I How do we distinguish two different notations for cokernel and coimage?

[elias001]

This post is a further clarification of this post here; Notation questions about kernel, cokernel, image and coimage

In that post, I asked about notations related difficulties about cokernel, coimage, image, kernel. I did some more research, and I am more clear on how to articulate what I am having trouble with.

If we have a map $f$ that maps from $A$ to $B$ and $f$ is a homomorphism, then the definition of cokernel and coimage are respectively $\frac{B}{\text{im }f},\frac{A}{\text{ker }f}$. We have the usual definition for kernel and image. Both $\frac{B}{\text{im }f},\frac{A}{\text{ker }f}$ are sets. But if we want to discuss properties of $f$ like surjectivity, injectivity, monomorphism, epimorphism using exact sequences, or in the context of additive category, why is it we can consider maps of the form $p:B\to \frac{B}{\text{im }f}$ cokernel of $f$, and similarly $k:\frac{A}{\text{ker }f}\to A$ to be the kernel of $f$? We end up having exact sequence of the following form: $\text{ker }f\xrightarrow{\text{ker }f} A\xrightarrow{f} B\xrightarrow{\text{coker }f} \text{coker }f,$ then one gets notation like $f\circ\text{ker }f=0,$ and $\text{coker }f\circ f=0$, and also, $\text{im }f=\text{coker }(\text{ker }f)=\text{coker }(\text{ker }f\to A)$ and $\text{coim }f=\text{ker }(\text{coker }f)=\text{ker }(B\to \text{coker }f).$

So when discussing cokernel, coimage of a map $f$, how does one know whether it is in terms of a map or a set?

Thank you in advance.

 

[Mark44]

So when discussing cokernel, coimage of a map f, how does one know whether it is in terms of a map or a set?

These co- terms are new to me, but kernel and image aren't. The kernel and image are sets, but they are attributes of the map. I'm reasonably sure the same is true for cokernel and coimage.

 

[fresh_42]

Have a look at my explanation here:

F

I Post in thread 'How to show $p(x)=g(x)x\pm 1\in\Bbb{Q}[x]$ is irreducible in $\Bbb{Q}_{\Bbb{Z}}[x]$?'

Jun 14, 2025

I'm not sure whether homological algebra/category theory is a good starting point for getting used to the concepts. E.g., I looked it up to avoid mistakes and found:

Let $f : X \longrightarrow Y$ be a morphism in an additive category. The image of $f$, denoted by $\operatorname{Im}(f)$, is defined as the kernel of the cokernel. Namely, if $(C, \pi : Y \longrightarrow C)$ is the cokernel of $f$, then $\operatorname{Im}(f)$ is defined to be the kernel of $\pi$. Similarly, coimage of $f$, denoted by $\operatorname{CoIm}(f)$, is the cokernel of the kernel.

What? Again...

• fresh_42

 

[martinbn]

So when discussing cokernel, coimage of a map f, how does one know whether it is in terms of a map or a set?

From the context. Just like all your recent questions, it depends on the context. If you work with a category where the objects are not sets, then you cannot use the definitions as sets. If you work with, say modules, then you can use either way. It depends what you are doing and which way is better for you.

 

[mathwonk]

Good question. I think the point is that both approaches are really the same when you remember what sort of objects you are dealing with.

I.e. the kernel of a map is not just an object, rather it is a SUB-object. I.e. Ker(f:X-->Y) is that subset of the module X consisting of those points p with f(p) = 0. Since it is defined as a subset of X, there is always understood a natural inclusion map from Ker(f) to X, so Ker(f), as naively defined, is naturally equipped with a(n injective) map Ker(f)-->X. Whether you write that map or not in the definition, the fact you are dealing with a subset of the module X, means you have that map available whenever you need it.

Now in abstract category theory, where the objects X may not even have elements, you have to use mapping notation to give clues as to the fact that your object Ker(f) is supposed to be a sub-object, i.e. you have to write
Ker(f)-->X; since you can't talk about the elements of Ker(f), in particular you can't point out they are all also elements of X, you have to show how to put ker(f) into X.
You also have to give some properties that contain the missing information about f, such as that for any map g:Z-->X such that fg:Z-->Y = 0, factors uniquely through the map ker(f)-->X, i.e. as g = (Z-->ker(f)-->X).

