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May 16th, 2018, 02:19 PM   #21
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Quote:
 Originally Posted by zylo modules are irrelevant, quotient rings are modules.
No, modules are different than rings. Two different mathematical objects.

Quote:
 Originally Posted by zylo R/I, R/J and (J:I)/I are groups (quotient rings)
Groups are not rings.

Zylo you're just throwing around words you don't understand. Pick up a pdf of a book on abstract algebra and start studying.

May 16th, 2018, 02:31 PM   #22
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Quote:
 Originally Posted by zylo Edit. Unless the set of all homeomorphisms R/I to R/J is an R-module (set with a particular structure) Why is the set of all homeomorphisms R/I to R/J an R module?
Yep, you got it. To see why this is plausible, first think about vector spaces: given vector spaces V and W over a field F, let S be the set of all linear maps from V to W. Then S is itself a vector space over F. It's clear what the "sum" of two linear maps means, and what it means to multiply a linear map by a scalar. Then checking the axioms are satisfied is a straightforward exercise.

Basically the same idea works with modules over a commutative ring (note: while R/I and R/J are rings in their own right, we're treating them as R-modules here. This is why there's a subscript "R" in the Hom notation).

Last edited by cjem; May 16th, 2018 at 02:33 PM.

May 17th, 2018, 06:38 AM   #23
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Quote:
 Originally Posted by zylo Ref (OP): Let R be a commutative ring with 1. If I, J $\displaystyle \triangleleft$ R then we define (J : I) = {x $\displaystyle \epsilon$ R | xI $\displaystyle \subset$ J} a) Show that (J : I) $\displaystyle \triangleleft$ R and J $\displaystyle \subset$ (J : I) b) Show that $\displaystyle Hom_{R}(R/I,R/J) \cong (J : I)/J$ via the map $\displaystyle \phi \mapsto \phi (1+I)$
So where we are at paradigmatically is:

Show that a set of things H with an operation is isomorphic to a set of things U with an operation:

Set of things H ($\displaystyle Hom_{R}(R/I,R/J)$): Set of homeomorphisms (mappings) from R/I to R/J with a + operation.
Set of things U, (J:I)/J with a + operation
Isomorphic mapping from H to U.

To establish the isomorphism you need to know the mapping from H to U
-------------------------------------------------------------------------------------
If h1 and h2 are elements of H, which map to elements u1 and u2 of U, then
h1+h2 maps to u1+u2 and u1+u2 maps to h1+h2.
-------------------------------------------------------------------------------------

Technically, it looks like all you need to know is what an element of H maps to in U. Is that what $\displaystyle \phi$ is all about?

WHAT is $\displaystyle \phi$?

Literally, $\displaystyle \phi \mapsto \phi (1+I)$ says map an element $\displaystyle \phi$ of H to the element $\displaystyle \phi$(I+1) of U.

EDIT: I+1 is not an element. It may mean i+1 where i is an element of I.

Last edited by zylo; May 17th, 2018 at 07:09 AM.

May 17th, 2018, 08:45 AM   #24
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Quote:
 Originally Posted by zylo EDIT: I+1 is not an element. It may mean i+1 where i is an element of I.
It's a coset.

https://en.wikipedia.org/wiki/Coset

 May 17th, 2018, 10:32 AM #25 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 I found coset (wiki), and its use to define quotient ring in http://www.maths.usyd.edu.au/u/de/AG...ra/pp22-51.pdf But I still don't understand how $\displaystyle \phi$ works in OP. What member of the set of homeomorphisms is associated with a member of {J:I)/I Excerpts from ggogling are below. $\displaystyle \cong$ isomorphic $\displaystyle \triangleleft$ ideal of $\displaystyle \mapsto$ maps to Definition Let R be a commutative ring. A subset I of R is an ideal if • If a, b ∈ I, then a + b ∈ I. • If a ∈ I and b ∈ R, then ab ∈ I. A quotient ring (also called a residue-class ring) is a ring that is the quotient of a ring A and one of its ideals a, denoted A/a. In general, a quotient ring is a set of equivalence classes where [x]=[y] iff x-y in a A module over a ring (R module) is a generalization of the notion of vector space over a field, wherein the corresponding scalars are the elements of an arbitrary given ring (with identity) and a multiplication (on the left and/or on the right) is defined between elements of the ring and elements of the module. In algebra, given a module and a submodule, one can construct their quotient module. This construction, described below, is analogous to how one obtains the ring of integers modulo an integer n, see modular arithmetic. It is the same construction used for quotient groups and quotient rings. Given a module A over a ring R, and a submodule B of A, the quotient space A/B is defined by the equivalence relation a ~ b if and only if b − a is in B, for any a and b in A. The elements of A/B are the equivalence classes [a] = { a + b : b in B }. The addition operation on A/B is defined for two equivalence classes as the equivalence class of the sum of two representatives from these classes; and in the same way for multiplication by elements of R. In this way A/B becomes itself a module over R, called the quotient module. In symbols, [a] + [b] = [a+b], and r·[a] = [r·a], for all a,b in A and r in R. In algebra, a module homomorphism is a function between modules that preserves module structures. Explicitly, if M and N are left modules over a ring R, then a function {\displaystyle f:M\to N} f:M\to N is called a module homomorphism or an R-linear map if for any x, y in M and r in R, f(x+y)=f(x)+f(y), f(rx)=rf(x).} f(rx)=rf(x). If M, N are right modules, then the second condition is replaced with f(xr)=f(x)r. The pre-image of the zero element under f is called the kernel of f. The set of all module homomorphisms from M to N is denoted by Hom_{R}(M, N). It is an abelian group (under pointwise addition) but is not necessarily a module unless R is commutative. ------------------------------------------------------------------------------ So at this point I don't even know if the OP is about quotient rings or modules. I have stepped into a morass of ambiguous definitions. I can see why those asian students who have to memorize a unique symbol for every word in their language do well at this. I note congruence, congruence classes, and congruence algebra are pretty straight forward until defined into oblivion. Reminds me of an organ concert I once went to. The organist performed a modern piece that was all over the organ almost at once. His manual dexterity was phenominal. After that, I imagine any organ students in the audience walked out and became day laborers.
May 17th, 2018, 11:46 AM   #26
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Quote:
 Originally Posted by zylo I have stepped into a morass of ambiguous definitions.
You have refuted your own original point.

