2.7 Special kinds of rings

Local ring

def: Local ring

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal {C}} \\ &\exists! \mathfrak m \in \mathfrak M_I(A) \\ \hline \\ &A \in \mathcal R^\mathcal L - \text{local ring} \end{align*}$$

Example: Field is a local ring

If $F \in \mathcal F$, we know that the only ideals of $F$ are $\{0\}$ and $F$, so the only maximal ideal is $\{0\}$

Example: $\Z_4$

The only non-trivial ideal in $\Z_4$ is $2\Z_4$, hence it's the only maximal ideal.

Proposition 2.7.1: Local ring criterion

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I := A \setminus A^* \\ \hline \\ &I \lhd_R A \implies A \in R^{\mathcal L}, I \in \mathfrak M_I(A) \end{align*}$$

Proof

Every proper ideal $J \lhd_R A, J \ne A$ should have only elements of $A/A^*$ by $(2.4.6)$. So $J \subseteq I$.

$\square$

Proposition 2.7.2: Generators for module over local ring

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal L} \\ &M \in \mathcal M_A \\ &\mathfrak m \in \mathfrak M_I(A) \\ &\phi: M \to M/\mathfrak mM, x \to x+\mathfrak mM \\ &X:=\{x_1, \ldots, x_n\} \subseteq M, \phi(X) \in \mathfrak B_{M_{A/\mathfrak m}}(M/\mathfrak mM) \\ \hline \\ &\begin{align*} &M/\mathfrak mM \in \mathcal M_{A/\mathfrak m} \hspace{0.5cm} \tag{a}\\ &\lang Y \rang_M = M \implies \lang \phi(Y) \rang_{M_{A/\mathfrak m}} = M/\mathfrak mM \hspace{0.5cm} \tag{b}\\ & \lang X \rang_M = M \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

a.

Note that $\forall x \in \mathfrak m:xM \subseteq \mathfrak mM \implies xM/\mathfrak mM=\{0\}$. Now $(x+\mathfrak m)\cdot(m+\mathfrak mM):= xm+\mathfrak mM$. Assume $x-x'\in \mathfrak m, m-m' \in \mathfrak mM$:

$$x'm' + \mathfrak mM = (x+\mathfrak m)(m+\mathfrak mM)+\mathfrak mM= \\ xm+\mathfrak mm+x\mathfrak mM+\mathfrak m^2M+\mathfrak mM=xm+\mathfrak mM$$

So scalar multiplication is well-defined. Now all module properties follows from the fact that $M \in \mathcal M_A$.

b.

Consider some $x \in M, Y = \{y_1, \ldots, y_n\}$. We know that $x = a_1y_1 + \ldots + a_ny_n$, so $\phi(x)=a_1y_1 + \ldots + a_ny_n + \mathfrak mM=(a_1+\mathfrak m)(y_1 + \mathfrak mM)+\ldots +(a_n+\mathfrak m)(y_n + \mathfrak mM)$. Since $\phi(M)=M/\mathfrak mM$, the set $\{y_1 + \mathfrak mM, \ldots, y_n+\mathfrak mM\}$ generates $M/\mathfrak mM$ as $A/\mathfrak m$-module.

c.

Define $N:=\lang X \rang_M$. Consider a natural homomorphism $\psi: N \to M \to M/\mathfrak mM$. Since $\phi(X) \in \mathfrak B_M(M/\mathfrak mM)$ we have $\psi(N)=\psi(\lang X \rang_M) = M/\mathfrak mM$.

Consider some $m \in M$ and $m+\mathfrak mM \in M/\mathfrak mM$. Since $\psi(N)=M/\mathfrak mM$ we know that $\exists n \in N: n+\mathfrak mM = \psi(n) = m + \mathfrak mM \implies m - n \in \mathfrak mM$. In other words $\forall m \in M: \exists n \in N, i \in \mathfrak m, m_2 \in M: m = n+im_2$ or $N+\mathfrak mM \supseteq M$.

On the other hand $N\subseteq M, \mathfrak mM \subseteq M \implies N+\mathfrak mM \subseteq M$ so:

$$N+\mathfrak mM = M$$

Now we can apply $(2.6.11)$ to finish the proof

$\square$

Radicals

def: Radical ideal

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \lhd_R A \\ \hline \\ &\sqrt I:=\{x \in A: \exists n \in \N: x^n \in I\} \end{align*}$$

Proposition 2.7.3: Radical ideal is ideal

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \lhd_R A \\ \hline \\ &\sqrt I \lhd_R A \end{align*}$$

Proof

First, note that $I \subseteq \sqrt I \implies I \ne \empty$.

Second consider some $x, y \in \sqrt I$. We know that $\exists a, b \in I: a = x^m, b = y^n$. If we examine $(x-y)^{m+n}$, we can see that it will be a sum of $c_ix^jy^k, j+k = m+n$ and $c_i \in A$. It's easy to see that either $j \ge m$ or $k \ge n$, so $c_ix^jy^k \in I$. Thus $x-y \in \sqrt I$.

Finally $\forall a \in A, x \in \sqrt I: \exists y \in I: x^n=y$ we have $(ax)^n = a^nx^n \in I$, so $ax \in \sqrt I$.

$\square$

Proposition 2.7.4: Nilradical is an intersection of all prime ideals

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ \hline \\ &\sqrt{\{0\}}=\bigcap_{P \in \mathfrak P_I(A)}P \end{align*}$$

Proof

Let's prove $\sqrt{\{0\}} \subseteq \bigcap_{P \in \mathfrak P_I(A)}P$

Consider arbitrary $a \in \sqrt{\{0\}}, P \in \mathfrak P_I(A)$. We want to prove that $a \in P$

$$a \in \sqrt{\{0\}} \implies \exists n \in \N: a^n = 0 \in P \implies \\ a^{n-1}\cdot a \in P \implies (a \in P) \vee (a^{n-1} \in P)$$

If $a \in P$ we're done, $a^{n-1} \in P$ otherwise, we're done by induction.

Now let's prove $\sqrt{\{0\}} \supseteq \bigcap_{P \in \mathfrak P_I(A)}P$

Assume $a \notin \sqrt {\{0\}}$. Consider the set of ideals $\Sigma:= \{I \lhd_R A: \forall n > 0: a^n \notin I\}$. Obviously $\{0\} \in \Sigma$. Similarly to $(2.3.5)$ we can prove that $\Sigma$ has a maximal element by inclusion that we denote by $\mathfrak m$.

