Ideals are generally not rings

(when rings are considered with identity)

Think about it if an ideal contained $1$ then it would have to be the whole ring

and are zero divisors closed under addition? Think back to the $\Bbb Z / 6 \Bbb Z$ example

Depends on your book

but I think most people go with subrings have identity

ideals are generally not subrings

And what does dummit say a ring is?

D&F defined subrings differently from any course I've taken on rings

fixed

in particular, it doesn't really ever make sense to look at subrings that don't contain 1, but on the other hand you rarely if ever want an ideal to contain 1

$6 \Bbb Z$ is an ideal, it does not contain $1$, so I don't know what dummit says a subring is

@PaulPlummer I'm pretty sure they'd claim that $6\mathbb{Z}$ is also a subring of $\mathbb{Z}$

Although I feel like you must be reading it wrong

If $Q=(x,y)$ and x is any Real number and y is any Complex number. Q is closed under addition, is there any example where a vector on the space v, and a scalar c, such that cv is not in Q?

I was given this problem.

I dont think so.

Because any point in this the space :/ There is no scalar multiple that can change that

@TheArtist depends on what field your scalars come from

are they real or complex?

What is the definition of ideal? @AlexClark

did you ever find a counterexample to "zero divisors form an ideal", @AlexClark

@SamuelYusim OH :O Haven't been mentioned. I assumed it to be real

in that case I believe what you said is true

@SamuelYusim Yes thanks :)

@SamuelYusim Yes they meant also complex :) I got it correct :)

@AlexClark No, that doesn't make sense, I agree.

I was struggling for a long time @SamuelYusim Thanks a lot for the support

yep

if it's real you're good and if it's complex just do c=i

@AlexC: any ideal that contains 1 is necessarily the whole ring, so that would be a rather silly requirement.

Hello friends.

@SamuelYusim yep ikr :)

Does it? Let me go see my copy.

@AlexC: I don't understand the statement "That's what D&F has"

D&F doesn't require that a subring contain 1

Subrings and ideals are distinct, usually. One usually requires subrings contain 1, so that they are rings themselves.

in particular a subring in D&F might have a different multiplicative identity than the whole ring, which is messed up and wrong

@SamuelYusim I don't believe you, how could that book become so popular with such a stupid definition?

because they do everything with the most generality you'll ever see in the literature, making it an amazing reference

Unless it says rings also don't have to have identity

yeah they do say that

Okay, so you lied @AlexClark And all this was for nothing

Btw, @AlexC, think about $\mathbb{Z}/6\mathbb{Z}$ if you want to give an answer to Mike's question.

You are a bad boy @AlexClark

@AlexClark: There is no issue with D&F's convention, other than that it's stupid. There are no internal consistency problems. Ideals are subrings according to D&F; according to most people they're not,.

I think that what D&F does is nice in the sense that it more closely resembles what goes on with groups, subgroups, and normal subgroups, but worse in the sense that it allows way too many of the worst counterexamples ever

@AlexClark And that is why I don't care for comments

Its taking me some time to soak in these limit points def of open sets etc I dont know why

it turns ring theory into topology

Yes they are ... Really hard

WHat do you mean by pictures@AlexC

Draw what limit point means, and see if you can see why it is related to whatever definition you know

@PaulP So first let me try to draw a neighbourhood

It says that the neighborhood of p is the set of all points q such that $d(p,q)<r$ for some r>0 right?@PaulPlummer

the set of points $q$

And it is not "the neighborhood" its a neighborhood

So if i am in lets say x-y plane the neighborhood will be a circle?

@AlexC it will be the set of all points x such that $x-0<2$ right?

Can you tell me I am really getting confused i dont know why?

So its just a line

An interval?

[0,2)

the negatives also

(-2,2)

A metric is basically the distance function satisfying three properties (which i will write if you ask me to)

$d(x,y)=|y-x|$

Hey @Alex

Curious, @AlexClark, did that ping you or AlexWertheim?

haha

I wonder if you just type @ if it pings everyone with special characters at the start of their name.

@AlexClark so is that your original gravitar or did you pick that one?

Laughing

Oh wait, it doesn't

Lol

I thought it did last time since there was already a noise associated with the first ping

Nope @Alex

@Kaj does though

Ted Shifrin always pings me as just @Stan

But the question still stands: did the above ping both you and Mr. Wertheim.