Now you see what happens with the cokernel too, I bet; i.e. the cokernel is not just an object, it is a QUOTIENT-object, namely Coker(f:X-->Y) is the quotient of the module Y by the submodule Im(f) of images f(p) under f, of elements of X. In particular, by its definition, Coker(f) is equipped with a surjection onto it from Y, i.e. Coker(f) = Y/Im(f), is the target of a surjective map Y-->Y/Im(f) = coker(f).
If the objects X,Y are abstract, without elements, then we have to give this surjective map Y-->Coker(f) explicitly as part of the data.
And we have to give some properties that capture its connection with f without mentioning elements; e.g. any map g:Y-->Z such that gf:X--Z = 0, factors uniquely through the map Y-->Coker(f), i.e. as g = (Y-->Coker(f)-->Z).

When working with concrete categories, like modules, as I do, you don't need the more abstract notation, since it does not add any information, and so I never use it. I.e. to me any map of modules f:X-->Y has associated to it two natural submodules, ker(f) in X, and Im(f) in Y, and hence also two natural quotient modules, X/ker(f) ≈ Im(f), and Y/Im(f) = Coker(f). (And I never use the terminology Coim(f) for X/ker(f).)

 

[martinbn]

When working with concrete categories, like modules, as I do, you don't need the more abstract notation, since it does not add any information, and so I never use it. I.e. to me any map of modules f:X-->Y has associated to it two natural submodules, ker(f) in X, and Im(f) in Y, and hence also two natural quotient modules, X/ker(f) ≈ Im(f), and Y/Im(f) = Coker(f). (And I never use the terminology Coim(f) for X/ker(f).)

Out of curiosity, when you work with sheaves how do you think of the image?

 

[mathwonk]

I'm not very good with sheaves. Off the top of my head, I recall that one progresses from presheaves to sheaves by "localizing". So for presheaves, the image is just the actual image (one image module for every open set). Then to define the sheaf image I believe you take, in each open set, every element that is locally an image....I will review these ideas and repost if I have better insight. Specifically, I recall a sheaf map is surjective iff it is surjective on "stalks", i.e. on germs of elements, at every point.
As an example, the exponential map from the sheaf O of holomorphic functions, to the sheaf O* of never-zero holomorphic functions is surjective, hence O* is the image of the map. In this case of course, on the complement of the origin in the complex numbers, the function z is of course not the exponential of any holomorphic function defined for all z≠0, but locally it is. (I need to re-read Serre's FAC.)
Sheaf cohomology gives you a measure of how far off locally surjective is from globally surjective, i.e. of the difference between the pre -sheaf image and the sheaf image. So for the complement X of the origin in the complex numbers C, the exact sequence of cohomology for the exponential map above, contains a short exact sequence (H^0(X,O)-->H^0(X,O*)-->H^1(X,Z)) =
Holo(X,C)-->Holo(X,C*)-->Z, where Z is the integers. The first (left-hand) map is exponentiation, and the second (right-hand, "coboundary") map is winding number of X around the origin of C* (which equals X, of course). I.e. a never vanishing holomorphic function on X has a logarithm iff it does not wind X around the origin of C*, iff it induces the trivial map on the fundamental group. (Of course winding number is actually defined-computed here by subtracting the difference in values of local logarithms.)

 

[martinbn]

I'm not very good with sheaves. Off the top of my head, I recall that one progresses from presheaves to sheaves by "localizing". So for presheaves, the image is just the actual image (one image module for every open set). Then to define the sheaf image I believe you take, in each open set, every element that is locally an image....I will review these ideas and repost if I have better insight. Specifically, I recall a sheaf map is surjective iff it is surjective on "stalks", i.e. on germs of elements, at every point.
As an example, the exponential map from the sheaf O of holomorphic functions, to the sheaf O* of never-zero holomorphic functions is surjective, hence O* is the image of the map. In this case of course, on the complement of the origin in the complex numbers, the function z is of course not the exponential of any holomorphic function defined for all z≠0, but locally it is. (I need to re-read Serre's FAC.)
Sheaf cohomology gives you a measure of how far off locally surjective is from globally surjective, i.e. of the difference between the pre -sheaf image and the sheaf image. So for the complement X of the origin in the complex numbers C, the exact sequence of cohomology for the exponential map above, contains a short exact sequence (H^0(X,O)-->H^0(X,O*)-->H^1(X,Z)) =
Holo(X,C)-->Holo(X,C*)-->Z, where Z is the integers. The first (left-hand) map is exponentiation, and the second (right-hand, "coboundary") map is winding number of X around the origin of C* (which equals X, of course). I.e. a never vanishing holomorphic function on X has a logarithm iff it does not wind X around the origin of C*, iff it induces the trivial map on the fundamental group. (Of course winding number is actually defined-computed here by subtracting the difference in values of local logarithms.)