Now let's prove that $\mathfrak m \in \mathfrak P_I(A)$. Consider $x, y \notin \mathfrak m$. Then the ideals $\mathfrak m+\lang x \rang_I$, $\mathfrak m+\lang y \rang_I$ are strict supersets of $\mathfrak m$ and so doesn't belong to $\Sigma$.

$$\exists n, m \in \N: a^m \in \mathfrak m + \lang x \rang_I, a^n \in \mathfrak m + \lang y \rang_I \implies \\ a^{m+n} \in \mathfrak m + \lang xy \rang_I \implies \mathfrak m + \lang xy \rang_I \notin \Sigma \implies \\ xy \notin \mathfrak m$$

So we proved $(x \notin \mathfrak m) \wedge (y \notin \mathfrak m) \implies xy \notin \mathfrak m$. Thus

$$xy \in \mathfrak m \implies (x \in \mathfrak m) \vee (y \in \mathfrak m)$$

And we proved that $\mathfrak m \in \mathfrak P_I(A)$. Next, $\mathfrak m \in \Sigma \implies \forall n > 0: a^n \notin \mathfrak m \implies a \notin \mathfrak m \implies a \notin \bigcap_{P \in \mathfrak P_I(A)}P$. So we proved $a \notin \sqrt {\{0\}} \implies a \notin \bigcap_{P \in \mathfrak P_I(A)}P$ or equivalently

$$a \in \bigcap_{P \in \mathfrak P_I(A)}P \implies a \in \sqrt{\{0\}}$$

$\square$

Corrolary 2.7.5: The radical of an ideal is the intersection of all prime ideals containing it

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \lhd_R A \\ \hline \\ &\sqrt{I}=\bigcap_{P \in \mathfrak P_I(A), I \subseteq P}P \end{align*}$$

Proof

We want to consider factor ring $A/I$ and natural homomorphism $\phi$ into it. Notice that:

$$r^n \in I \iff r^n+I = I \iff (r+I)^n = I$$

So under $(2.3.12)$ $\sqrt I \mapsto \sqrt {\{0 + I\}}$. We know by $(2.7.4)$ that the latter is the intersection of all prime ideals in $A/I$. And by $(2.3.12)$ these ideals are mapped to prime ideals in $A$ containing $I$.

$\square$

Proposition 2.7.6: Radical power inclusion

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal N} \\ &I \lhd_R N \\ \hline \\ &\exists m \in \N: (\sqrt I)^m \subseteq I \end{align*}$$

Proof

Since $\sqrt I$ is finitely generated, assume $\sqrt I=\lang a_1, \ldots, a_n \rang_I$. Then $\forall i: \exists n_i \in \N: a_i^{n_i} \in I$. Define $m:=\sum_{i=1}^k(n_i-1)+1$. Then $(\sqrt I)^m$ is generated by the products $a_1^{r_1}\cdot \ldots \cdot a_n^{r_n}$ with $\sum r_i = m$. By the definition of $m$, $\exists r_i \ge n_i$, so each product will be in $I$. Thus $(\sqrt I)^m \subseteq I$.

$\square$

Ideal quotients

def: Ideal quotient

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I, J \lhd_R A \\ &a \in A \\ \hline \\ &(I:J):= \{x \in A: xJ \subseteq I\} \\ &(I:a) := (I:\lang a \rang_I) \end{align*}$$

Proposition 2.7.7: Ideal quotient is ideal

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I, J \lhd_R A \\ \hline \\ &(I:J) \lhd_R A \end{align*}$$

Proof

$$0 \in (I:J) \implies (I:J) \ne \empty$$ $$\forall x, y \in (I:J), z \in J: (x - y)z = \underbrace{xz}_{\in I}-\underbrace{yz}_{\in I} \in I \implies \\ x-y \in (I:J)$$ $$\forall x \in (I:J), z \in J, k \in A: k\underbrace{xz}_{\in I} \in I \implies kx \in (I:J)$$

$\square$

Primary ideals

def: Primary ideal

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \lhd_R A, I \ne A \\ &\forall x, y \in A: xy \in I \implies \exists n \in \N: (x \in I) \vee (y^n \in I) \\ \hline \\ &I \in \sqrt \mathfrak P_I(A) \end{align*}$$

Proposition 2.7.8: Radical of a primary is prime

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \in \sqrt{\mathfrak P}_I(A) \\ \hline \\ &\begin{align*} &\sqrt I \in \mathfrak P_I(A) \tag{a}\\ &\forall J \in \mathfrak P_I(A), I \subseteq J \implies \sqrt I \subseteq J \hspace{1cm} \tag{b} \\ \end{align*} \end{align*}$$

Proof

a.

$$\forall x, y \in A: xy \in \sqrt I \implies \exists m \in \N: (xy)^m \in I \overset{I \in \sqrt{\mathfrak P}_I(A)}\implies \\ (x^m \in I) \vee (y^{nm} \in I) \implies (x \in \sqrt I) \vee (y \in \sqrt I)$$

b.

Follows from $(2.7.5)$

$\square$

Proposition 2.7.9: Ideal with maximal radical is primary

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &I \lhd_R A \\ &\sqrt I \in \mathfrak M_I(A) \\ \hline \\ &I \in \sqrt \mathfrak P_I(A) \end{align*}$$

Proof

Сonsider a factor ring $A/I$. By $(2.3.12)$ $\sqrt I \mapsto \sqrt {\{0 + I\}}$. Since $\sqrt I \in \mathfrak M_I(A)$, and there are no ideals containing $\sqrt I$, then there are no ideals containing $\sqrt {\{0 + I\}}$ in $A/I$. Moreover, by $(2.7.5)$ $\sqrt {\{0 + I\}}$ is the intersection of all prime ideals, so it's the only prime ideal in $A/I$ and hence maximal.

Since $A/I$ is local, each element of $A/I$ is either in $\sqrt {\{0 + I\}}$ or a unit. Each zero divisor would be in $\sqrt {\{0 + I\}}$, otherwise if $xu=0 \implies x uu^{-1}=0 \implies x = 0$.

$$\forall x, y \in A: xy \in I \implies xy+I = I \implies \\ (x+I)(y+I)=I \implies y+I \in \sqrt{\{0+I\}} \implies \\ \exists n \in \N: (y + I)^n = y^n + I = I \implies y^n \in I$$

$\square$

Proposition 2.7.10: m-primary ideal criteria

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal N} \\ &I \lhd_R N \\ &\mathfrak m \in \mathfrak M_I(N) \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} & \sqrt I = \mathfrak m, I\in \sqrt{\mathfrak P}_I(N) \hspace{1cm} \tag{a}\\ & \sqrt I = \mathfrak m \hspace{1cm} \tag{b}\\ & \exists n \in \N: \mathfrak m^n \subseteq I \subseteq \mathfrak m \tag{c}\\ \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

Obvious

$(b) \implies (a)$

Follows from $(2.7.9)$

$(b) \implies (c)$

Follows from $(2.7.6)$

$(c) \implies (b)$

$\sqrt {\mathfrak m^n} = \{x \in A: \exists m: x^m \in \mathfrak m^n\}$. Obviously $\mathfrak m \subseteq \sqrt {\mathfrak m^n}$ if we assume $m:=n$. On the other hand $\mathfrak m^n \subseteq \mathfrak m \implies \sqrt{\mathfrak m^n} \subseteq \sqrt \mathfrak m \overset{(2.7.5)}\subseteq \mathfrak m$. So

$$\mathfrak m = \sqrt {\mathfrak m^n} \subseteq \sqrt I \subseteq \sqrt \mathfrak m = \mathfrak m$$

