I ##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?

[elias001]

TL;DR Summary

Question about online answer about a question concerning a proposition concerning $A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}$

I asked online questions about Proposition 2.1.1:

Question

For the proof of the Proposition 2.1.1 at the end of the Background section below, the text make use of Proposition 3.5 preceding Definition 6.6. Is it possible to use the classic method of the first isomorphism theorem by constructing a surjective homomorphic map and finding the kernel of the map. Basically, we let:

$m,s\in A$, and $\bar{(\frac{m}{s})}=\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}$

then $\frac{A_\mathfrak{p}}{\mathfrak{a}_\mathfrak{p}}=\{\bar{(\frac{m}{s})}\mid m\in A, s\not\in P\}=\{\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}\mid m\in A, s\not\in P\}$

and since $(\frac{A}{\mathfrak{a}})_{\frac{\mathfrak{p}}{\mathfrak{a}}}$ denotes the localization of the quotient ring $\frac{A}{\mathfrak{a}}$ at the ideal $\frac{\mathfrak{p}}{\mathfrak{a}}$. Also

let $\bar{m}, \bar{s}\in \frac{A}{\mathfrak{a}}$ where $\bar{m}=m+\mathfrak{a}, \bar{s}=s+\mathfrak{a}$. So

$(\frac{A}{\mathfrak{a}})_{\frac{\mathfrak{p}}{\mathfrak{a}}}=\{\frac{\bar{m}}{\bar{s}}\mid \bar{s}\not\in \frac{\mathfrak{p}}{\mathfrak{a}}\}=\{\frac{m+\mathfrak{a}}{s+\mathfrak{a}}\mid s+\mathfrak{a} \not\in \frac{\mathfrak{p}}{\mathfrak{a}} \}$

We define the map: $f:A_\mathfrak{p}\to (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}$ as $\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}\mapsto \frac{m+\mathfrak{a}}{s+\mathfrak{a}}$

with $\mathrm{ker}f\;=\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}=\mathfrak{a}_\mathfrak{p}=\{\frac{m}{s}\mid m\in \mathfrak{a}, s\not\in\mathfrak{p}\}$.

Also since $\mathfrak{a}_\mathfrak{p}$ is an ideal extension, then

$\mathfrak{a}_\mathfrak{p}=\{\sum_i \frac{m_i}{s_i}\mid m_i\in \mathfrak{a}, s_i\not\in\mathfrak{p}\}=\{\frac{\sum_{j} m_j\left(\prod_{i \neq j} s_i\right)}{\prod_i s_i}\mid m_j\in \mathfrak{a}, s_i\not\in\mathfrak{p}\}$.

One thing I don't know how to deal with is how do I account for the map $\mathfrak{q}\mapsto \mathfrak{q}A_{\mathfrak{p}}$ in the statement of the theorem in my proof?


Background

The following is taken from

Algebra Vol 2 by Luthar and Passi

3.5 Proposition. Let $\mathfrak{a}$ be an ideal of $A$, and $\bar{S}$ be the image of $S$ in $\bar{A}=A/\mathfrak{a}$. Then there exists a unique isomorphism $\eta$ of $S^{-1}A/S^{-1}\mathfrak{a}$with$\bar{S}^{-1}\bar{A}$which maps$(\mathfrak{a}/s)^{-1}$to$\bar{\mathfrak{a}}/\bar{s}##.

Definition 6.6. The ring $(A-\mathfrak{p})^{-1}A$ is a local ring and is called the $\textit{localization of$A$at its prime ideal}$ $\mathfrak{p}$ be denoted by $A_\mathfrak{p}$, and it is defined as $$A_\mathfrak{p}=\{a/s\mid a\in A, s\not\in\mathfrak{p}\}$$

If $\mathfrak{a}$ is an ideal of $A$, it is customary to denote the extension of $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}$ of $\mathfrak{a}$ in $A_\mathfrak{p}$ by $\mathfrak{a}_\mathfrak{p}$. Thus $\mathfrak{a}_\mathfrak{p}$ consists of elements of the form $\frac{a}{s}$, $a$ in $\mathfrak{a}$ and $s\not\in \mathfrak{p}$.

2.1.1 Proposition. The ring $A_\mathfrak{p}$ is a local ring with maximal ideal $\mathfrak{p}A_\mathfrak{p}$. As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$. For any ideal $\mathfrak{a}$ of $A$, contained in $\mathfrak{p}$, we have

$$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.$$

In particular, $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is isomorphic to the field of fractions of $A/\mathfrak{p}$.

The answer I got is the following:

The isomorphism $(S/\mathfrak{a})^{-1}(A/\mathfrak{a})\cong S^{-1}A/S^{-1}\mathfrak{a}$ in Prop 3.5 is proved by the authors using the First Isomorphism Theorem, so applying it to the case $S=A-\mathfrak{p}$ as in Prop 2.1.1 is essentially the proof you wanted.


As for the map $\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}$, this is the map from prime ideals of $A$ disjoint from $A-\mathfrak{p}$ to prime ideals of $A_\mathfrak{p}$ given by extension of ideals along the localization map $A\to A_\mathfrak{p}$ as in Prop 1.3.10 and Prop 3.3. As stated in the propositions, this is a bijection with inverse given by the contraction map along $A\to A_\mathfrak{p}$, which is used to prove the statement in Prop 2.1.1 about maximality of $\mathfrak{p} A_\mathfrak{p}$ in $A_\mathfrak{p}$ and the statement immediately following it.

I have some questions about the answer I got.

When the person answering says:

As for the map $\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}$, this is the map from prime ideals of $A$ disjoint from $A-\mathfrak{p}$ to prime ideals of $A_\mathfrak{p}$ given by extension of ideals along the localization map $A\to A_\mathfrak{p}$ as in Prop 1.3.10 and Prop 3.3. As stated in the propositions, this is a bijection with inverse given by the contraction map along $A\to A_\mathfrak{p}$, which is used to prove the statement in Prop 2.1.1 about maximality of $\mathfrak{p} A_\mathfrak{p}$ in $A_\mathfrak{p}$ and the statement immediately following it.

$1.$

Is the map $\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}$ from $A\setminus \mathfrak{p}\to A_\mathfrak{p}$?

But I don't understand what the author meant for the rest of the sentence in mathematical notation:

....given by extension of ideals along the localization map $A\to A_\mathfrak{p}$ as in Prop 1.3.10 and Prop 3.3

$2.$

In the next statement where the author says:

As stated in the propositions, this is a bijection with inverse given by the contraction map along $A\to A_\mathfrak{p}$, which is used to prove the statement in Prop 2.1.1 about maximality of $\mathfrak{p} A_\mathfrak{p}$ in $A_\mathfrak{p}$ and the statement immediately following it.

How is $A\to A_\mathfrak{p}$ this a contraction map? and

How is it "used to prove the statement in Prop 2.1.1 about maximality of $\mathfrak{p} A_\mathfrak{p}$ in $A_\mathfrak{p}$ and the statement immediately following it"

$3.$

In the second sentence of Proposition 2.1.1, what does the phrase: "runs once through the prime ideals..." precisely in mathematical notation.

As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$.

Thank you in advance.

 

[elias001]

There are a few two more questions I forgot to also ask:

5.

For 3.5 Proposition, the map $f:A_\mathfrak{p}\to (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}$ I am suppose to use to show $(S/\mathfrak{a})^{-1}(A/\mathfrak{a})\cong S^{-1}A/S^{-1}\mathfrak{a}$, is it suppose to be $\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}\mapsto \frac{m+\mathfrak{a}}{s+\mathfrak{a}}$ or is it suppose to be instead:
$\frac{m}{s}\mapsto \frac{m+\mathfrak{a}}{s+\mathfrak{a}}$ then show that it is surjective with kernel $S^{−1}a$

6.

In the text Introduction to Commutative Algebra and Algebraic Geometry by Ernst Kunz, Chapter 3 section 4, pg 82

He states:

A submodule $U\subset M$ is always contained in the kernel of the composite mapping
$$M\xrightarrow{i}M_S\xrightarrow{\epsilon}\frac{M_S}{U_S},$$
if $i$ and $\epsilon$ are the canonical mappings. By the universal properties of the residue module $M/U$ and of the module of fractions $(M/U)_S,$ an $R_S-$linear mapping is induced:
$$\rho:(\frac{M}{U})_S\to \frac{M_S}{U_S},\quad \rho(\frac{m+U}{s})=\frac{m}{s}+U_S.$$
Rule 4.15. (Permutability of forming the residue class module and the module of fractions ) $p$ is an isomorphism.

Proof. $\rho$ is obviously surjective. We know that $\text{Ker}(\rho)=0.$ If $\rho(\frac{m+U}{s})=0,$ then $\frac{m}{s}\in U_S,$ so there are elements $u\in U,s'\in S,$ with $\frac{m}{s}=\frac{u}{s'},$ and so there is an $s''\in S$ with $s''(s'm-su)=0.$ It follows that
$$\frac{m+U}{s}=\frac{s''s'm+U}{s''ss'}=\frac{s''su+U}{s''s's}=0.$$
Rule 4.16. Let $I$ be an ideal in $R$ and $S'$ the image of $S$ in $R/I.$ The canonical mapping
$$\rho:(\frac{R}{I})_{S'}\to \frac{R_S}{I_S},\quad \rho(\frac{r+I}{s+I})=\frac{r}{s}+I_S$$
is a ring isomorhism.

This is proved like 4.15

Compare to 3.5 Proposition, does it matter when showing isomorphism for the context of localization commuting with quotient, that one uses either of the map:

$$\frac{r+I}{s+I}\mapsto \frac{r}{s}+I_S,\quad\text{ or } \frac{r}{s}+I_S\mapsto \frac{r+I}{s+I}.$$

7.

