May 15th, 2018, 08:15 AM  #11  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
Obviously this question isn't really appropriate for someone that hasn't already had a fair amount of experience with rings and ideals  it would normally take until the ~2nd/3rd year of an undergraduate degree to get to this stage. Note: there is some nontrivial work to be done in answering question (b). But the specific issue Mona had with this question (that I was replying to) really is a one line argument that falls straight out of the definition; I find it hard to believe someone working at Mona's level would struggle with it unless they just hadn't given it a proper attempt. Quote:
For example, when I first came across quotient groups, I found them very confusing and abstract. I could answer questions involving them, but I'd say it took at least 6 months before I really understood what they were all about. Within another half a year, they had basically become second nature to me. Once you get to this point, the basic theory of quotients in other algebraic structures (showing they are welldefined, isomorphism theorems, etc.) also becomes very natural and obvious.  
May 15th, 2018, 09:32 AM  #12  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Or are you saying the OP shoildn't ask the question until she has had years of experience with the thread?  
May 15th, 2018, 11:36 AM  #13  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
In fact, she had managed to do the only conceptually challenging part of the problem: coming up with the formula for her map $h: (J:I) \to \text{Hom}_R(R/I, R/J)$. The rest is just applying quick standard arguments to see that everything works out: 1) Check $h$ welldefined 2) Check $h$ is an $R$module homomorphism 3) Check the kernel of $h$ is $J$ 4) Check the resulting map $(J:I)/J \to \text{Hom}_R(R/I, R/J)$ is inverse to the map $\text{Hom}_R(R/I, R/J) \to (J:I)/J$ induced by the map given in the question. Note that Mona only claimed to have difficulty with 1). _____________________ 1) Suppose $r + I = r' + I$. Then $r  r' \in I$, so for all $x \in (J:I)$, we have $x(rr') \in J$, i.e. $xr + J = xr' + J$. 2) Let $x, x' \in (J:I)$, $r, s \in R$. Then $h_{x+x'}(r+I) = (x+x')r + J = (xr + J) + (x'r + J) = h_x(r+I) + h_{x'}(r+I)$ and $h_{sx}(r+I) = (sx)r + J = s(xr) + J = s \cdot h_x(r+I)$. 3) Suppose $x \in (J:I)$ is such that $h_x$ is the zero map. Then $x + J = 1 \cdot x + J = h_x(1+I) = J$, so $x \in J$. Conversely, if $x \in J$, then $h_x(r+I) = xr + J = J$, so $h_x$ is the zero map. Hence $h$ a homomorphism $h': (J:I)/J \to \text{Hom}_R(R/I, R/J)$ such that $h'_{x+J} = h_x$ for all $x \in (J:I)$. C 4) Let $g: \text{Hom}_R(R/I, R/J) \to (J:I)/J$ be given by $g(\varphi) = \varphi(1+I) + J$. If $\varphi \in \text{Hom}_R(R/I, R/J)$, then for all $r \in R$, we have $h'_{g(\varphi)}(r+I) = g(\varphi) \cdot r + J = \varphi(1 + I) \cdot r + J = \varphi(r + I) + J = \varphi(r + I) + J$, so $h'_{g(\varphi)} = \varphi$. On the other hand, $x \in (J:I)$, then $g(h'_{x+J}) = h'_{x+J}(1+I) + J = x + J$. So $h'$ and $g$ are inverses.  
May 15th, 2018, 02:16 PM  #14  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
But just out of curiousity, what is h? What does x↦hx mean? what is $\displaystyle Hom_{g}( R/I, R/J)$? Does R/I designate a quotient ring? What is a quotient ring?  