It taking to much time to upload i cant show you my ball@AlexClark Well is this ball closed

Oh there will be a limit point which wont be in the ball right @alexC

Ah so that means i will get a circle but basically points on the boundary

Not the line

Yes got it thanks!!

Yes @AlexC what will happen in 4 dim (I understand three it will be a sphere but not the boundary)

Okay

But in 4 dim(Only difference

SO if it has to be an closed Ball then it should also contain the boundary

@AlexClark The definition of zero divisor :P

@AlexC So in the given ball which i made any point except the boundary will be an isolated point right?

I was using the definition of geodesics for my definition of zero divisors. No wonder I couldn't get anywhere.

It's no worries

@anon that'll get you into trouble

common error

It will

Many days

SO any point in this neighborhood is a limit point because every neighborhood of that point will contain a point x such that it is in the open ball right?@AlexC

@AlexClark how should i prove it?

What what? I really didnt get your last statement@AlexC

Oh ok

IMAO

Got it

So we can conclude that all points in a neighborhood are limit points right @AlexC

So will the same apply to a closed set

Can we go with the same proof?

Let me assume its not a limit point

We dont need that

$d(0,2)$ if we take the standard metric is $|(2,0)-(0,0)|\leq2$

@AlexClark Yes

So every neighborhood of $(2,0)$ will contain the point 2 and hence a limit point

Am i right or horribly wrong@AlexC

Oh ya

So what should i do

Yes I can think of a picture of what you say

But how to prove in words

@AlexClark Yes

Yes i can think it like this IF i make the neighbourhood arbitrary small it will still have a point x because the neighborhood should contain all points $\leq2$ and because of this there will always be such a x

And proper. I am guessing these are the definitions given, not sure why you are asking me. I am probably not more reliable than Dummit @AlexClark

Oh okay

Its not a big deal, just was wondering why you were asking me if the definitions are right

Alright see yah

Well i am still not happy i dont know why

@Paul Are you free?

Sort of. What is it? @Rememberme

Well it says that the neighborhood of a point should be less than or equal to 2 SO i take (2,0) so basically the neighborhood will be $d((2,0),x)\leq 2$ which is basically $|x-(2,0)|\leq2$ Now can i use the modulus properties and say about the range of x which will show that the neighborhood will always contain a point less than 2 even if i make it arbitrary small @PaulPlummer

That is i have taken the standard metric

@Paulp

Make what abritrarily small? And what does it mean for a point in the plane to be less than 2? @Rememberme

Make the neighborhood arbitrary small....

You can show that there is a point in the ball by giving an example...

So you want to change the "$2$"?

Thats what i am trying to do I am trying to show that (2,0) is a limit point by showing that even if i make the neighborhood arbitrary small it will still have a point in the closed set

A limit point of what set?

chat.stackexchange.com/transcript/message/21713948#21713948 @Paul

It should not be to difficult to show that, If you draw a picture you should "know" what the smart choice of points should be. Then you just have to specify mathematically what those points are. Say I had an ball of radius $\epsilon$ around $(2,0)$, is there a point you can thing of that is within $\epsilon$ distance from that point? @Rememberme

I can think of a picture and yes it is true but i cannot write it in words@pAUL

Do an example, do a couple example, that should end up being the blueprint for what you will end up writing @Remembermer, maybe even try a "simpler" version where you are just looking at the real line instead of the plane (a ball around the point $2$, and show it is a limit point of $[0,2][$) @Rememberme

@Rememberme can you think of elements approaching 2 from the left on the real line?

you absolutely should be able to visualize this situation

Also I know you are just beginning analysis, struggling with how to but these concepts into words is part of the point. Doing examples can really help get at what you need to show, or how you would show it, if you can't quite figure out how to formalize it. @Rememberme

Hm. Could a user retag a closed question so as to unilaterally reopen it (via gold badge) and then rollback the retag?

the other direction works

just makes you a dickhead

You can instantly reopen any question closed as a duplicate that was originally asked with a tag you have a gold badge for.

meta.stackoverflow.com/questions/254589/… So I don't think so

And this unambiguously confirms that meta.stackoverflow.com/questions/254589/… @anon

@Mike $K$ be your knot in $S^3$, $X = S^3 - K$ being the complement and $\widetilde{X}$ the infinite cyclic cover. $\Sigma_n$ be branched $n$-sheeted covers of $S^3$ branched at $K$ (i.e., is a cover away from $K$). Nice fact : $\tilde{H}_\bullet(\widetilde{X}) \hookrightarrow \varprojlim H_\bullet(\Sigma_n)$.

