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13:00
A nontrivial example is the following, having defined K[X] you can define K[X,Y] as (K[X])[Y], that is polynomials in Y with coefficents in K[X] (and inductively you can define the ring of polynomials in any number of finite variables)
nor is sqrt(5)
how about the field over the surreals?
oops edited
what's this notation? $S \sqcup T$ for two sets $S, T$
I haven't defined polynomials over a ring though @AlessandroCodenotti
@AlexKChen yes it is
13:00
ignoring that it's too big to be a set
@GFauxPas disjoint union, i.e. you treat same things in S and T as "different" things
@LeakyNun Ok, then ?
@LeakyNun It's not like the definition is any different from polynomials over a field
the formal construction is $S \times \{0\} \cup T \times \{1\}$ @GFauxPas
@AlexKChen now the basic constants and functions of a ring (and of a field) is {0,1,+,*} right
ah, that makes sense
13:01
we need them to exist for every ring
and field
and then we impose additional properties
field -> Inverse need to exist
@LeakyNun like ?
@AlexKChen like inverse needs to exist, addition is commutative, etc
context: the topological space $\mathbb R \sqcup \mathbb R' / \sim$ with $x \sim y$ unless $x = 0, y = 0'$
13:02
OK
@GFauxPas ah, the line with two origins
yes
its interesting
Was it false ?
If you read somewhere that $A=B\sqcup C$ it means that $A=B\cup C$ and $B\cap C=\varnothing$
it has all the properties of a manifold expect Hausdorffness
13:03
thanks Leaky et al
@AlexKChen so, a homomorphism between two rings <R,+,*> and <R',+',*'> is a mapping f:R->R' that preserves the 2 constants and 2 functions
yes thats what I was looking up, interesting non-Hausdorfness examples
:)
So basically a renaming ?
@AlexKChen it doesn't have to be bijective
But must be injective ?
13:04
no
it doesn't have to be injective either
it's just any old function
OK, so I don't understand it then. Give me an example.
consider Z -> Z/2Z
that maps an integer to its parity
Oh ok
So a homemorphism is that for all x, y in R, we have f(x)+f(y) = f(x+y) and f(xy) = f(x)f(y) ?
yes, and f(0)=0 and f(1)=1
and f(1)=1
13:07
for the purposes now, a homomorphism is any function that preserves those things, a monomorphism is an injective homomorphism, an epimorphism is a surjective homomorphism, and an isomorphism is a bijective homomorphism
and @MatheinBoulomenos I know what you would say
let's just not confuse him
Come on :/
so $\bigsqcup S_i$ would be defined either iteratively or by multiplying by $\{0\}, \{1\}, \{2\},\ldots$ I guess?
why not just use "surjective homomorphism"?
Surely you don't expect me to understand all of them ?
@GFauxPas yes
13:08
makes sense :)
@AlexKChen they are just three terms that have easy definitions
you already know what a homomorphism is
you know what injective, surjective, and bijective mean
and per the suggestion of Mathein, let's not use the term "epimorphism"
so a monomorphism is an injective homomorphism and an isomorphism is a bijective homomorphism
do you get it?
yeah
sure
for every element k in K, there is a homomorphism from K[X] to K called the evaluation homomorphism, which basically evaluates each polynomial with k
13:10
I got homomorphism and homeomorphism confused in my class once and my question to the professor didnt make sense and I felt silly :(
@GFauxPas that's unfortunate
ok then
so if you consider K=Q and k=1/2
From my experience, feeling silly seems to be an unavoidable part of learning mathematics
10
then the homomorphism maps x^2+4 to 17/4
13:11
OK great
now we'll step away from field and focus on rings
we say that m divides n if there is a number k such that n=mk
and we call m a divisor of n
now, a unit is a divisor of 1
13:15
ok
the units are closed under multiplication and inverse
huh?
oh ok
if x and y are units, so is xy and x^-1
@LeakyNun are you assuming your ring is commutative? You have to be careful with left-divisor, left-unit etc. in general
and the identity is a unit
13:16
ya sure
@MatheinBoulomenos yes
we only need comm.rings for this context
ok then
don't get me wrong: there are very useful rings that are not commutative
ok so how this is related with capelli's theorem ?
