I Localising a non integral domain at a prime

[elias001]

TL;DR Summary

How does one determine all the equivalence classes and how many there are in the case of localising a non integral domain at a prime?

The two screenshots below are taken from the book Fundamentals of Modern Algebra A global Perspective by: Robert G Underwood

Screenshot 1

Screenshot 2

The two screenshots above concerns localizing a non integral domain at a prime; in the second screen shot, the example given is $(Z_6)_3$ which denote the multiplicative closed set $Z_6\times\{1,3\}=\{(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(0,3),(1,3),(2,3),(3,3), (4,3),(5,3)\},$ with the equivalence relation $\sim$ define as $(a,b)\sim(c,d)\text{ iff for some } s\in S \text{ for which }s(ad-bc)=0.$ and the equivalence classes in $(Z_6)_3:$ $\{\frac{0}{1}, \frac{1}{1}\}$.

What I would like to know are the following:

1. How did the author determine that $\{\frac{0}{1}, \frac{1}{1}\}$ are the only equivalence classes? I mean the reason $\frac{0}{3}$ is not an elements of the equivalence classes listed because we are in $\pmod 6?$

2. Also, for any other a large non integral domain integer ring $n$ along with large prime $p$, say if we have $(Z_n)_p$, where $n$ is in the couple hundred trillion, and $p$ has couple hundred billion digits, are there algorithms that can systemically determine all of its equivalence classes, and also upper and lower bounds for how many there are?

Thank you in advance.

 

[martinbn]

0/3 and 0/1 are in the same class.

 

[elias001]

@martinbn ops you are right. Are there ways to show that $\{0/1,1/1\}$ are the only equivalent classes in the localisation?

 

[martinbn]

@martinbn ops you are right. Are there ways to show that $\{0/1,1/1\}$ are the only equivalent classes in the localisation?

Yes, you check that any other a/b is equivalient to one of these two.

 

[elias001]

@martinbn I should have asked what about the case $\frac{1}{3}$, but $\frac{1}{3}\sim \frac{1}{1}.$ Actually, the way the set $S,$ is it defined in general as $S=\{p^n: n\geq 0\}$ and is always equal to $\{1,p\}$ for $p-$prime?

Also, just to make sure I understand the example properly, if I want to determine the set of equivalence class for $(\Bbb{Z}_n)_p$, where $n$ is some large composite number, say $n=100$, and $p=41$, then I would have the set of equivalence classes $\{\frac{0}{1},\frac{1}{1}\},$ and $\Bbb{Z}_100\times \{1,41\}=\{(0,1),(1,1),(2,1),\ldots, (99,1),(0,41),(1,41),(2,41),\ldots,(99,41)\}$?

 

[mathwonk]

Note: a/1 ≈ 0/1 iff 3a = 0 in Z/6. Hence the kernel of the map Z/6-->S^-1(Z/6) is {0,2,4} and the image is {0/1, 1/1}. Hence every element of the localized ring is equivalent to one of 0/1, 0/3, 1/1, or 1/3. But 0/1 and 0/3 are equivalent and 1/1 ≈ 1/3 since 3.(3.1-1.1)= 0 in Z/6.

In your last example, note 41 is not a zero divisor in Z/100, so the map from Z/100 to the localization is injective. That's a start. Thus the first 100 of your fractions are all different, and so are the last 100. But some of the first 100 are equivalent to some of the last 100. Do you see how many?

Or you could just ask yourself what the units are in Z/100, and use the Proposition 2.44,(ii), that you cite.
Hint: what is (41).(61) ?

General advice: (more below): When they give us a theorem, that is a potential tool, and we need to pause, read it, and think about what it can be useful for. You were asking how many elements are in a certain localization. Prop. 2.44(i) gives a criterion for when S^-1.R is at least as large as R, and 2.44(ii) is a criterion for when R and S^-1.R are the same size. These criteria are ideally suited to answer the question you posed on localizing Z/100.

The example they gave on localizing Z/6 is a case where the criteria do not seem to apply, but actually they can be tweaked until they do apply. Namely since the kernel of the map Z/6-->S^-1.Z/6 is {0,2,4}, the result seems to be that we are essentially localizing the quotient (Z/6)/{0,2,4} ≈ Z/2 = {0,1}, at the multiplicative set {1,3}/{0,2,4} = {1} which is a unit in Z/2, so the result, again by 2.44(ii) is isomorphic to just Z/2 = {0,1}.
This technique of first forming the quotient of R by the ideal of elements which are annihilated by elements of S, always allows us to get into the situation of 2.44(i), where the natural map is injective, but not always in the situation of 2.44(ii). (A general principle is that forming quotients "commutes with localization".)
So in this example, all our questions are answered by, just thinking about and using, the tool given in the proposition.

As exercise, you might try to compute the result of localizing Z/100 at powers of 2, or powers of 3, or powers of 5, or powers of 6. (my answers: Z/25, Z/100, Z/4, Z/25.)

 

[mathwonk]

aside: (preachy commentary, hopefully useful;)
By the way, please forgive me, but from your questions, I have the impression you have not studied a first course of abstract algebra, and yet are attempting to read books that seem to be for second courses in the topic. (That book by Underwood looks very advanced and very condensed even for that audience. The book of Arbib seems even less useful.) I recommend first going carefully through a more introductory book, such as Abstract Algebra, by Ted Shifrin, or maybe (based only on reviews, as I do not own it) A book of abstract algebra by Charles Pinter.
It seems hard to find a good elementary book on abstract algebra, but the very far out categorical ones you are referencing are to me not useful at all for early learning in the subject. One needs first to acquire the habit of looking for structure in the objects you study, and how to compare them, i.e. looking at homomorphisms between them. It is of little use to memorize the categorical definition of a kernel, and then when trying to understand the relation between a ring and its localization, not even to ask oneself what is the kernel of the map from R to S^-1.R. I.e. the first job is not to ask what is the most abstract possible definition of a kernel, but what is the concept useful for: namely it tells you how big is the image of a map, exactly what you were asking above.
When studying localization, the basic result is a characterization of existence of homomorphisms, i.e. any ring map f:R--->A that maps the multiplicative set S into a set of units of A, induces a unique ring map S^-1.R-->A, sending x/y to f(x)/f(y). In particular, if S consists of units of R, the identity map R--->R induces a unique map S^-1.R--->R, inverting the natural map R-->S^-1.R, which proves your Prop. 2.44(ii) above.
If and when you need a book aimed at graduate students, I recommend Dummitt and Foote as a more understandable and helpful alternative to Underwood. I agree also with martinbn and would expand his suggestion to include a recommendation to go through a book of elementary number theory. There is such a book by Ethan Bolker that uses abstract algebra to treat elem. number theory, but I don't know how helpful it is. There are also well regarded introductory books on number theory by Underwood Dudley, David Burton, and Charles Van den Eynden. These of course are just suggestions, as only you know your needs best.

