My Math Forum Isomorphism

 Abstract Algebra Abstract Algebra Math Forum

 May 17th, 2018, 08:20 PM #31 Senior Member   Joined: Aug 2012 Posts: 2,201 Thanks: 646 ps -- Zylo here is a more straightforward response to your question, assuming that by saying, "Wrong" that is simply Zylo-speak for, "I don't understand, can you please explain it?" Why sure, I'll be happy to! You are correct that the group operation is not necessarily ordinary multiplication. For example in the group of invertible 2x2 matrices, the group operation is matrix multiplication. In the group of permutations on 3 letters, the operation is composition of permutations. In the group of integers, the operation is ordinary addition. In the group of nonzero positive real numbers, the operation is ordinary multiplication. Now, we all agree I hope that whatever a group's operation is, we can classify every group as either satisfying the relation $xy = yx$ for all $x$ and $y$ in the group; or not. In the former case, we call the group Abelian and we typically write the operation as $+$. If we write a group's operation as $gh$ or $g \cdot h$, we call that "multiplicative notation." If we write the group's operation as $+$, we refer to this as "additive notation." We are not implying that the group operation is necessarily ordinary multiplication or addition. Rather, in this context "multiplicative notation" and "additive notation" simply refer to whether we notate the group operation $gh$ or $g + h$. By convention we use additive notation for Abelian groups and multiplicative notation for non-Abelian groups. And since modules are generally required to have commutative addition (remember that modules are like vector spaces over a ring instead of a field), the additive notation is appropriate. And this notational convention extends to cosets. Hope this is helpful. Last edited by Maschke; May 17th, 2018 at 08:26 PM.
 May 18th, 2018, 07:39 AM #32 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 From wiki "In mathematics, a group is an algebraic structure consisting of a set of elements equipped with an operation that combines any two elements to form a third element and that satisfies four conditions called the group axioms, namely closure, associativity, identity and invertibility." Which is what I said in last post, when you insisted on calling gh multiplication. But I can sympathize, there is just so much to remember. Still waiting for an explanation of $\displaystyle \phi$ in OP b). Perhaps cjem, who claimed the solution was obvious in 3 secs, might condescend to explain it.
May 18th, 2018, 09:41 AM   #33
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 311
Thanks: 109

Math Focus: Number Theory, Algebraic Geometry
Quote:
 Originally Posted by zylo From wiki "In mathematics, a group is an algebraic structure consisting of a set of elements equipped with an operation that combines any two elements to form a third element and that satisfies four conditions called the group axioms, namely closure, associativity, identity and invertibility." Which is what I said in last post, when you insisted on calling gh multiplication. But I can sympathize, there is just so much to remember.
He said that it's multiplicative notation, not that the operation is multiplication... That, when you write the group operation as "gh", you're writing it in the same way you write multiplication, even though it might not actually be multiplication.

Quote:
 Originally Posted by zylo Still waiting for an explanation of $\displaystyle \phi$ in OP b). Perhaps cjem, who claimed the solution was obvious in 3 secs, might condescend to explain it.
For the third time, I only claimed the specific issue Mona had with question (b) could be resolved that easily.

Anyway, the map in the OP works as follows:

Take any element $\varphi$ in $\text{Hom}_R{R/I,R/J}$ (recall: this means $\varphi$ is an $R$-module homomorphism from $R/I$ to $R/J$). The identity element of $R/I$ is the coset $1 + I$. Since $1 + I$ is an element of $R/I$, we can apply $\varphi$ to it to get an element $\varphi(1+I) \in R/J$. The map in the question simply takes $\varphi$ to this value $\varphi(1+I)$.

___________

Anyway, I don't see what you're really trying to achieve, here. It would be much more beneficial for you to work from the basics of group theory and ring theory and build your way up. You'll end up with a much better overall understanding, and things will actually make sense to you rather than feeling like a bunch of abstract definitions.

Last edited by cjem; May 18th, 2018 at 09:47 AM.