You claimed that everything in math is obvious because it's one definition built on another, therefore an intelligent person should be able to untangle any complicated problem by drilling down to the definitions.

This is in fact true in principle.

But in practice, one acquires experience and context as one learns. You learn what a quotient is in beginning abstract algebra; and by the time you get to a more advanced class in ring theory, you have become so familiar with the concept of quotients that you no longer remember that there was a time you struggled with the concept.

You have discovered for yourself that context and experience, gained over months and years of studying and doing problems, are how you learn math. The fact that in theory you could drill it all down to a mechanical unwinding of definitions is true but irrelevant to how we understand things.

And your ethnic slur is most definitely not appreciated here. Not by me, anyway.

 May 17th, 2018, 12:57 PM #27 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 In certain cultures you grow up having to do a lot of memorization. That's not an ethnic slur. Calling it an ethnic slur is defamation. I'm sure the organist spent years developing his technigue. But enough politics. You quote wiki: gH = { gh : h an element of H } is the left coset of H in G with respect to g Given that, how do you conclude that (1+I) is a coset? (1+I)=1I , and NumberGroup is a unique designation that tips off a coset? So back to OP b): What does $\displaystyle \phi \mapsto \phi$(1+I) mean? Presumably $\displaystyle \phi$ is a member of HomR(), ie, a Homeomorphism. Why is $\displaystyle \phi$(1+I) a member of (J:I)/I
May 17th, 2018, 01:25 PM   #28
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Quote:
 Originally Posted by zylo I'm sure the organist spent years developing his technigue.
Are you playing with your organ again?

Quote:
 Originally Posted by zylo You quote wiki: gH = { gh : h an element of H } is the left coset of H in G with respect to g
I haven't quoted Wiki. You quoted Wiki. I did give you the link to a Wiki page but earlier I suggested that you get a book on abstract algebra. I reiterate that suggestion.

Quote:
 Originally Posted by zylo Given that, how do you conclude that (1+I) is a coset? (1+I)=1I , and NumberGroup is a unique designation that tips off a coset?
gH is multiplicative notation. 1 + I is additive notation. The addition operation in modules is commutative, so the additive notation is used. Multiplicative notation is typically used for arbitrary groups, which need not be commutative. But when the group operation is commutative (such groups are called Abelian after Niels Henrik Abel) the additive notation is used by convention.

How would anyone know that convention? By studying basic abstract algebra instead of frustratingly trying to work backwards from Wikipedia.

Quote:
 Originally Posted by zylo So back to OP
Better to start with the ring of integers mod 5 and the corresponding $\mathbb Z$-module. Once you grok that the rest will be much more straightforward. You have to learn to walk before you can run, contrary to your earlier claim that simply powering through a tower of definitions is sufficient.

By the way a module is like a vector space except that the scalars live in a ring instead of a field. A ring is an algebraic system in which you can add, subtract, and multiply, like the integers. A field is a ring in which you can also divide, like the rationals or the reals.

Last edited by Maschke; May 17th, 2018 at 01:31 PM.

May 17th, 2018, 02:59 PM   #29
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Quote:
 Originally Posted by Maschke gH is multiplicative notation.1 + I is additive notation.
Both wrong

gH is simply a symbol which stands for coset, according to wiki:
"gH = { gh : h an element of H } is the left coset of H in G with respect to g"
where gh is not multiplication but a generic relation g*h, which could be std addition, multiplication, or some other relation between elements.

There is no such thing as sum of 1 plus a group. If it's a coset, it means iI, as defined above.

Still no explanation of $\displaystyle \phi$.

May 17th, 2018, 03:03 PM   #30
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Quote:
 Originally Posted by zylo Both wrong gH is simply a symbol which stands for coset, according to wiki: "gH = { gh : h an element of H } is the left coset of H in G with respect to g" where gh is not multiplication but a generic relation g*h, which could be std addition, multiplication, or some other relation between elements. There is no such thing as sum of 1 plus a group. If it's a coset, it means iI, as defined above.
Good God Zylo I'm trying to help you. Are you here to learn or to argue?

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