$\square$

Dimension

def: Krull dimension

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ \hline \\ &\dim A:=\sup_n\{P_0 \subset P_1 \subset \ldots \subset P_n, P_i \in \mathfrak P_I(A)\} \end{align*}$$

def: Height of a prime ideal

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &P \in \mathfrak P_I(A) \\ \hline \\ &\text{ht}(P):=\sup_n\{P_0 \subset P_1 \subset \ldots \subset P_n=P, P_i \in \mathfrak P_I(A)\} \end{align*}$$

Artin rings

def: Artin ring

$$\begin{align*} &\sphericalangle \\ &T \in \mathcal R^{\mathcal C} \\ &I_i \lhd_R T, I_1 \supseteq I_2 \supseteq \ldots \implies \exists n \in \N: \forall k > n: I_k = I_n \ \\ \hline \\ &T \in \mathcal R^{\mathcal A} \end{align*}$$

Proposition 2.7.11: Artin ring criterion

$$T \in \mathcal R^{\mathcal A} \iff T \in \mathcal R^{\mathcal N}, \dim T = 0$$

Proof

We will not prove this theorem here as it requires a number of technical propositions which we want to avoid for brevity. For details see see Michael Atiyah "Introduction To Commutative Algebra", Theorem 8.5

Proposition 2.7.12: Nilradical is nilpotent

$$\begin{align*} &\sphericalangle \\ &T \in \mathcal R^{\mathcal A} \\ \hline \\ &\exists n \in N: (\sqrt {\{0\}})^n = \{0\} \end{align*}$$

Proof

By the definition of Artin rings: $\exists n \in N: I: = (\sqrt {\{0\}})^n = (\sqrt {\{0\}})^{n+1} = \ldots$. We need to prove $I = 0$. Assume not and let $\Sigma:= \{J \lhd_R T:IJ\ne\{0\} \}$. $I \in \Sigma \implies \Sigma \ne \empty$. Let $K$ be the minimal by inclusion element in $\Sigma$.

Obviously $IK \ne \{0\} \implies \exists x \in K: xI \ne \{0\}$. But $\lang x \rang_I \subseteq K$ and $\lang x \rang_I \in \Sigma$ so by minimality of $K$ we have $\lang x \rang_I = K$. Next $(xI)I=xI^2 = xI \ne \{0\}$. So similarly by minimality $xI = \lang x \rang_I$. So $\exists y \in I: x = xy$. And so we have:

$$x = xy = xy^2 = \ldots = xy^k = \ldots$$

Next, $y \in I = (\sqrt {\{0\}})^n \subseteq \sqrt {\{0\}}$. So for some $k$ we have $y^k = 0$ and consequently $x = 0$ which contradicts to the definition of $x$.

$\square$

Proposition 2.7.13: Ideals in Artin local ring

$$\begin{align*} &\sphericalangle \\ &T \in \mathcal R^{\mathcal A} \cap \mathcal R^{\mathcal L}\\ &\mathfrak m \in \mathfrak M_I(T) \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} &I \lhd_R T \implies (\exists x \in T: \lang x \rang_I = I) \wedge (\exists n \in \N: I = \mathfrak m^n) \hspace{0.5cm} \tag{a}\\ &\exists x \in T: \lang x \rang_I = \mathfrak m \hspace{0.5cm} \tag{b}\\ & \dim_{T/\mathfrak m}\mathfrak m/\mathfrak m^2 \leq 1 \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

Obvious

$(b) \implies (c)$

Follows from $(2.7.2.b)$ by assuming $M:=\mathfrak m \in \mathcal M_T$

$(c) \implies (a)$

If $\dim_{T/\mathfrak m}\mathfrak m/\mathfrak m^2 = 0$ then $\mathfrak m = \mathfrak m^2$. Assuming $I:=\mathfrak m, M:=\mathfrak m \in \mathcal M_T$ (note that $T \in \mathcal R^{\mathcal N}$ and so $\mathfrak m$ is finitely generated) and applying $(2.6.10)$ we have $\mathfrak m=0$. So every ideal is $\{0\}$.

If $\dim_{T/\mathfrak m}\mathfrak m/\mathfrak m^2 = 1$ then by $(2.7.2.c)$ $\mathfrak m = \lang x \rang_I$. Consider $I \lhd_R \mathfrak m, I \ne \{0\}, I \ne \mathfrak m$. Since $T \in \mathcal R^{\mathcal A}$, we know that $\dim T = 0$, so $\mathfrak m$ is the only prime ideal and so by $(2.7.4)$ $\mathfrak m = \sqrt {\{0\}}$. Thus by $(2.7.12)$ $\exists n: \mathfrak m^n = 0$.

This implies that

$$\exists r: I \subseteq \mathfrak m^r, I \nsubseteq \mathfrak m^{r+1}$$

and thus $\exists y \in I, \exists t \in T: y=tx^r, y \notin \lang x^{r+1} \rang_I$. So $tx^r \notin \lang x^{r+1} \rang_I \implies t \notin \lang x \rang_I=\mathfrak m$ and so by $(2.7.1)$ $t \in T^*$. So $x^r=t^{-1}y \in I$.

Finally $x^r \in I \implies \mathfrak m^r = \lang x^r \rang_I \subseteq I \subseteq \mathfrak m^r \implies I = \lang x^r \rang_I = \mathfrak m^r$

$\square$.

Proposition 2.7.14: Maximal ideals filtration in Noetherian local ring

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal N} \cap R^{\mathcal L} \\ &\mathfrak m \in \mathfrak M_I(N) \\ \hline \\ &\text{Either } \forall n: \mathfrak m^n \ne \mathfrak m^{n+1} \text{ or } (\exists n: \mathfrak m^n = 0) \wedge (N \in \mathcal R^{\mathcal A}) \end{align*}$$

Proof

Assume $\exists n: \mathfrak m^n = \mathfrak m^{n+1}$. By $(2.3.4)$ $\mathfrak m^n$ is an ideal, it's contained in a sole maximal ideal $\mathfrak m$ and it's also $N$-module which is finitely generated since $N \in R^{\mathcal N}$. So by Nakayama's lemma $(2.6.10)$ taking $M:=\mathfrak m^n \in \mathcal M_N, I:=\mathfrak m$ we have $\mathfrak m^n = 0$.

Consider some $P \in \mathfrak P_I(N)$. We know $\mathfrak m^n=\{0\} \subseteq P$. Next, by $(2.7.5): \sqrt P = P$. So we have $\mathfrak m \subseteq \sqrt{\mathfrak m^n} \subseteq \sqrt P = P \subseteq \mathfrak m \implies P = \mathfrak m$. So the only prime ideal in $A$ is $\mathfrak m$ and so $\dim A = 0$ and $N \in \mathcal R^{\mathcal A}$.