Lastly, here is the corrected laTex of

3.5 Proposition Let $\mathfrak{a}$ be an ideal of $A$, and $\bar{S}$ be the image of $S$ in $\bar{A}=A/\mathfrak{a}$. Then there exists a unique isomorphism $\eta$ of $S^{-1}A/S^{-1}\mathfrak{a}$ with $\bar{S}^{-1}\bar{A}$ which maps $(\mathfrak{a}/s)^{-1}$ to $\bar{\mathfrak{a}}/\bar{s}$.

 

[elias001]

@fresh_42 Ok the passage from the first post is as follows:


3.5 Proposition.Let $\mathfrak{a}$ be an ideal of $A$, and $\bar{S}$ be the image of $S$ in $\bar{A}=A/\mathfrak{a}$. Then there exists a unique isomorphism $\eta$ of $S^{-1}A/S^{-1}\mathfrak{a}$ with $\bar{S}^{-1}\bar{A}$ which maps $(\mathfrak{a}/s)^{-1}$ to $\bar{\mathfrak{a}}/\bar{s}.\\\\$

Definition 6.6. The ring $(A-\mathfrak{p})^{-1}A$ is a local ring and is called the $\textit{localization of A at its prime ideal}$ $\mathfrak{p}$ be denoted by $A_\mathfrak{p}$, and it is defined as $$A_\mathfrak{p}=\{a/s\mid a\in A, s\not\in\mathfrak{p}\}$$

If $\mathfrak{a}$ is an ideal of $A$, it is customary to denote the extension of $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}$ of $\mathfrak{a}$ in $A_\mathfrak{p}$ by $\mathfrak{a}_\mathfrak{p}$. Thus $\mathfrak{a}_\mathfrak{p}$ consists of elements of the form $\frac{a}{s}$, $a$ in $\mathfrak{a}$ and $s\not\in \mathfrak{p}.\\\\$


2.1.1 Proposition. The ring $A_\mathfrak{p}$ is a local ring with maximal ideal $\mathfrak{p}A_\mathfrak{p}$. As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$. For any ideal $\mathfrak{a}$ of $A$, contained in $\mathfrak{p}$, we have$\\\\$

$$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.$$

In particular, $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is isomorphic to the field of fractions of $A/\mathfrak{p}\\\\$.


$\textbf{The questions 1 to 6 in that post boils down to:}$ if the following four parts are correct? Then there is a question from a passage from Ernst Kunz's text after. $\\\\$

The first one is just checking correctness of notations:$\\\\$

$\textbf{First,}$ are the following notations correct:$\\\\$

$\mathfrak{a}_\mathfrak{p}:\\\\$

$\mathfrak{a}_\mathfrak{p}=\{\frac{a}{s}\mid a\in \mathfrak{a}, s\not\in \mathfrak{p}\}\\\\$.

$A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}:\\\\$

$A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p} = \left\{ \frac{b}{s} + \mathfrak{a}_\mathfrak{p} \mid b \in A, s \in A \setminus \mathfrak{p} \right\},\\\\$

where $\frac{b_1}{s_1} + \mathfrak{a}_\mathfrak{p} = \frac{b_2}{s_2} + \mathfrak{a}_\mathfrak{p}$ if $\exists t \in A \setminus \mathfrak{p}$ such that $t(b_1 s_2 - b_2 s_1) \in \mathfrak{a}.\\\\$

$(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}:\\\\$

$(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}} = \left\{ \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \mid b \in A, s \in A \setminus \mathfrak{p} \right\},\\\\$

where $\frac{b_1 + \mathfrak{a}}{s_1 + \mathfrak{a}} = \frac{b_2 + \mathfrak{a}}{s_2 + \mathfrak{a}}$ if $\exists t \in A \setminus \mathfrak{p}$ such that $t(b_1 s_2 - b_2 s_1) \in \mathfrak{a}.\\\\$

$\mathfrak{p} / \mathfrak{a}:\\\\$

$\mathfrak{p} / \mathfrak{a} = \{ b + \mathfrak{a} \mid b \in \mathfrak{p} \}.\\\\$

$\textbf{Second,}$ the phrase: "runs once through the prime ideals" in the statement: As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$ mean?:$\\\\$

$\textbf{-}$ There is a bijective correspondence between the set of prime ideals $\mathfrak{q} \subseteq \mathfrak{p}$ in $A$ and the prime ideals of $A_\mathfrak{p}.$ Translating into mathematical notations, we have:

$$\{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\} \to \{\text{prime ideals of } A_\mathfrak{p}\}.$$

More pendantically precise, does it mean:$\\\\$

For a commutative ring $A$ with unity, a prime ideal $\mathfrak{p} \subset A$, and the localization $A_\mathfrak{p} = S^{-1}A$ with $S = A \setminus \mathfrak{p}$, there exists a bijection:$\\\\$

with the forward map:$\\\\$

$$f: \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\} \to \{\text{prime ideals of } A_\mathfrak{p}\}, \quad f(\mathfrak{q}) = \mathfrak{q}A_\mathfrak{p} = \left\{ \frac{a}{s} \in A_\mathfrak{p} \mid a \in \mathfrak{q}, s \in S \right\},$$

defined as: $f: \mathfrak{q} \mapsto \mathfrak{q}A_\mathfrak{p},\\\\$

and it's inverse map is:$\\\\$

$$f^{-1}: \{\text{prime ideals of } A_\mathfrak{p}\} \to \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}, \quad f^{-1}(\mathfrak{P}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P}\}$$

defined as: $f^{-1}: \mathfrak{P} \mapsto \{ a \in A \mid \frac{a}{1} \in \mathfrak{P}\}.\\\\$

The composition of the two maps are: $\\\\$

$f \circ f^{-1}:\\\\$

$(f \circ f^{-1})(\mathfrak{P}) = f(\{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}A_\mathfrak{p} = \mathfrak{P}=\text{id}_{\{\text{prime ideals of } A_\mathfrak{p}\}}.\\\\$

and $\\\\$

$f^{-1} \circ f:\\\\$

$(f^{-1} \circ f)(\mathfrak{q}) = f^{-1}(\mathfrak{q}A_\mathfrak{p}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{q}A_\mathfrak{p}\} = \mathfrak{q}=\text{id}_{\{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}}.\\\\$

$\textbf{Third, showing:}\; A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.\\\\$

We use the first isomorphism theorem to show:

$$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.$$

Using the following well defined surjective homormorphism map (can be easily shown):$\\\\$

$g: A_\mathfrak{p} \to (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}, \quad g\left( \frac{b}{s} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}\\\\$

with its kernel defined as:$\\\\$

$\ker(g) = \left\{ \frac{b}{s} \in A_\mathfrak{p} \mid \frac{b + \mathfrak{a}}{s + \mathfrak{a}} = 0 \right\} = \left\{ \frac{b}{s} \mid \exists t \in A \setminus \mathfrak{p}, tb \in \mathfrak{a} \right\} = \mathfrak{a}_\mathfrak{p}.\\\\$

By the first isomorphism theorem, $A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p} \cong (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}.\\\\$

$\textbf{Fourth,}\\\\$

Can I define the inverse of $g$ to be:$\\\\$

$g^{-1}: (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}} \to A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}, \quad g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p}.?\\\\$

I am not sure that $\ker(g) =\mathfrak{a}_\mathfrak{p},$ means that it has a trivial kernel? $\\\\$

$\textbf{Because if it doesn, then we have}\\\\$

$g \circ g^{-1}:\\\\$

$(g \circ g^{-1})\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = g\left( \frac{b}{s} + \mathfrak{a}_\mathfrak{p} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}= \text{id}_{(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}}.\\\\$

and $\\\\$

$g^{-1} \circ g:\\\\$

$(g^{-1} \circ g)\left( \frac{b}{s} \right) = g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p}= \text{id}_{A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}}.\\\\$

For Kunz's passage:$\\\\$

in his text Introduction to Commutative Algebra and Algebraic Geometry, Chapter 3 section 4, pg 82$\\\\$

He states:$\\\\$

A submodule $U\subset M$ is always contained in the kernel of the composite mapping
$$M\xrightarrow{i}M_S\xrightarrow{\epsilon}\frac{M_S}{U_S},$$

if $i$ and $\epsilon$ are the canonical mappings. By the universal properties of the residue module $M/U$ and of the module of fractions $(M/U)_S,$ an $R_S-$linear mapping is induced:
$$\rho:(\frac{M}{U})_S\to \frac{M_S}{U_S},\quad \rho(\frac{m+U}{s})=\frac{m}{s}+U_S.$$

Rule 4.15. (Permutability of forming the residue class module and the module of fractions ) $p$ is an isomorphism.$\\\\$

Proof. $\rho$ is obviously surjective. We know that $\text{Ker}(\rho)=0.$ If $\rho(\frac{m+U}{s})=0,$ then $\frac{m}{s}\in U_S,$ so there are elements $u\in U,s'\in S,$ with $\frac{m}{s}=\frac{u}{s'},$ and so there is an $s''\in S$ with $s''(s'm-su)=0.$ It follows that
$$\frac{m+U}{s}=\frac{s''s'm+U}{s''ss'}=\frac{s''su+U}{s''s's}=0.$$

Rule 4.16. Let $I$ be an ideal in $R$ and $S'$ the image of $S$ in $R/I.$ The canonical mapping
$$\rho:(\frac{R}{I})_{S'}\to \frac{R_S}{I_S},\quad \rho(\frac{r+I}{s+I})=\frac{r}{s}+I_S$$
is a ring isomorhism.$\\\\$

This is proved like Rule 4.15. $\\\\$

Compare to 3.5 Proposition, does it matter when showing isomorphism for the context of localization commuting with quotient, that one uses either of the map:

$$\frac{r+I}{s+I}\mapsto \frac{r}{s}+I_S,\quad\text{ or } \frac{r}{s}+I_S\mapsto \frac{r+I}{s+I}?$$

 

[fresh_42]

This post contains more unwritten claims than written reasons. You have to prove these statements, not just list them. If I shall correct them, then show what you did to support these claims.