By the way, since you say elsewhere you are studying math on your own, may I suggest books by great masters as the best for providing insight, such as Euclid's Elements, Euler's Elements of Algebra, and Gauss' Disquisitiones? For modern or abstract algebra, one of the first texts, by Van der Waerden, preferably an early edition, like https://www.amazon.com/Modern-Algebra-Vol-Van-Waerden/dp/B000Q9W1CA?tag=pfamazon01-20
is still one of the best.

But, even if you decide to consult some of these books, (recalling the advice of Samuel Johnson), I would actually recommend you just read whatever appeals to you, since you have that luxury. At least, that's what I do.

 

[fresh_42]

As an add-on to what @mathwonk has said, you can always find lecture notes online. My proposal for search keys would be: "Introduction to Abstract Algebra + pdf", possibly dropping "Introduction to" and/or "abstract". Only "Algebra" and "+pdf" are important. Actually, many universities have English texts on their web servers these days. You don't even have to buy a book.

Another good advice, a bit hidden in what @mathwonk has said is to take the integers and calculate examples. Start with numbers less than ten.

 

[elias001]

@mathwonk sorry for my late reply here. I am only looking at Robert Underwood's text only for the section on localization, specifically the example he provides. it was really difficult to find a decent example beyond just $\Bbb{Z}_2$. Also, I find that the more advanced the concepts I learn in abstract algebra, the less examples are used to illustrate the new concepts. Also I am going through books specifically on introduction to commutative ring theory and module theory. I feel that is much better since the pacing is slower in those books. Dummit and Foote is a good source for exercises, but the presentation feel overwhelming and many of the exercises has zero motivations behind them or there is zero mention if they are important in the overall theory or was it an exercise based on some journal article.

 

[elias001]

@mathwonk I still have more questions about the example in Robert G Underwood's text, with $Z_6$ as a non-integral domain, localise at $3$, $3$ is a non-nilpotent element.

Question 1. Given a non integral domain integer ring, $Z_n$, $n$ is composite, then it can have unique factorization, $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ where all $p_i$s are prime integers and all $e_i$s are greater than or equal to zero.

When one says that one wants to localise $Z_n$ at a prime, does whatever prime say $q$, one is discussing have to satisfy the following criterias:

1) $q$ has to come from one of the primes $p_i,$ which came from the unique factorization of $n$

2) $q$ is a non-nipotent element of $Z_n$

3) $q$ has to be one of the zero divisor of $n$

4) After completing steps 1 to 3, what are the equivalence classes?

So lets say we have $n=232792560=2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17.$.


So for $Z_{232792560}$, we want to localise at a prime, by forming the set $S=\{a^n:n\geq 0\},$ $a$ can only be one of $2,3,5,7,11,13,17,$ and also as long as $a$ is non-nilpotent.

After that, would the equivalence classes of $(Z_{232792560})_a$ still be $\{\frac{0}{1}, \frac{1}{1}\}?$

 

[elias001]

@fresh_42, @mathwonk my post #10 was more questions after reading your post #6. specifically

In your last example, note $41$ is not a zero divisor in $Z_{100}$, so the map from $Z_{100}$ to the localization is injective. That's a start. Thus the first $100$ of your fractions are all different, and so are the last $100$. But some of the first $100$ are equivalent to some of the last $100$. Do you see how many?

In post #6 and #7, you said:

$\text{post: 6}$

Note: $\frac{a}{1} \sim \frac{0}{1}$ iff $3a = 0$ in $Z_6$. Hence the kernel of the map $Z_6\to S^{-1}(Z_6)$ is $\{0,2,4\}$ and the image is $\{\frac{a}{1}, \frac{0}{1}\}$. Hence every element of the localized ring is equivalent to one of $\frac{0}{1}, \frac{0}{3}, \frac{1}{1},$ or $\frac{1}{3}$. But $\frac{0}{1}$ and $\frac{0}{3}$ are equivalent and $\frac{1}{1}\sim \frac{1}{3}$ since $3\cdot(3\cdot 1-1\cdot 1)= 0$ in $Z_6$.

In your last example, note $41$ is not a zero divisor in $Z_{100}$, so the map from $Z_{100}$ to the localization is injective. That's a start. Thus the first $100$ of your fractions are all different, and so are the last $100$. But some of the first $100$ are equivalent to some of the last $100$. Do you see how many?

Or you could just ask yourself what the units are in $Z_{100}$, and use the Proposition 2.44,(ii), that you cite.
Hint: what is $(41)\cdot(61)$?

$(41)\cdot (61)\equiv 1 {\pmod{100}}$. For $\text{gcd}(a,100)=1, a=1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39,41,43,47,49,51,53,57,59,61,63,79,81,83,87,89,91,93,97,99$

General advice: (more below): When they give us a theorem, that is a potential tool, and we need to pause, read it, and think about what it can be useful for. You were asking how many elements are in a certain localization. Prop. 2.44(i) gives a criterion for when $S^{-1}\cdot R$ is at least as large as $R$, and 2.44(ii) is a criterion for when $R$ and $S^{-1}\cdot R$ are the same size. These criteria are ideally suited to answer the question you posed on localizing $Z_{100}$.

