my hs friernds who hate math
#groups-rings-fields
solemn rain
upper pivot
you need to think hard and long
woven delta
How long do you spend on individual problems
upper pivot
before finding the tricks
solemn rain
idk
woven delta
Yeah
upper pivot
try for atleast 30 mins
solemn rain
it took me like 30 seconds to find the iso that i told john about
and john told me its trash
and then i just gae up and asked heree
solemn rain
yea shamrock told me same buit
its just hard to believe that
i will get it
like its not there at all boys
upper pivot
just try spending more time on problems
solemn rain
the trick that is
solemn rain
yea but i think
upper pivot
i do
solemn rain
๐
woven delta
I once spent 2 weeks on a single problem
upper pivot
i mean you dont know that unless you try a problem lol
solemn rain
maybe ith elps? idk
latent anvil
math is hard
solemn rain
as fuck
woven delta
^
solemn rain
really its so hard
latent anvil
I don't think I've ever spent 2 weeks on a problem
latent anvil
Most problems I'm working on are for classes, where I have 1 week to do them
woven delta
I mean if it weren't for quarantine I wouldn't have
solemn rain
i am like genuinely terrified of how math can be difficult in like
grrad school
lmfao
woven delta
And also it wasn't like constant work
latent anvil
Oh yeah I mean it never is
solemn rain
is it normal
that you guys go from literally 0 ideas to solve
a probleml
to the right idea?
latent anvil
like I've spent days working on a problem, but really I mean I've spent days with it in the back of my mind
mo2men it like, happens sometimes
woven delta
There are some problems that I actually spent days on like almost nonstop
But research problems
woven delta
Yeah
latent anvil
why did no one tell me this
upper pivot
yeah im curious too
solemn rain
i forgot how to do like
change of basis stuff
in transformations
do i need to review that
woven delta
But anyway mo2men
solemn rain
okay liquid im doing lin algebra uptill jordan
latent anvil
As least I get to draw pretty pictures
woven delta
You are far to early to be stressing about these things
solemn rain
i am just stressing about problem solving
i once heard that like the purpose of UG is just to teach u how to solve problems
not the specific conent
which was weird for me
woven delta
Eventually you will be good at problem solving
latent anvil
I mean I feel like some of the specific content is also pretty important
woven delta
Problem solving for research is very different than regular problem solving
upper pivot
you dont need rudin for LA?
solemn rain
i meant like
for learning math
ig there is a reason ug starts with analysis
and lin alg
upper pivot
i mean its comp math
also like, same principle
just be willing to spend hours on a problem lol
solemn rain
cool boys
can i use another li nalgebra text?
rather than HK
i want something new
upper pivot
i mean id advice sticking with it
woven delta
Yeah stick to what you're doing now
@solemn rain the last thing you want to do is switch around and get nothing done
(see NC)
shy bluff
Question about this
Part b
I dont' really know how to show that the normalizer of T is PT?
Like I see that the normalizer is gTg^{-1}, and that because T is diagonal it.. has a basis that's just made up of elementary basis vectors, and that it must be equal to something in the permutation group becaus ethe permutation group is simply a ... permutation of said basis vectors?
But I don't know how to show that?
I tried expanding it out but I'm not really getting anywhere?
shy bluff
Also like PT is just going to be the set containing all the products of permutation matrices with all the products of diagonal matrices right?
shy bluff
I dont' know how to show that the normalizer of T is PT
Like I see that the elements of PT are the elements of the normalizer but I don't know how to prove it
woven delta
- show congugating by a permutation matrix gives you another diagonal matrix. Therefore PT is inside the normalizer
- Show that the matrices which normalize the invertible diagonal matrices are exactly the ones which map basis elements to nonzero multiples of other basis elements
shy bluff
And by "T is normal in PT" you mean "show that tPTt^{-1} = PT"?