$A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p} = \left\{ \frac{b}{s} + \mathfrak{a}_\mathfrak{p} \mid b \in A, s \in A \setminus \mathfrak{p} \right\},\\\\$

where $\frac{b_1}{s_1} + \mathfrak{a}_\mathfrak{p} = \frac{b_2}{s_2} + \mathfrak{a}_\mathfrak{p}$ if $\exists t \in A \setminus \mathfrak{p}$ such that $t(b_1 s_2 - b_2 s_1) \in \mathfrak{a}.\\\\$

Let's see. If $\frac{b_1}{s_1} + \mathfrak{a}_\mathfrak{p} = \frac{b_2}{s_2} + \mathfrak{a}_\mathfrak{p}$ then $\frac{b_1s_2-b_2s_1}{s_1s_2}\in \mathfrak{a}_\mathfrak{p}.$ If we write the elements $\mathfrak{a}_\mathfrak{p}=\left\{\frac{a}{t}\,|\,a\in \mathfrak{a}\, , \,t\in S\right\}$ then
$$
\dfrac{b_1s_2-b_2s_1}{s_1s_2}=\dfrac{a}{t} \Longrightarrow t\left(b_1s_2-b_2s_1\right)=(s_1s_2)a\in \mathfrak{a}.
$$
You should have made this step. It uses that $\mathfrak{a}$ is an ideal!

$(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}:\\\\$

$(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}} = \left\{ \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \mid b \in A, s \in A \setminus \mathfrak{p} \right\},\\\\$

where $\frac{b_1 + \mathfrak{a}}{s_1 + \mathfrak{a}} = \frac{b_2 + \mathfrak{a}}{s_2 + \mathfrak{a}}$ if $\exists t \in A \setminus \mathfrak{p}$ such that $t(b_1 s_2 - b_2 s_1) \in \mathfrak{a}.\\\\$

I haven't checked, but I guess this is correct. You should have done the calculation.

$\textbf{Second,}$ the phrase: "runs once through the prime ideals" in the statement: As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$ mean?:$\\\\$

$\textbf{-}$ There is a bijective correspondence between the set of prime ideals $\mathfrak{q} \subseteq \mathfrak{p}$ in $A$ and the prime ideals of $A_\mathfrak{p}.$ Translating into mathematical notations, we have:

$$\{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\} \to \{\text{prime ideals of } A_\mathfrak{p}\}.$$

Firstly, I would use the notation $\stackrel{1:1}{\longleftrightarrow },$ and secondly, note the set on the right as what it is claimed to be:
$$
\left\{\mathfrak{q'}\subseteq A_\mathfrak{p} \,|\, \mathfrak{q'} \text{ is prime}\right\}
$$
Now, this claim has to be proven, namely that any prime ideal $\mathfrak{q'}\subseteq A_\mathfrak{p}$ is of the form $\mathfrak{q'}=\mathfrak{q}A_\mathfrak{p}$ for some prime ideal $\mathfrak{q}\subseteq \mathfrak{p}\subseteq A.$

More pendantically precise, does it mean:$\\\\$

For a commutative ring $A$ with unity, a prime ideal $\mathfrak{p} \subset A$, and the localization $A_\mathfrak{p} = S^{-1}A$ with $S = A \setminus \mathfrak{p}$, there exists a bijection:$\\\\$

with the forward map:$\\\\$

$$f: \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\} \to \{\text{prime ideals of } A_\mathfrak{p}\}, \quad f(\mathfrak{q}) = \mathfrak{q}A_\mathfrak{p} = \left\{ \frac{a}{s} \in A_\mathfrak{p} \mid a \in \mathfrak{q}, s \in S \right\},$$

defined as: $f: \mathfrak{q} \mapsto \mathfrak{q}A_\mathfrak{p},\\\\$

and it's inverse map is:$\\\\$

$$f^{-1}: \{\text{prime ideals of } A_\mathfrak{p}\} \to \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}, \quad f^{-1}(\mathfrak{P}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P}\}$$

defined as: $f^{-1}: \mathfrak{P} \mapsto \{ a \in A \mid \frac{a}{1} \in \mathfrak{P}\}.\\\\$

The composition of the two maps are: $\\\\$

$f \circ f^{-1}:\\\\$

$(f \circ f^{-1})(\mathfrak{P}) = f(\{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}A_\mathfrak{p} = \mathfrak{P}=\text{id}_{\{\text{prime ideals of } A_\mathfrak{p}\}}.\\\\$

and $\\\\$

$f^{-1} \circ f:\\\\$

$(f^{-1} \circ f)(\mathfrak{q}) = f^{-1}(\mathfrak{q}A_\mathfrak{p}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{q}A_\mathfrak{p}\} = \mathfrak{q}=\text{id}_{\{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}}.\\\\$

See my previous comment. I wouldn't accept this as a proof. You should use elements, i.e., prime ideals, not sets, i.e., the spectra. It is a bit too handwavy for my taste. Say, I don't want to add all missing arguments in order to check whether this is true.

You need to show that $\operatorname{Spec}(A_\mathfrak{p})=\left\{\mathfrak{q}A_\mathfrak{p}\,|\,\mathfrak{q}\subseteq \mathfrak{p} \text{ and }\mathfrak{q}\subseteq A\text{ is prime }\right\}.$ Then you must show that your mappings $f,f^{-1}$ actually map between these sets. The equations $f\circ f^{-1}=\operatorname{id}_{\operatorname{Spec}(A_\mathfrak{p})}$ and $f^{-1} \circ f =\operatorname{id}_{\{ \mathfrak{q} \subseteq A \,|\, \mathfrak{q} \text{ is prime and } \mathfrak{q} \subseteq \mathfrak{p}\} }$ are only the last step.

$\textbf{Third, showing:}\; A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.\\\\$

It is essential that $S=A\setminus \mathfrak{p}$ and $\mathfrak{a}\subseteq \mathfrak{p}.$ This should be mentioned!

We use the first isomorphism theorem to show:

$$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.$$

Using the following well defined surjective homormorphism map (can be easily shown):$\\\\$

$g: A_\mathfrak{p} \to (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}, \quad g\left( \frac{b}{s} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}\\\\$

with its kernel defined as:$\\\\$

$\ker(g) = \left\{ \frac{b}{s} \in A_\mathfrak{p} \mid \frac{b + \mathfrak{a}}{s + \mathfrak{a}} = 0 \right\} = \left\{ \frac{b}{s} \mid \exists t \in A \setminus \mathfrak{p}, tb \in \mathfrak{a} \right\} = \mathfrak{a}_\mathfrak{p}.\\\\$

By the first isomorphism theorem, $A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p} \cong (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}.\\\\$

Same here. This requires a lot more explanations and equations, beginning with why the elements of $(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}$ look like the way you claim they look like. I don't think you can use the isomorphism theorem. I think that this is exactly what has to be shown, namely, that the isomorphism theorem still holds for localizations. You have to focus on elements, not sets.

$\textbf{Fourth,}\\\\$

Can I define the inverse of $g$ to be:$\\\\$

$g^{-1}: (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}} \to A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}, \quad g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p}.?\\\\$

I am not sure that $\ker(g) =\mathfrak{a}_\mathfrak{p},$ means that it has a trivial kernel? $\\\\$

I don't know. There is too much missing in your statements to give a reasonable answer. Once you figured out how and why elements look like they do, you will probably see what the kernel has to be.

$\textbf{Because if it doesn, then we have}\\\\$

$g \circ g^{-1}:\\\\$

$(g \circ g^{-1})\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = g\left( \frac{b}{s} + \mathfrak{a}_\mathfrak{p} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}= \text{id}_{(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}}.\\\\$

and $\\\\$

$g^{-1} \circ g:\\\\$

$(g^{-1} \circ g)\left( \frac{b}{s} \right) = g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p}= \text{id}_{A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}}.\\\\$

For Kunz's passage:$\\\\$

in his text Introduction to Commutative Algebra and Algebraic Geometry, Chapter 3 section 4, pg 82$\\\\$

He states:$\\\\$

A submodule $U\subset M$ is always contained in the kernel of the composite mapping
$$M\xrightarrow{i}M_S\xrightarrow{\epsilon}\frac{M_S}{U_S},$$

if $i$ and $\epsilon$ are the canonical mappings. By the universal properties of the residue module $M/U$ and of the module of fractions $(M/U)_S,$ an $R_S-$linear mapping is induced:
$$\rho:(\frac{M}{U})_S\to \frac{M_S}{U_S},\quad \rho(\frac{m+U}{s})=\frac{m}{s}+U_S.$$

Please don't use the fraction symbol if you mean quotient modules.

Rule 4.15. (Permutability of forming the residue class module and the module of fractions ) $p$ is an isomorphism.$\\\\$

No. He says quotient module, not module of fractions. These are different things.

Proof. $\rho$ is obviously surjective. We know that $\text{Ker}(\rho)=0.$ If $\rho(\frac{m+U}{s})=0,$ then $\frac{m}{s}\in U_S,$ so there are elements $u\in U,s'\in S,$ with $\frac{m}{s}=\frac{u}{s'},$ and so there is an $s''\in S$ with $s''(s'm-su)=0.$ It follows that
$$\frac{m+U}{s}=\frac{s''s'm+U}{s''ss'}=\frac{s''su+U}{s''s's}=0.$$

Rule 4.16. Let $I$ be an ideal in $R$ and $S'$ the image of $S$ in $R/I.$ The canonical mapping
$$\rho:(\frac{R}{I})_{S'}\to \frac{R_S}{I_S},\quad \rho(\frac{r+I}{s+I})=\frac{r}{s}+I_S$$
is a ring isomorhism.$\\\\$

This is proved like Rule 4.15. $\\\\$

Compare to 3.5 Proposition, does it matter when showing isomorphism for the context of localization commuting with quotient, that one uses either of the map:

$$\frac{r+I}{s+I}\mapsto \frac{r}{s}+I_S,\quad\text{ or } \frac{r}{s}+I_S\mapsto \frac{r+I}{s+I}?$$

Yes, I think so. The problem here is probably the well-definedness of these mappings because we deal with residue classes.