The example they gave on localizing $Z_6$ is a case where the criteria do not seem to apply, but actually they can be tweaked until they do apply. Namely since the kernel of the map $Z_6\to S^{-1}\cdot Z_6$ is $\{0,2,4\}$, the result seems to be that we are essentially localizing the quotient $\frac{Z_6}{\{0,2,4\}} \cong Z_2 = \{0,1\}$, at the multiplicative set $\frac{\{1,3\}}{\{0,2,4\}}= \{1\}$ which is a unit in $Z_2$, so the result, again by 2.44(ii) is isomorphic to just $Z_2 = \{0,1\}$.
This technique of first forming the quotient of $R$ by the ideal of elements which are annihilated by elements of $S$, always allows us to get into the situation of 2.44(i), where the natural map is injective, but not always in the situation of 2.44(ii). (A general principle is that forming quotients "commutes with localization".)
So in this example, all our questions are answered by, just thinking about and using, the tool given in the proposition.

As exercise, you might try to compute the result of localizing $Z_{100}$ at powers of $2$, or powers of $3$, or powers of $5$, or powers of $6$. (my answers: $Z_{25}, Z_{100}, Z_4, Z_{25}$.)


$\text{post:7}$

It is of little use to memorize the categorical definition of a kernel, and then when trying to understand the relation between a ring and its localization, not even to ask oneself what is the kernel of the map from $R$ to $S^{-1}\cdot R$. I.e. the first job is not to ask what is the most abstract possible definition of a kernel, but what is the concept useful for: namely it tells you how big is the image of a map, exactly what you were asking above.
When studying localization, the basic result is a characterization of existence of homomorphisms, i.e. any ring map $f:R\to A$ that maps the multiplicative set $S$ into a set of units of $A$, induces a unique ring map $S^{-1}\cdot R\to A$, sending $\frac{x}{y}$ to $\frac{f(x)}{f(y)}$. In particular, if $S$ consists of units of $R$, the identity map $R\to R$ induces a unique map $S^{-1}\cdot R\to R$, inverting the natural map $R\to S^{-1}\cdot R$, which proves your Prop. 2.44(ii) above

What does the following notations mean: $\frac{Z_6}{\{0,2,4\}},\frac{\{1,3\}}{\{0,2,4\}}$? Also, for $Z_{100}$ and since $100=2^2\cdot 5^2$, $Z_{100}$ is only allow to be localised at $2,5$? I mean we want $a\cdot p\equiv 0 {\pmod {100}}$, where $p=2$, or $5$

 

[fresh_42]

What does the following notations mean: $\frac{Z_6}{\{0,2,4\}},\frac{\{1,3\}}{\{0,2,4\}}$?

Here, it is quotient building (I think): $\mathbb{Z}_6/\{0,2,4\}$ after the kernel of the ring homomorphism that was mentioned. It was another way to look at $S^{-1}R.$ This notation is not very common, at least I saw it for the first time. I would avoid it. Either we have a quotient ring in which case you should write $R/I$ or $R/\bigl\langle \ldots \bigr\rangle ,$ or we have a localization where we introduce quotients, in which case you should write it as $S^{-1}R,$ or in case $S=R\setminus P$ with a prime ideal $P\triangleleft R,$ you can write $R_P.$ The horizontal line is misleading, in my opinion.

Also, for $Z_{100}$ and since $100=2^2\cdot 5^2$, $Z_{100}$ is only allow to be localised at $2,5$? I mean we want $a\cdot p\equiv 0 {\pmod {100}}$, where $p=2$, or $5$

We need a multiplicative closed set $S$ to build $S^{-1}R$ and further require $1\in S.$

Now let us assume that $P\triangleleft R$ is a prime ideal. Then we define $S=R\setminus P.$ Since $1\not\in P$ we have $1\in S.$ How about the multiplicative closure? Say, we have $a,b\in S.$ This means that $a\not\in P$ and $b\not\in P.$ Assume that $a\cdot b\not\in S.$ Then $a\cdot b\in P.$ By the definition of a prime ideal, this means that $a\in P$ or $b\in P$ (or both). However, we already know that neither of them is in $P.$ Hence, our assumption $a\cdot b\not\in S$ is false, i.e., $a\cdot b\in S$ and $S$ is multiplicative closed.

This means that whenever we have a prime ideal, then its complement set in a ring is a multiplicative closed set and may serve for a localization. In your example of $R=\mathbb{Z}_{100}$ we have a prime ideal $5\mathbb{Z}/100\mathbb{Z}\triangleleft \mathbb{Z}_{100}$ and we can localize with the multiplicative closed set $S=\{a\in \mathbb{Z}_{100}\,|\,a\not\in 5\mathbb{Z}\}.$ This gives us all quotients with a denominator that is not divisible by $5.$ Same with $2\mathbb{Z}/100\mathbb{Z}\triangleleft \mathbb{Z}_{100}$ which results in all odd denominators.

Note that you cannot combine these two since $10\mathbb{Z}/100\mathbb{Z}\triangleleft \mathbb{Z}_{100}$ is not a prime ideal. Prime ideals are a natural source to define multiplicative closed sets by their set-theoretical complement.

 

[elias001]

@fresh_42 in reply 11, when I asked you about the notations: $\frac{Z_6}{\{0,2,4\}} \cong Z_2 = \{0,1\}, frac{\{1,3\}}{\{0,2,4\}}= \{1\}.$ This $\frac{Z_6}{\{0,2,4\}} \cong Z_2$ is due to $\lvert\frac{Z_6}{\{0,2,4\}} \rvert=\frac{\lvert Z_6\rvert}{\lvert \{0,2,4\}\lvert }=\frac{6}{3}=2?$ But what about for $\frac{\{1,3\}}{\{0,2,4\}}?$

Also for the stated $\textbf{Example:}$ in Underwood's text, of $\Bbb{Z}_6$ at $S=\{3^n:n\geq 0\}=\{1,3\}$ (in our notation: this is $(\Bbb{Z}_6)_3)$ with all its quotient set elements $\Bbb{Z}_6\times \{1,3\}=\{(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(0,3),(1,3),(2,3),(3,3),(4,3),(5,3)\},$ with the equivalence relation $\sim$ and its equivalence classes in $(\Bbb{Z}_6)_3: \{\frac{0}{1},\frac{1}{1}\};$

Lets say we have $(\Bbb{Z}_n)_p$ where $n$ is composite and $p$ is prime. What I want to know is when we want to localize $\Bbb{Z}_n$ at a prime integer $p$. I have some questions about generalizing the example.