And by 1), you mean conjugation a diagonal matrix by a permutation matrix correct?
woven delta
Yeah
1 is obvious because of what you said
But you can do it concretely also
With lots of indices
woven delta
Okay so the diagonal matrices are the ones so that for each basis element e_i, D(e_i) is r_i e_i for some nonzero constant r_i
shy bluff
Yes
woven delta
Okay now suppose a permutation matrix maps e_i to e_j
shy bluff
And the permutation matrix is just [sigma(1) .... sigma(n)] right?
woven delta
Then $Q^{-1}DQ(e_i) = Q^{-1}D(e_{\sigma(i)}) = Q^{-1}(r_{\sigma(i)}e_{\sigma(i)}) = r_{\sigma(i)} e_i$
And that's it
That actually wasn't so bad
cloud walrusBOT
Liquid:
woven delta
Yeah
The matrix associated to sigma
Wow I actually was able to write those indices on mobile
And didn't make any mistakes
Does that make sense @shy bluff
shy bluff
I think so
Um
I'll try writing it out aggain but this time using what you gave me and see if I get it
Thank you!
woven delta
๐
Oh my step 2 is bad
Let me change that
Okay @shy bluff fixed my step 2
The step 2 I wrote before was just step 1 lol
I basically described PT in a nicer way
shy bluff
oooh ok I gget what you did now!
You went and showed it using the standard basis vectors instead of only using the matrices
Makes sense!
Thank you!
shy bluff
Question, is this a valid function?
$$\phi: PT/T \rightarrow T$$ such that $\phi(a) = \phi(ptT) = p$ for some $a \in PT/T, p \in P, t \in T$?
cloud walrusBOT
Liria ^(;,;)^:
tribal pasture
You have the output to be p, an element of P, whereas the codomain is T. This should raise a problem
@shy bluff
shy bluff
Wait I meant to P not to T
Sorry
Uh $$\phi:PT/T \rightarrow P$$ is what I meant I think
cloud walrusBOT
Liria ^(;,;)^:
tribal pasture
Then this seems to be okay
shy bluff
Okie thank you!
kindred mist
Let A be a ring, let M be an A-module, and let ${M_i}_{i \in S}$ be a family of submodules of M. How do we distinguish (notationally) between
cloud walrusBOT
๐๐ธ๐พ๐ท๐ฝ๐ช๐ซ๐ต๐ฎ:
kindred mist
${\sum_{i}(m_i) | m_i \in M_i}$ and the direct sum ${(m_{i}, ... ) | m_i \in M_i}, because I have seen these both written as $\sum_i (M_i)$ ?
kindred mist
Thank you!
for some reason Hungerford uses \sum for direct sum, but using \oplus definitely makes more sense to me
upper pivot
i think that notation is used in some places
and the sum coincides with \oplus if for example nothing intersects
(and we take the sums in M)
kindred mist
so if the M_i are all disjoint, yeah that makes sense
coincides as in isomorphic as modules right (I think I can prove that, just making sure I get what you mean)?
upper pivot
yeah
kindred mist
thanks!
latent anvil
It's not if they're all disjoint
(or more properly if they all intersect at the identity)
Take A = Z, M = Z/2Z (+) Z/2Z, and M_1, M_2, M_3 the district nontrivial subgroups, so M_1 = Z/2Z (+) 0, M_2 = 0 (+) Z/2Z, and M_3 = {(0,0), (1, 1)}
these sum to the whole group and the pairwise intersections are trivial
cloud walrusBOT
shamrock:
upper pivot
oh yeah i see
oops i was thinking of the case with 2 submodules and didnt think far ahead
latent anvil
Yeah np I've made this mistake before
I changed a problem set at the last minute and put the incorrect statement on it
which is extra yikes
upper pivot
rip
broken basin
Might be a weird question, but has anyone ever thought about developing an algebra for emotions?
It seems a monoidal structure exists. e.g. insecurity โ fear = jealousy
Also, see this standard wheel of emotions
Seems like there are natural inverses, and therefore could constitute a group?
scarlet estuary
er
i guess you could find a monoid that corresponds to some "chart" of emotion "combinations", if you wanted?
just take an existing monoid that satisfies your properties and rename everything
i'm not sure why you'd do this.