Kunz's Chapter III.4 is full of definitions and some conventions. I suggest working through them to understand all his rules. This is quite a bit of work. There are no shortcuts to see all these rules without examining the elements. E.g., if I were to try to understand where the universal property kicked in before rule 4.15, I would definitely need some paper and thoughts.

 

[elias001]

@fresh_42 in post 1, i quoted an answer from MSE it says:

'The isomorphism $(S/\mathfrak{a})^{-1}(A/\mathfrak{a})\cong S^{-1}A/S^{-1}\mathfrak{a}$ in Prop 3.5 is proved by the authors using the First Isomorphism Theorem, so applying it to the case $S=A-\mathfrak{p}$ as in Prop 2.1.1 is essentially the proof you wanted.

As for the map $\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}$, this is the map from prime ideals of $A$ disjoint from $A-\mathfrak{p}$ to prime ideals of $A_\mathfrak{p}$ given by extension of ideals along the localization map $A\to A_\mathfrak{p}$ as in Prop 1.3.10 and Prop 3.3. As stated in the propositions, this is a bijection with inverse given by the contraction map along $A\to A_\mathfrak{p}$, which is used to prove the statement in Prop 2.1.1 about maximality of $\mathfrak{p} A_\mathfrak{p}$ in $A_\mathfrak{p}$ and the statement immediately following it.'

How close is what that person says compare to what i wrote?

The thing is he stated the map for proposition 2.1 1 that i need to apply the first isomorphism theorem to is: $\frac{m}{s}\mapsto \frac{m+\frak{a}}{s+\frak{a}}$. Also the statement where it uses the phrase: "runs once through the prine odeals of...." It feels extremely sloppy and i had a hard time trying to figure out what is meant ny it. Also according to the MSE author who gave the answer when he says there is the bijection.. Is the bijection I have given correct?

 

[fresh_42]

Is the bijection I have given correct?

I don't know. You left out too many steps. This makes it impossible to answer your question without doing the calculations myself. I already listed what is missing.

1.) What are the elements in $A_\mathfrak{p}$?

$A_\mathfrak{p}=\left\{\left.\dfrac{a}{s}\,\right|\,a\in A \wedge s\not\in \mathfrak{p} \right\}.$

2.) What are the ideals in $A_\mathfrak{p}?$

If $\mathfrak{q'}\subseteq A_\mathfrak{p}$ is an ideal, then, setting $S=A\setminus\mathfrak{p},$ it has to hold that
$$
\dfrac{q_1}{s_1}-\dfrac{q_2}{s_2}=\dfrac{q_1s_2-q_2s_1}{s_1s_2}\in\mathfrak{q'} \wedge \dfrac{a}{s}\cdot\dfrac{q_1}{s_1}=\dfrac{aq_1}{ss_1}\in \mathfrak{q'}
$$
Let's set $\mathfrak{q}:=\left\{q\in A\,\left|\,\dfrac{q}{s}\in \mathfrak{q'}\text{ for some }s \in S\right.\right\}.$ Setting $s=s_1=s_2=1$ shows that $q_1-q_2 \in \mathfrak{q}$ and $aq_1\in \mathfrak{q}$ which proves that $\mathfrak{q}\subseteq A$ is an ideal. Hence all ideals $\mathfrak{q'}$ of $A_\mathfrak{p}$ look like $\mathfrak{q}_\mathfrak{p}$ where $\mathfrak{q}$ is an ideal of $A.$

3.) What are the prime ideals $\mathfrak{q'}$ of $A_\mathfrak{p}?$

It means that
$$
\dfrac{a_1}{s_1}\cdot \dfrac{a_2}{s_2}\in \mathfrak{q'} \Longrightarrow \dfrac{a_1}{s_1}\in\mathfrak{q'}\vee \dfrac{a_2}{s_2}\in \mathfrak{q'}
$$
which can be rephrased as
$$
a_1\cdot a_2\in \mathfrak{q} \Longrightarrow a_1\in \mathfrak{q} \vee a_2\in \mathfrak{q}
$$
since all elements of $\mathfrak{q'}$ look like this by our definition of $\mathfrak{q}.$ But this condition means that $\mathfrak{q}\subseteq A$ is a prime ideal.

4.) If $\mathfrak{q'}=S^{-1}\mathfrak{q}\subseteq A_\mathfrak{p}$ is a prime ideal in $A_\mathfrak{p},$ is $\mathfrak{q}$ contained in $\mathfrak{p}?$

Assume $\mathfrak{q}\not\subseteq \mathfrak{p}.$ Then there is an element $q\in \mathfrak{q}$ which is not in $\mathfrak{p}.$ But that means, $q\in S$ by definition of $S.$ This means $\dfrac{q}{q}=1_{A_\mathfrak{p}}\in \mathfrak{q'}.$ However, prime ideals are proper ideals, so they cannot contain the unit element. Hence, $q\in \mathfrak{p}$ and all numerators of elements of $\mathfrak{q'}$ are contained in $\mathfrak{p}.$

Finally, we have all the ingredients for our isomorphism:
$$
g\, : \,\operatorname{Spec}(A_\mathfrak{p}) \longrightarrow \left\{\mathfrak{q}\in \operatorname{Spec}(A)\,|\,\mathfrak{q}\subseteq \mathfrak{p}\right\}
$$
defined by $g(\mathfrak{q'})=\mathfrak{q}$ where $q:=\left\{q\in A\,\left|\,\dfrac{q}{s}\in \mathfrak{q'}\text{ for some }s \in S\right.\right\}.$ By 2.) - 4.) we have such a mapping $f$ between those sets.

6.) Is $g$ bijective?

$g$ is injective per construction. If two prime ideals $\mathfrak{q'}_1$ and $\mathfrak{q'}_2$ of $A_\mathfrak{p}$ lead to the same set $q:=\left\{q\in A\,\left|\,\dfrac{q}{s}\in \mathfrak{q'}\text{ for some }s \in S\right.\right\}$ then they have to be included in each other. It remains to show that for every prime ideal $\mathfrak{q}\subseteq A$ with $\mathfrak{q}\subseteq \mathfrak{p},$ $S^{-1}\mathfrak{q}\subseteq A_\mathfrak{p}$ is a prime ideal. If $\dfrac{a_1}{s_1}\cdot \dfrac{a_2}{s_2}\in \mathfrak{q'}$ then $\dfrac{a_1a_2}{s_1s_2}\in \mathfrak{q'}$ and $a_1a_2\in \mathfrak{q}.$ Since $q\subseteq A$ is prime, we have $a_1\in \mathfrak{q}$ or $a_2\in \mathfrak{q}$ which means $\dfrac{a_1}{s_1}\in \mathfrak{q'}$ or $\dfrac{a_2}{s_2}\in \mathfrak{q'}$ and $\mathfrak{q'}=S^{-1}\mathfrak{q}\subseteq A_\mathfrak{p}$ is a prime ideal. Our function $g$ is thus surjective, and therefore bijective.

7.) You have chosen a different direction with $f\stackrel{?}{=}g^{-1}$ and defined
$$
f\, : \,\left\{\mathfrak{q}\in \operatorname{Spec}(A)\,|\,\mathfrak{q}\subseteq \mathfrak{p}\right\}\longrightarrow \operatorname{Spec}(A_\mathfrak{p})
$$
by defining $f(\mathfrak{q})=\mathfrak{q}\cdot A_\mathfrak{p}.$ This requires showing that $\mathfrak{q}\cdot A_\mathfrak{p}$ is a prime ideal. I do not see this. The elements of $\mathfrak{q}A_\mathfrak{p}$ are $\dfrac{q\cdot a}{s}$ with any $a\in A\, , \,q\in \mathfrak{q},$ and $s\in S=A\setminus \mathfrak{p}.$ We have to prove that
$$
q_1\dfrac{a_1}{s_1} \cdot q_2\dfrac{a_2}{s_2}=\dfrac{q_1a_1q_2a_2}{s_1s_2}\in \mathfrak{q}A_\mathfrak{p}
$$
implies $\dfrac{q_1a_1}{s_1}\in \mathfrak{q}A_\mathfrak{p}$ or $\dfrac{q_2a_2}{s_2}\in \mathfrak{q}A_\mathfrak{p}.$ Why is this the case? But it's late, maybe I have overlooked something. Nevertheless, it has to be proven.

 

[elias001]

@fresh_42 in post 1, I quoted the two propositions from Luthar's 4. volumes text, only passages i was had difficulty understanding. I was trying to understand notations. Both propositions were provided with proofs. Would it help if I attached a pdf of those pages of the book where the propositions were from.

Also I am still finish updating the group action post in my DM to you, i am half way down.

 

[fresh_42]

@fresh_42 in post 1, I quoted the two propositions from Luthar's 4. volumes text, only passages i was had difficulty understanding. I was trying to understand notations. Both propositions were provided with proofs. Would it help if I attached a pdf of those pages of the book where the propositions were from.

It is not important what someone wrote; it is important what you, or in this case, I, understand. You basically claimed that $\mathfrak{q}A_\mathfrak{p}=\mathfrak{q}_\mathfrak{p}.$ Now that I think about it after some sleep, this is correct.

But all these little remarks I made contribute to our understanding. Skipping them makes a book slimmer and is part of what has to be done when studying a text. Those conclusions that I broke down into little pieces are what it means to read a math book. Reading alone isn't sufficient - even with an eidetic memory.

And on the internet, skipping those steps makes a post harder to read because someone has to make the effort and explain why the claims are true. And it is not a good idea to expect this from your readers.

Also I am still finish updating the group action post in my DM to you, i am half way down.

I wish you would post more short texts and questions instead of long ones, which I meanwhile measure in meters! It always takes an hour to even fight through them, let alone give a reasonable answer.

 

[martinbn]

@elias001 I would suggest that you explicitly state where and how you use AI in your posts. @fresh_42 is putting a lot of effort and time to help you understand, but if you are just copying AI generated text, then copy his reply back to the AI and then repeat, it is a waste of his time. He could just as well have that conversation with the AI, it would be faster. Are you interested in learning math or are you doing something else here?