$\square$

Integral closures

def: Integral element, Integral closure

$$\begin{align*} &\sphericalangle \\ &A, B \in \mathcal R^{\mathcal {C}} \\ & A \subseteq_R B \\ & s \in B: \exists p \in A[x], LC(p)=1: p(s)=0 \\ \hline \\ &s \in \mathfrak I_B(A) - s \text{ is integral over }A \text{ in } B \\ &\mathfrak I_B(A) - \text{integral closure of } A \text{ in } B \end{align*}$$

def: Integrally closed ring in a superring

$$\begin{align*} &\sphericalangle \\ &A, B \in \mathcal R^{\mathcal {C}} \\ & \mathfrak I_B(A) = A \\ \hline \\ &A \text{ is integrally closed in } B \end{align*}$$

def: Integrally closed ring

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal {C}} \\ & \mathfrak I_{\mathfrak F(A)}(A) = A \\ \hline \\ &A \text{ is integrally closed} \end{align*}$$

Proposition 2.7.15: Integral element criteria

$$\begin{align*} &\sphericalangle \\ & S, B \in \mathcal R^{\mathcal {C}} \\ & B \subseteq_R S \\ & s \in S \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} & s \in \mathfrak I_S(B) \tag{a}\\ & \exists L \subseteq B[s], |L| < \infty, \lang L \rang_{M_B}=B[s] \hspace{1cm} \tag{b} \\ & \exists M: B[s] \subseteq M, M \subseteq_R S, \exists L: |L| < \infty: \lang L\rang_{M_B} = M \hspace{0.5cm} \tag{c} \\ &\exists M \in \mathcal M^F_{B[s]}, \exists L: |L| < \infty: \lang L\rang_{M_B} = M \hspace{0.5cm} \tag{d} \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

$$s^n+b_{n-1}s^{n-1}+\ldots+b_0 =0 \implies \\ s^n = -b_{n-1}s^{n-1}-\ldots-b_0 \\ s^{n+1} = -b_{n-1}s^{n}-b_{n-2}s^{n-1}-\ldots-b_0s= \\ -b_{n-1}(-b_{n-1}s^{n-1}-\ldots-b_0)-b_{n-2}s^{n-1}- \ldots-b_0s \\ \ldots \\ L:=\{1, \ldots, s^{n-1}\}$$

$(b) \implies (c)$

$$M:= B[s]$$

$(c) \implies (d)$

$$M:=B[s]$$

Note that $\forall x \in B[s]: xB[s]=0 \implies x \cdot 1 = 0 \implies x = 0$

$(d) \implies (a)$

Consider theorem $(2.6.12)$ and take $\phi_s: M \rightsquigarrow M, x \mapsto sx$, $I:= B$. First note that $\phi_s(M) = sM \subseteq M = BM$ since $M \in \mathcal M_{B[s]}$. So by $(2.6.12)$ we have:

$$\phi_s^n + \phi_{b_1}\phi_s^{n-1} + \ldots + \phi_{b_n} = 0, b_i \in B \implies \\ \forall m \in M: s^nm + b_1s^{n-1}m + \ldots + b_nm=0, b_i \in B \implies \\ (s^n + b_1s^{n-1}+ \ldots + b_n)m=0$$

Since $M \in \mathcal M^F_{B[s]}$ is faithful we have:

$$s^n + b_1s^{n-1}+ \ldots + b_n = 0$$

$\square$

note

To see why faithfulness of module in a clause $d$ matter - consider $B = \Z[x], S = \Z, s = x, M = \Z[x]/\lang x \rang_I$.

First note that $M \cong_R \Z$. $M = \lang 1 \rang_{M_\Z} \in \mathcal M_{\Z}$, $M \in \mathcal M_{\Z[x]}$. It will be not faithful in $M_{\Z[x]}$ as $\forall m \in M:x \cdot m = 0$. So it satisfies all properties of $(d)$ except faithfulness and obviously $x \notin \mathfrak I_{\Z[x]}(\Z)$.

Corrolary 2.7.16: Integral extensions are finitely generated modules

$$\begin{align*} &\sphericalangle \\ & A, B \in \mathcal R^{\mathcal {C}} \\ & A \subseteq_R B \\ & x_1, \ldots, x_n \in \mathfrak I_B(A) \\ \hline \\ & \exists L: |L| \lt \infty, \lang L \rang_{M_A}=A[x_1, \ldots, x_n] \end{align*}$$

Proof

By induction: when $n=1$ follows from $(2.7.15.b)$.

When $n>1$ define $A_k:=A[x_1, \ldots, x_k]$. Then $A_n=A_{n-1}[x_n]$. Since $x_n \in \mathfrak I_{A_n}(A_{n-1})$ by $(2.7.15.b)$ $A_n$ is a finitely generated $A_{n-1}$-module. Since in turn $A_{n-1}$ is finitely generated over $A$, we finish the proof.

$\square$

Proposition 2.7.17: Integral closure is a ring

$$\begin{align*} &\sphericalangle \\ &A, B \in \mathcal R^{\mathcal C} \\ &A \subseteq_R B \\ \hline \\ &\mathfrak I_B(A) \in \mathcal R^{\mathcal C} \end{align*}$$

Proof

Consider $x, y \in \mathfrak I_B(A)$. By $(2.7.16)$ $A[x, y]$ is a finitely generated $A$-module. $A[x+y], A[xy] \subseteq A[x,y] \subseteq_R B$ by $(2.7.15.c)$ $x+y, xy \in \mathfrak I_B(A)$.

$\square$

Valuation rings

def: Valuation ring

$$\begin{align*} &\sphericalangle \\ &V \in \mathcal R^{\mathcal {ID}} \\ &\forall x \in \mathfrak F(V)^*: (x \in V) \vee (x^{-1} \in V) \\ \hline \\ &V \in \mathcal R^{\mathcal V} - \text{valuation ring} \end{align*}$$

Proposition 2.7.18: Valuation ring properties

$$\begin{align*} &\sphericalangle \\ &V \in \mathcal R^{\mathcal {V}} \\ \hline \\ &\begin{align*} &V \in \mathcal R^{\mathcal L} \hspace{0.5cm} \tag{a}\\ &\forall B: V \subseteq_R B \subseteq_R \mathfrak F(V) \implies B \in \mathcal R^{\mathcal V} \hspace{0.5cm} \tag{b}\\ &\mathfrak I_{\mathfrak F(V)}(V) = V \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

Define $I:=V \setminus V^*$. Let's check that $I \lhd_R V$:

$$0 \in I \implies I \ne \empty$$ $$a \in V, x\in I \implies x \notin V^* \implies x^{-1} \notin V \implies \\ a(ax)^{-1} \notin V \implies (ax)^{-1} \notin V \overset{V \in \mathcal R^{\mathcal V}}\implies ax \in V \setminus V^* \implies ax \in I$$ $$\forall x, y \in I \setminus \{0\}: (xy^{-1} \in V) \vee (yx^{-1} \in V) \implies \\ \begin{cases} xy^{-1} \in V \implies x-y = \underbrace{(1-xy^{-1})}_{\in V}\cdot y \in I \\ yx^{-1} \in V \implies x-y = \underbrace{(1-yx^{-1})}_{\in V} \cdot x \in I \end{cases} \\ \forall x \in I: x=x+0=0+x \in I$$

Thus we proved $I \lhd_R V$ and by $(2.7.1)$ $V \in \mathcal R^{\mathcal L}$.

b.