I probably should have been clearer. I know all this, my question was more about if you have any specific way to think about it that suits you.

 

[mathwonk]

not really, sounds as if you know more about them than I do. I have struggled all my life to grasp the algebra side of algebraic geometry. In fact I went into it partly for the challenge, since I had little intuition for algebra compared to geometry, but I hoped the geometry could help me understand the algebra.

 

[elias001]

@martinbn and @mathwonk thank you for both of your replies. I just want to know if I am looking cokernel/coimage as maps instead of as them being sets, what can I say to inform/signal to the reader that I am doing so. My understanding is that we can talk about cokernel in the context of abeluan groups as either as sets or as maps/morphisms.

 

[mathwonk]

that is a problem if you just use the word. I guess you'll need to either write out the map or just explain yourself. e.g. f:X-->Y implies ker(Y-->Coker(f)) = im(f), rather than just ker(Coker(f)) = im(f).

 

[elias001]

@mathwonk So if I have a map $f:A\to B$, where $A, B$ are some mathematical objects: groups, rings, modules, vector space, etc, and $f$ is a map has certain properties: continuity, invertible, or merely injective/surjective, homomorphic, non-linear, etc. According to linear algebra, there are four fundamental subspaces: kernel, cokernel, image, coimage, and there are calculations for determining each of the four respective matrix. Also, one can talk about them the map $f$ using exact sequences. But even from the view point of exact sequence, the notion of cokernel and coimage as maps and as sets, how does one communicate to the reader whether one is talking about sets or maps for either cokernel or coimage. I am not even talking about sub-objects, sub-quotient objects in terms of concrete or abstract categories or even using the language of homological algebra. I think I should learn how to sort out how to signal to the reader in concrete cases first. I think in the language of category theory, there are much more terms or theorem that one can cite to signal to the reader whether you are talking about sets or about maps.

 

[fresh_42]

My recommendation is to close your book about homological algebra for now and study abstract algebra instead. And outside of homological algebra, those terms are always sets. The morphism may define those sets, but that's it. You have to understand those sets before you head on chasing diagrams. You can go back to homological algebra when you encounter your first chain complex, but not sooner.

 

[martinbn]

What are you writing? Which reader do you want to comunicate to?

 

[elias001]

@fresh_42 what about when I am studying modules, can I safely avoid talking about cokernel or coimage as morphism? I did exercises in chain complex in John Fraeligh's A first course in abstract algebra 7th edition. It is worse than symbol pushing. It feel so abstract, divorce from any concrete content.

@mathwonk i should have mentioned when I am both reading or writing, as in a homework assignment. Also, I hope I never have to leave the land of concrete categories. It was really strange when I was doing all those proofs about 3 by 3, 4 by 4 and 5 by 5 and snake lemmas. I learn to proof them as exercises. I have zero appreciate for them. As I told @fresh_42 everything feels so abstract.

 

[fresh_42]

@fresh_42 what about when I am studying modules, can I safely avoid talking about cokernel or coimage as morphism?

Yes. You will need kernels and images, not their dual objects. Sure, quotients will be necessary, but it's better to think of them as quotients rather than cokernels and coimages. These are just words.

 

[elias001]

@fresh_42 as an example of what I am talking about of all the feeling of all of it being mere abstractions: i have the following two sequences of maps

1. $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$

2. $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$

I had to ask how to how the second sequence is short exact. It was strange because I can do a lot of those fancy lemmas that I mentioned, but when it comes concrete examples, I fell flat on my face mathematically speaking.