May 15th, 2018, 03:38 PM  #15  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
$\begin{align*} f: \; &\mathbb{Z} \to \mathbb{Z} \\ &x \mapsto x^2 \end{align*}$ But there's another way this notation is used. Say I wanted to refer to the function from the integers to the integers sending each element to its cube. I could write this out formally as above, or I could simply write the function as $x \mapsto x^3$. Note that this notation is only really useful when there's a nice formula for the function, such as in these examples. For an example that combines these two uses of the notation, let $\mathbb{N}$ be the set of natural numbers, and let $B$ be the set of functions from the integers to themselves. Notice that a function $f$ from $\mathbb{N}$ to $B$ sends natural numbers to functions. So for each natural number $n$, $f(n)$ is itself a function  I might now want to give a formula for this function. Our new notation lets you do this in quite a neat way. Say I write: $\begin{align*} f: \; &\mathbb{N} \to B \\ &n \mapsto (x \mapsto x^n) \end{align*}$ This tells you that for each natural number $n$, $f(n)$ is the function sending each integer $x$ to its $n$th power, $x^n$. Returning to the question at hand, for now just consider $(J:I)$ to be some set. $\text{Hom}_R(R/I, R/J)$ is a set consisting of (some certain) functions from $R/I$ to $R/J$. So $h$ is a function sending elements of a set to functions. For each $x$ in $(J:I)$, what $h$ sends $x$ to some function from $R/I$ to $R/J$. Which function does it send it to? It sends it to the function $h_x$, which is defined by $h_x(r+I) = xr+J$. We could write this in our new notation: $\begin{align*} f: \; &(J:I) \to \text{Hom}_R(R/I, R/J) \\ &x \mapsto (r+I \mapsto xr+J) \end{align*}$ _____________________ As for explaining what $R/I$ means (yes, it's the quotient of a ring by an ideal) and what $Hom_R(R/I, R/J)$ is, would you mind describing your background so I can pitch my response? For example, do you know much group theory (e.g. homomorphisms, normal subgroups, quotients) or linear algebra (e.g. given vector spaces $V$ and $W$, do you know that the set of all linear maps $V \to W$ can itself be viewed as a vector space?) Last edited by cjem; May 15th, 2018 at 03:50 PM.  
May 15th, 2018, 11:36 PM  #16 
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630  Excellent question. This is a perfect example of what it means for a subject to be at a certain level in math. When math majors take their first course in abstract algebra, they are taught about groups, which have one associative operation with an identity and inverses. Students are introduced to a construction called quotient groups, and at first it takes a while for students to grok the concept. Quotient groups are actually very simple (though they are not simple groups!) Imagine the integers, and the proper subset consisting of the multiples of 5. The multiples of five are not only a subset, they turn out to be a subgroup of the integers; and when you "take the quotient" of the integers by the multiples of 5, you get ... drum roll ... the plain old integers mod 5. So quotienting, also known casually as "modding out," is a generalization of the way in which we derive the integers mod 5 from the integers. Now the students are taught various theorems about quotient groups, like the "first isomorphism theorem," and the "second isomorphism theorem." These names are standard and everyone knows them by those names. At first this is all abstract symbol pushing, the students don't actually understand what a quotient group is, they just push the symbols the best they can. Still in your first algebra class, you finish up learning the very basics of groups, and they introduce something call rings. A ring is a set with two operations, one of which distributes over the other. The integers are the most basic example of a ring. You learn about quotient rings, and now this is the second time you've seen the idea. You gain a little more understanding. In topology they have quotient topologies. You keep seeing this idea in different settings. Any time you want to consider two objects to be "the same except for some attributes we don't care about at the moment," you can quotient out by the attributes you don't care about. In the case of the integers mod 5, you are calling two integers "the same" if they are the same after integer division by 5. If you take the closed unit interval [0.1] and mod out {0,1}, you are saying that you regard 0 and 1 as the same. What do you get then, topologically? You get a circle. A line segment with its endpoints identified. So you keep seeing this idea of modding out, and without even realizing it you gain instinctive understanding. You no longer recall what it's like to not grok quotients. Someone says "quotient" and you have this large context of what it means. It becomes like breathing, it become second nature. And you no longer remember what it's like to not get it. So even though IN THEORY you can break down the definition to its logical components, people are not machines. People learn by accumulated experience; not by mechanically following chains of logic. To wrap this all up, if you consider the integers as a ring and not just a group, then the analogy of a subgroup is NOT a subring  this is a very subtle point that confuses students  but something called an ideal. And in the ring of integers, the multiples of 5 are an ideal of the integers, and the quotient ring is ... drum roll again ... the integers mod 5. Hope this helped. Last edited by Maschke; May 15th, 2018 at 11:45 PM. 