(1) By the fundamental theorem of arithmetic, $n$ factors as $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, the prime integer $p$ can only come from one of the elements from the list $\{p_1,p_2,\ldots p_k\}?$

(2) because $\Bbb{Z}_n$ is not an integral domain, then from (1), if the prime integer $p$ can only come from one of the $p_i$s, then $p_i\cdot (p_1^{e_1}p_2^{e_2}\cdots p_i^{e_i-1}\cdots p_k^{e_k})=0$ and $p_i, p_1^{e_1}p_2^{e_2}\cdots p_i^{e_i-1}\cdots p_k^{e_k}$ are considered as zero divisors?

(3) as to the set $S$, the general case would be $S=\{p_i^n:n\geq 0\}=\{1,p_i\}$, where $p_i$ is a prime integer.

(4) for the general case of $\Bbb{Z}_n\times \{1,p_i\}$, we would have $\Bbb{Z}_n\times \{1,p_i\} =\{(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),\ldots (n-1,1), (0,p_i),(1,p_i),(2,p_i),(3,p_i),(4,p_i),(5,p_i),\ldots,(n-1,p_i)\}.$ Is this correct?

(5) Would the set of equivalence classes be still: $\{\frac{0}{1},\frac{1}{1}\}$ and only those two equivalence classes given a specific $p_i$?

 

[fresh_42]

@fresh_42 in reply 11, when I asked you about the notations: $\frac{Z_6}{\{0,2,4\}} \cong Z_2 = \{0,1\}, frac{\{1,3\}}{\{0,2,4\}}= \{1\}.$ This $\frac{Z_6}{\{0,2,4\}} \cong Z_2$ is due to $\lvert\frac{Z_6}{\{0,2,4\}} \rvert=\frac{\lvert Z_6\rvert}{\lvert \{0,2,4\}\lvert }=\frac{6}{3}=2?$ But what about for $\frac{\{1,3\}}{\{0,2,4\}}?$

I am still not sure what this horizontal line actually means. You said quotient building. This makes sense for
$\mathbb{Z}_6/\mathbb{Z}_3\cong \mathbb{Z}_2$ because $\mathbb{Z}_6\cong \mathbb{Z}_3\times \mathbb{Z}_2$ is a direct product, and factoring one leaves the other one remaining. I wouldn't write $\mathbb{Z}_3=\{0,2,4\}$ but ok.

You can write the group (!) isomorphism as follows
$$
\mathbb{Z}_2\times \mathbb{Z}_3=\{0,1\}\times \{0,1,2\}=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}\stackrel{(*)}{\cong} \mathbb{Z}_6=\{0,1,2,3,4,5\},
$$
or multiplicatively
$$
\mathbb{Z}_2\times \mathbb{Z}_3=\{1,a\}\times \{1,b,b^2\}=\{1,b,b^2,a,ab,ab^2,\}\stackrel{(**)}{\cong} \mathbb{Z}_6=\{1,c,c^2,c^3,c^4,c^5\}
$$
where $a^2=b^3=c^6=1.$ I leave it as an exercise to figure out how the group (!) mappings in $(*)$ and $(**)$ work.

However, $frac{\{1,3\}}{\{0,2,4\}}= \{1\}$ doesn't make any sense. You can only factor along subgroups, and $\{0,2,4\}$ is obviously not included in $\{1,3\}.$ So this horizontal line means likely something else, which I don't know, since I've never seen such a notation before. A guess would be $S^{-1}R=\dfrac{R}{S},$ not the quotient but the localization with $S=\{1,3\}.$ But then, $S$ should be in the denominator, not the numerator like $S^{-1}R=\dfrac{R}{S}.$

Also for the stated $\textbf{Example:}$ in Underwood's text, of $\Bbb{Z}_6$ at $S=\{3^n:n\geq 0\}=\{1,3\}$ (in our notation: this is $(\Bbb{Z}_6)_3)$ with all its quotient set elements $\Bbb{Z}_6\times \{1,3\}=\{(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(0,3),(1,3),(2,3),(3,3),(4,3),(5,3)\},$ with the equivalence relation $\sim$ and its equivalence classes in $(\Bbb{Z}_6)_3: \{\frac{0}{1},\frac{1}{1}\};$

This is very confusing, and I don't understand it. If you take the multiplicatively closed set $S= \{1,3\} \subseteq \{0,1,2,3,4,5\}$ as rings, where $3\cdot 3=3,$ then shouldn't we get twelve elements in $S^{-1}\mathbb{Z}_6$, for each possible new denominator six elements? Maybe there are only two equivalence classes left, but I don't see why without any calculations.

Please don't write this as $R_p$ if there are specific integers involved.

Lets say we have $(\Bbb{Z}_n)_p$ where $n$ is composite and $p$ is prime. What I want to know is when we want to localize $\Bbb{Z}_n$ at a prime integer $p$. I have some questions about generalizing the example.