(i'd also assume "an algebra" here doesn't mean a literal algebra, since "emotions", however ill-defined the differences between them may be, certainly dont fit the structure of an algebra)
half nebula
Hey could I have some help with these 3 problems? I just have no idea how to really start them outside of 2 which I have fine. It's really just 1 and 3.
As for 1, I found this online but I don't fully understand what it's saying.
dreamy anchor
We need to ensure that phi(x + y) = phi(x) + phi(y) for all choices of x and y in Z_24 when getting output in Z_18, and the easiest way is to work from the identity because its properties are the nicest (and that it always has to be in the mapping; the image of Z_18 is in fact a subgroup). They're using Fermat's Little Theorem/Euler's Theorem to say that anything to the 18th power is going to be the same as (congruent to) itself mod 18, and then deducing that, for example, x^6 can't be e (otherwise it's no longer a [in this case cyclic] subgroup).
They play with a few more possibilities until they get that bottom part of text where if everything works out, then the power y^k that works as homomorphism if and only if 18 divides 6k, which gives that they have below
half nebula
Okay that makes sense. Where did they get y^k from though?
Like how did they know to use that?
And further, what actually is y and what is k?
Oh wait never mind. I see what y is.
dreamy anchor
For 3, the image is just by taking all those classes and seeing their result mod 15. [a] x [a] = [a^2 mod 15] in this case. Those a that map to [1], the multiplicative identity, are what gives us the kernel.
half nebula
Wait hold on I still don't get 1 yet. Like where do they get y^6 from?
dreamy anchor
I believe in general it's easiest to work with the lcm of the two moduli because that (6 here) can be raised to a power in either side to yield the identity
You could pick a number like 12 and produce the same argument, but more complicated
half nebula
But the identity for Z18 is x^18 and for Z24 it's x^24 right?
How is 6 the identity?
dreamy anchor
It's not, they're arguing there that the for <e, x^6, x^12, x^18> to be a subgroup which must be, x^6, x^12, and x^18 must either be all the identity (trivial homomorphism, so phi(x) = x^6 = x^12 = x^18 = e) or all not be the identity
half nebula
I get how x^18 is the identity but how is x^6 and 12 the identity?
dreamy anchor
For the trivial case y^6k gives the identity, but this is just one unique homorphism formula
When they're all not is when we produce the case work that they show
half nebula
Sorry I guess I'm still confused on that. What's a trivial case and what's a homomorphism formula?
dreamy anchor
A homomorphism formula is just phi(x) = some function of x
The trivial case is my term for when the subgroup is just the identity
You may have heard the "trivial group" be just the identity
half nebula
Okay I get that now. So how is y^6k the trivial case? Because 6 goes into both 18 and 24?
dreamy anchor
y^6k has two possibilities:
1.) For any integer k, y^6k is the identity, so e = y^6 = y^-6 = y^12 = y^-12 = ... so our image aka subgroup of Z_18 is just {e}
2.) Using case 1, we argue that now they all must be different. But by the properties of a homomorphism giving the image of something from Z_18, each 6k must divide 18. This is where we get the "multiple of 3" idea.
half nebula
But how is y^6 the identity in case 1? And how is it equal to y^12?
Or is that the point that it's not the identity?
dreamy anchor
Because lcm(24, 18) = 6, we put our attention to the subgroup {e = x^24, x^6, x^12, x^18}. If all the elements are equal, clearly e = e^2, etc., so y^6 = y^12 = (y^6)^2, etc.
So for the top case, we're really just finding out what the trivial homomorphism has the form of.
