by all the $s_m$ equals $R$. Why is that so? – CookieMaster Jul 13 '15 at 14:58
@CookieMaster If not this ideal is contained in a maximal ideal, say $n$. Then what about $s_n$? – user26857 Jul 13 '15 at 15:07
ah I see. So $s_n$ would be element of $n$ which is a contradiction of the choice of $s_n$. Thus the ideal of generated by all $s_n$ equals R. This leads to $Rx \in R$ especially $1x=x \in R$ is this conclusion correct? – CookieMaster Jul 13 '15 at 17:23
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@CookieMaster Correct. – user26857 Jul 13 '15 at 20:21

Show that $R = \bigcap_mR_m$ whenever $R$ is an integral domain

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