May 18th, 2018, 12:06 PM   #34
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

Quote:
 Originally Posted by cjem .................................................. .......... Anyway, the map in the OP works as follows: Take any element $\varphi$ in $\text{Hom}_R{R/I,R/J}$ (recall: this means $\varphi$ is an $R$-module homomorphism from $R/I$ to $R/J$). The identity element of $R/I$ is the coset $1 + I$. Since $1 + I$ is an element of $R/I$, we can apply $\varphi$ to it to get an element $\varphi(1+I) \in R/J$. The map in the question simply takes $\varphi$ to this value $\varphi(1+I)$. .................................................. .........
From http://www.maths.usyd.edu.au/u/de/AG...ra/pp22-51.pdf
A/I = { I + a | a ∈ A } , ***
the group of cosets of I ,with addition defined by, for a, b ∈ A
(I + a) + (I + b) = I + (a + b)
Further A/I forms a ring by defining, for a, b ∈ A ,
(I + a) (I + b) = I + (a b)

*** This makes sense, considering I+a doesn't (what is sum of group and element of group), in light of definition of coset from wiki:
gH = { gh : h an element of H } is the left coset of H in G with respect to g,
Hg = { hg : h an element of H } is the right coset of H in G with respect to g.
In which case (I + a) $\displaystyle \equiv$ Ia makes sense.

Then (I+1)(I+a) = I + a and I+1 is indeed an identity element of A/I

Just checking you weren't pissing in my ear and telling me it was raining.

What applying the identity element of (R/I) to a homeomorphism means, and whether it's an element of (J:I)/I, I don't see. But since I, J, and (J:I)/J are Ideals (sub-rings) of R and therefore with the same operations, I can see twilight through the trees.

The only thing I still can't grasp is the operation of a coset on a homeomorphism (function). But let it go.

Many thanks for your expertise and patience.

Why do I want to know? I'm interested in the basic structure of things. Like where does money come from. Every country in the world is massively in debt. But whom do they owe the money too? Surely there must be some country or organization that has a massive credit balance equal to everybody else's debt. You can understand it without getting into all the details of Abstract Banking. But that's a beer for another time and place.

May 19th, 2018, 02:44 PM   #35
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 311
Thanks: 109

Math Focus: Number Theory, Algebraic Geometry
Quote:
 Originally Posted by zylo What applying the identity element of (R/I) to a homeomorphism means, and whether it's an element of (J:I)/I, I don't see. But since I, J, and (J:I)/J are Ideals (sub-rings) of R and therefore with the same operations, I can see twilight through the trees. The only thing I still can't grasp is the operation of a coset on a homeomorphism (function). But let it go.
$\varphi$ is a function that takes elements of $R/I$ to elements of $R/J$. So, for every element $A$ of $R/I$, we have a corresponding element $\varphi(A)$ of $R/J$. For example, one element of $R/I$ is the coset $1 + I$, so we have a corresponding element $\varphi(1 + I)$ in $R/J$.

The map $\varphi \mapsto \varphi(1+I)$ simply sends each homomorphism $\varphi: R/I \to R/J$ to the corresponding element $\varphi(1+I)$ of $R/J$.

A few things:
1) The word here is homomorphism, not homeomorphism (homeomorphisms are basically "isomorphisms" of topological spaces - they are not really to do with rings/modules/groups.)

2) $(J:I)/J$ is not an ideal of $R$, not is it a ring. However, it is an ideal of $R/J$ and it is also naturally an $R$-module.

3) Ideals are not in general subrings. In fact, the only ideal of a ring $R$ that is also a subring is $R$ itself.

So what is an ideal? Well, firstly, it must be an (additive) subgroup $I$ of your ring $R$. Given just this condition, we can certainly form the quotient group $R/I$. (Recall: the elements of this group are cosets $x + I$, for $x \in R$, with operation $(x + I) + (y + I) = (x + y) + I$). But we don't just want a group, we actually want $R/I$ to be a ring in some canonical way. The only natural multiplication operation to try would be: the product of the cosets $x + I$ and $y + I$ is the coset $xy + I$.

But now we run into a problem: this might not be well-defined. Indeed, if $x + I = x' + I$ and $y + I = y' + I$, there is no guarantee that we'd have $xy + I = x'y' + I$. In other words, the product of two cosets might change depending on how we choose to write them down! This is a big problem. For example, in integer multiplication, I could write the number $3$ as $3$, or as $2 + 1$. It's important that I get the same result if I calculate $3 \times 6$ compared to if I calculate $(2+1) \times 6$. If I take the same two inputs, even if I write them differently, I need the output to be the same!