I guess this should tell us something about Alexander polynomial of $K$, as the thing at the right hand side is really Cech homology of the solenoidal space $\varprojlim \Sigma_n$.

And Cech stuff are generally easier to handle.

Anyway, here's the paper if you're interested.

Can a polynomial only have one term?

@plebian It's called a monomial

Ok.

@BalarkaSen sup

why does $N/M \cong \Bbb{Z}$ imply that $N\cong M \oplus \Bbb{Z}$?

oh, hello, @iwriteonbananas

well, consider an exact sequence $0 \to M \to N \to \Bbb Z \to 0$

yeah

we want a section $\Bbb Z \to N$.

a section is a homomorphism that's a right inverse to the map $N \to \Bbb{Z}$?

yes.

ok

the section can be set up by sending $1$ to anything in $N$ which maps to $1$ after composing with the third map

and then extending $\Bbb Z$-linearly.

we know that we can do that since the third map is surjective

Well @BalarkaSen if I have to prove that a given point is a limit point then how should I go on doing it?

you begin by showing that every sequence in your set clutters up around the point you want to prove to be the limit point

not the kernel, no.

my bad

ok

where did we use that $M\to N$ is injective?

@BalarkaSen I have to prove that^

Then what should I do??

@Rememberme: write out the definition of "limit point"

If (x/y)=z + r/y... does that imply that z is a whole number?

why doesn't the same argument work for $$0\to M \to N \to \Bbb{Z}/2 \to 0$$? define a section $\Bbb{Z}/2 \to N$ by $1\mapsto$ some element in $N$ that maps to $1$ under the third map

this works because the third map is surjective

@BalarkaSen ok

@plebian what are $x$, $y$, and $r$?

@robjohn

@iwriteonbananas think about it

@robjohn a point p is called a limit point if every neighborhood of p contains a point q such that $p\neq q$ and $q\epsilon E$

@robjohn x y are numbers and r is the rest

@Rememberme can you show that is true in the case given?

@BalarkaSen i don't know. why does the existence of a section not imply $N \cong M\oplus \Bbb{Z}/2$ in this case?

@Remember what are the neighborhoods of $(2, 0)$ that you can think of?

draw a picture. it'll help.

@iwriteonbananas 'cause your section isn't well-defined.

I know @BalarkaSen a picture really helped me and also I was able to prove it using the picture but I am not able to put it into words

@BalarkaSen why not?

you're sending $a \in \Bbb Z/2$ to something in $N$ which slithers back to $a$ via the third map (which is possible, as the third map is surjective). but you also need this map to be a homomorphism, and it's not necessary that image of $a$ under the section, $s(a) \in N$, is a 2-torsion :)

it must hold that $s(a) + s(a) = s(a + a) = s(0) = 0$, i mean.

and an example where existence of such a map is impossible is $N = \Bbb Z/4$, which has no element of order 2.

i'll be right back.

yes

oh

but then

@BalarkaSen I sent you a message on MMF, could you please check it? I didn't know you could be contacted here.

oh ok

MMF?@octatoan

if we have an exact sequence of $\Bbb{Z}/2$-modules $$0\to M \to N \to \Bbb{Z}/2 \to 0$$, the section IS well defined though

mymathforum.com

I don't go there anymore, @octatoan. was the message you sent had anything to do with math? if so, you can just ask it here.

otherwise, i am not bothering to have a look

@BalarkaSen well anyways what I did with the help of @AlexClark hint I tried making the neighborhood arbitrarily small and see what happened when I made the neighbourhood arbitrarily small even then there were some points in the neighbourhood which were present in the closed set (with the help of a diagram) but I am not happy a d I want to put it in words can you help me or give me some kind of a direction for it

@iwriteonbananas k, guess it is.

Hello @TedShifrin

How are you

So what should I do @BalarkaSen

well, for any $\epsilon > 0$, an $\epsilon$-ball around $x$ in $\Bbb R^2$ always intersects your set $S$ (prove this rigorously). so that should be enough of a proof.

@BalarkaSen Well, I was wondering how you learn math, since I'm in sort of the same place as you (both geographically and age-wise), and taking classes at a local college isn't an option (because Kolkata), so knowing what works for you might also help me.

pick up a book and start studying

that's what worked for me.

Same here