 

[elias001]

@martinbn I am replying back to him this afternoon. There was several definitions I did not forget to type out from Luthar's text and a few things I did not typed out from different MSE posts. Also, I just got out of bed a short while ago.

Also, I am hoping besides homological algebra, and this topic, other topics in algebra will have more concrete examples. That way I don't always have to pester LLM to do searches for me.

 

[elias001]

@fresh_42 Here is another answer I got from MSE:

From the above, you know that there is a map:$\\\\$

$$\{\text{Ideals of }A\} \rightarrow \{\text{Ideals of }A_{\mathfrak{p}}\}$$
which sends $\mathfrak{a}$ to $\mathfrak{a}A_{\mathfrak{p}}.\\\\$

What happens when you restrict the source to ##\textit{prime}## ideals of $A$, do you find ##\textit{prime}## ideals of $A_{\mathfrak{p}}$? The answer is no. For instance, take $\Bbb Z_{(2)}$ the localization of $\Bbb Z$ at $(2)$, then the image of the prime ideal $(3) \subseteq \Bbb Z$ is in fact $(3) \Bbb Z_{(2)}= \Bbb Z_{(2)}$ as a whole, which is not prime. That sucks.$\\\\$

However, if we restrict even more to $\textit{prime ideals of}$ ##A## ##\textit{contained inside}## $\mathfrak{p}$, then the image of the map above does land inside prime ideals of $A_{\mathfrak{p}}:\\\\$

$$\{\text{Prime ideals of }A \text{ contained in }\mathfrak{p}\}
\rightarrow \{\text{Prime ideals of }A_{\mathfrak{p}}\}.$$

As before, this sends a prime ideal $\mathfrak{q}$ (which lives inside the prime ideal $\mathfrak{p}$) to the prime ideal $\mathfrak{q}A_{\mathfrak{p}}$ (which then trivially lives inside the prime ideal $\mathfrak{p}A_{\mathfrak{p}}).\\\\$

The end statement simply tells you that once you have made this restriction (and corestriction), this map is a bijection$\\\\$

Prime ideals of the localization $A_{\mathfrak{p}}$ are in one-to-one correspondence with the prime ideals of $A$ contained inside $\mathfrak{p}$, and the correspondence is given by the map above.$\\\\$

 

[fresh_42]

Also, I am hoping besides homological algebra, and this topic, other topics in algebra will have more concrete examples.

The idea is simple (cp. Kunz, example III.4.18):

Algebra:

If $\mathfrak{p}\subseteq A$ is a prime ideal, then $S=A\setminus \mathfrak{p}$ is a multiplicative set, so we have a localization, which means we can introduce denominators $S.$ Moreover, $A/\mathfrak{p}$ is an integral domain, and as such possesses a quotient field. What is the quotient field of $A/\mathfrak{p}$?

$\mathfrak{p}A_\mathfrak{p}\subseteq A_\mathfrak{p}$ is a maximal ideal because of the bijection above. This means, $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is a field. This field is exactly the quotient field of $A/\mathfrak{p}$ for which we used the isomorphism of Proposition 3.5. This is the background of all of this.

Topology:

This is the more important application. For each subset $E\subseteq A$ we define $V(E):=\left\{\mathfrak{p}\in \operatorname{Spec}(A)\,|\,E\subseteq \mathfrak{p}\right\}.$ Note that this is one of the sets of our bijection $f$ for the case $E=\mathfrak{q}.$ The essential part now is that the sets $V(E)$ fulfill the axioms of closed sets in a topology. (Prove it!) The resulting topology is called Zariski topology, and the topological space is the spectrum or prime spectrum of $\operatorname{Spec}(A)$ of $A.$ I haven't worked out what this means for localizations and quotient spaces, but Proposition 3.5 provides the tools to do this.

 

[elias001]

@fresh_42 I asked in another post about notations:$\\\\$
$\textbf{Definition 1.}$ The ring $(A-\mathfrak{p})^{-1}A$ is a local ring and is called the $\textit{localization of}$ $A$ $\textit{at its prime ideal}$ $\mathfrak{p}$ be denoted by $A_\mathfrak{p}$, and it is defined as $$A_\mathfrak{p}=\{a/s\mid a\in A, s\not\in\mathfrak{p}\}$$

If $\mathfrak{a}$ is an ideal of $A$, it is customary to denote the extension of $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a} \;(*)$ of $\mathfrak{a}$ in $A_\mathfrak{p}$ by $\mathfrak{a}_\mathfrak{p}$. Thus $\mathfrak{a}_\mathfrak{p}$ consists of elements of the form $a/s$, $a$ in $\mathfrak{a}$ and $s\not\in \mathfrak{p}.\\\\$

$\textbf{Definition 2.}$ for Extension of ideals/extended ideals$\\\\$

Let $f:M\to N$ be a ring homomorphism. If $I\triangleleft M$, define the extension $I^e$ of $I$ with respect to $f$ to be$\\\\$

$$I^e=\langle f(I)\rangle \text{(ideal generated in }N)$$

$$=\{\sum_{i=1}^{n}y_if(x_i)\mid n\geq 1,\forall i(y_i\in N,x_i\in I)\}\;(**)$$



Questions for the MSE post:$\\\\$

I would like to match the definition of extended ideal in the context of localization for $(*)$ in $\textbf{Definition 1.}$ to the more general one $(**)$ in $\textbf{Definition 2.}$. For $(*)$ we have $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}$. I am guessing in the context of $(**)$, we can define a surjective homomorphic map $\phi:A\to A_\mathfrak{p}$ for $(*)$ with $A=M$, and $A_\mathfrak{p}=N$ corresponding to the $f$ map in $(**)$ along with both $\mathfrak{a},\mathfrak{p}$ are ideals of $A$. Then the ideal $\mathfrak{a}$ in $(*)$ corresponds to the ideal $I$ in $(**)$ and $\phi(\mathfrak{a})$ corresponds to $f(I)$ in $(**)$. But then I don't know how to show that $(*)$ one is a special case of $\sum_{i=1}^{n}y_if(x_i)$?

MSE answer:

According to definition ($**$), every element of $\mathfrak{a} A_{\mathfrak{p}}$ is of the form $\sum \frac{a_i b_i}{s_i}$ with $a_i \in \mathfrak{a}, b_i \in A$, and $s_i \in A \setminus \mathfrak{p}$. We may assume $b_i = 1$, since $a_ib_i \in \mathfrak{a}$ again. Any such linear combination can be expressed as a single fraction by taking common denominators:
$$
\sum_i \frac{a_i}{s_i} = \frac{\sum_{j} a_j\left(\prod_{i \neq j} s_i\right)}{\prod_i s_i},$$
noting that the numerator lies in $\mathfrak{a}$ and the denominator lies in $A \setminus \mathfrak{p}$ by multiplicativity. Hence this element also satisfies definition $(*)\\\\$.

 

[fresh_42]

@fresh_42 Here is another answer I got from MSE:

Crossover discussions on different platforms don't make much sense, be it LLMs or MSE, or even different textbooks or lecture notes. This only confuses. I suggest concentrating on one question about one source at a time. If not, then you have to deal with several notations, conventions, styles, and intentions. You will be so busy handling those that you lose your target of an actual understanding.

 

[elias001]

@fresh_42 I thought it would make it easier for you, since the questions originated from MSE. The stuff I asked you in post 3, a lot of it overlaps with what I asked in the MSE posts or were already defined in Luthar's text. The MSE posts were mainly about asking for clarifications about them and I know it comes across that a lot of the statements seems to be I have not proven them, since they seem to be appearing out of nowhere. I am just worry about the phrase "run once through...", whether the isomorphism in the proposition 2.1.1 can be shown using the first isomorphism theorem and does there exists an inverse map of the surjective homormorphism map, $q?\\\\$ Lastly, I really dislike the notation $S^{-1}A.$ This $S^{-1}$ notation is driving me bananas.

Also, here is:$\\\\$

$\textbf{Definition 6.6.}$ The ring $(A-\mathfrak{p})^{-1}A$ is a local ring and is called the $\textit{localization of}$ $A$ $\textit{at its prime ideal}$ $\mathfrak{p}$ be denoted by $A_\mathfrak{p}$, and it is defined as $$A_\mathfrak{p}=\{a/s\mid a\in A, s\not\in\mathfrak{p}\}$$

If $\mathfrak{a}$ is an ideal of $A$, it is customary to denote the extension of $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}$ of $\mathfrak{a}$ in $A_\mathfrak{p}$ by $\mathfrak{a}_\mathfrak{p}$. Thus $\mathfrak{a}_\mathfrak{p}$ consists of elements of the form $a/s, a$ in $\mathfrak{a}$ and $s\not\in \mathfrak{p}\\\\$.