13:19
@MatheinBoulomenos all rings are commutative!
@AlessandroCodenotti nonsense
46 mins ago, by Leaky Nun
how much time do you have?
I mean what does K[alpha][x] means ?
45 mins ago, by Alex K Chen
As many as you can give me :)
@MatheinBoulomenos matrices don't exist
13:19
@AlexKChen it just means (K[alpha])[X]
Ok, but what is K[alpha] ?
I know what K[x] is
K[x] is the ring of polynomials in x over K
x is just a formal symbol
you tuold me that
(it isn't even an object, poor x)
so alpha is the ring of polynomials in alpha over K
13:21
now use the symbol alpha instead of x, and you have K[alpha]
wait
sorry, K[alpha] means something different there
What is that
do you still have as much time as I can give you?
about 15 minutes
i initially didn't thought htis would take so much time
13:22
lmao
there are much to learn about rings and fields
I don't want to just give you everything that is directly related to the statement without the background knowledge
so close to irreducible polynomials though
exactly
@AlexKChen would you let me continue my path
now a unit is a divisor of 1
13:24
now, a ring is a domain if there are no pair of non-zero elements whose product is 0
thats the defination
i.e. xy=0 implies x=0 or y=0
@AlexKChen yes
(after this lengthy 15 minutes I won't do any summary, just bookmark the conversation yourself)
Sure, Im very grateful to you explaining all this ! :)
13:25
a non-zero non-unit element in a domain is called irreducible if it is not the product of any non-zero non-unit elements
Originally asked artofproblemsolving.com/community/c7h1565464_capellis_theorem but nobody explained it.
@LeakyNun ok
a non-zero non-unit element p in a domain is called prime if for every x and y, p divides xy implies p divides x or p divides y
sorry cat
ok
@AlexKChen Nice cat.
Knew how to step all the keys first before pressing Enter.
13:27
now <R,+,*> is a subring of <R',+',*'> if R is a subset of R' and + is inherited from +' and * is inherited from *'
what is "inherition" ?
i.e. x + y = x +' y for every x and y in R
and likewise for *
@AlexKChen Inheritance is the noun.
if K is a subring of L and b is an element of L
then K[b] is the smallest subring of L containing K and b
13:29
lemme think for a bit
Q is a subring of C
sqrt2 is an element of C
so Q[sqrt2] is the smallest subring of C containing Q and sqrt2
i.e. {a+bsqrt2 | a,b in Q}
you told the example before
Q[cbrt2] would be {a+bcbrt2+ccbrt4 | a,b,c in Q}
13:31
Yeah of course
Q[e] would be isomorphic to Q[X]
what is e ?
euler's constant
elements in Q[e] are polynomials in e
How "
Q[e] would be isomorphic to Q[X]" makes sense ? One is polynomial and one is ring, right ?
because e is transcendental
@AlexKChen no, they are both rings
Q[X] is the ring of all polynomials in X over Q
Q[e] is the smallest subring of C that contains Q and e
13:33
Wait a bit I'm confused
Obviously x^3-x-1 is a element in Q[X]
yes
and e^3-e-1 is an element in Q[e]
(if a ring contains Q and e, it must contain e^2, so e^3, so -e, so e^3-e, so e^3-e-1)
Oh yeah
They're isomorphic
13 mins ago, by Alex K Chen
about 15 minutes
2 minutes to go
<F,+,*> is a subfield of <F',+',*'> if F and F' are fields and F is a subring of F'
13:36
OK
that's every word in Capelli's theorem's statement defined
49 secs ago, by Leaky Nun
<F,+,*> is a subfield of <F',+',*'> if F and F' are fields and F is a subring of F'
11 mins ago, by Leaky Nun
a non-zero non-unit element in a domain is called irreducible if it is not the product of any non-zero non-unit elements
15 mins ago, by Alex K Chen
about 15 minutes
Sure thakns, no need to search older messages
that's time's up :P
Can I ask you one more quesiton (should take around 2~3 minute)
go ahead
13:38
What does Q[a_1, a_2] means ?