To find any other forms/formulae, we now know they'd have to be different, and that's where our argument comes into play that each 6k divide 18
We now try to find other cases where y^k = phi(x)^k gives us the identity
We get that k is a multiple of 3, so phi(x)^3, phi(x)^6, phi(x)^9.
half nebula
But wait the lcm(24, 18) is 72.
half nebula
Ooooooh okay
dreamy anchor
12 and 15 fail because they don't divide 18, which is the max order of any element in Z_18
Now I sleep
half nebula
Okay cool. And what you typed for 3 is all I need to know right?
dreamy anchor
For 3 they might want the image also as the name of a group
So whatever you get after checking 1, 2, 4, 7, 8, 11, 13, and 14, maybe there's a name for that group of integer classes left doing a -> a^2
Besides that, yes
half nebula
Okay.
lyric plover
Are there generators of (R,+)?
dreamy anchor
Equivalent would be "is there a smallest real number?"
oblique river
there's a generating set of (R,+)
yes, the smallest generating set of (R,+) is uncountable
there aren't any single generators of R
woven delta
Any generating set must have exactly the same Cardinality as R itself
lyric plover
Wouldn't {0} be a trivial generating set?
hot lake
no because the subgroup of R generated by {0} is {0} and {0} is not R
prime oriole
So like R is a commutative ring with 1.
And R has exactly one maximal ideal
Need to show that set of all non- units forms an ideal .
I'm getting stuck, how do i show that this is closed under addition
oblique river
if a and b are two nonunits
you need to use the fact that the ideals (a) and (b) are both contained in the maximal ideal m
also sorry zoph haha
i saw you typing just as I started
mild laurel
I just started typing too its fine
prime oriole
Ok I'll try thanks
red imp
is it a thing that every permutation polynomial on a finite field gives you a different permutation of that field
mild laurel
If I'm understanding you right, no, the polynomials x and x^p would give you the same permutation for example
@red imp on F_p
red imp
oh I meant to say polynomials of degree less than p-1
mild laurel
Yeah, since if two polynomials gave the same permutations, then you could subtract them, and you'd have a polynomial of degree less than p - 1 with p roots, which implies that its 0
mild laurel
This is a pretty common idea that's used a lot
red imp
yeah but I'm a noob to algebra so it's like black magic
mild laurel
You maybe should think a bit more about why a polynomial of degree n has at most n roots over a finite field
This isn't super obvious, since for example the polynomial x^2 + x over Z/6Z has 4 roots
red imp
okay that's the main issue I've been having in my algebra class
what the hell does a quotient ring look like
I have zero intuition for them
is there a resource that's like "quotient rings for dummies"
mild laurel
well, I guess the way that I'd explain it for groups is like
6Z (the multiples of 6) is a subgroup of Z
or replace subgroup with ideal
So you have {..., -12, -6, 0, 6, 12, ...} = 6Z
but you also have translates of this subgroup/ideal in some sense
red imp
oh so the group analogue of an ideal is just a subgroup?
mild laurel
so you can consider like 1 + 6Z { ..., -11, -5, 1, 7, 13, ...}
Yeah
(well, maybe normal subgroup)
red imp
oh yeah
mild laurel
And so you have 6 of these translates
0 + 6Z, 1 + 6Z, 2 + 6Z, 3 + 6Z, 4 + 6Z, 5 + 6Z
and you can ask if you can put a group/ring structure on these 6 elements, that kind of reflects the group/ring structure of Z
In other words, you want (n + 6Z) + (m + 6Z) = (n + m) + 6Z
so like (1 + 6Z) + (2 + 6Z) = 3 + 6Z
red imp
ah yes I have seen how addition is defined in these rings
mild laurel
So your quotient ring is just the 6 translates
red imp
:o
mild laurel
and you can put corresponding addition/multiplication operations on these 6 cosets they're normally called
red imp
so Z/6Z just has 6 elements and that's it?
mild laurel
yeah
red imp
well that has been far more complicated in my mind, that simplifies it a bunch
appreciate it
mild laurel
I mean, in this case, you can of course think about it as just the integers mod 6, and this translates to the cosets of 6Z
red imp
is R/{0} isomorphic to R
mild laurel
as rings? groups?
red imp
rings
mild laurel
How are you putting a ring structure on R/ {0}?