Also, $\mathfrak{p}/\mathfrak{a}=\{x+\mathfrak{a}\mid x\in\mathfrak p\}\subseteq R/\frak{a}$

 

[elias001]

@fresh_42 for the isomorphism in proposition 2.1.1, the surjective homomorphism map is:

5. Proving the Isomorphism$\\\\$

Define the ring homomorphism:$\\\\$

$g: A_\mathfrak{p} \to (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}, \quad g\left( \frac{b}{s} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}.\\\\$

$\textbf{Well-defined}:$ If $\frac{b_1}{s_1} = \frac{b_2}{s_2}$, then $\exists t \in S$ such that $t(b_1 s_2 - b_2 s_1) = 0$, so $b_1 s_2 - b_2 s_1 \in \mathfrak{a}$, and $\frac{b_1 + \mathfrak{a}}{s_1 + \mathfrak{a}} = \frac{b_2 + \mathfrak{a}}{s_2 + \mathfrak{a}}.\\\\$

$\textbf{Homomorphism}:\\\\$

$g\left( \frac{b_1}{s_1} + \frac{b_2}{s_2} \right) = g\left( \frac{b_1 s_2 + b_2 s_1}{s_1 s_2} \right) = \frac{(b_1 s_2 + b_2 s_1) + \mathfrak{a}}{(s_1 s_2) + \mathfrak{a}} = \frac{b_1 + \mathfrak{a}}{s_1 + \mathfrak{a}} + \frac{b_2 + \mathfrak{a}}{s_2 + \mathfrak{a}},\\\\$

$g\left( \frac{b_1}{s_1} \cdot \frac{b_2}{s_2} \right) = g\left( \frac{b_1 b_2}{s_1 s_2} \right) = \frac{(b_1 b_2) + \mathfrak{a}}{(s_1 s_2) + \mathfrak{a}} = \frac{b_1 + \mathfrak{a}}{s_1 + \mathfrak{a}} \cdot \frac{b_2 + \mathfrak{a}}{s_2 + \mathfrak{a}}.\\\\$

$\textbf{Surjectivity}:$ For $\frac{b + \mathfrak{a}}{s + \mathfrak{a}} \in (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}$, $g\left( \frac{b}{s} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}}.\\\\$

$\textbf{Kernel}:\\\\$

$\text{ker }(g) = \left\{ \frac{b}{s} \in A_\mathfrak{p} \mid \frac{b + \mathfrak{a}}{s + \mathfrak{a}} = 0 \right\} = \left\{ \frac{b}{s} \mid \exists t \in A \setminus \mathfrak{p}, tb \in \mathfrak{a} \right\} = \mathfrak{a}_\mathfrak{p}.\\\\$


By the first isomorphism theorem, $A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p} \cong (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}.\\\\$

The inverse map $g$ is:$\\\\$

$g^{-1}: (A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}} \to A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}, \quad g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p}.\\\\$

Compositions:$\\\\$

$g \circ g^{-1}:\\\\$

$(g \circ g^{-1})\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = g\left( \frac{b}{s} + \mathfrak{a}_\mathfrak{p} \right) = \frac{b + \mathfrak{a}}{s + \mathfrak{a}},\\\\$

so $g \circ g^{-1} = \text{id}_{(A / \mathfrak{a})_{\mathfrak{p} / \mathfrak{a}}}.\\\\$


$g^{-1} \circ g:\\\\$

$(g^{-1} \circ g)\left( \frac{b}{s} \right) = g^{-1}\left( \frac{b + \mathfrak{a}}{s + \mathfrak{a}} \right) = \frac{b}{s} + \mathfrak{a}_\mathfrak{p},\\\\$

so $g^{-1} \circ g = \text{id}_{A_\mathfrak{p} / \mathfrak{a}_\mathfrak{p}}.\\\\$

 

[elias001]

@fresh_42 When the statement:$\\\\$

"As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}".\\\\$
The forward map is:$\\\\$

is translated into:

$f: \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\} \to \{\text{prime ideals of } A_\mathfrak{p}\}, \quad f(\mathfrak{q}) = \mathfrak{q}A_\mathfrak{p} = \left\{ \frac{a}{s} \in A_\mathfrak{p} \mid a \in \mathfrak{q}, s \in S \right\}.\\\\$

The inverse map is:$\\\\$

$f^{-1}: \{\text{prime ideals of } A_\mathfrak{p}\} \to \{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}, \quad f^{-1}(\mathfrak{P}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P}\}.\\\\$

3. Bijective Maps and Compositions$\\\\$

The map $f: \mathfrak{q} \mapsto \mathfrak{q}A_\mathfrak{p}$ and its inverse $f^{-1}: \mathfrak{P} \mapsto \{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}.\\\\$

Compositions:$\\\\$

$f \circ f^{-1}:\\\\$

$(f \circ f^{-1})(\mathfrak{P}) = f(\{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{P} \}A_\mathfrak{p} = \mathfrak{P},\\\\$

so $f \circ f^{-1} = \text{id}_{\{\text{prime ideals of } A_\mathfrak{p}\}}.\\\\$

$f^{-1} \circ f:\\\\$

$(f^{-1} \circ f)(\mathfrak{q}) = f^{-1}(\mathfrak{q}A_\mathfrak{p}) = \{ a \in A \mid \frac{a}{1} \in \mathfrak{q}A_\mathfrak{p} \} = \mathfrak{q},\\\\$

so $f^{-1} \circ f = \text{id}_{\{\mathfrak{q} \subseteq A \mid \mathfrak{q} \text{ is prime}, \mathfrak{q} \subseteq \mathfrak{p}\}}.\\\\$

You said in post 4, i have to show: $\operatorname{Spec}(A_\mathfrak{p})=\left\{\mathfrak{q}A_\mathfrak{p}\,|\,\mathfrak{q}\subseteq \mathfrak{p} \text{ and }\mathfrak{q}\subseteq A\text{ is prime }\right\}.$ The thing is in vol 2 of Luthar's text, at the point where I quote localization and prime ideal, $\text{Spec }A$ has not gotten mentioned at all. Are there equivalent or alternative things I can shown. Those four volues all concentrates on undergraduate abstract algebra. I am guessing he is considering the concept of $\text{Spec }A$ to be beyond undergraduate abstract algebra.

 

[fresh_42]

I only used $\operatorname{Spec}(A)$ because it was shorter than writing "the set of all prime ideals of $A,$" especially when posing additional assumptions like $\mathfrak{q}\subseteq \mathfrak{p}$ on those ideals. And you had already used it elsewhere. It was a shortcut, not the topological space.

This is what happens when you mix different sources. One author writes it out, I used it as a shortcut, and another one speaks of the Zariski topology.

Since we can only write things on the internet, we first have to agree on a setup that makes the terms we use unambiguous. If we deal with many posts from different sources, from different authors, and even different subjects, all we get is a mess.

 

[elias001]

@fresh_42 here is the proof of Proposition 2.1.1 according to Luthar's text.$\\\\$

Proposition 3.5: Let $\mathfrak{a}$ be an ideal of $A$, and let $\bar{S}$ be the image of $S$ in $\bar{A}=A/\mathfrak{a}$. Then there exists a unique isomorphism $\eta$ of $S^{-1}A/S^{-1}\mathfrak{a}$ with $\bar{S}^{-1}\bar{A}$ which maps $(a/s)^{-1}$ to $\bar{a}/\bar{s}.\\\\$

Proposition 1.3.3: If $\varphi: A\to A'$ is a surjective homomorphism of commutative rings, thenin the 1-1 correspondence between the ideals of $A'$ and those of $A$ which $\mathfrak{n}=\text{ker }(\varphi),$ the prime (resp: maximal) ideals of $A'$ correspond with prie (resp: maximal) ideals of $A$ which contain $\mathfrak{n}.\\\\$

Proposition 1.3.10: Let $S$ be a multiplicative subset of thte commutative ring $A.$ In the correspondence of Proposition 1.3.3 the prime ideals of $A'=S^{-1}A$ correspond with the prime ideals of $A$ which are disjoint from $S,$ while the maximal ideals of $A'$ correspond with those ideals of $A$ which are maximally disjoint from $S.\\\\$

Proof of Proposition 2.1.1: Being disjoint from $S=A-\mathfrak{p}$ is obviously the same as being contained in $\mathfrak{p}.$ Therefore (by Proposition 1.3.10), $\mathfrak{q}\mapsto \mathfrak{q}A_\mathfrak{p} (=S^{-1}\mathfrak{q})$ is a 1-1 correspondence between the prime ideals of $A,$ contained in $\mathfrak{p}$, and the prime ideals of $A_\mathfrak{p}.$ In particular, $\mathfrak{p}A_\mathfrak{p}$ is the unique maximal ieal of $A_\mathfrak{p}.$ Therefore $A_\mathfrak{p}$ is a local ring with maximal ideal $\mathfrak{p}A_\mathfrak{p}$ If $\mathfrak{a}$ is an ideal of $A$, contained in $\mathfrak{p}$, then the image $\bar{S}$ of $S=A-\mathfrak{p}$ in $A/\mathfrak{a}$ is the complement in $A/\mathfrak{a}$ of the prime ideal $\mathfrak{p}/\mathfrak{a}$ of $A/\mathfrak{a}.$ Therefore by Proposition 3.5, $$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}=S^{-1}A/S^{-1}\mathfrak{a}\cong(A/\mathfrak{a}-\mathfrak{p}/\mathfrak{a})^{-1}A/\mathfrak{a}=(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}.$$
Finally taking, taking $\mathfrak{a}=\mathfrak{p},$ and noticing that $\mathfrak{a}_\mathfrak{p}=\mathfrak{p}A_\mathfrak{p},$ we have $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\cong (A/\mathfrak{p})_\mathfrak{0}=$ the field of fractions of the integral domain $A/\mathfrak{p}.\\\\$


I am not sure what maximally disjoint from $S$ mean in the end of the statement for Proposition 1.3.10, My impression is that Luthar is making a lot of effort by discussing about topics from advanced commutative algebra without directly mentioning anything from advanced commutative algebra. Am I correct? Also in the proof of proposition 2.1.1, Luthar doesn't mention what the inverse map is suppose to be.

 

[fresh_42]

@fresh_42 here is the proof of Proposition 2.1.1 according to Luthar's text.$\\\\$
I am not sure what maximally disjoint from $S$ mean in the end of the statement for Proposition 1.3.10, My impression is that Luthar is making a lot of effort by discussing about topics from advanced commutative algebra without directly mentioning anything from advanced commutative algebra. Am I correct?

No, it's way simpler. If $\mathfrak{q}\subseteq A'=S^{-1}A=A_\mathfrak{p}$ is a prime ideal, then $S\cap \mathfrak{q'}=\emptyset.$ I don't know why he uses the word maximal here; it is unnecessary. The maximal prime ideal with that property is $\mathfrak{q'}=\mathfrak{p}.$

I assume that he considers arbitrary multiplicative sets $S$ which are not necessarily of the form $A\setminus \mathfrak{p}.$ In this case, we still need $\mathfrak{q'}$ to be an ideal and $S\cap \mathfrak{q'}=\emptyset.$ So the maximal elements of these sets have to be a) still an ideal. b) disjoint from $S,$ and c) maximal among all ideals with this property. It is not clear whether there is always only one such ideal, as is the case with $S=A\setminus \mathfrak{p}.$ So you have to phrase it carefully as being maximal among the possibilities.