Does it mean
(Q[a_1])[a_2] ?
where a_i is roots of some integral polynomial
yes
I hesitated to say yes before your last statement
So "The smallest subring of C that contains Q, a_1 and a_2 ?"
yes
ah, it's yes even without your last assumption
Thanks a ton, Leaky nun for you help !
@user21820 oh hey I didn't even notice that it was you (I knew you talked, but I didn't know it's you)
@AlexKChen thanks for letting me waste your time lol
13:41

Leaky nun explain terms for Capelli's criterion

1 hour ago, 1 hour 13 minutes total – 329 messages, 7 users, 4 stars

Bookmarked 8 secs ago by Alex K Chen

you included quite a few irrelevant messages though
you take that back, those messages heard you
Could you tell me what kind of topic in mathematics discussing the number of ways representing a number N as a finite sum of 1, 2, ... up to r where r<=N?
For example, N=2

2 = 1 + 1
2= 2

So there are 2 ways only.
Partition is the keyword you want to google
In number theory and combinatorics, a partition of a positive integer n, also called an integer partition, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. (If order matters, the sum becomes a composition.) For example, 4 can be partitioned in five distinct ways: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1 The order-dependent composition 1 + 3 is the same partition as 3 + 1, while the two distinct compositions 1 + 2 + 1 and 1 + 1 + 2 represent the same partition 2 + 1 + 1. A summand in a partition is also...
13:51
Ugh. Why can't we use one word to mean only one thing?
Good, thanks
because there are more things than words
pigeon-holed
Now we know that $\Bbb{R}[x]/(x^2+1)$ is isomorphic to $\Bbb{C}$. This fact is well understood but what I dont understand is why cant we similarly construct $\Bbb{R}$ by quotienting out $\Bbb{Q}[x]$.
I mean I do understand that why we cant do it, because like in the case of $\Bbb{C}$ we have $i$ and every complex number corresponds to a tuple of real numbers and one of them having the term $i$ associated with it, but for the reals we dont have any such $i$, that is there is no such unifying factor. But can this idea be made rigorous ?
This doesn't work for several reasons. First, $\Bbb R$ is uncountable, but $\Bbb Q[x]$ (and thus any quotient thereof) is countable
How about a partition with restriction as I mentioned above.
For example, N=4 and the partition can only use 1, 2, 3 but not 4.
13:54
@MatheinBoulomenos Yeah thats true.
$\Bbb R$ also contains transcendental numbers like $e$ or $\pi$ which are another reason why we can't realize it as a quotient of $\Bbb Q[x]$
Also, if we have a nonzero polynomial $f$, then $\operatorname{dim}_{\Bbb Q}\Bbb Q[x]/(f)=\operatorname{deg}(f)$, but $\Bbb R$ is infinite-dimensional over $\Bbb Q$
But to come back to your question, yes, it's true that one can show that $\Bbb R$ is not a quotient of $\Bbb Q[x]$ by proving that $\Bbb R \neq \Bbb Q[\alpha]$ for all $\alpha \in \Bbb Q$
cantor's approach is much better
people back down didn't know how to count
@ArtificialStupidity there's only one partition that uses 4
this comes up in other situations, for example, if one considers $\Bbb F_p(\sqrt[p]{x},\sqrt[p]{y})$ as an extension of $\Bbb F_p(x,y)$
@LeakyNun ok. lets take N=10 and we can only use 1,2, 3, 4 , how many ways are there?
that's the coefficient of x^10 in (x+x^2+x^3+x^4)^10
14:02
@LeakyNun Nice hint! Thank you!