I assume that $p\,|\,n$ so that $n\mathbb{Z}\subseteq p\mathbb{Z}$ and we can build $p\mathbb{Z}/n\mathbb{Z}$ such that $p\mathbb{Z}_n \subseteq \mathbb{Z}_n$ is an ideal. You should prove that it is a prime ideal in $\mathbb{Z}_n.$

(1) By the fundamental theorem of arithmetic, $n$ factors as $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, the prime integer $p$ can only come from one of the elements from the list $\{p_1,p_2,\ldots p_k\}?$

Yes, because we need an ideal in $\mathbb{Z}_n=\mathbb{Z}(n\cdot\mathbb{Z}$ and ideals look like $I/n\mathbb{Z}$ where $I\subseteq \mathbb{Z}$ is an ideal in $\mathbb{Z}$ that contains $n\cdot \mathbb{Z}.$ Now, $\mathbb{Z}$ is a principal ideal domain, so all ideals are principal ideals and $I=a\cdot \mathbb{Z}.$ So what does it mean, that $n\cdot \mathbb{Z}\subseteq a\cdot \mathbb{Z}?$ It means, that $a$ has to divide $n.$ In the case $a=p$ is our prime number, we need $p\,|\,n$ which means $p$ is one of these prime factors of $n.$

(2) because $\Bbb{Z}_n$ is not an integral domain, then from (1), if the prime integer $p$ can only come from one of the $p_i$s, then $p_i\cdot (p_1^{e_1}p_2^{e_2}\cdots p_i^{e_i-1}\cdots p_k^{e_k})=0$ and $p_i, p_1^{e_1}p_2^{e_2}\cdots p_i^{e_i-1}\cdots p_k^{e_k}$ are considered as zero divisors?

Yes.

(3) as to the set $S$, the general case would be $S=\{p_i^n:n\geq 0\}=\{1,p_i\}$, where $p_i$ is a prime integer.

This is not the general case. Say we have $n=24$ and the multiplicative set generated by $3.$ This set contains $\{1=3^0,3=3^1,9=3^2\}.$ So higher powers are possible as long as they don't exceed $n.$

(4) for the general case of $\Bbb{Z}_n\times \{1,p_i\}$, we would have $\Bbb{Z}_n\times \{1,p_i\} =\{(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),\ldots (n-1,1), (0,p_i),(1,p_i),(2,p_i),(3,p_i),(4,p_i),(5,p_i),\ldots,(n-1,p_i)\}.$ Is this correct?

See my previous remark.

(5) Would the set of equivalence classes be still: $\{\frac{0}{1},\frac{1}{1}\}$ and only those two equivalence classes given a specific $p_i$?

I don't know what your equivalence relation is. I assume we are still talking about $S^{-1}R,$ a localization, and not quotients, which I thought you meant by that horizontal line when I began my answer.

This means we consider $R\times S$ and define
$$
(r,s)\sim (r',s') \Longleftrightarrow (rs'-r's)\cdot t=0 \text{ for some }t\in S.
$$
Take $R=\mathbb{Z}_{24}$ and $S=\{1,3,9\}.$ Then there are $72$ elements in $R\times S.$

First, start with the example $\mathbb{Z}_6\times \{1,3\}$ with $12$ elements and figure out which are equivalent and, then do the same for $\mathbb{Z}_{24}\times \{1,3,9\}.$

Maybe you were right and the example $\mathbb{Z}_6\times \{1,3\}$ has indeed only two equivalence classes, but I do not see it without calculations.

 

[elias001]

@fresh_42 for question 1., which came from reply 11. I am only going by what mathwonk said in post 6:

The example they gave on localizing Z/6 is a case where the criteria do not seem to apply, but actually they can be tweaked until they do apply. Namely since the kernel of the map Z/6-->S^-1.Z/6 is {0,2,4}, the result seems to be that we are essentially localizing the quotient (Z/6)/{0,2,4} ≈ Z/2 = {0,1}, at the multiplicative set {1,3}/{0,2,4} = {1} which is a unit in Z/2, so the result, again by 2.44(ii) is isomorphic to just Z/2 = {0,1}.

Why can I not use $R_p$ when integers are involved? You make it sound like I am about to commit some mathematical deviant act. Should I go to a confession with a math priest to ask for absolution?

Given what your corrections are at (3) about $S=\{p_i^m: m\geq 0\}=\{1,p_i,p_i^2,\ldots,p_i^m\}$, would (4) be changed to
$\Bbb{Z}_n\times \{1,p_i,p_i^2,\ldots,p_i^m\}=\{(0,1),(1,1),(2,1),\ldots,(n-1,1),(0,p_i),(1,p_i),(2,p_i),\ldots,(n-1,p_i),(0,p_i^2),(1,p_i^2),(2,p_i^2),\ldots,(n-1,p_i^2),\ldots,(0,p_i^m),(1,p_i^m),(2,p_i^m),\ldots,(n-1,p_i^m)\}?$

For (5), I am referring to localization when asking about the number of equivalence classes. I thought there must be some convenient ways to list and count them, since they there are repeats. For large composite $n$ and localization at a large prime $p$, How does one list them all, make sure there are no repeats, and give a number for counting the number of equivalence classes.

 

[fresh_42]

@fresh_42 for question 1., which came from reply 11. I am only going by what mathwonk said in post 6:

... which I didn't understand either. The linguistic difficulty is that $R\longrightarrow R/I$ is called a quotient and a localization $S^{-1}R$ is also intended to establish "quotients" in the arithmetical sense that the elements of $S$ become denominators. So, I guess I stumbled upon the double meaning of the word quotient.

However, $S=\{0,2,4\}$ doesn't contain the element $1.$ and with $0$ in it, it hardly is a multiplicative closed set, at least not in the usual sense.

Why can I not use $R_p$ when integers are involved? You make it sound like I am about to commit some mathematical deviant act. Should I go to a confession with a math priest to ask for absolution?