Obviously $\mathfrak F(V)=\mathfrak F(B)$.

$$\forall x \in \mathfrak F(B), x \ne 0 \implies x \in \mathfrak F(V) \implies \\ (x \in V) \vee (x^{-1} \in V) \implies \\ (x \in B) \vee (x^{-1} \in B)$$

c.

Obviously $V \subseteq \mathfrak I_{\mathfrak F(V)}(V)$. Let's prove $V \supseteq \mathfrak I_{\mathfrak F(V)}(V)$.

Fix some $x \in \mathfrak I_{\mathfrak F(V)}(V)$. We have

$$x^n + v_1x^{n-1} + \ldots + v_n = 0$$

If $x \in V$ then we're done. Otherwise $x^{-1} \in V$. Multiplying the equation above by $x^{1-n}$ we have:

$$x = -v_1-v_2x^{-1}-\ldots-v_nx^{1-n} \in V$$

$\square$

def: Discrete valuation

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &v: F^* \to \Z\\ &v(xy)=v(x)+v(y) \\ &v(x+y) \ge \min(v(x),v(y)) \\ \hline \\ &v \in \mathcal V(F) - \text{discrete valuation} \end{align*}$$

def: Valuation ring of a field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &v \in \mathcal V(F) \\ &F_v^+:=\{x \in F: v(x) \ge 0\} \\ \hline \\ &F_v^+ - \text{valuation ring of a field } F \end{align*}$$

def: Discrete valuation ring

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal {ID}} \\ &\exists v \in \mathcal V(\mathfrak F(B)) \\ & B = \mathfrak F(B)_v^+ \\ \hline \\ &B \in \mathcal R^{\mathcal {DVR}} \end{align*}$$

Proposition 2.7.19: Discrete valuation ring is a valuation ring

$$V \in \mathcal R^{\mathcal {DVR}} \implies V \in \mathcal R^{\mathcal V}$$

Proof

First note that $v(1) = v(1\cdot 1)=v(1)+v(1) \implies v(1) =0$. Additionally $0= v(1) = v(xx^{-1})=v(x)+v(x^{-1})$

Consider some $x \in \mathfrak F(V)^*$. If $v(x) \ge 0$ then $x \in V$ and we're done. If $v(x)<0$ then $v(x^{-1})=-v(x) > 0$, so $x^{-1} \in V$.

$\square$

Proposition 2.7.20: Discrete valuation ring is a local noetherian domain of dimension 1

$$\begin{align*} &\sphericalangle \\ &D \in \mathcal R^{\mathcal {DVR}} \\ \hline \\ &\begin{align*} & D \in R^{\mathcal L}\hspace{0.5cm} \tag{a}\\ & D \in R^{\mathcal N}\hspace{0.5cm} \tag{b}\\ & \dim D = 1 \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

Exercise

Proposition 2.7.21: Discrete valuation ring criteria

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal N} \cap \mathcal R^{\mathcal L} \cap \mathcal R^{\mathcal {ID}} \\ &\dim A = 1 \\ &\mathfrak m \in \mathfrak M(A) \\ &F:=A/\mathfrak m \in \mathcal F \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} & A \in \mathcal R^{\mathcal {DVR}} \hspace{0.5cm} \tag{a}\\ & \mathfrak I_{\mathfrak F(A)}(A)=A \hspace{0.5cm} \tag{b}\\ & \exists x \in A: \lang x \rang_I = \mathfrak m \hspace{0.5cm} \tag{c}\\ & \dim_F \mathfrak m / \mathfrak m^2=1 \tag{d}\\ & \forall I \lhd_R A, I \ne \{0\}, I \ne A: \exists n \in \N: I = \mathfrak m^n \hspace{0.5cm} \tag{e}\\ & \exists x \in A: \forall I \lhd_R A, I \ne \{0\}, I \ne A: \exists n \in \N: I = \lang x^n \rang_I \hspace{0.5cm} \tag{f}\\ \end{align*} \end{align*}$$

Proof

Before we start proving, let's make the following observations.

If we take arbitraty ideal $\forall I \lhd_R A, I \ne \{0\}, I \ne A$ then by $(2.7.5)$ we have

$$\sqrt{I}=\bigcap_{P \in \mathfrak P_I(A), I \subseteq P}P$$

Note that $I \subseteq \mathfrak m$ because $\mathfrak m$ is the only maximal ideal in the ring. Next $A \in \mathcal R^{\mathcal {ID}} \implies A/\{0\} \in R^{\mathcal {ID}} \implies \{0\} \in \mathfrak P_I(A)$. Since $\dim A = 1$, every prime ideal $\ne \{0\}$ must be maximal. And since $A \in \mathcal R^{\mathcal L}$, we have exactly one maximal ideal $\mathfrak m$. So the only prime ideals in the ring are $\{0\}$ and $\mathfrak m$. And so $\sqrt{I} = \mathfrak m$. And so by $(2.7.10)$:

$$\begin{align*} & \exists n \in \N: \mathfrak m^n \subseteq I \tag{I}\\ \end{align*}$$

Next, we cannot have some $n \in \N$ such that $\mathfrak m^n = \mathfrak m^{n+1}$. Because if that's the case, by $(2.7.14)$ we have $A \in \mathcal R^{\mathcal A}$ so $\dim A = 0$. But we know that $\dim A = 1$. So

$$\begin{align*} & \forall n \in \N: \mathfrak m^n \ne \mathfrak m^{n+1}\tag{II}\\ \end{align*}$$

$(a) \implies (b)$

By $(2.7.19)$ we have $A \in \mathcal R^{\mathcal {V}}$. So by $(2.7.18.c)$ $\mathfrak I_{\mathfrak F(A)}(A)=A$

$(b) \implies (c)$

Take some $a$:

$$a \in \mathfrak m, a \ne 0$$

If $\lang a \rang_I = \mathfrak m$ then we're done. Otherwise, by $(\text{I})$ there exists some $n$:

$$\mathfrak m^n \subseteq \lang a \rang_I, \mathfrak m^{n-1} \nsubseteq \lang a \rang_I$$

Let's choose some $b$:

$$b \in \mathfrak m^{n-1}, b \notin \lang a \rang_I$$

And define $x$:

$$x:= \frac{a}{b}$$

Note that $x^{-1} \notin A$ (since $b \notin \lang a \rang_I$ and $x^{-1}$ has $a$ in denominator) and so

$$x^{-1} \notin \mathfrak I_{\mathfrak F(A)}(A)$$

We know that $\mathfrak m \in \mathcal M_{A}$. Additionally, $\mathfrak m^n \ne \mathfrak m^{n+1} \implies \mathfrak m \ne \{0\}$. So if there's an element $z \in A, z \ne 0: z\mathfrak m = 0$ then $\exists w \in A, w \ne 0: zw=0$. But that contradicts to the fact that $A \in \mathcal R^{\mathcal {ID}}$. So we proved $\mathfrak m \in \mathcal M^F_{A}$.