 

[fresh_42]

@fresh_42 as an example of what I am talking about of all the feeling of all of it being mere abstractions: i have the following two sequences of maps

1. $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$

2. $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$

I had to ask how to how the second sequence is short exact. It was strange because I can do a lot of those fancy lemmas that I mentioned, but when it comes concrete examples, I fell flat on my face mathematically speaking.

You have to look at the homomorphisms. How are they defined? How do they act on specific elements of the rings. Once you know this, you can define their images and kernels. Exactness means the image of one equals the kernel of the next. This means to show that two sets are equal. To show equality of sets, say $A=B$ we usually show $A\subseteq B$ and $B\subseteq A.$

Short exactness means that the left homomorphism is injective, i.e., its kernel is $\{0\},$ and that the right homomorphism is surjective, i.e., it hits all available ring elements.

Comment: I assumed rings. It also works for groups, but then we would have written $\{e\}$ or $1$ at the boundaries and not $\{0\}.$

 

[martinbn]

@fresh_42 what about when I am studying modules, can I safely avoid talking about cokernel or coimage as morphism? I did exercises in chain complex in John Fraeligh's A first course in abstract algebra 7th edition. It is worse than symbol pushing. It feel so abstract, divorce from any concrete content.

@mathwonk i should have mentioned when I am both reading or writing, as in a homework assignment. Also, I hope I never have to leave the land of concrete categories. It was really strange when I was doing all those proofs about 3 by 3, 4 by 4 and 5 by 5 and snake lemmas. I learn to proof them as exercises. I have zero appreciate for them. As I told @fresh_42 everything feels so abstract.

Are you stadying on your own or is this a course?

 

[elias001]

@martinbn I am studying on my own. I went through two thirds of the book Arrows, structures and functors by Arbib and Manes. I stopped at the section after natural transformation in the chapter about functors. I think i need to know more math for the section onwards adjoint functors. I also gone through Thomas Scott Blyth's old module theory text on my own. I also learned about direct and inverse limits in the setting of abelian groups. I find i cannot picture the concepts as well as say in real and complex analysis.

 

[mathwonk]

Your post #17 brings up an important point I meant to mention sooner. namely, as fresh says, it is important to know what the maps are. In your sequence 2, only one map is possible, namely multiplication by p. [oops, maybe also multiplication by any multiple of p? Either way the image and cokernel are the same.]
But in sequence 1, there are several possible maps with different images, e.g. inclusion onto the first factor Z/p+{0}, or inclusion onto the second factor {0}+Z/p, or even mapping [1] to ([1],[1]).
[In fact every non trivial element of Z/p+Z/p seems to have order p, so you could map [1] to any non zero element. So there are apparently p^2-1 different injective maps, having p+1 different images groups, it seems.]
In these 3 different cases you get different images, even though all are isomorphic to Z/pZ. I.e. just writing what you have does not display what the image is, nor what the cokernel is fully. I.e. the cokernel is not just Z/pZ, although it is isomorphic to it. Rather the cokernel is either (Z/pZ+Z/pZ)/({0}+Z/pZ) or something else, depending on what the maps are. In particular just writing an arrow does not specify a map fully. I.e. whether or not you define a cokernel to be a map, the map it represents must be detectable from the data you give.

I think of direct and inverse limits as generalizations of unions and cartesian products from set theory and analysis. In module theory they generalize direct sums and direct products. In fact, as I recall, then they are actually quotients of direct sums, or submodules of direct products. (That's in the usual case where the index set is a directed set, but as I vaguely recall the index can be more general, such as maybe a category! Indeed I still have class notes, from a class I myself taught!, detailing how to define direct limits of functors, as functors adjoint to the "constant functor" construction, but I can't easily understand them myself, which tells you something about how useful these abstractions are, to me at least.)

 

[elias001]

@fresh_42 When you said

My recommendation is to close your book about homological algebra for now and study abstract algebra instead. And outside of homological algebra, those terms are always sets. The morphism may define those sets, but that's it. You have to understand those sets before you head on chasing diagrams.