red imp
I dunno I just thought of the ideal {0} and wondered what R/{0} was gonna be like
mild laurel
Oh, that's quotient
red imp
trying to gain some familiarity
mild laurel
Yeah, it is
red imp
ohkay cool
mild laurel
I mean, going back to our example
Z/{0}
its translates are basically just {1}, {2}, {n}
mild laurel
You mean like Z[x] modulo a polynomial right
red imp
yeah I think so
mild laurel
The other way to think about quotients is that
In like Z/6Z, you can think of sending all of 6Z to 0
so that in the new ring Z/6Z, everything in the ideal 6Z becomes zero
And the same thing happens in Z[x]/(p(x))
you're essentially stating that p(x) = 0
So like, for Z[x]/(x^2)
you're stating that x^2 = 0
So x^3 = 0 as well, since you can multiply both sides by x
or x^n = 0 for any n > 2
so you're only left with linear polynomials in some sense
red imp
woah
mild laurel
or, as a more typical example, if you have Z[x]/(x^2 + x + 1)
you're stating that x^2 + x + 1 = 0
red imp
okay yeah I've been missing that "equivalent to 0" throughout this whole class
mild laurel
or that x^2 = - x - 1
so you can take like x^3 = x(x^2) = x(-x - 1) = -x^2 - x = x + 1 - x = 1
Yeah, its partially because you can think of ideals (or normal subgroups) as kernel of homomorphisms
gilded citrus
or if you want something that looks sort of like that 6, you can have Z[x, y] with the ideal generated by x^2, xy and y^3 for example
somewhat analogous to 3*2=0
red imp
I'm trying to get how that works
so that ideal is the smallest ideal that contains x^2, xy and y^3?
mild laurel
That's one possible definition yeah
red imp
there's another definition of ring generation?
mild laurel
for example <x,y> = {x f + y g | f,g \in Z[x,y] }
kind of all the "linear" combinations of x and y in some sense
upbeat juniper
the 2 defns are equivalent
prime oriole
No thats just F(x) -{0}
steady axle
hmm better to use \ for that
prime oriole
Thank you
steady axle
what is deg(fg)?
steady axle
try to make final argument more formal maybe
prime oriole
Oh that was rough i was just solving
Like if i need to show that 5 is prime in Z[โ2].
I know something like norm function can be used
cloud walrusBOT
bertwit:
steady axle
do you understand why this helps?
prime oriole
I think i do
steady axle
๐
steady axle
i am not sure what ur saying. but its because R/I is integral domain iff I is prime ideal
prime oriole
Yes i have
steady axle
good
norm can be used to prove that a given element is irreducible
for ufd that implies prime
cloud walrusBOT
bertwit:
steady axle
and ofc euclidean domain imples ufd
prime oriole
Then if i prove Z[โ2] is Euclidean domain then the element i want to prove as prime
No no
Um i actually got confused
I need to show that Z[โ2] /(5) is I.D
Then it is proven by implication
solid drum
Why would modding out by (5) show that the original ring is Euclidean?
rugged moss
https://cdn.discordapp.com/attachments/328208536029102081/721853355936186478/unknown.png
HI was told to move this here ๐
olive mirage
But yes, one group has an element of an order the other does not
so they cannot be isomorphic because isomorphisms preserve order
[Of course what that order is, I'm leaving to you]
But for example, if I was asked to show that D_12 and A_4 are not isomorphic, I might note that the rotation in D_12 has order 6, but A_4 has no elements of order 6
scarlet estuary
different (tangentially related) meanings of order
the order of a group is, indeed, the number of elements of that group
the order of an element of a group is how many times you need to apply the group operation to that element to get the identity
i.e. the order of $a$ is the smallest positive integer $n$ such that $a^n = e$
cloud walrusBOT
Namington:
scarlet estuary
i'm surprised you're doing this sort of problem but haven't heard of this concept before - maybe you know it by another name?
half nebula
Hey so for this problem, I already have the answer I think. All we need is that first part. We just need to prove that it is a homomorphism. Is that not just the phi(a+b)=phi(a)+phi(b)?