@MatheinBoulomenos okay, I see thanks
I obviously meant $\alpha \in \Bbb R$ not $\Bbb Q$
Yea or else it would be meaningless to call it an extension
the rationals and the reals have the same birthday though
go figure
@LeakyNun Hey! I just dropped in because on Main the chat-room came up and said you were talking in there. =)
14:06
@LeakyNun The coef of x^10 is always 1.
Working on a two part problem and the first part is to show that for any $A \in M_2(\Bbb{Z})$, $(x+A)(x-A)=x^2 - A^2 \in M_2(\Bbb{Z})[x]$. If I can say that $Ax=xA$, then this problem is trivial (note $x$ is not a vector but an indeterminate). Am I missing something or does $Ax=xA$ hold trivially (by definition)? How do I justify this?
$Ax=xA$ does hold by definition
@MatheinBoulomenos Thanks! That's what I figured.
For a ring (not necessarily commutative) $A$, $A[x]$ denotes the polynomial ring where it is assumed that the variable $x$ commutes with all elements from $A$
14:39
Hello. What are some examples of smooth local injections which are not immersions?
dense line in the torus
the figure eight map R --> R^2
What do you mean by 'dense line in the torus'? Also, I thought the figure eight map was a generic example of an immersion. Could you please elaborate?
@BalarkaSen you're giving examples of immersions that are not embeddings
oh i misread
injections which are not immersions are super easy. consider x --> x^3, R --> R
x --> x^(2n+1) for any n
Thank you. I don't know how to phrase this formally, but are there any "other" examples? Ones that are not given by "stopping", i.e by vanishing derivative.
14:50
for higher dimensions you kill a subspace by reducing rank
it's stopping on a dimension
and that's all that can happen, right?
Yes, by the very definition of the word "immersion", and how rank of a matrix works...
yep. Ok, thank you very much
15:03
why matrix determinant is maximum when it’s diagonal elements are least?
15:16
It’s not. A multiple of the n-by-n identity matrix $cI_n$ has determinant $c^n$ which is increasing with $c$
You can find examples where increasing a diagonal element decreases the determinant, but it’s by no means true in general @Fawad
15:35
@Semiclassical when entries are only 0 and 1
Can we relate it from volume of parallelopiped?
Considering 3by3 matrix
Hadamard's maximal determinant problem, named after Jacques Hadamard, asks for the largest determinant of a matrix with elements equal to 1 or −1. The analogous question for matrices with elements equal to 0 or 1 is equivalent since, as will be shown below, the maximal determinant of a {1,−1} matrix of size n is 2n−1 times the maximal determinant of a {0,1} matrix of size n−1. The problem was posed by Hadamard in the 1893 paper in which he presented his famous determinant bound and remains unsolved for matrices of general size. Hadamard's bound implies that {1, −1}-matrices of size n have ...
15:53
Do I understand this error term correctly? I have a function $f(n)=g(n)+O(e^{-c\sqrt{\log n}})$. This means that $\exists K\in\Bbb R^+$ such that $f(n)\in\left[g(n)-Ke^{-c\sqrt{\log n}},g(n)+Ke^{-c\sqrt{\log n}}\right]$, correct?
16:16
paths in topology are not functions right?
they are just maps
@ManolisLyviakis A path in a topological space $X$ is a continuous function $[0,1]\to X$
so a loop is not a path
its just a map
What do you mean with "map"? Usually it's synonymous with function
for one input it can give more than 1 values
that is just a map not a function
a circle is a map not a function
a loop is definitely a path then
$f:[0,1]\to X$ continuous with $f(0)=f(1)$ is a loop
16:19
but a loop gives 2 values for one input
no, it gives the same output for 2 different inputs
A loop function does not necessarily mean that there are multiple outputs for a single input.