You can, it is only confusing, and notations are meant to be clear, not misleading. E.g., if you write $\left(\mathbb{Z}_6\right)_3$ then it is unclear what the $3$ stands for. It is not even clear whether you speak about the ring $\mathbb{Z}_6$ or the group $\mathbb{Z}_6.$

I know the notations $S^{-1}R$ or $R\times S/\sim$ or in case $S=R\setminus P$ where $P\subseteq R$ is a prime ideal, the notation $R_P.$ But here, we have no prime ideal, but $S=\{s^n\,|\,n\in \mathbb{N}_0\}$ instead. So you abbreviate $S$ with $s$ and right $R_S=R_s$ which is not the usual notation. Wikipedia suggests in that case where $S$ are the powers of a single element to write $S^{-1}R=R\left[\frac{1}{s}\right]=R\left[s^{-1}\right]$ or $S^{-1}R=R[x]/\bigl\langle sx-1 \bigr\rangle.$

This would lead in our case to $\mathbb{Z}_6\left(\frac{1}{3}\right)$ or $\mathbb{Z}_6[x]/\bigl\langle 3x-1 \bigr\rangle.$ This is way better than $\left(\mathbb{Z}_6\right)_3$ where the second index isn't self-explaining. We already have a quotient (factor) ring $\mathbb{Z}_6=\mathbb{Z}/6\cdot\mathbb{Z}$ in the first index. So what does ${}_3$ in the second index mean, $3\cdot \mathbb{Z}/6\cdot \mathbb{Z}$?
I haven't checked whether these two are the same ring, nor did you provide a proof. Hence, it is not self-explanatory and at best a statement that needs to be proven. $\mathbb{Z}_6\left[\frac{1}{3}\right]$ is better, since it describes what we did: take the ring $\mathbb{Z}_6$ and adjoint the element $1/3.$ It shows that we speak of rings and not of groups, and that we added the element $1/3$ with the rule $(1/3)^2=1/3.$

It is hard enough to cope with all the different notations in all your different sources, since every author, including the posts here at PF has their own preferences.

Given what your corrections are at (3) about $S=\{p_i^m: m\geq 0\}=\{1,p_i,p_i^2,\ldots,p_i^m\}$, would (4) be changed to
$\Bbb{Z}_n\times \{1,p_i,p_i^2,\ldots,p_i^m\}=\{(0,1),(1,1),(2,1),\ldots,(n-1,1),(0,p_i),(1,p_i),(2,p_i),\ldots,(n-1,p_i),(0,p_i^2),(1,p_i^2),(2,p_i^2),\ldots,(n-1,p_i^2),\ldots,(0,p_i^m),(1,p_i^m),(2,p_i^m),\ldots,(n-1,p_i^m)\}?$

Yes, but nobody would write that. Plus, it isn't clear how $m$ is defined. I suggest $R\left[\frac{1}{p_i}\right]$ because that is what the brackets $\left[\cdot\right]$ mean. And you don't have to bother equivalence classes, since the new ring inherits the arithmetic rules from $R.$

If $R$ is an integral domain, the equivalence relation is particularly easy:
$$
(r,s^n)\sim (r',s^m)\Longleftrightarrow \left(rs^m-r's^n\right)s^k=0 \Longleftrightarrow \left(rs^m-r's^n\right)=0 \Longleftrightarrow rs^m=r's^n
$$
because $s^k\neq 0$ for all $k$ and $R$ is an integral domain, which allows us to "divide" by $s^k.$
Things are a bit more complicated if $R$ is not an integral domain, since $s^k$ could be a zero divisor, so that we may not cancel it.

For (5), I am referring to localization when asking about the number of equivalence classes. I thought there must be some convenient ways to list and count them, since they there are repeats. For large composite $n$ and localization at a large prime $p$, How does one list them all, make sure there are no repeats, and give a number for counting the number of equivalence classes.

See my previous remark.

Let's see what my example says. We have $R=\mathbb{Z}_{24}$ and $S=\{3^n\,|\,n\in \mathbb{N}_0\}=\{1,3,3^2\}$ since $3^3=27\equiv 3\pmod{24}$ and we run in circles by all other powers. The equivalence relation is defined by
\begin{align*}
\left(a,3^n\right)\sim \left(b,3^m\right) &\Leftrightarrow \left(3^ma-3^nb\right)3^k=0 \;\text{ for some }\;k\in \mathbb{N}_0\\
&\Leftrightarrow 3^{m+k}a-3^{n+k}b=0 \;\text{ for some }\;k\geq 0\\
&\Leftrightarrow 3^{m+k}a=3^{n+k}b\;\text{ for some }\;k\geq 0
\end{align*}
This is a bit nasty to check for all possible values $k,m,n\in \{0,1,2\}$ and $a,b\in \mathbb{Z}_{24}.$

So let's take the other approach by looking at $\mathbb{Z}_{24\left[\frac{1}{3}\right]}.$ Since $\frac{1}{3}\cdot \frac{1}{3}=\frac{1}{3}$ we have all polynomials $a+\frac{1}{3}b$ of degree one and coefficients from $\mathbb{Z}_{24}.$ This gives us $24$ possibilities for $a$ and $24-(24/3)=16$ possibilities for $b,$ namely the elements that are not divisible by $3.$ This makes $40$ equivalence classes total. But do we still have repetitions among those elements?

I suggest you think about that question. Hint: consider the cases $b_i\equiv 1\pmod{3}$ and $b_j\equiv 2\pmod{3},$ i.e., all four possible pairs $\frac{1}{3}b_i\stackrel{?}{\sim}\frac{1}{3}b_j.$

And of course, we still do not have a formal proof why $\{s^k\,|\,k\in \mathbb{N}_0\}^{-1}\mathbb{Z}_n =\mathbb{Z}_n\left[\frac{1}{s}\right].$

 

[elias001]

@fresh_42 Here is what I got about how to generalize the example from Underwood's text. Before I begin, I don't like the notation $S^{-1}A$. Does this part $S^{-1}$ of the notation denote elements in it to mean $\frac{1}{s}\in S^{-1}, s\in S?$

Some definitions.$\\\\$

The $p$-adic valuation $v_p(n)$ is defined as the highest exponent $e$ such that $p^e$ divides $n$.$\\\\$ Formally:$\\\\$

$v_p(n) = \max \{ e \geq 0 : p^e \mid n \}.\\\\$

For $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, we have $v_{p_i}(n) = e_i$. In the context of $(\mathbb{Z}_n)_p$, if $p = p_i$, then $e = v_p(n) = e_i$, and the number of equivalence classes is: $p^{v_p(n)}.\\\\$