Next, since $b \in A, b \ne 0$ and $\mathfrak m \in \mathcal M^F_{A}$ we have $b \mathfrak m \ne \{0\} \implies x^{-1} \mathfrak m \ne \{0\}$. Now if we assume that $x^{-1}\mathfrak m \subseteq \mathfrak m$ then $\mathfrak m \in \mathcal M^F_{A[x^{-1}]}$. Also since $A \in \mathcal R^{\mathcal N}$ we know that $\mathfrak m$ is a finitely generated $A$-module so by $(2.7.15.d)$ we would have $x^{-1} \in \mathfrak I_{\mathfrak F(A)}(A)$ which is a contradiction. So we proved:

$$x^{-1}\mathfrak m \nsubseteq \mathfrak m$$

Notice that $b \in \mathfrak m^{n-1} \implies b \mathfrak m \subseteq \mathfrak m^n \subseteq \lang a \rang_I = aA$ and so $x^{-1}\mathfrak m = \frac{b}{a}\mathfrak m \subseteq \frac{aA}{a}=A$ so

$$x^{-1}\mathfrak m \subseteq A$$

$x^{-1}\mathfrak m \nsubseteq \mathfrak m$, $x^{-1}\mathfrak m \subseteq A$ and $\mathfrak m-$ maximal ideal implies

$$x^{-1}\mathfrak m = A \implies \mathfrak m = Ax = \lang x\rang_I$$

On the final note, it's sort of not obvious that $x = \frac{a}{b} \in A$. Hoewever it is, since it generates $\mathfrak m \subseteq A$.

$(c) \implies (d)$

Assuming $M:=\mathfrak m \in \mathcal M_A$, by $(2.7.2.b)$ we have $\dim_F \mathfrak m / \mathfrak m^2 \leq 1$. By $(\text{II})$ we have $\mathfrak m \ne \mathfrak m^2$ and so $\dim_F \mathfrak m / \mathfrak m^2 = 1$.

$(d) \implies (e)$

Consider

$$I \lhd_R A, I \ne \{0\}, I \ne A$$

By $(\text{I})$ we have

$$\exists n \in \N: \mathfrak m^n \subseteq I$$

Consider a factor ring $A/\mathfrak m^n$. Remember that we have 2 prime ideals in $A$, namely $\{0\}$ and $\mathfrak m$. Under the correspondence theorem $(2.3.12)$ the only proper ideal in $A/\mathfrak m^n$ is $\phi(\mathfrak m)$ which is prime so $\dim A/\mathfrak m^n=0$ and $A/\mathfrak m^n \in \mathcal R^{\mathcal A} \cap \mathcal R^{\mathcal L}$.

Next denote $\phi: A \rightsquigarrow_{M_A} A / \mathfrak m^n$ - natural module homomorphism. If $n=1$ then $\phi(\mathfrak m)=\phi(\mathfrak m^2)=\{0\}$, so $\dim_F(\phi(\mathfrak m)/\phi(\mathfrak m^2)) = 0$. If $n\ge2$ then we can apply third isomorphism theorem for modules $(2.6.6)$ assuming $M:=\mathfrak m, N: = \mathfrak m^2, K:=\mathfrak m^n$ so $\mathfrak m/ \mathfrak m^2 \cong \frac{\mathfrak m / \mathfrak m^n}{\mathfrak m^2 / \mathfrak m^n} = {\phi(\mathfrak m)}/{\phi(\mathfrak m^2)}$ and conclude that $\dim_F(\phi(\mathfrak m)/\phi(\mathfrak m^2)) = 1$.

So we have $\dim_F(\phi(\mathfrak m)/\phi(\mathfrak m^2))\le 1$. Now we can apply $(2.7.13)$ and get that $\phi(I) = \phi(\mathfrak m)^k$ and so by taking $\phi^{-1}$ we have $I = \mathfrak m^k$.

$(e) \implies (f)$

By $(\text{II})$ $\mathfrak m \ne \mathfrak m^{2}$, take $x \in \mathfrak m \setminus \mathfrak m^2$. We know that $\exists r: \lang x \rang_I = \mathfrak m^r$. Since $x \notin \mathfrak m^2$ it follows that $r = 1$ and $\lang x \rang_I = \mathfrak m$. Finally $\forall I \lhd_R A: I = \mathfrak m^k = \lang x^k \rang_I$

$(f) \implies (a)$

First note that $\lang x \rang_I = \mathfrak m$. Otherwise if $\lang x^k \rang_I = \mathfrak m$ then $\lang x \rang_I \supseteq \lang x^k \rang_I = \mathfrak m$. Since $x^k \notin A^*$ and so $x \notin A^*$ then $\lang x \rang_I = \mathfrak m$. So we established:

$$\lang x \rang_I = \mathfrak m$$

Next, by $(\text{II})$ we have $\lang x^k \rang_I \ne \lang x^{k+1} \rang_I$. Consider some $a \in A \setminus A^* \cup \{0\}$, it follows that $\exists! k \in \N: \lang a \rang_I = \lang x^k \rang_I$. Define $v(a):=k$. For $a' \in A^* \cup \{0\}$ define $v(a')=0$. Next extend $v$ to $\mathfrak F(A)$ by defining $v(\frac{a}{b})=v(a)-v(b)$.

Let's prove that this valuation is well-defined.

First consider $v$ defined on $A$

Consider some $a, b \in A, a \in A^*$. Then $\lang ab \rang_I = \lang b \rang_I$ and so $v(ab)=v(b) = 0 + v(b) = v(a)+v(b)$.

Next, $v(a)=0, v(b) \ge 0, v(a+b) \ge 0 \implies v(a+b)\ge 0 = \min(v(a), v(b))$.

Finally we need to prove valuation properties for elements $a, b \in A \setminus A^*$. Assume $\lang a \rang=\lang x^k \rang$, $\lang b \rang=\lang x^n \rang$. Then $\lang ab \rang=\lang x^{n+k} \rang$ and we have $v(ab) = v(a) + v(b)$.

Next, by $(2.4.6)$ $\exists u, v \in A^*: a = ux^k, b=vx^n$. Assume $k \leq n$, $\lang a + b \rang_I=\lang ux^k + vx^n \rang_I=\lang x^k(1+u^{-1}vx^{n-k}) \rang_I \subseteq \lang x^k \rang_I$. So we can see that $v(a+b) \ge k = \min(k, n) = \min(v(a), v(b))$.

