# Suppose that $R$ is a one-dimensional normal Noetherian local ring. Then the maximal ideal $m_R$ is principal

Theorem 11.2 (Matsumura's Commutative Ring theory) gives us equivalent conditions for a ring $R$ to be considered a DVR. I was stuck while reading the proof of $(4) \implies (3)$,

(3) $R$ is a Noetherian local ring, dim $R > 0$ and the maximal ideal $m_R$ is principal;

(4) $R$ is a one-dimensional normal Noetherian local ring

Let $K$ be the field of fractions of $R$ and let $m$ be the maximal ideal of $R$. Let $x \in m - m^2$ and let $y \in R$ such that $xR : y = m$. We have $a = yx^{-1}$.

Question 1

Suppose that $am \subset m$. How does this imply that $a$ is integral over $R$ by Theorem 2.1?

Theorem 2.1 Suppose that $M$ is an $A$-module generated by $n$ elements, and that $\phi \in$ Hom$_A (M,M)$; let $I$ be an ideal of $A$ such that $\phi(M) \subset IM$. Then there is a relation of the form $$\phi^n + b_1 \phi^{n-1} + ... + b_{n-1} \phi + b_n = 0$$

Since $m$ is an ideal of $R$, it is definitely an $R$-module. It is finitely generated, since $R$ is Noetherian. We can define $\phi : m \rightarrow m$ by $r \mapsto ar$ for all $r \in m$. But then, this only gives $$a^nr + b_1 a^{n-1}r + ... + b_{n-1} ar + b_n = 0$$

We cannot get $$a^n + b_1 a^{n-1} + ... + b_{n-1} a + b_n = 0$$

Since $1 \not\in m$. Then how does this theorem imply that $a$ is integral over $R$?

Question 2

The definition of "normal" for Matsumura is that for any prime ideal "p" of a ring $R$, we have that $R_p$ is integrally closed. How does this imply that $R$ itself is integrally closed?

Thanks in advance

• For the question 2, you should know that being integrally closed is a local property on $\textrm{Spec}(A)$. I tried to find a reference for it (as it is well-known) and came across this : proofwiki.org/wiki/Integrally_Closed_is_Local_Property which made me laugh SO MUCH. ;-) – ujsgeyrr1f0d0d0r0h1h0j0j_juj Mar 4 '15 at 12:45
• @RobertGreen Looks like the margin wasn't big enough to fit the full proof? ;) – rschwieb Mar 4 '15 at 12:52
• @rschwieb Fermat was a mathematician, despite of the margin situation. ;-) Here we had someone who IMO realized that (s)he didn't know how to prove (3)=> (1) or (2) only when (s)he arrived to the point where (s)he had to show it... – ujsgeyrr1f0d0d0r0h1h0j0j_juj Mar 4 '15 at 12:57
• Ok, I will answer to prove question at least. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Mar 4 '15 at 13:02

## 1 Answer

For your question 1, you made a mistake, for theorem 2.1 will imply that $a^nr + b_1 a^{n-1} + ... + b_{n-1} ar + b_n r = 0$, there's an $r$ in front of $b_n$, because $\Phi^0 = \textrm{Id}_M$. From there, do you see how to conlude ?

For the question 2, here you go :

Lemma 1. Let $\rho : A \to B$ be a ring morphism making $B$ an $A$ algebra and $S$ be a multiplicatively closed subset of $A$. Let $C$ be the integral closure of $A$ in $B$. Then (the image of) $S^{-1} C$ is the integral closure of (the image of) $S^{-1}A$ in $S^{-1}B$. One says that localization commutes with integral closure.

Proof. First of all that (the image of) $S^{-1} C$ is integral over $S^{-1}A$ (I drop the image of's from here) is easy. Now let $x\in S^{-1}B$ a root of a monic polynomial $P$ in $S^{-1}A[T]$. Let $s\in S$ be the product of "denominators" of $x$ and coefficients of $P$. Then $sx$ verifies a monic relation over $A$, so that it is in $C$ by definition of $C$, so that $x$ is in $S^{-1}C$. $\square$

Proposition. Let $A$ be a domain. The following are equivalent :

1. $A$ is integrally closed ;
2. For all $\mathfrak{p}\in\textrm{Spec}(A)$ the localization $A_{\mathfrak{p}}$ is integrally closed ;
3. For all $\mathfrak{p}\in\textrm{Spec}(A)$ that is maximal, the localization $A_{\mathfrak{p}}$ is integrally closed.

Proof. As $A$ is a domain, everything (localizations $A_{\mathfrak{p}}$'s included) will take place in $K = \textrm{Frac}(A)$. Then $\textrm{Frac}(A_{\mathfrak{p}}) = K$ for all $\mathfrak{p}\in\textrm{Spec}(A)$, so that the lemma shows that (1) implies (2). No comment one (2) implies (3). Now suppose (3). Note $C$ the integral closure of $A$ in $K$. By the lemma, the morphism (the inclusion) $C\to A$ induces an isomorphism $C_{\mathfrak{p}}\to A_{\mathfrak{p}}$ for $\mathfrak{p}\in\textrm{Spec}(A)$ that is maximal. The conclusion results from the following lemma. $\square$

Lemma 2. Let $\varphi : M\to N$ be an morphisms of $A$-modules. The following are equivalent :

1. $\varphi$ is an isomorphism ;
2. $\varphi_{\mathfrak{p}} : M_{\mathfrak{p}}\to N_{\mathfrak{p}}$ for all $\mathfrak{p}\in\textrm{Spec}(A)$ ;
3. $\varphi_{\mathfrak{p}} : M_{\mathfrak{p}}\to N_{\mathfrak{p}}$ for all $\mathfrak{p}\in\textrm{Spec}(A)$ that is maximal.

Proof. Results immediately from the more general follwoing lemma. $\square.$

Lemma 3. Let $A$ be a ring and $S$ be a multiplicatively closed subset of $A$. Then $M\mapsto S^{-1}M$ is an exact functor.

Proof. $S^{-1} M \simeq M \otimes_A S^{-1} A$ functorialy so that $M\mapsto S^{-1}M$ is already right exact as "tensoring is a right exact functor". To conclude, one just has to show that $M\mapsto S^{-1}M$ sends injective maps to injective maps. So let $\varphi : M\to N$ be injective and let $\frac{m}{s}$ be in the kernel of $S^{-1}\varphi$, which means that $\frac{\varphi(m)}{\varphi(s)} = 0$ in $S^{-1}N$, which means that you can find an $s'\in S$ such that $s'\varphi(m) = 0$, which means $\varphi(s'm) = 0$, and $\varphi$'s injectivity implies that $s'm = 0$, which implies that $\frac{m}{s} = \frac{1}{ss'} s'm = 0$ in $S^{-1} M$. $\square$

Remark. The proposition answers to your question 2.