Can you elaborate more on what you meant by understanding those four objects more as sets. To me they are either sets or quotient sets. There is a definition in category theory where one talks about kernel or cokernel pair along with a triangle looking commutative diagram. The thing is, when hitting the topic of theory of modules, the topic of exact sequences comes up in Dummit & Foote. in fact, any textbook solely about Rings and modules, or just theory of modules have the topic of exact sequences, and the technique of diagram chasing is not far behind. I have the feeling exact sequences are a topic that prepare students for homological algebras/algebraic topology. Also I understand that the theory of spectral sequences comes sometime after that. Another thing is depending on the text being used, basic category theory up to the yoneda lemma and adjoint functors also introduced. Which going back to our four candidates of discussions, I understand that depending on which concrete category (groups, rings, modules, etc) one is discussing, image, coimage, cokerenl might not exists. If that is the case, how does one prove that it does not exists. I hope i am not asking something unreasonable.

 

[fresh_42]

Say, we have a morphism $f\, : \, A\rightarrow B.$ What we really need in algebra are $\operatorname{ker}(f)\, , \,A/\operatorname{ker}(f)\, , \,\operatorname{im}(f)$ and rarely $B/\operatorname{im}(f).$ There is no need to call the quotients cokernel or coimage other than as a (rarely used) abbreviation.

Relevant exact sequences are usually short exact sequences, namely
$$
\operatorname{ker}(f) \rightarrowtail A \twoheadrightarrow A/\operatorname{ker}(f) .
$$
This is brief for
$$
\{0\}\stackrel{\subseteq }{\rightarrow} \operatorname{ker}(f) \stackrel{\iota}{\rightarrow} A \stackrel{\pi_f}{\rightarrow } A/\operatorname{ker}(f) \stackrel{0}{\rightarrow} \{0\}.
$$
Exactness means that the image of one morphism is the kernel of the following morphism. Let's check this:

1. $\operatorname{im}(\subseteq )=\{0\}=\operatorname{ker}(\iota)$ since $\iota\, : \,\operatorname{ker}(f) \rightarrowtail A$ is the injective embedding from the kernel of $f$ into $A$ and has therefore a trivial kernel. The tail of the arrow indicates injectivity. This proves exactness at the first object $\operatorname{ker}(f).$
   
   
2. $\operatorname{im}(\iota) = \operatorname{ker}(f)=\operatorname{ker}(\pi_f)$ since $\pi_f \, : \, A\twoheadrightarrow A/\operatorname{ker}(f)$ is the surjective projection onto the quotient $A/\operatorname{ker}(f)$ and has therefore the kernel of the object we built the quotient with. The double head of the arrow indicates surjectivity. This proves exactness at the second object $A.$
   
   
3. $\operatorname{im}(\pi_f)= A/\operatorname{ker}(f)=\operatorname{ker}(0)$ since the right most morphism $0$ maps everything onto $\{0\}.$ This proves exactness at the third object $A/\operatorname{ker}(f).$

The central question of this short exact sequence is whether we can find a morphism $g\, : \,A/\operatorname{ker}(f) \rightarrow A$ such that $\pi_f\circ g =\operatorname{id}_{A/\operatorname{ker}(f)}.$ This leads to the concept of semidirect products and direct products. Such a morphism $g$ does not automatically exist. If it exists, then it makes the quotient object $A/\operatorname{ker}(f)$ a subobject of $A.$ If such a morphism $g$ exists, we say that the short exact sequence splits.

This is what is important in algebra. The sets are:

1. $\operatorname{ker}(f)=\left\{a\in A\,|\,f(a)=0\right\}$ or whatever the neutral element is. E.g., in the category of (multiplicative) groups, we have $\operatorname{ker}(f)=\left\{a\in A\,|\,f(a)=1\right\}.$
   
   
2. $\operatorname{im}(f)=\left\{b \in B\,|\,b=f(a)\text{ for some }a\in A\right\}.$
   
   
3. $A/\operatorname{ker}(f)=\left\{a+\operatorname{ker}(f)\,|\,a\in A\right\}$ is a set of sets, the equivalence classes $a+\operatorname{ker}(f).$ The equivalence relation is defined as $a\sim a' \Longleftrightarrow a-a'\in \operatorname{ker}(f).$ Again in the category of (multiplicative) groups, this changes to $A/\operatorname{ker}(f)=\left\{a\cdot\operatorname{ker}(f)\,|\,a\in A\right\}$ and $a\sim a' \Longleftrightarrow a\cdot a'^{-1}\in \operatorname{ker}(f).$ This is also the reason why we use to write abelian groups with an addition and non-abelian groups with a multiplication. Commutativity makes a significant difference here, and rings, algebras, or vector spaces have a commutative addition.