Now $v$ defined on $\mathfrak F(A)$

First, let's prove that it's well-defined. Consider $\frac{a}{b}=\frac{c}{d} \implies ad = bc$:

$$v(a)+v(d)=v(ad)=v(bc)=v(b)+v(c) \\ v(\frac{a}{b}) = v(a) - v(b) = v(c) - v(d) = v(\frac{c}{d}) \\$$

Next let's prove valuation properties:

$$v(\frac{a}{b}\frac{c}{d})=v(\frac{ac}{bd})=v(ac)-v(bd)= \\ v(a)+v(c)-v(b)-v(d)=v(a)-v(b)+v(c)-v(d)= \\ v(\frac{a}{b})+v(\frac{c}{d})$$ $$v(\frac{a}{b}+\frac{c}{d})=v(ad+bc)-v(bd) \ge \min(v(ad), v(bc)) -v(bd) = \\ =\min(v(ad)-v(bd), v(bc)-v(bd)) = \min(v(\frac{a}{b}), v(\frac{c}{d}))$$

So we just proved that $v$ is a valid valuation on $\mathfrak F(A)$.

Finally, we need to prove that $A=B:=\{a \in \mathfrak F(A): v(a) \ge 0\}$. $A \subseteq B$ by definition of $v$ that requires every element of $A$ to have non-negative valuation. Now assume $\frac{a}{b} \in B \implies v(\frac{a}{b}) \ge 0$. This implies $v(a) \ge v(b) \implies \lang a\rang_I \subseteq \lang b\rang_I \implies a \in \lang b\rang_I \implies a = bx, x \in A \implies \frac{a}{b} \in A$.

$\square$.

Proposition 2.7.22: Discrete valuation ring has a unique discrete valuation up to a scaling factor

$$\begin{align*} &\sphericalangle \\ &D \in \mathcal R^{\mathcal {DVR}} \\ &v_1, v_2 \in \mathcal V(\mathfrak F(D)) \\ \hline \\ &\exists n_1, n_2 \in \N: \forall a \in D: \frac{v_1(a)}{n_1}=\frac{v_2(a)}{n_2} \end{align*}$$

Proof

From $(2.7.20)$ we know that $D \in \mathcal R^{\mathcal L} \cap R^{\mathcal N} \cap R^{\mathcal {ID}}$ and $\dim D = 1$. Thus we can use all the properties of $(2.7.21)$.

Let $x$ be a generator from $(2.7.21.f)$. Define

$$n_1:=v_1(x), n_2:=v_2(x)$$

Note that:

$$v_i(x^r)=rv_i(x)$$

Consider some $a \in D$, we know that $\lang a \rang_I=\lang x^r \rang_I$, so $n_2v_1(a) = n_2rv_1(x)=n_2rn_1=n_1rn_2=n_1rv_2(x)=n_1v_2(a)$.

$\square$

def: Uniformizer

$$\begin{align*} &\sphericalangle \\ &D \in \mathcal R^{\mathcal {DVR}} \\ &\lang x \rang_I \in \mathfrak M_I(D)\\ \hline \\ &x - \text{uniformizer} \end{align*}$$

note

If we define valuation of uniformizer to be equal to 1 then this valuation is unique

Alternatively, if for any valuation there's an element with $v(x)=1$ then all these valuations are the same.

Localizations

def: Ring of fractions

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal ID} \\ &S: \{1\} \subseteq S \cdot S \subseteq S \subseteq B \\ \hline \\ &S^{-1}B:=\{\frac{b}{s}: b \in B, s \in S\} \end{align*}$$

note

Remember, $\frac{a}{s}$ is defined as fractions equivalence relation class.

note

$S:=B \setminus 0 \implies S^{-1}B = \mathfrak F(B)$

Proposition 2.7.23: Complement of a prime ideal is multiplicatively closed

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal ID} \\ &I \lhd_R B \\ \hline \\ &I \in \mathfrak P_I(B) \iff \{1\} \subseteq (B \setminus I)^2 \subseteq B \setminus I \subseteq B \end{align*}$$

Proof

$\implies$

First, let's prove that $(B \setminus I)^2 \subseteq B \setminus I$. Consider $x, y \in B \setminus I$ or equivalently $x, y \notin I$:

$$(x \notin I) \wedge (y \notin I) \overset{I \in \mathfrak P_I(B)}\implies xy \notin I \implies \\ xy \in B \setminus I \\$$

Next we prove that $\{1\} \subseteq (B \setminus I)^2$:

$$I \in \mathfrak P_I(B) \overset{(2.4.6)}\implies 1 \notin I \implies 1 \in B/I \implies 1 \in (B/I)^2$$

$\impliedby$

First note that if $xy \notin B \setminus I$ then either $x \notin B \setminus I$ or $y \notin B \setminus I$. Otherwise if they are both in $B \setminus I$ then by $(B \setminus I)^2 \subseteq B \setminus I$ we'll have $xy \in B \setminus I$ which is a contradiction. So we have:

$$\forall x, y \in I: xy \in I \implies xy \notin B \setminus I \implies \\ (x \notin B \setminus I) \vee (y \notin B \setminus I) \implies (x \in I) \vee (y \in I) \\$$ $$\{1\} \subseteq B \setminus I \implies 1 \notin I \implies I \subset B$$

$\square$

def: Ring localization

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal ID} \\ &P \in \mathfrak P_I(B) \overset{(2.7.23)}\implies \{1\} \subseteq (B\setminus P)^2 \subseteq B\setminus P \subseteq B \\ \hline \\ &B_P:=(B \setminus P)^{-1}B \\ \end{align*}$$

Proposition 2.7.24: Ring localization is a local ring

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal {ID}} \\ &P \in \mathfrak P_I(B) \\ \hline \\ &B_P \in \mathcal R^{\mathcal {L}} \end{align*}$$

Proof

Consider a subset $I \subseteq B_P$:

$$I:=\{\frac{a}{s}: a \in P, s \in B\setminus P\}$$

Let's prove $I \lhd_R B_P$:

$$I\ne \empty \\ \forall \frac{a}{s}, \frac{b}{t} \in I: \frac{a}{s} -\frac{b}{t}= \frac{\overbrace{at-bs}^{\in P}}{\underbrace{st}_{\in B\setminus P}} \in I\\$$ $$\forall \frac{b}{t} \in B_P, \frac{a}{s} \in I: \frac{b}{t}\frac{a}{s} = \frac{ab}{st} \in I$$

Next, let's note that $B_P^* = \{\frac{a}{s}: a \in B\setminus P, s \in B\setminus P\}$, so $I = B_P \setminus B_P^*$. By $(2.7.1)$ we finish the proof.