$\textbf{Note:}$ The valuation $v_p(n)$ determines the size of the $p$-primary component, directly giving the number of classes.$\\\\$

Definition of the $p-$primary Component.$\\\\$

The $p$-primary component of $\mathbb{Z}_n$ is the localization $(\mathbb{Z}_n)_p$, isomorphic to $\mathbb{Z}_{p^{v_p(n)}}$, consisting of fractions $\frac{a}{b}$ with $b \in S = \{1, p, \ldots, p^{e-1}\}$. It isolates the part of $\mathbb{Z}_n$ associated with the prime $p.\\\\$

Note that the $p$-primary component $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}$ has order $p^e$, where $e = v_p(n)$. The number of equivalence classes is thus:$.\\\\$

$p^{v_p(n)}$ directly tied to the size of the $p$-primary component.$\\\\$

We generalize the example of $\mathbb{Z}_6)_3$ to the general case of localization of $\mathbb{Z}_n)_p$ in several steps, and we provide two computational examples to illustrate the steps.$\\\\$

3. Generalization of the set $S.\\\\$

The set $S$ in the example of $\mathbb{Z}_6)_3$ is $S = \{3^n : n \geq 0\} = \{1, 3\}$ in $\mathbb{Z}_6$. For the general case of $(\mathbb{Z}_n)_p$, we define:$\\\\$

$S = \{p^k : k \geq 0\} \subset \mathbb{Z}_n,\\\\$

where $p$ is a prime dividing $n$, and exponents $k$ are taken modulo the ring $\mathbb{Z}_n$. Since $n = p^e m$ with $\gcd(p, m) = 1$ and $e = v_p(n), p_i$ (the $p$-adic valuation of $n$), we have $p^e \equiv 0 \pmod{n}$. Thus:$\\\\$

$S = \{p^0, p^1, p^2, \ldots, p^{e-1}\} = \{1, p, p^2, \ldots, p^{e-1}\},\\\\$

since $p^e \equiv 0 \pmod{n}$, and higher powers do not produce new elements. The set $S$ is multiplicatively closed because $1 \in S$ (for $k = 0$) and the product $p^k \cdot p^l = p^{k+l} \in S$ if $k + l < e$, or $p^{k+l} \equiv 0 \notin S$, but only non-zero elements are included.$\\\\$

2. Then we list the elements of $\mathbb{Z}_n \times S\\\\$

The set $\mathbb{Z}_n \times S$ is the Cartesian product:$\\\\$

$\mathbb{Z}_n \times S = \{(a, s) : a \in \mathbb{Z}_n, s \in S\},\\\\$

where $\mathbb{Z}_n = \{0, 1, 2, \ldots, n-1\}$ and $S = \{1, p, p^2, \ldots, p^{e-1}\}$. Thus:$\\\\$

$\mathbb{Z}_n \times S = \{(a, p^k) : a = 0, 1, \ldots, n-1, k = 0, 1, \ldots, e-1\}.\\\\$

The number of elements is $|\mathbb{Z}_n| \cdot |S| = n \cdot e$, since $|S| = e.\\\\$

3. We then list the equivalence classes in $(\mathbb{Z}_n)_p\\\\$

The equivalence relation$\\\\$ on $\mathbb{Z}_n \times S$ is defined as $(a, b) \sim (c, d)$ if there exists $s \in S$ such that:

$s (ad - bc) \equiv 0 \pmod{n}.\\\\$

The localization $(\mathbb{Z}_n)_p = S^{-1} \mathbb{Z}_n$ is the set of equivalence classes $\frac{a}{b}$, where $(a, b) \in \mathbb{Z}_n \times S$. To list the equivalence classes, note that $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}$, where $e = v_p(n)$. Thus, the equivalence classes correspond to residues $\{0, 1, \ldots, p^e - 1\}$ in $\mathbb{Z}_{p^e}.\\\\$

To find representatives, consider the map $\lambda:\mathbb{Z}_n \to (\mathbb{Z}_n)_p, a \mapsto \frac{a}{1}$. The equivalence classes are:

$$\{\frac{a}{1} : a = 0, 1, \ldots, p^e - 1\}$$

where $\frac{a}{1} = \frac{c}{d}$ if there exists $s = p^k \in S$ such that $p^k (ad - c) \equiv 0 \pmod{n}$. Since $n = p^e m$, this implies $ad \equiv c \pmod{p^e}.\\\\$

4. We then eliminate the repeats of equivalence c'lasses$\\\\$

To eliminate repeats, choose representatives $\frac{a}{1}$ for $a = 0, 1, \ldots, p^e - 1$. For any $(c, d) \in \mathbb{Z}_n \times S$, find an $a$ for which:$\\\\$

$\frac{c}{d} = \frac{a}{1} \implies \exists s = p'^k \in S \text{ such that } p^k (c - ad) \equiv 0 \pmod{n}.\\\\$

Since $n = p^e m$, this means $c \equiv ad \pmod{p^e}$. Solve for $a \equiv c d^{-1} \pmod{p^e}$, where $d = p^k$ is invertible modulo $p^e$ (since $\gcd(p^k, p^e) = p^k$). We calculate the inverse via employing the extended Euclidean algorithm. The steps can be summarized in a$\\\\$

$\textbf{systematic processes:}\\\\$

1'. For each $(c, p^k) \in \mathbb{Z}_n \times S$, compute $d^{-1} \pmod{p^e}.\\\\$

2'. Compute $a \equiv c d^{-1} \pmod{p^e}.\\\\$

3'. The class of $(c, p^k)$ is $\frac{a}{1}.\\\\$

This ensures each class is represented exactly once by $\frac{a}{1}, a = 0, 1, \ldots, p^e - 1.\\\\$

$\textbf{Note:}$ The choice of representatives modulo $p^e$ avoids duplication by mapping all pairs to a unique residue.$\\\\$

5. Making sure the elements in the list is exhaustive$\\\\$

The list $\{\frac{0}{1}, \frac{1}{1}, \ldots, \frac{p^e - 1}{1}\}$ is exhaustive because:$\\\\$