This is the environment and the sets that are commonly used in abstract algebra. Long exact sequences and duality occur if we consider other objects like chain or simplicial complexes. These, however, are not suited to learn the basis of algebra. They are relatively easy once you understand what's going on within the realms I just described: short exact sequences, and the three sets I defined.

 

[martinbn]

@martinbn I am studying on my own.

In that case why don't you study something else? Algebra can be a bit abstract an unmotivated. But if you study fist other parts mathematics it can become easier. For example if you study some algebraic number theory and/or algebraic geometry, commutative algebra becomes much easier to study.

 

[elias001]

@martinbn for the second exact sequence above $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$, in the second map for the portion, $0\xrightarrow{f_1}\mathbb{Z}/p\mathbb{Z}\xrightarrow{f_2} \mathbb{Z}/p^2\mathbb{Z}$, would the map be defined as $f_2(n+\mathbb{Z}/p\mathbb{Z})=np+\mathbb{Z}/p^2\mathbb{Z}$ or can I write it as $f_2(n+\mathbb{Z}/p\mathbb{Z})=n+\mathbb{Z}/p^2\mathbb{Z},$ where $n=0,1,2,\ldots ,p^2-1$?

 

[martinbn]

@martinbn for the second exact sequence above $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$, in the second map for the portion, $0\xrightarrow{f_1}\mathbb{Z}/p\mathbb{Z}\xrightarrow{f_2} \mathbb{Z}/p^2\mathbb{Z}$, would the map be defined as $f_2(n+\mathbb{Z}/p\mathbb{Z})=np+\mathbb{Z}/p^2\mathbb{Z}$ or can I write it as $f_2(n+\mathbb{Z}/p\mathbb{Z})=n+\mathbb{Z}/p^2\mathbb{Z},$ where $n=0,1,2,\ldots ,p^2-1$?

The second guess is not well defined, so it cannot be that.

 

[elias001]

@martinbn the second one is not well defined because it won't allow $n$ to go beyond $n>p$ in the range?

 

[fresh_42]

Well definition means that you cannot map one element to two different targets. This is usually no problem if the elements are singletons. Here, however, we have sets as elements, the equivalence classes. The same sets can have different representatives and they may not be mapped differently.

Example:

The ring $\mathbb{Z}_5$ can be viewed as $\mathbb{Z}/5\mathbb{Z}$ and therefore as the set of equivalence classes $\left\{0+5\mathbb{Z},1+5\mathbb{Z},2+5\mathbb{Z},3+5\mathbb{Z},4+5\mathbb{Z}\right\}.$ But for example $3+5\mathbb{Z} = 8+5\mathbb{Z}.$ A well defined function $f$ on $\mathbb{Z}/5\mathbb{Z}$ requires that $f(3+5\mathbb{Z})=f(8+5\mathbb{Z}).$ It may not depend on the representative you choose for a class.

If you write $\mathbb{Z}_5=\left\{0,1,2,3,4\right\}$ which by the way should be better written as $\mathbb{Z}_5=\left\{\bar 0,\bar 1,\bar 2,\bar 3,\bar 4\right\}$ then we still have $\bar 3=\bar 8$ and the function may not distinguish between them.

 

[martinbn]

@martinbn the second one is not well defined because it won't allow $n$ to go beyond $n>p$ in the range?

Because if you take a different representative of the same class, it will not.be mapped to the same element.

 

[elias001]

@fresh_42 @martinbn if we have $f(n+5\mathbb{Z})=n+5^2\mathbb{Z}$, then for $n=7$, then both $7+5\mathbb{Z}, 2+5\mathbb{Z}$ maps to $7+5^2\mathbb{Z}$. So you two are right.

