Mathematics - 2023-07-24

Thomas Finley

02:20

math.stackexchange.com/questions/4741183/…

@robjohn Can you please help me delete this question?

@XanderHenderson If you don't mind, you might do it as well!

The reason is, I think this question is no good and much trivial. I was little hasty in putting it up. Excuse me for that.

But in addition, I feel it's cluttering things up.

shintuku

03:16

if $t_1, \dots, t_n$ are algebraically dependent, then $k[t_1, \dots, t_n]$ can be shortened to $k[t_{i_1}, \dots, t_{i_k}]$ with $i_1, \dots, i_k \in \{1, \dots, n\}$ and $k<n$, right?

robjohn

@ThomasFinley since there is an answer, it would be disrespectful to those who’ve spent time answering to delete the question.

Thomas Finley

@robjohn oh! I see. You do made a good point. Then leave it.

copper.hat

03:35

I spent some time answering a question yesterday, even interacting with the OP and after it was answered they deleted it. Which is irritating, since the point it to help others.

one potato two potato

reopen it by force

copper.hat

04:16

@onepotatotwopotato how does one re open by force?

robjohn

Use the Force, Luke!

Thomas Finley

@copper.hat I understand

0

Q: How can they say, " If $R$ is a ring, and $q$ a positive integer, then $\underbrace {a+a+a+\cdots +a}_{\text{q times}}=a.q$"?

I was studying Ring Theory and I think I have a couple of basic doubts. First of all, let me write the definition of rings. It is, A set R is called an (associative) ring if there are two binary operations defined as addition "+" and multiplication ".", which satisfies the following: If $a,b\in...

ring-theory

Help needed in here!

copper.hat

It is just notation for $a+a+\cdots + a$.

one potato two potato

@copper.hat I once saw ted reopen the answered question by asking moderators in here.

copper.hat

The mods are on strike atm, so I didn't want to ask. I did bring it up, but did not want to put them in an awkward spot.

one potato two potato

04:28

oh

Thomas Finley

@copper.hat But that pops up a connection between two arbitrary operations defined formerly, out of nowhere. This seems weird

Any idea?

copper.hat

Really, it is just notation, you are reading too much into it. You can write $\underbrace {a+a+a+\cdots +a}_{\text{q times}}$ instead, but it will become wearisome.

Thomas Finley

@copper.hat Did you check out the complete reference in my updated post, i.e

0

Q: How can they say, " If $R$ is a ring, and $q$ a positive integer, then $\underbrace {a+a+a+\cdots +a}_{\text{q times}}=q.a$"?

I was studying Ring Theory and I think I have a couple of basic doubts. First of all, let me write the definition of rings. It is, A set R is called an (associative) ring if there are two binary operations defined as addition "+" and multiplication ".", which satisfies the following: If $a,b\in...

ring-theory

It seems they use it prove something important without a justification of it's legality!

copper.hat

I don't want to wade through it all. If you do not like the notation, use another one. The $\cdot$ above is not the ring multiplication operator.

Given that the comments have mentioned that many times, I am not sure what you are hoping for in terms of resolution? I'm not sure I get the issue other than a mild abuse of notation?

There is nothing to be proved, it is purely a notational device.

Thomas Finley

04:45

@copper.hat You are misunderstanding me. I agree with you. But he uses it as a complete ring multiplication operator to prove that if $a.q=0$ (here, . is a ring multiplication operator) , then $a\neq 0$ and $q\neq 0$ might be possible.

If you would've checked out the notation, you will know what I am talking about.

copper.hat

I have no idea what a complete ring multiplication operator is.

Jesus, git a grip, will ya

Thomas Finley

@copper.hat just a ring operator.

@copper.hat cant

copper.hat

I can repeat myself, but will add nothing new. Yes, there are finite rings for which $qa=0$. And...

Thomas Finley

@copper.hat Allow me to try one last time. Let us assume $\underbrace_{a+a+...+a}_{\text{q times}}=q\times a.$ We needed to prove that $q.a=0$ not, $ q\times a=0$ but it appears they showed the later instead of the former.

shintuku

alternatively, as someone has mentioned, all rings are Z-algebras. so if $n \in \mathbb Z$ is not an element of your ring read $n \cdot r$ as $f(n)\cdot r$ where $f$ is the only homomorphism from Z to your ring. e.g. read $3 \cdot r$ as $f(3)r = f(1+1+1)r = (f(1)+f(1)+f(1))r = r+r+r$

Thomas Finley

04:53

@shintuku homomorphism is not mentioned up until the part I am in. So, I can't interpret it, like that. Sorry :?)

shintuku

all you need to know is that $z \cdot r$ with $z \in \mathbb Z$ and $r \in R$ is well defined, even when $z$ is not an element of $R$, and it is well defined without needing a new operation other than those you have

you'll learn why it is well defined when you get to homomorphisms

so there's a reason it works, and you want the reason it works, but you need the homomorphisms to understand it

one potato two potato

accept as a definition.

robjohn

@copper.hat you can always ask. If you ask a striking mod, they will just strike you with their sign.

copper.hat

@robjohn If it is a negative sign it just might put my mood in a more positive direction...

robjohn

Most signs held by strikers are usually negative.

copper.hat

05:10

The product would work in a mind altering way...