- $\textbf{Surjectivity:}$ Every $(c, p^k) \in \mathbb{Z}_n \times S$ is equivalent to some $\frac{a}{1}$, as shown above.$\\\\$

- $\textbf{Injectivity:}$ If $\frac{a}{1} = \frac{b}{1}$, then there exists $p^k \in S$ such that $p^k (a - b) \equiv 0 \pmod{n}$, implying $a \equiv b \pmod{p^e}$, so $a = b$ in $\{0, 1, \ldots, p^e - 1\}.\\\\$

Since $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}$, and $\mathbb{Z}_{p^e}$has exactly $p^e$ elements, the list covers all classes.$\\\\$

6. The number of equivalence classes$\\\\$

The number of equivalence classes in $(\mathbb{Z}_n)_p$ is the order of the ring $\mathbb{Z}_{p^e}$, which is: $p^e, \text{ where } e = v_p(n).\\\\$

This is due to the isomorphism $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}.\\\\$

$\textbf{Note:}$ The formula $p^{v_p(n)}$ counts the elements in the $p$-primary component.$\\\\$

Showing $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}.\\\\$

7. Functions $f$ and $f^{-1}$ for the Isomorphism $(\mathbb{Z}_n)_p \cong \mathbb{Z}_{p^e}\\\\$

Define the isomorphism $f: (\mathbb{Z}_n)_p \to \mathbb{Z}_{p^e}$ and its inverse:$\\\\$

$f\left( \frac{a}{b} \right) = a b^{-1} \pmod{p^e}, \quad b = p^k \in S,\\\\$

where $b^{-1}$ is the inverse of $b$ modulo $p^e$. The inverse is:$\\\\$

$f^{-1}(c) = \frac{c}{1}, \quad c \in \mathbb{Z}_{p^e}.\\\\$

$\textbf{Computing compositions:} f \circ f^{-1},f^{-1} \circ f\\\\$

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

$f(f^{-1}(c)) = f\left( \frac{c}{1} \right) = c \cdot 1^{-1} \equiv c \pmod{p^e},\\\\$

so $f \circ f^{-1} = \text{id}_{\mathbb{Z}_{p^e}}.\\\\$

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

$f^{-1}\left( f\left( \frac{a}{b} \right) \right) = f^{-1}(a b^{-1}) = \frac{a b^{-1}}{1} = \frac{a}{b}= \text{id}_{(\mathbb{Z}_n)_p},\\\\$

because $\frac{a b^{-1}}{1} = \frac{a}{b}$ in $(\mathbb{Z}_n)_p$, as $b (a b^{-1} \cdot 1 - a \cdot 1) = b (a b^{-1} - a) = a - a = 0.\\\\$

8. $\textbf{Computational examples:}\\\\$

$\textbf{Examples for: }n = 232792560\\\\$

We compute $n$'s factorization:$\\\\$

The factorization of $232792560$ is $232792560 = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 47 \cdot 2207.\\\\$

$\mathbf{p = 2: }\\\\$

- For $p = 2, e = v_2(n) = 4$, so $S = \{2^0, 2^1, 2^2, 2^3\} = \{1, 2, 4, 8\}$. Then:$\\\\$

$\mathbb{Z}_{232792560} \times S = \{(a, s) : a = 0, 1, \ldots, 232792559, s \in \{1, 2, 4, 8\}\},\\\\$

with $232792560 \cdot 4 = 931170240$ elements.$\\\\$

- For $p = 2, e = 4$, so $p^e = 2^4 = 16$. The classes are $\{\frac{0}{1}, \frac{1}{1}, \ldots, \frac{15}{1}\}$ in $\mathbb{Z}_{16}.\\\\$

- And, $e = v_2(n) = 4$, and so there are $2^4 = 16$ classes.$\\\\$

- Also, $p = 2, e = 4, p^e = 16\\\\$.

Let $\frac{a}{b} = \frac{5}{2}: q\left( \frac{5}{2} \right) = 5 \cdot 2^{-1} \pmod{16}.\\\\$

Compute $2^{-1} \pmod{16}: 2 \cdot 8 = 16 \equiv 0$, but $2 \cdot 9 = 18 \equiv 2 \pmod{16}$, so no inverse exists unless restricted to odd numbers. For in $\mathbb{Z}_{16}$, we use the extended Euclidean algorithm. Also we verify the representative class:$\\\\$

$q^{-1}(5) = \frac{5}{1}.\\\\$

$\mathbf{p = 3: }\\\\$

- For $p = 3, e = v_3(n) = 2$, so $S = \{1, 3\}$. Then:$\\\\$

$\mathbb{Z}_{232792560} \times S = \{(a, s) : a = 0, 1, \ldots, 232792559, s \in \{1, 3\}\},$$\\\\$

with $232792560 \cdot 2 = 465585120$ elements.$\\\\$

- And, $p = 3, e = 2$, so $p^e = 3^2 = 9$. The classes are $\{\frac{0}{1}, \frac{1}{1}, \ldots, \frac{8}{1}\}$ in $\mathbb{Z}_9.\\\\$

- Also, $p = 3, e = v_3(n) = 2$, and so there are $3^2 = 9$ classes.$\\\\$

$q^{-1}(5) = \frac{5}{1}.\\\\$

- Lastly, $p = 3, e = 2, p^e = 9, \frac{a}{b} = \frac{5}{3}:\\\\$

$q\left( \frac{5}{3} \right) = 5 \cdot 3^{-1} \pmod{9}.\\\\$

Since $3 \cdot 3 = 9 \equiv 0$, the inverse is: $3 \cdot 6 = 18 \equiv 0$, but compute modulo 9 correctly later.$\\\\$

 

[elias001]

@fresh_42 i forgot to clarify, if points 4. to point 7., look ok. If anything you think are missing proof justifications, I am not worry about that. I would like your opinion on if the steps were there, does the steps used sound plausible especially on the elimination of repeated representative classes and the use of p-adic valuation map for counting the number of equivalence classes.