 

[elias001]

@martinbn Isn't it supposed to be if i know basic commutative algebra, it would make studying either Algebraic number theory and Algebraic geometry easier.

 

[fresh_42]

Isn't it supposed to be if i know basic commutative algebra, it would make studying either Algebraic number theory and Algebraic geometry easier.

In my opinion, yes, but you don't need the five or nine lemmas for it. The rings $\mathbb{Z}_n$ are the standard examples in many cases, so it's necessary to understand them. Rings are a generalization of the integers, and modules are a generalization of vector spaces. So it helps to study those foundations before the general case, and if it were for having a reservoir of examples.

 

[fresh_42]

@fresh_42 @martinbn if we have $f(n+5\mathbb{Z})=n+5^2\mathbb{Z}$, then for $n=7$, then both $7+5\mathbb{Z}, 2+5\mathbb{Z}$ maps to $7+5^2\mathbb{Z}$. So you two are right.

The other way around: $7+5\mathbb{Z}, 2+5\mathbb{Z}$ are different elements in $\mathbb{Z}_{25}$ but identical in $\mathbb{Z}_5.$ So it maps the same element to two different targets, which is not allowed for a function. That means in technical terms "$f$ isn't well defined".

We only have a (surjective) homomorphism $\mathbb{Z}_{25} \longrightarrow \mathbb{Z}_5$ by mapping $n+25\mathbb{Z}$ to $(n\pmod{5})+5\mathbb{Z}.$

 

[elias001]

@fresh_42 in linear algebra, for matrices, one can calculate the equivalent of its kernel and cokernel. I have seen cokernel amd exact sequences being discussed in a few linear algebra texts. Some of those texts includes category theory. Also $\Bbb{Z}-$modules are not vector spaces. it seems i can't always rely on vector spaces as a crutch for examples.

 

[elias001]

@fresh_42 i think I got a bit confused when I thought about $np$ instead of $n$.

 

[fresh_42]

Also I have never encountered a module that is not a vector space.

I bet you have. Take any abelian group $\left(G,+\right).$ Then we can define
$$
z\cdot g=\begin{cases}
\underbrace{-g-g-\ldots -g}_{-z\text{ times }}&\text{ if }z<0\\
0&\text{ if }z=0\\
\underbrace{g+g+\ldots +g}_{z\text{ times }}&\text{ if }z>0
\end{cases}
$$
and $G$ becomes a natural $\mathbb{Z}$-module.

Also, every ideal in a ring is a module over that ring.

 

[elias001]

@fresh_42 i edited my post. I said with modules over the integers, I can't always rely on vector spaces as a crutch for sources of examples.

 

[martinbn]

@martinbn Isn't it supposed to be if i know basic commutative algebra, it would make studying either Algebraic number theory and Algebraic geometry easier.

Yes, but for an introductory course on either of those you don't need much to start with and a lot is usually covered in the text.

 

[elias001]

@martinbn I have never taken any introductory course in abstract algebra. I am at a point at Dummit and Foote where the examples being presented are from things that are beyond the introductory mark.

 

[martinbn]

@martinbn I have never taken any introductory course in abstract algebra. I am at a point at Dummit and Foote where the examples being presented are from things that are beyond the introductory mark.

You said that you are studying it on your own, not as a part of a course. My point is, in that case if you find it unmotivated study something else.

 

[elias001]

@martinbn I am also studying on my own so that I know how to do all the exercises in the first 14 chapters in Dummit and Foote since that will be the textbook my university will be using in their introductory abstract algebra course. Fourteen chapters of raw unadulterated abstract algebra served and forced fed to the students at an overdosed speed in the span of less than nine months. I have met students who fail that class. One professor said anyone she has seen that fell behind in that course never manages to catch up. In another occasion, she told me the material in Dummit and Foote was very basic as far as abstract algebra is concerned. I am not in a position to judge what she meant by being basic is suppose to mean.

 

[elias001]

@martinbn I think the two volumes of Samuel and Zariski were what Courant's two volumes of Differential and Integral calculus to the subject of Calculus and Analysis. Thee is no similar text for beginning level abstract algebra. The written style for both two-volumes were written in such s way the authors were giving a lecture to the reader in person.