An example of where this actually goes wrong if we just have a subgroup of a ring rather than an ideal: take $R = \mathbb{Q}$ to be the ring of rational numbers, and $I = \mathbb{Z}$ to be the subgroup of integers. Let's consider the cosets $\frac{1}{2} + \mathbb{Z}$ and $\frac{1}{3} + \mathbb{Z}$; their "product" should then be $\frac{1}{6} + \mathbb{Z}$. But we could also write the coset $\frac{1}{2} + \mathbb{Z}$ as $-\frac{1}{2} + \mathbb{Z}$ (note that this is the same coset, since the difference between $\frac{1}{2}$ and $-\frac{1}{2}$ is $1$, which is an element of $\mathbb{Z}$.) But now if we take the product of $-\frac{1}{2} + \mathbb{Z}$ with $\frac{1}{3} + \mathbb{Z}$, we'd get $-\frac{1}{6} + \mathbb{Z}$. But now this is actually a different coset to $\frac{1}{6} + \mathbb{Z}$ (as the difference between $\frac{1}{6}$ and $-\frac{1}{6}$ is $\frac{1}{3}$ which is not in $\mathbb{Z}$.) So in this case, the product of the same cosets was different depending on how we wrote them down! This would be like if $(1+2) \times 6$ was not equal to $3 \times 6$ in integer multiplication!

This is why ideals are not just subgroups $I$ of our ring $R$, but we also require them to satisfy the clever condition that $rx \in I$ for all $r \in R, x \in I$. This isn't just some random abstract thing mathematicians decided, it's precisely the condition that makes multiplication in $R/I$ work! Now I could show you why this is the condition that makes the multiplication work (it's just a 1 or 2 line proof), but I feel like it'd be more beneficial to you if you had a try yourself first. I'll be happy to help if you get stuck, though.

Last edited by cjem; May 19th, 2018 at 03:00 PM.

May 19th, 2018, 02:47 PM   #36
Senior Member

Joined: Aug 2012

Posts: 2,201
Thanks: 646

Quote:
 Originally Posted by cjem 3) Ideals are not in general subrings. In fact, the only ideal of a ring $R$ that is also a subring is $R$ itself. So what is an ideal?
Why are normal subgroups the thing you mod out by in groups; and ideals (not subrings) are the things you mod out by in rings?

The insight is that you mod out by kernels of homomorphisms. The kernel of a group homomorphism is a normal subgroup; and the kernel of a ring homomorphism turns out to be an ideal, and not a subring.

In case anyone ever wondered why we mod out by ideals and not subrings, that's why.

May 19th, 2018, 03:14 PM   #37
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 311
Thanks: 109

Math Focus: Number Theory, Algebraic Geometry
Quote:
 Originally Posted by Maschke Why are normal subgroups the thing you mod out by in groups; and ideals (not subrings) are the things you mod out by in rings? The insight is that you mod out by kernels of homomorphisms. The kernel of a group homomorphism is a normal subgroup; and the kernel of a ring homomorphism turns out to be an ideal, and not a subring. In case anyone ever wondered why we mod out by ideals and not subrings, that's why.
Yes, this is the more conceptual way of viewing things. I was planning on basing my rant on this perspective, but I figured I'd first see how Zylo coped with: ideals are precisely the subgroups for which we can actually define a natural ring structure on R/I.

May 20th, 2018, 10:52 AM   #38
Senior Member

Joined: Oct 2009

Posts: 755
Thanks: 261

Quote:
 Originally Posted by Maschke Why are normal subgroups the thing you mod out by in groups; and ideals (not subrings) are the things you mod out by in rings? The insight is that you mod out by kernels of homomorphisms. The kernel of a group homomorphism is a normal subgroup; and the kernel of a ring homomorphism turns out to be an ideal, and not a subring. In case anyone ever wondered why we mod out by ideals and not subrings, that's why.
Actually, in entire mathematics, we mod out with equivalence relations. This happens very transparently in topology, but not so in group theory or ring theory.

But also in group or ring theory, we mod out by certain equivalence relations. Indeed, we mod out by congruence relations.