Also in the proof of proposition 2.1.1, Luthar doesn't mention what the i-nverse map is suppose to be.

Well, he mentions the bijection on the element level $\mathfrak{q}\stackrel{1:1}{\longleftrightarrow } \mathfrak{q}A_\mathfrak{p}.$ All you need to show is that the prime ideals in $A_\mathfrak{p}$ look this way.

 

[martinbn]

Borcherds has lectures on algebraic geometry that might be of interest. He is following Hartshorn's book, but he draws very good pictures. Here is the one on localization.

Localization

 

[elias001]

@martinbn hey I just want to apologize if I upset you earlier or awhile back. Anyways, you asked for examples from LLMs. So for the notations appearing in proposition 2.1.1 and definitions appearing before it. I asked the for the set builder notations again and also for an explanation for each one. $\\\\$

Also @fresh_42

Here they are: $\\\\$

$\mathbf{1. A_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in A, s \in A \setminus \mathfrak{p} \}$, where elements are identified under the equivalence relation $\frac{a}{s} = \frac{b}{t}$ if there exists $u \in A \setminus \mathfrak{p}$ such that $u(at - bs) = 0.\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
$A_\mathfrak{p}$ is the localization of the ring $A$ at the prime ideal $\mathfrak{p}$. It consists of fractions where the numerator is any element of $A$ and the denominator is an element of $A$ not in $\mathfrak{p}$, with equivalence defined by the given relation to account for the multiplicative set $A \setminus \mathfrak{p}.\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
Let $A = \mathbb{Z}$, $\mathfrak{p} = (5)$, the prime ideal generated by 5. Then $A \setminus \mathfrak{p} = \mathbb{Z} \setminus \{ \ldots, -10, -5, 0, 5, 10, \ldots \} = \{ \ldots, -4, -3, -2, -1, 1, 2, 3, 4, \ldots \}$.
Thus, $A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in \mathbb{Z}, s \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$, with $\frac{a}{s} = \frac{b}{t}$ if there exists $u \not\equiv 0 \pmod{5}$ such that $u(at - bs) = 0$. $\\\\$
For example, $\frac{10}{3}$, $\frac{15}{2}$, $\frac{0}{1}$ are elements, and $\frac{10}{3} = \frac{20}{6}$ since, for $u = 1 \not\equiv 0 \pmod{5}$, we have $1 \cdot (10 \cdot 6 - 20 \cdot 3) = 1 \cdot (60 - 60) = 0.\\\\$

$\mathbf{2. (A - \mathfrak{p})^{-1}A:}\\\\$$\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$(A - \mathfrak{p})^{-1}A = \{ \frac{a}{s} \mid a \in A, s \in A \setminus \mathfrak{p} \}$, with the same equivalence relation as above.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is another notation for the localization $A_\mathfrak{p}$, where $A - \mathfrak{p} = A \setminus \mathfrak{p}$ is the multiplicative set used for localization. It is identical to $A_\mathfrak{p}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
As above, for $A = \mathbb{Z}$, $\mathfrak{p} = (5)$, we have $(A - \mathfrak{p})^{-1}A = \{ \frac{a}{s} \mid a \in \mathbb{Z}, s \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$.
For example, $\frac{7}{2}$, $\frac{-3}{4}$, with $\frac{7}{2} = \frac{14}{4}$ since $1 \cdot (7 \cdot 4 - 14 \cdot 2) = 1 \cdot (28 - 28) = 0.\\\\$

$\mathbf{3. \mathfrak{p}A_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$\mathfrak{p}A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in \mathfrak{p}, s \in A \setminus \mathfrak{p} \}$, with the equivalence relation $\frac{a}{s} = \frac{b}{t}$ if there exists $u \in A \setminus \mathfrak{p}$ such that $u(at - bs) = 0.\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the ideal of $A_\mathfrak{p}$ generated by $\mathfrak{p}$. It consists of fractions where the numerator is an element of the prime ideal $\mathfrak{p}$, and the denominator is in $A \setminus \mathfrak{p}$. This is the maximal ideal of the local ring $A_\mathfrak{p}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
For $A = \mathbb{Z}$, $\mathfrak{p} = (5)$, we have $\mathfrak{p} = \{ \ldots, -10, -5, 0, 5, 10, \ldots \}$. Thus, $\mathfrak{p}A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in (5), s \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$. $\\\\$
For example, $\frac{5}{3}$, $\frac{10}{2}$, $\frac{-15}{7}$ are elements, since $5, 10, -15 \in (5)$. Also, $\frac{5}{3} = \frac{10}{6}$ as shown above.$\\\\$

$\mathbf{4. \mathfrak{q}A_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$\mathfrak{q}A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in \mathfrak{q}, s \in A \setminus \mathfrak{p} \}$, with the equivalence relation as above.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the ideal of $A_\mathfrak{p}$ generated by a prime ideal $\mathfrak{q} \subseteq \mathfrak{p}$ of $A$. It consists of fractions with numerators in $\mathfrak{q}$ and denominators in $A \setminus \mathfrak{p}$. The proposition states that as $\mathfrak{q}$ varies over prime ideals contained in $\mathfrak{p}$, $\mathfrak{q}A_\mathfrak{p}$ covers all prime ideals of $A_\mathfrak{p}.\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
Let $\mathfrak{q} = (0)$, the zero ideal (a prime ideal in $\mathbb{Z}$), and $\mathfrak{p} = (5)$. Then $\mathfrak{q}A_\mathfrak{p} = \{ \frac{0}{s} \mid s \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \} = \{ 0 \}$, since $\frac{0}{s} = 0$ in $A_\mathfrak{p}$. This is the zero ideal in $A_\mathfrak{p}.\\\\$

$\mathbf{5. \mathfrak{a}_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} \mid a \in \mathfrak{a}, s \in A \setminus \mathfrak{p} \}$, with the equivalence relation $\frac{a}{s} = \frac{b}{t}$ if there exists $u \in A \setminus \mathfrak{p}$ such that $u(at - bs) = 0$.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the ideal of $A_\mathfrak{p}$ generated by an ideal $\mathfrak{a} \subseteq \mathfrak{p}$ of $A$. It consists of fractions with numerators in $\mathfrak{a}$ and denominators in $A \setminus \mathfrak{p}$. This is often denoted $\mathfrak{a}A_\mathfrak{p}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
Let $\mathfrak{a} = (10)$, $\mathfrak{p} = (5)$. Since $10 = 2 \cdot 5$, we have $\mathfrak{a} \subseteq \mathfrak{p}$. Then $\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} \mid a \in (10), s \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$.
For example, $\frac{10}{3}$, $\frac{20}{7}$, $\frac{-30}{1}$ are elements, since $10, 20, -30 \in (10).\\\\$

$\mathbf{6. A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} + \mathfrak{a}_\mathfrak{p} \mid a \in A, s \in A \setminus \mathfrak{p} \}$, where $\frac{a}{s} + \mathfrak{a}_\mathfrak{p} = \frac{b}{t} + \mathfrak{a}_\mathfrak{p}$ if $\frac{a}{s} - \frac{b}{t} \in \mathfrak{a}_\mathfrak{p}$, i.e., there exists $c \in \mathfrak{a}$, $u \in A \setminus \mathfrak{p}$ such that $\frac{a}{s} = \frac{b}{t} + \frac{c}{u}$.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the quotient ring of $A_\mathfrak{p}$ by the ideal $\mathfrak{a}_\mathfrak{p}$. Elements are cosets of fractions in $A_\mathfrak{p}$, with equivalence defined by the ideal membership in $\mathfrak{a}_\mathfrak{p}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
Using $\mathfrak{a} = (10)$, $\mathfrak{p} = (5)$, we have $A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$, $\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} \mid a \in (10), s \not\equiv 0 \pmod{5} \}$.$\\\\$
The quotient is $A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} + \mathfrak{a}_\mathfrak{p} \}$. For example, $\frac{1}{3} + \mathfrak{a}_\mathfrak{p}$, $\frac{2}{7} + \mathfrak{a}_\mathfrak{p}$.$\\\\$$\\\\$
Cosets are equal if their difference is in $\mathfrak{a}_\mathfrak{p}$: $\frac{1}{3} + \mathfrak{a}_\mathfrak{p} = \frac{11}{3} + \mathfrak{a}_\mathfrak{p}$ since $\frac{11}{3} - \frac{1}{3} = \frac{10}{3} \in \mathfrak{a}_\mathfrak{p}$.$\\\\$

$\mathbf{7. A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} = \{ \frac{a}{s} + \mathfrak{p}A_\mathfrak{p} \mid a \in A, s \in A \setminus \mathfrak{p} \}$, where $\frac{a}{s} + \mathfrak{p}A_\mathfrak{p} = \frac{b}{t} + \mathfrak{p}A_\mathfrak{p}$ if $\frac{a}{s} - \frac{b}{t} \in \mathfrak{p}A_\mathfrak{p}$, i.e., there exists $c \in \mathfrak{p}$, $u \in A \setminus \mathfrak{p}$ such that $\frac{a}{s} = \frac{b}{t} + \frac{c}{u}$.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the quotient of $A_\mathfrak{p}$ by its maximal ideal $\mathfrak{p}A_\mathfrak{p}$. The proposition states it is isomorphic to the field of fractions of $A/\mathfrak{p}$, which is a field since $\mathfrak{p}$ is prime.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
For $\mathfrak{p} = (5)$, $\mathfrak{p}A_\mathfrak{p} = \{ \frac{a}{s} \mid a \in (5), s \not\equiv 0 \pmod{5} \}$. The quotient is $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p} = \{ \frac{a}{s} + \mathfrak{p}A_\mathfrak{p} \}$.
For example, $\frac{1}{3} + \mathfrak{p}A_\mathfrak{p}$, $\frac{6}{7} + \mathfrak{p}A_\mathfrak{p}$. $\\\\$
Since $A/\mathfrak{p} = \mathbb{Z}/(5) \cong \mathbb{Z}/5\mathbb{Z}$, the field of fractions is $\mathbb{Z}/5\mathbb{Z}$ itself (as it’s a field). Thus, cosets correspond to elements in $\mathbb{Z}/5\mathbb{Z}$, e.g., $\frac{5}{3} + \mathfrak{p}A_\mathfrak{p} = 0 + \mathfrak{p}A_\mathfrak{p}$ since $\frac{5}{3} \in \mathfrak{p}A_\mathfrak{p}$.$\\\\$