Thomas Finley

@onepotatotwopotato check my updated post, please.

2

Q: Isn't an integral domain implies automatically that it is of characteristic zero?

I was reading about Ring Theory from the book Topics in Algebra by I.N Herstein. I encountered the following definitions: Definition 1: If $a\neq 0$ is in a commutative ring $R,$ such that there exists a $b(\neq 0)\in R$ such that $ab=0$, (,where $0$ is the zero element of $R$), then $a$ is cal...

abstract-algebra ring-theory definition

copper.hat

I think you are confusing the ring multiplication and the natural addition again.

Soumik Mukherjee

Every ring has an underlying additive group structure, so $q\cdot a=0$ implies that the order of $a$ divides $q$

Thomas Finley

@SoumikMukherjee I would better say, $a+a....+a=0$

Why is $q\times a$ same as $q\cdot a$ ?

@copper.hat Sorry, it makes no sense to me now.

Soumik Mukherjee

@ThomasFinley That is just the notation

copper.hat

05:21

@ThomasFinley Noah gave an example of a ring that has no zero divisors but $2.1 = 0$.

Thomas Finley

@SoumikMukherjee But then can you elaborate why using this notation is justified to establish that "If "." is a multiplication operator in R, it might very well be possible for q\neq 0$ and $a\neq 0$ satisfying, $qa=0$"". But they conclude it from the fact, that $a\times q=0$ ?

But $\times$ is not the same as $\cdot$

copper.hat

i need mathjax, i'm too tired to parse the above

Thomas Finley

@copper.hat I think that example proves it and the proof given in herstein imho is completely wrong, bogus, unnecessary and irritating.

I should move on satisfied with the example you quoted.

Ted Shifrin

No, Herstein is a well-respected expert on algebra. I dismiss you as bogus and irritating.

Soumik Mukherjee

$q$ is not an element of the ring, so $qa=0$ does not prove the existence of a zero divisor

copper.hat

05:25

I don't see what it is about Herstein that it bothering you. It seems consistent with the above.

shintuku

@ThomasFinley $\{r \in R: r = q \times a\} \subsetneq \{r \in R: r = q \cdot a\}$

Thomas Finley

@copper.hat I am sorry, but I am dumb enough to not get you. Better, if you are interested, post an answer to quench my thirst.

Otherwise, Herstein is the villain.

shintuku

@shintuku forgot my quantifiers: $r \in R: \exists q \exists a$ blahblah

Soumik Mukherjee

Btw, in the text it is written by juxtaposition, not by $\cdot$

So where are you finding that "relation between the operations is established"?

Thomas Finley

@SoumikMukherjee herstein says, let us write a.b as ab at the beginning

copper.hat

05:36

Sorry, but I don't see what is bothering you so much. One operation is a fundamental ring operation (multiplication), the other a notational convenience. That is it.

Thomas Finley

@copper.hat Ok, are you saying all the places of occurences of say ab or xy or pq or blah blah anything like this, is not the same as ring operation of a and b i.e ab, p and q i.e pq and so on?

If this it, my problem is resolved.

But is it really so simple?

copper.hat

I believe so.

Thomas Finley

What do you believe? This, chat.stackexchange.com/transcript/message/64037924#64037924?

copper.hat

05:51

Is there something inconsistent in what I have been saying since 21.16 pst?

Vitalie Ghelbert

Problem of Logic

Argument A is the reason that imply statement S is true, otherwise S is false.
If we can demonstrate that exist argument B equivalent with A,
A became non-reason argument and neither A or B not imply S is true, therefore S is false.

A is "0/2 = 0"
B is "0/3 = 0"
S is "0 is even".

Is S true or false?

🌹

— Apologies for any confusion in my previous responses.
Let's reevaluate the logic based on the demonstration:
Argument A implies statement S is true.
A: 0/2 = 0 implies S: "0 is even."

copper.hat

Huh?

Thomas Finley

@copper.hat My answer in 22.16 is "no." But consider adding these in your answer that all these are used differently and so on in the whole piece of text. Not only in that particular line but everywhere in the picture it's used differently. I have edited the question appropriately so that you can answer it and I can accept it. Finally, closing the issue once and for all. I wont write it myself for you explained it, so it's all yours.

@copper.hat now what?

I said you may.

Ofc, you might not.

copper.hat

To be honest, I don't understand the issue or why you are getting so spikey about it.

And I am going to sleep now. Good luck.

Thomas Finley

@copper.hat gdnight

Sweet dreams ;)

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