Like this is the version from the book. Our homework version is the same thing but without everything past "Prove that phi is a homo."
golden pasture
the group operation of $\mathbb C^\times$ is multiplication
cloud walrusBOT
ariana:
golden pasture
so prove ฯ(ab)=ฯ(a)ฯ(b)
mild laurel
Woah I just assigned this problem as homework
golden pasture
wts lol
half nebula
Oh dang cool XD Do you teach at University of South Carolina?
snow flint
dox dox dox dox dox nvm
golden pasture
zoph already doxxed themselves as an anime girl
mild laurel
Not quite
upper pivot
lol coincidence then kek
golden pasture
yea
snow flint
i mean to soulgiver not zoph
half nebula
I can send you the pdf for our homework. I am curious if he jsut got it from some sort of teacher resource or something off the internet <->
Anyways though, would it just be proving that phi(a+bi)=phi(a)+phi(bi)?
half nebula
But then what is the a and b replacing?
solemn rain
phi: C --> R
where C is the multiplicative
gr
group
and R is the same
so you would want to prove that
phi(ab) = phi(a)phi(b)
where a and b are in C
if u know basic complex numbers
thats just the magnitude of acomplex number
half nebula
Oh wait so is it literally just like phi(ab)=a^2b^2 then?
half nebula
So what is b?
solemn rain
any complex number?
b = f+ji for some f and j in R
arbitrary complex numbers
and the operation defined on this group is multiplication
half nebula
Sorry I feel really dumb for asking this but what is a complex number? I just haven't heard that term before.
solemn rain
so u want to show that this map preserves this operation
complex numbers are numbers of the form x+yi
for x and y in R
and i^2=-1
R is real numbers
half nebula
Oh so imaginary numbers.
half nebula
Okay so i isn't another variable.
half nebula
Okay yeah I remember that.
half nebula
Okay so a and b are both in C which means they're both in the form x+yi (but they can be different so we have a=x+yi and b=g+hi or some other variables. Correct?
solemn rain
nothing says they must be the same
u have to pick totally arbitrary
complex numbers
and show that phi preserves the structure
ie a homoo
half nebula
So it would be phi((x+yi)(g+hi))=phi(x+yi)phi(g+hi)
half nebula
Sorry mistype.
solemn rain
cool
so yea
for any x y g h in R
show that this is true
and then find image and kernel
half nebula
I don't need to do that for mine so no worries on that. He said we don't have to learn that for right now.
solemn rain
and describe the fibers geometrically
half nebula
I just need to prove it's a homo.
half nebula
I mean sure, I don't mind. But I also wanna get this homework done asap because I am already super tired.
half nebula
Thank you ^^
solemn rain
np
half nebula
Well my book is called Beachy and Blair and itโs in 3.7 for mine.
Also I feel like my math is off.
Sorry for my horrible handwriting but this is what Iโm getting for this part.
But for the other part phi(a)phi(b), I got this.
dreamy anchor
Erm
solemn rain
tbh i cant understand
can u just write it out
it should be clear if u just write it
in text here
half nebula
I can it'll take me a second though.
solemn rain
tyt
dreamy anchor
You're trying to show that these two things end up being equal:
phi((a + bi)(c + di)) = phi((ac - bd) + (ad + bc)i) = (ac - bd)^2 + (ad + bc)^2
phi((a + bi)(c + di)) = phi(a + bi)phi(c + di) = (a^2 + b^2)(c^2 + d^2)
So your algebra hopefully shows this
half nebula
I was 1 variable off but yes.
I'm trying to show that (ac - bd)^2 + (ad + bc)^2=(a^2 + b^2)(c^2 + d^2)
Okay never mind. It's right.
It was that one variable.
sterile garden
Here \mu is the mobius function. Can someone help me to understand how the last step in this equality comes to be?
mild laurel
take n = 6
I can iterate over the divisors like d = 1,2,3,6
but n/d is 6,3,2,1
which is the exact same set of numbers
latent anvil
What would happen if f failed to be irreducible in K[x]?