$\mathbf{8. \mathfrak{p}/\mathfrak{a}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$\mathfrak{p}/\mathfrak{a} = \{ p + \mathfrak{a} \mid p \in \mathfrak{p} \}$, where $p + \mathfrak{a} = q + \mathfrak{a}$ if $p - q \in \mathfrak{a}$.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the quotient of the ideal $\mathfrak{p}$ by the ideal $\mathfrak{a} \subseteq \mathfrak{p}$ in the ring $A$. It is an ideal in the quotient ring $A/\mathfrak{a}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
Let $\mathfrak{a} = (10)$, $\mathfrak{p} = (5)$. Since $(10) \subseteq (5)$, we have $\mathfrak{p}/\mathfrak{a} = \{ p + (10) \mid p \in (5) \}$.
Elements are cosets like $5 + (10)$, $0 + (10)$. Since $5 - 0 = 5 \notin (10)$, these are distinct, but $5 + (10) = 15 + (10)$ since $15 - 5 = 10 \in (10)$. In $A/\mathfrak{a} = \mathbb{Z}/(10) \cong \mathbb{Z}/10\mathbb{Z}$, this ideal is generated by $5 + (10)$.$\\\\$

$\mathbf{9. (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}:}\\\\$

$\textbf{Set Builder Notation:}\\\\$$\\\\$
$(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}} = \{ \frac{a + \mathfrak{a}}{s + \mathfrak{a}} \mid a \in A, s \in A \setminus \mathfrak{p} \}$, with $\frac{a + \mathfrak{a}}{s + \mathfrak{a}} = \frac{b + \mathfrak{a}}{t + \mathfrak{a}}$ if there exists $u \in A \setminus \mathfrak{p}$ such that $(a + \mathfrak{a})(t + \mathfrak{a}) = (b + \mathfrak{a})(s + \mathfrak{a})$ in $A/\mathfrak{a}$, i.e., $at - bs \in \mathfrak{a}$.$\\\\$

$\textbf{Explanation:}\\\\$$\\\\$
This is the localization of the quotient ring $A/\mathfrak{a}$ at the prime ideal $\mathfrak{p}/\mathfrak{a}$ of $A/\mathfrak{a}$. Elements are fractions with numerators and denominators as cosets, where the denominator coset $s + \mathfrak{a}$ has $s \notin \mathfrak{p}$.$\\\\$

$\textbf{Numerical Example:}\\\\$$\\\\$
For $\mathfrak{a} = (10)$, $\mathfrak{p} = (5)$, we have $A/\mathfrak{a} = \mathbb{Z}/(10)$, $\mathfrak{p}/\mathfrak{a} = (5)/(10) = \{ 0 + (10), 5 + (10) \}$. Thus, $A \setminus \mathfrak{p} = \{ s \in \mathbb{Z} \mid s \not\equiv 0 \pmod{5} \}$, and $(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}} = \{ \frac{a + (10)}{s + (10)} \mid a \in \mathbb{Z}, s \not\equiv 0 \pmod{5} \}$.$\\\\$
For example, $\frac{1 + (10)}{3 + (10)}$, $\frac{2 + (10)}{7 + (10)}$. Equivalence holds if, e.g., $\frac{1 + (10)}{3 + (10)} = \frac{11 + (10)}{3 + (10)}$ since $1 \cdot 3 - 11 \cdot 3 = 3 - 33 = -30 \in (10)$.$\\\\$

$\textbf{Computation for Isomorphism:}\\\\$$\\\\$
The proposition states $A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p} \cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}$. For $\mathfrak{a} = (10)$, $\mathfrak{p} = (5)$, we verify: $\\\\$$\\\\$
- Left: $A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}$ has elements $\frac{a}{s} + \mathfrak{a}_\mathfrak{p}$, where $\mathfrak{a}_\mathfrak{p} = \{ \frac{a}{s} \mid a \in (10), s \not\equiv 0 \pmod{5} \}$. $\\\\$$\\\\$
- Right: $(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}} = (\mathbb{Z}/(10))_{(5)/(10)}$ consists of fractions in $\mathbb{Z}/10\mathbb{Z}$ with denominators not in $\{ 0 + (10), 5 + (10) \}$, i.e., $s + (10)$ where $s \not\equiv 0, 5 \pmod{10}$.$\\\\$
The isomorphism maps $\frac{a}{s} + \mathfrak{a}_\mathfrak{p} \to \frac{a + (10)}{s + (10)}$, preserving operations, as verified by checking addition and multiplication modulo the ideals.$\\\\$

 

[elias001]

@martinbn Also thanks for the reference to Borcherds.

I did found two friendly reference:

1. Introduction to algebraic geometry by Justin R. Smith

2. A first course in algebraic geometry and algebraic varieties by: Flaminio Flamini

3. A course in commutative algebra by: Gregor Kemper

4. Commutative algebra through exercises by: Andrea Bandini, Patrizia Gianni, Enrico Sbarra.

5. A term of commutative algebra by: Steven Kleiman and Allen Altman


Truth be told, for anyone who is going to any math field where it has the adjective "algebraic" attached to it, chances are one has to take a course in algebraic geometry. That is the equivalent of all the pre-medical students who want to study clinical medicine has to take that dreaded organic chemistry course.

 

[elias001]

@fresh_42 when you say Luthar mentioned the bijections on the elements level
$\mathfrak{q}\stackrel{1:1}{\longleftrightarrow } \mathfrak{q}A_\mathfrak{p},$ I find that when it comes to algebra, the more details to unobfuscate the notations. The better and more convenient it is whenever i need to do proofs. So I say it is better if we know what the inverse map is suppose to be. I mean why not, via two compositions, we get to show bijection which is much less writing than having to show one to one and onto.

 

[fresh_42]

@fresh_42 when you say Luthar mentioned the bijections on the elements level
$\mathfrak{q}\stackrel{1:1}{\longleftrightarrow } \mathfrak{q}A_\mathfrak{p},$ I find that when it comes to algebra, the more details to unobfuscate the notations. The better and more convenient it is whenever i need to do proofs. So I say it is better if we know what the inverse map is suppose to be. I mean why not, via two compositions, we get to show bijection which is much less writing than having to show one to one and onto.

You have to go on the element level and show that $f\circ f^{-1}=1$ and $f^{-1}\circ f=1.$ Either that, or injectivity and surjectivity. In both cases, we need elements, not sets.

 

[mathwonk]

fresh, I suspect you are helping train a bot.

 

[fresh_42]

fresh, I suspect you are helping train a bot.

I have Kunz's book, and his citations were one-to-one, even the letters he used. Only mine is in German and his is in English, so the page numbers don't match. His PM sounded like a human being. But you are right, I'm still looking for an approach to change his way of learning, esp. his use of LLMs or various sources at the same time. Self-study is not easy if you have no control mechanisms.

 

[elias001]

@fresh_42 Okay, let try to show: $\operatorname{Spec}(A_\mathfrak{p})=\left\{\mathfrak{q}A_\mathfrak{p}\,|\,\mathfrak{q}\subseteq \mathfrak{p} \text{ and }\mathfrak{q}\subseteq A\text{ is prime }\right\}$ per your suggest. We have the set builder notation equality from the other post for reference: $x \in \text{Spec } R:=\{ x : x \in \text{Spec } R \} = \{ \mathfrak{p}: \mathfrak{p} \subset R, \mathfrak{p} \text{ is a prime ideal in }R\}.\;\;(*)$ We partially modify it to be $x \in \text{Spec } A_\mathfrak{p}:=\{ x : x \in \text{Spec } A_\mathfrak{p} \} = \{ \mathfrak{p}: \mathfrak{p} \subset A_\mathfrak{p}, \mathfrak{p} \text{ is a prime ideal in }A_\mathfrak{p}\}.$ I suspect this is only the start.

@martinbn as per your suggestion in this thread, post #2 about getting a book that is native to my language. Well English has become the equivalent of my first language. Anyways, I am using Elements of Mathematics Algebra Chapters 1-3 by the collective name Bourbaki, or what I like to call it using "super saiyan ultra master super ultra ocd beyond austistic level anal pedantic language".

 

[elias001]

@fresh_42 what about this: $x \in \text{Spec } A_\mathfrak{p}:=\{ x : x \in \text{Spec } A_\mathfrak{p} \} = \{ \mathfrak{q}A_\mathfrak{p}:\mathfrak{q}\subseteq \mathfrak{p} , \mathfrak{p} \subset A_\mathfrak{p}, \mathfrak{p} \text{ is a prime ideal in }A_\mathfrak{p}\}$, having written that out, do I need to prove the statement: $\operatorname{Spec}(A_\mathfrak{p})=\left\{\mathfrak{q}A_\mathfrak{p}\,|\,\mathfrak{q}\subseteq \mathfrak{p} \text{ and }\mathfrak{q}\subseteq A\text{ is prime }\right\}$?

 

[fresh_42]

Do you even read what I post? See post #6 points 2.)-4.)

 

[elias001]

@fresh_42 your post #6 points 2.)-4.) is where you try to shown that statement?

 

[fresh_42]

Yes.