$\square$

def: Ring localization at point

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &B_a:=\{a^n, n \in \N \cup \{0\}\}^{-1}B \end{align*}$$

Proposition 2.7.25: Ring localizations intersection = Ring

$$\begin{align*} &\sphericalangle \\ & B \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &\begin{align*} &B \cong \bigcap_{M \in \mathfrak M_I(B)}B_M \hspace{1cm} \tag{a}\\ &B \cong \bigcap_{P \in \mathfrak P_I(B)}B_P \hspace{1cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

Before we begin the proof, let's consider an ideal $I$ and homomorphism: $\phi: B \to B_I, b \mapsto \frac{b}{1}$. It's easy to check that it's indeed a homomorphism with $\ker \phi = \{0\}$, so $B \cong \phi(B)$. In the proof we'll assume that $B \subseteq B_I$ for any ideal $I$. Strictly speaking this would mean that $\phi(B) \subseteq B_I$ but we'll just write $B$ instead thinking about fractions with $1$ denominator.

a.

Obviously

$$\forall M \in \mathfrak M_I(B): B \subseteq B_M \subseteq \mathfrak F(B) \implies B \subseteq \bigcap_{M \in \mathfrak M_I(B)}B_M \subseteq \mathfrak F(B)$$

Now let's prove $B \supseteq \bigcap_{M \in \mathfrak M_I(B)}B_M$. Fix some $b \in \bigcap_{M \in \mathfrak M_I(B)}B_M$ and consider an ideal quotient $I_b:=\{x \in B: xb \in B\}$. Note it's very similar to $(B: b)$ and we can prove very similarly to $(2.7.7)$ that $I_b \lhd_R B$.

Note that $b \in B \iff 1 \in I_b \iff I_b = B$. So all we need to prove is $I_b = B$. Assume otherwise $I_b \ne B$:

$$(2.3.5) \implies \exists M \in \mathfrak M_I(B): I_b \subseteq M \\ b \in \bigcap_{M \in \mathfrak M_I(B)}B_M \implies b \in B_M$$

If $b = \frac{r}{s}$ that means that $r \in B, s \in B \setminus M$. On the other hand $sb = \frac{r}{1} \in B$ so $s \in I_b \subseteq M \implies s \in M$. This is a contradiction.

So we proved $I_b = B \implies b \in B$. Since $b$ was arbitrary in $\bigcap_{M \in \mathfrak M_I(B)}B_M$, we proved $B \supseteq \bigcap_{M \in \mathfrak M_I(B)}B_M$.

b.

Each prime ideal $P$ is contained in some maximal ideal $M$. Because $B_P \supseteq B_M$, the intersection by all prime ideals contained in $M$ is exactly $B_M$. So $\bigcap_{P \in \mathfrak P_I(B)}B_P = \bigcap_{M \in \mathfrak M_I(B)}B_M$ and using $(a)$ we finish the proof.

$\square$

Graded rings

def: Graded ring

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal C} \\ &\forall n \in \N \cup \{0\}: B_n \subseteq_{G,+} B \\ &\forall m,n \in \N \cup \{0\}: B_mB_n \subseteq B_{m+n} \\ &B=\bigoplus_{n\ge 0}B_n \\ \hline \\ &B \in \mathcal R^{\mathcal {GR}}, B - \text{graded ring} \end{align*}$$

note

Since $B_0 B_0 \subseteq B_0$, it follows $B_0 \subseteq_R B$ and $B_0B_n \subseteq B_n$ so $B_n \in \mathcal M_{B_0}$.

Example: Graded ring of polynomials

The example that we'll often use is graded ring of polynomials $R[x_1, \ldots, x_k ]$ where $R_n$ is a group of polynomials of degree $n$. Note that each polynomial is $R_n$ is a homogeneous degree $n$ and $R_n = \lang \{x_1^{r_1}\ldots x_k^{r_k}, r_1 + \ldots + r_k = n\} \rang_G$.

def: Graded subring

$$\begin{align*} &\sphericalangle \\ &B \in \mathcal R^{\mathcal {GR}} \\ &A \subseteq_R B \\ & A = \bigoplus_{n\ge 0}(A \cap B_n) \\ \hline \\ &A \subseteq_{gr} B, A - \text{graded subring of }B \end{align*}$$

def: Graded homomorphism

$$\begin{align*} &\sphericalangle \\ &A, B \in \mathcal R^{\mathcal {GR}} \\ & \phi: A \rightsquigarrow_R B \\ & \phi(A_n) \subseteq B_n \\ \hline \\ & \phi: A \rightsquigarrow_{gr} B \\ \end{align*}$$

def: Associated graded ring

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal {C}} \\ &I \lhd_R A\\ &gr_I(A):=\bigoplus_{n=0}^{\infty}I^n/I^{n+1} \\ &\cdot: gr_I(A) \times gr_I(A) \to gr_I(A), \\ &(x+ I^{n})(y + I^{m}) \mapsto xy+I^{m+n}\\ \hline \\ &gr_I(A) - \text{associated graded ring} \end{align*}$$

note

In this definition we assume $I^0 := A$

Proposition 2.7.26: Associated graded ring for maximal ideal

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal N} \cap \mathcal R^{\mathcal L} \\ &\dim N = d \\ &\mathfrak m \in \mathfrak M_I(N) \\ &F:= N / \mathfrak m \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} & gr_\mathfrak{m}(N) \cong_{gr} F[x_1, \ldots, x_d]\hspace{0.5cm} \tag{a}\\ & \dim_F \mathfrak m/\mathfrak m^2 = d \hspace{0.5cm} \tag{b}\\ & \exists m_1, \ldots, m_d \in N: \lang m_1, \ldots, m_d \rang_I=\mathfrak m \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

The strict proof will require some additional propositions which we'd like to avoid proving. So we'd rather give a semi-formal proof.

$(a) \implies (b)$

Denote $P:=F[x_1, \ldots, x_d]$, note that it's a ring but we make it as part of a graded ring homomorphism. That means that we assume natural grading by polynomial degree. And so $\mathfrak m/\mathfrak m^2 \cong P_1$. $P_1$ is linear polynomials with $d$ indeterminates over the field $F$ so $\dim_F \mathfrak m/\mathfrak m^2 =\dim_F P_1= d$.

$(b) \implies (c)$

Consider $(2.7.2)$ and assume $M:=\mathfrak m \in \mathcal M_A, I:=\mathfrak m$

$(c) \implies (a)$

This part is for reference only, as it uses some notions like algebraic independence which we not covered yet and some propositions we have not intent to prove here. For more details see Michael Atiyah "Introduction To Commutative Algebra", Proposition 11.22.

We can make an isomorphism $\phi: F[m_1, \ldots, m_d] \to gr_\mathfrak{m}(N)$. It splits a polynomial in $m_1, \ldots, m_d$ into homogeneous parts and then each homogeneous part of degree $n$ is mapped into $\mathfrak m^n / \mathfrak m^{n+1}$. We can finish the proof using auxiliary fact that if some elements generate maximal ideal (or more generally $\mathfrak m$-primary ideal) then they are algebraically independent over $F = A / \mathfrak m$, so they will be intederminates in this ring as stated in $(a)$.