oh actually I think I have an example where f isn't irreducible. K = Q and F = Q(i, 2^(1/13)), and f(x) = x^10 + 4
sorry I was assuming F/K was Galois for some reason
prime oriole
Is 1+3i a prime element in Z[i]
prime oriole
@knotty mason its not a prime element right
latent anvil
[Q(i, 2^(1/13)) : Q] = [Q(i, 2^(1/13)) : Q(2^(1/13))] [Q(2^(1/13)) : Q]
mild laurel
No, it's not
knotty mason
correct
prime oriole
@knotty masonokay thank you
latent anvil
First one is 2 because Q(2^(1/13)) is contained in R, second is 13 since x^13 - 2 is irreducible by Eisenstein
mild laurel
(1 + 3i) = (1 + i)(2 + i)
prime oriole
@mild laurel okay
latent anvil
it's possible that x^10 + 4 factors but I don't think so
it's irreducible mod 11, which you can check by brute force
latent anvil
oh yeah sorry
the point is that Q(i, 2^(1/13)) is a degree 26 extension of Q over which the irreducible degree 10 polynomial f(x) = x^10 + 4 factors
yes
x^10 + 4 = (x^5 - 2i)(x^5 + 2i)
oh for some reason I didn't want f to have a root but I don't think that matters
you could take f = x^10 + 1
why are you deleting your posts?
upbeat juniper
I believe that if a is a root of f(x), then f(x) irreducible in K[x] means K(a) (the composite of F(a) and K) has degree 10 over K
but this is not necessarily the case since 10 and 26 are not coprime
latent anvil
this is true, but it also wouldn't tell us much if they were coprime
it could factor without having a root in F
(as in the x^10 + 4 example)
upbeat juniper
your example doesn't seem to contradict my point though
based on your example, if we let a be a root of x^5-2i then [K(a) : K] = 5 but [F(a) : F] = 10
and I didn't assume the root was in F (or even K)
if they were coprime, [K(a) : K] would equal [F(a) : F] so f(x) would be irreducible in K[x]
prime oriole
mild laurel
Remember that a^41 = a
kindred rivet
warm pawn
solemn rain
the ideal generated by 105 and x^2+2x+2
knotty mason
do you have a charcterization of maximal ideals in Z[x]?
knotty mason
I'm not sure if you can do that here, can you?
knotty mason
all maximal ideals of Z[x] are of a specific form
solemn rain
yea
knotty mason
@warm pawn do you have a definition of maximal ideal?
solemn rain
me?
@warm pawn every maximal ideal is in the form (p,f(x)) where p is prime
and f is irreducible
in mod p
think of this
half nebula
Hey so I just want to be sure I'm going this right. I need to check if this is a homomorphism.
I'm on d right now and I made 2 matrixes,
(a b)
(c d)
represented by x
(f g)
(h k)
represented by y.
And phi(x)phi(y)=abfg
So it's not a homomorphism right?
half nebula
Lol I wasn't sure how else to do it XD
dawn kiln
do you know latex?
half nebula
Actually I could do it in excel and screenshot it and not really.
Like I know some things but not matrixes.
dawn kiln
$\begin{pmatrix}
1 & 2 \
3 & 4
\end{pmatrix}$
half nebula
Oh okay.
dawn kiln
you can format it to make it a bit more readable
half nebula
How do I do phi?
cloud walrusBOT
maximwebb:
dawn kiln
$\phi$, or $\Phi$
cloud walrusBOT
maximwebb:
hot lake
or even $\varphi$
cloud walrusBOT
Zef Klop ๐ ๐ฟ ๐ป:
dawn kiln
or even $\mathbb{\phi}$
cloud walrusBOT
maximwebb:
dawn kiln
maybe not that last one lol
half nebula
So for c. I did $\phi(
\begin{pmatrix}
a & b
c & d
\end{pmatrix}
\begin{pmatrix}
f & g
h & k
\end{pmatrix}
)$
cloud walrusBOT
Soulgiver831:
dawn kiln
double backslash
dawn kiln
$\phi \left(
\begin{pmatrix}
a & b\
c & d
\end{pmatrix}
\begin{pmatrix}
f & g\
h & k
\end{pmatrix}
\right)$
hot lake
lol the map in d) isn't even well-defined
cloud walrusBOT
maximwebb:
$\phi \left(
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
f & g\\
h & k
\end{pmatrix}
\right)$
Oooooh okay
dawn kiln
if you look at the tex, discord covers up the backslashes
half nebula
Okay so I have to do, for c, $\phi(x+y)=\phi(x)+\phi(y)$ right?