It could mean that the function is cyclic
i dont get it
a loop always gives at least for one input 2 values
hey @BalarkaSen
do you know about locally free sheaves?
I just have a simple question
Consider the function $f:[0,1]\to\Bbb R^2$ defined by $f(t)=(\cos 2\pi t,\sin 2\pi t)$, do you agree that this is a function and that its image in $\Bbb R^2$ is a circle?
16:24
its image is not a circle unleess it goes 0-2π
@AlessandroCodenotti: you might use $f:[0,2\pi]\to\Bbb R^2$...
the half circe is a function
@ManolisLyviakis fixed it
@ManolisLyviakis: the whole circle is a function; which value of $t$ forces $f(t)$ to produce more than one value?
yes
you are right
no input gives multiple values
16:31
If I have a function $f(n)=e^n+O(n)$, can I say that there exists a real value $K$ such that $e^n-Kn\leq f(n)\leq e^n+Kn$?
Based on the answer here math.stackexchange.com/a/1067625/86846 I'm going to say "yes" and move along...
Is it known that there are always two primes between $\sqrt{n} $ and $n+1$?
If $n\ge 16$ then I think you can use Bertrand's Postulate...
16:49
i need to prove that the set of classes of hopotoic functions from X-->[0,1] has only one element
but i just started algebraic topology
so im kinda stuck my intuition doesnt really help
im trying to imagine the possible functions but since my X is an arbitrary topological space i cant xD
maybe i can say all of them are homotopic through the straight line homotopy?
Hi all; I'm doing a Number Theory question, and if someone could offer a hint, that would be greatly appreciated;
The question is: Find the sum of the perfect square divisors of the smallest integer with exactly 6 perfect square divisors.
since [o,1] is Kinda like $\mathbb{R}$
@BalarkaSen can you let me know when you come back I want to discuss some small issue with you
My method so far has just been listing out combinations of exponents that are multiples of 2, but obviously that hasn't been working out too well
@Adeek Ask. I can't say I would be able to answer.
16:56
@BalarkaSen it is just very small issue so in the why do we require the base space of vector bundle to be connected in order to prove isomorphism of the category of locally free sheaves and vector bundles.
is it because the rank will not be constant otherwise ?
You have answered your own question.
okay yeah :P @BalarkaSen thanks
17:42
Does anyone here know the rule whereby log(1 / (1 + math.e**-x)) == -log(1 + math.e ** -x)? I can see the truth equivalence but wasn't sure what rule to apply to get from the former to the latter...
negative exponent rule: 1/a=a^(-1)
power rule for logs: log(x^b)=b log(x)
alternatively
quotient rule for logs: log(x/y)=log(x)-log(y)
fact about logs: log(1)=0
but one may say the quotient rule comes from using the negative exponent rule, power rule, and product rule
@anon thank you for your note. My little motor is computing...
hey, has anybody heard of the question -- What is the probability that a triangle formed by selecting three points on a circumference of a circle at random, contains the center?
if so, I've got a quick question about my solution
what's your question
@anon I wanted to say thanks again--this is perfect :)
17:52
np
well i compute the probability to be integral from 0 to pi of (theta/2pi) dtheta
twice that quantity ^
and divide by 2pi again afterwards?
theta/2pi talks about the probability of the third point in the arc formed by theta, but you want to pick the second point (defined by theta) uniformly at random which means you gotta divide by 2pi for that too
yes, why
i have intuition that i ought to interpret [0, 2pi] as the unit interval [0, 1]
but i cant justify it
ohhhhhh
duh
the uniform distribution on [a,b] is the usual length-measure divided by (b-a). for instance, for the uniform distribution on [0,pi], the probability of getting in the interval [pi/3,2pi/3] will be (2pi/3-pi/3)/(pi-0)
thank you.
17:55
mmhmm
there's another argument using symmetry (and no integral calculations) that generalizes to 4 points on a sphere and tetrahedra (and higher dimensions)

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