Conversation started Dec 29, 2017 at 12:27.
Dec 29, 2017 12:27
HI everybody
What doe this means in elementary terms:
Let K be a subfield of C and f,g $\in$ K[x]. Let z be a complex root of f and assume f is irreducible in K[x] and g(x)-z is irreducible in K[z][x]. Then f(g(x)) is irreducible in K[x]
@jublikon $x$ is fixed
let $t=h/x$
@AlexKChen do you know what a field is?
Nope very sorry (I know what finite field is though, courtsey you)
where does that question come from?
statement
Problems from the book, by Gabriel Dospeniscu and Titu Andreescu
is it from a course?
Dec 29, 2017 12:32
This is from chapter 21 of that book, called Cohn or something irreducibilty criteria
That is a book for preparing for mathematical olympiads
I'm uploading a picture, you'll see
how much time do you have?
As many as you can give me :)
so we are familiar (let's say we are) with the real numbers and complex numbers
we know many of their properties
but one might wonder, what properties are retained with other types of numbers
Dec 29, 2017 12:35
Like ?
so we might want to define other types of numbers
this is an abstract setting: we want it to work for all kinds of numbers
firstly we might want to define what a number is
we don't actually define what a number is
but a set of numbers, when they satisfy certain properties, we call them a ring
What properties ?
if they satisfy other properties, we call them a field
so let's say we have a set of numbers
we aren't just concerned with the numbers though; they need to interact
and we want them to interact using the two basic operations: addition and multiplication
Can I interrupt a bit ?
Dec 29, 2017 12:38
I know a bit about integers modulo prime and very little about finite fields.
So no need to introduce them again.
Please tell what happens after that.
are you sure?
OK, so you can give analogies with integers modulo p and inverses modulo p. I know them well.
if you like it then you shoulda put a ring on it
@AlexKChen I'm doing this in the abstract setting
it works with every possible kind of numbers
OK, sure, then go on.
Dec 29, 2017 12:40
so we have a set of things we call numbers, as well as two binary functions we call addition and multiplication
ok
what's binary function btw ?
let our set be called R
then a binary function is a function of the type R^2 -> R
must be commutative ?
Dec 29, 2017 12:42
now we want those numbers and the two functions to satisfy some properties
we want the two functions to be associative and commutative
we want multiplication to distribute over addition
what does that mean ?
"distribute over addition" ?
that means, (x+y)*z = x*z+y*z
Dec 29, 2017 12:44
we want additive identity and multiplicative identity to exist, i.e. a number we call 0 such that x+0=x for all x, and a number we call 1 such that x*1=x for all x
we want negatives to exist, i.e. for every x there is y such that x+y=0
we call those numbers together with the addition and multiplication a ring
"the [set] forms a ring under [addition] and [multiplication]"
@AlessandroCodenotti buongiorno
Dec 29, 2017 12:46
Hi
@AlexKChen do you get it?
what does "buongiorno" mean ?
it means good morning in italian
oh ok
So that's the defination of ring
now if multiplicative inverses exist for every non-zero element, we call them a field
Dec 29, 2017 12:47
so integers modulo composites are ring, and modulo primes are field ?
that you can spell buongiorno is more impressive...assuming that's how you spell it
@AlexKChen yes
and every field is a ring
of course
now let's say we have a field K
Dec 29, 2017 12:49
we can define polynomials over K
a polynomial is defined to be something of the form a_n x^n + ... + a_1 x^1 + a_0 x^0
they are not functions. they are just formal objects
where n is a natural number
"they are just formal objects" I don't understand
Gimme an example
I thought polynomials were functions
formal (adj)
of or concerned with outward form or appearance as distinct from content.
"I don't know enough about art to appreciate the purely formal qualities"
@AlexKChen but here we are not considering them as functions
think of em as combinatorial objects....at least, i usually see them as physical objects with lots of little pieces
Dec 29, 2017 12:51
@LeakyNun we're treating them (like say) circles ?
nah, like legos
if we treat them as functions, we start running into problems like "why is 0 and x^2+x different polynomials over Z/2Z if they agree at every point in the domain?"
because a function is uniquely determined by its domain, codomain, and the image of each point in the domain
OK, so we can treat them as tuple of numbers: (Coeff of x^n, Coeff of x^n-1, ..., constant coeff) ?
precisely
now we can add polynomials together
we can also multiply polynomials together
and you can verify that the set of the polynomials over K forms a ring under polynomial addition and polynomial multiplication
ok
yeah of course
Dec 29, 2017 12:53
we call that ring K[X]
X is numbers from ?
no
K is a field
we denote by K[X] the ring of monovariate polynomials over K under addition and multiplication
"polynomials in X"
(polynomials with one variable)
only blue legos
Give me an example except common things (polynomials in R, Q, integers modulo P)
of what
Dec 29, 2017 12:56
a nontrivial example of a polynomial in K[x]
you need K to exist for K[X] to exist
It's just an ordered n-tuple of elements of K
if K is Z/2Z, K[X] contains x^3 + x + 1
the coefficients are only 0 and 1
I wrote "except common things (polynomials in R, Q, integers modulo P)"
Dec 29, 2017 12:58
but you don't seem to understand that K[X] depends on K
No I seem to understand.
let's say K is Q[sqrt(2)], i.e. {a+bsqrt(2) | a,b in Q}
I was asking for a nontrivial example of K
yeah that
Z/2Z is trivial? you take that back
Do you know complex analysis? The meromorphic functions on a domain form a field
Dec 29, 2017 12:59
So this is a good polynomial:
it heard you
(2+sqrt{2})x^3 + sqrt{2} ?
no
2+sqrt3 isn't in K
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)
Dec 29, 2017 13:00
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
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
Dec 29, 2017 13:00
@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
we need them to exist for every ring
and field
Dec 29, 2017 13:02
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 $x = 0, y = 0'$
@GFauxPas ah, the line with two origins
Dec 29, 2017 13:02
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
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
Dec 29, 2017 13:03
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 ?
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.
Dec 29, 2017 13:05
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
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 :/
Dec 29, 2017 13:07
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
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
Dec 29, 2017 13:08
yea
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
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
Dec 29, 2017 13:10
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
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
Dec 29, 2017 13:12
ok
and we call m a divisor of n
now, a unit is a divisor of 1
the units are closed under multiplication and inverse
Dec 29, 2017 13:15
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
ya sure
@MatheinBoulomenos yes
we only need comm.rings for this context
Dec 29, 2017 13:17
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 ?
@MatheinBoulomenos all rings are commutative!
@AlessandroCodenotti nonsense
Dec 29, 2017 13:19
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
@AlexKChen it just means (K[alpha])[X]
Ok, but what is K[alpha] ?
I know what K[x] is
Dec 29, 2017 13:20
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
now use the symbol alpha instead of x, and you have K[alpha]
wait
sorry, K[alpha] means something different there
What is that
Dec 29, 2017 13:21
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
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
Dec 29, 2017 13:23
exactly
@AlexKChen would you let me continue my path
now a unit is a divisor of 1
now, a ring is a domain if there are no pair of non-zero elements whose product is 0
thats the defination
Dec 29, 2017 13:24
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 ! :)
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
Dec 29, 2017 13:26
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.
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" ?
Dec 29, 2017 13:28
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
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
Dec 29, 2017 13:30
aha
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}
Yeah of course
Q[e] would be isomorphic to Q[X]
Dec 29, 2017 13:32
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
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)
Dec 29, 2017 13:35
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'
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
Dec 29, 2017 13:37
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
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
Dec 29, 2017 13:39
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
 
Conversation ended Dec 29, 2017 at 13:40.