May 16th, 2018, 08:44 AM  #17  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
I only ask about things I can't find definitions of. R/I is a quotient ring: Couldn't find a coherent definition. From: Quotient Ring  from Wolfram MathWorld "A quotient ring (also called a residueclass ring) is a ring that is the quotient of a ring A and one of its ideals a, denoted A/a." But quotient in this context is undefined and not referenced, which is unusual for Wolfram. Another source began by introducing cosets and I lost interest. Maschke tried to explain it to me, and now that I compare it to "residue class ring" it may turn out to be a simple concept. If a quotient ring is a defined subset of the elements of a {commutative?) ring, that already anchors it for me. Somebody has defined a subset of a commutative ring and investigated the implications. If they saw it as the solution to a vexing problem, I would be impressed. If they just did it for something new to investigate the implications of, fine, it's something new to play with. In this context, a quotient ring is a new subset constructed by a defined operation on a set and a subset. These are very basic, simple, primal concepts. How about a subset composed, by a rule, from elements of a ring, ideal, and quotient ring. The Asians would love it. $\displaystyle \mapsto$ maps to and $\displaystyle \triangleleft$ is an ideal of. But $Hom_R(R/I, R/J)$ is a stumper, simply because I don't know what to google for $Hom_R$. If you could describe it in words that I can look up, that would be fine. And what does $\displaystyle \cong$ say in the context of OP question b. I'm guessing isomorphic to Last edited by zylo; May 16th, 2018 at 08:58 AM.  
May 16th, 2018, 09:13 AM  #18  
Senior Member Joined: Aug 2012 Posts: 2,156 Thanks: 630  Quote:
The elements of the integers mod 5, namely 0, 1, 2, 3, 4 with addition and multiplication taken mod 5, are not a subset of the integers. Each element is an equivalence class of integers. See https://en.wikipedia.org/wiki/Quotient_group  
May 16th, 2018, 11:10 AM  #19  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
When googling these words, it might help to keep the example of vector spaces in mind so that the definitions don't seem so abstract. $R$modules are kind of like vector spaces where the scalars need only come from a ring rather than a field (in fact, if $R$ is a field, then an $R$module is exactly the same thing as a vector space over $R$). Similarly, homomorphisms of $R$modules are the analogue of linear maps; it's quite common to even call them $R$linear maps. With regards to the last sentence in the above paragraph, recall that the set of linear maps between two vector spaces can itself be seen as a vector space  it's a similar sort of thing going on here. Yes, the $\cong$ means isomorphic (as $R$modules).  
May 16th, 2018, 03:14 PM  #20  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
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 $\displaystyle \phi \mapsto \phi (1+I)$ modules are irrelevant, quotient rings are modules. R/I, R/J and (J:I)/I are groups (quotient rings) which have a defined "addition." A homeomorphic map from R/I to R/J makes sense. The set S of all maps from R/I to R/J makes sense. An isomorphic map from S to (J:I)/I makes NO* sense: What structure of S is being preserved? What does $\displaystyle \phi$ map an element of S to? Edit. Unless the set of all homeomorphisms R/I to R/J is an Rmodule (set with a particular structure) Why is the set of all homeomorphisms R/I to R/J an R module? Last edited by zylo; May 16th, 2018 at 03:23 PM.  

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