An equivalence relation $\sim$ is called a congruence relation on a group $G$ iff $g\sim g'$ and $h\sim h'$ implies $gh\sim g'h'$ and $g^{-1} \sim (g')^{-1}$.
Congruence relations are the only kind of relations such that the quotient $G/\sim$ is a group again.

Now the relation with normal subgroups is the following:
1) If $N$ is normal in $G$ then
$$x\sim y~\Leftrightarrow~xy^{-1}\in N$$
is a congruence relation.
2) Conversely, if $\sim$ is any congruence relation then
$$N = \{x\in G~\vert~x\sim e\}$$
is a normal subgroup and we have $x\sim y$ if and only if $xy^{-1}\in N$.

A similar thing happens in rings.

Now, your observation that we mod out with kernels of homomorphisms is correct in rings and groups, but it is morally wrong. Indeed, take a look at the abstract algebraic structure of a lattice. It is a set $L$ with two operations $\wedge$ and $\vee$ and satisfying the following axioms:
1) $x\vee y = y\vee x$ and $x\wedge y = y\wedge x$
2) $(x\vee y)\vee z = x\vee (y\vee z)$ and $(x\wedge y)\wedge z = x\wedge (y\wedge z)$
3) $x\vee (x\wedge y) = x$ and $x\wedge (x\vee y) = x$
4) $x\vee x = x$ and $x\wedge x = x$.

( A lattice is actually "the same" as a partially ordered set where every finite subset has a supremum and an infimum)

Now, how to define quotients of lattices? Homomoprhisms still make sense, but how will you make sense of kernels? It doesn't make immediate sense here.
Congruence relations will still make sense though as equivalence relations $\sim$ satsifying $a\sim x$ and $b\sim y$ implies $a\wedge b\sim x\wedge y$ and $a\vee b\sim x\vee y$. And quotients with respect to congruence relations still go through.
(Incidentally, one can define the kernel of a homomorphism, but the result wil again be a congruence relation: define $x\sim y$ if and only if $f(x) = f(y)$.)

The branch of algebra that investigates this is called universal algebra.

 July 20th, 2018, 10:04 AM #39 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Thought I'd take a shot at part a) of OP from definitions: Let R be a commutative ring with 1. If I,J $\displaystyle \triangleleft$ R, then define (J:I) = {x ∈ R| xI $\displaystyle \subset$ J). a) Show (J:I) $\displaystyle \triangleleft$ R and J $\displaystyle \triangleleft$ (J:I) xI $\displaystyle \subset$ J if x+i=j, i ∈ I and j ∈ J (J:I) = {j - i | i ∈ I and j ∈ J} 1) (J:I) $\displaystyle \triangleleft$ R closed: (j-i) + (j'-i') = (j+j') - (i + i'), ∈ (J:I). additive identity: i=0, j=0 additive inverse: (-j) + i Ideal: r(j-i)=rj=ri. rj ∈ J and ri ∈ I because I and J are ideals. 2) J $\displaystyle \triangleleft$ (J:I) J is a sub-group of (J:I) If i=0, (J:I) = {j-0 | 0 ∈ I and j ∈ J} J is an ideal of (J:I) if jr ∈ J for j ∈ J and r ∈ (J:I): j(j'-i') = jj' - ji' and ii' ∈ J and ji' ∈ I because I is an ideal of R
July 20th, 2018, 10:43 AM   #40
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 124

Quote:
 Originally Posted by Maschke Why are normal subgroups the thing you mod out by in groups; and ideals (not subrings) are the things you mod out by in rings?
G/N is a unique partition if aN=Na, otherwise aN and Na lead to different partitions.

Google "ben1994", select playlists/groups/quotient groups. Same applies for an Ideal in its capacity as a subgroup: playlists/Rings/quotient rings

Or directly:

The series is phenomenal.

Suggest you set playback speed to 1.5

kernel of a homomorphism is just an example of an ideal, not a definition.

Last edited by skipjack; July 20th, 2018 at 02:55 PM.

 Tags isomorphism

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mathbalarka Abstract Algebra 2 November 4th, 2012 10:53 PM tiger4 Abstract Algebra 2 July 2nd, 2012 02:17 AM jpav Abstract Algebra 6 July 11th, 2011 06:00 AM mia6 Linear Algebra 1 November 10th, 2010 08:31 AM just17b Abstract Algebra 4 December 18th, 2007 07:57 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top