cloud walrusBOT
Soulgiver831:
hot lake
if you want to show that phi is a morphism then yes you have to show that this is true
half nebula
Okay cool.
hot lake
the implicit group laws of M2(R) and R are addition
half nebula
Almost done with my homework, on the last problem now and I kinda get it but I'm not fully sure. I think it has something to do with proving the two groups are abelian first maybe.
sour plume
Nah, the groups don't need to be abelian. What's your definition of a normal subgroup?
half nebula
A subgroup that is normal I guess XD Sorry if I misunderstand.
sour plume
What does "normal" mean, then?
There's a lot of different ways to define the term "normal subgroup", so it's good to know which one your lecture is working with
half nebula
A normal subgroup means that for g in G and n in N then gng^-1 is in N. That is, when N is a normal subgroup of G.
hot lake
what a coincidence, you have to show that for every g in G and h in H there are some h1 or h2 in H such that things
half nebula
such that things?
sour plume
Yeah, so so this kind of definition fits very nicely to the problem that we have; the problem tells us "for every g in G and h in H, find some h_1 and h_2 in H with some properties". The definition of a normal subgroup tells us that for every g in G and h in H, we find that ghg^-1 fulfils some property
So you don't have to go that far, just play around with the definitions a bit
hot lake
I mean, you have to show a statement of the form "forall g in G and forall h in H there is h1 in G such that ....", and the normality of H in G says "forall g in G and forall h in H, ... is in H"
it's almost the same
hot lake
well you have to figure out how to get h1 and h2 from what the normality of H gives you
stone fulcrum
Let's say I gave you a g and an h. Then h1 is an element such that
gh = h1g
Natural question, does that even exist?
half nebula
Well that would have to mean h1 is equal to h right?
hot lake
um no ?
stone fulcrum
But what makes you say that?
half nebula
Or wait that would only be if it is abelian.
I'm still getting used to the mindset that ab doesn't always equal ba.
stone fulcrum
Well, what is h1 equal to?
half nebula
Some element in H
half nebula
Well you just divide both sides by g right?
stone fulcrum
In a sense yes. But we don't really talk "division" in groups. Instead we talk about "multiplying by the inverse"
stone fulcrum
Yup lol. Can you finish the logic from there?
half nebula
I think so ^^
kindred mist
Silly question but I can't get a clear answer from the google,
Let F be a field, am I correct that F(x) := Frac(F[x])?
ty
worldly galleon
smart
chilly ocean
no u
solemn rain
what axioms do left group actions have to satisify
shy bluff
shy bluff
So we want to go and show that e dot x = x right?
solemn rain
yea thats the first axiom
shy bluff
So we need (g dot f)(e) = .... What does that equal?
solemn rain
so the group G acts on X and Y
solemn rain
so define the group action g* f(x) = g* f(g^-1*x)
for x in X
yea
and i think g^-1*x is the action of G on X
shy bluff
Wait question, the second dot, g^{-1} * x, is that dot the group product or is that actually the action?
shy bluff
Ok so it is again an action
shy bluff
Yea idk
Wait so what I'm tyring to show is that if g = e
Then we just get x back?
Like uh if g = e, then we have f(x) = f(x) lol
solemn rain
show that g*f(x) = f(x) for g being the identity in G
shy bluff
Ist' it specifically for g = e*?
shy bluff
Oh
solemn rain
e being the identity
shy bluff
Yea
shy bluff
Issok
shy bluff
Oh, ok so we're assuming that we already have a left action of G on sets X and Y
solemn rain
yea G acts on X and Y
solemn rain
g^-1*x is the action on X
g*f(g^-1 * x ) is the action on Y
f(g^-1 *x ) is in Y
as f:X-->Y
got it?
shy bluff
Yea I think so
solemn rain
cool
shy bluff
Lemme type it up and I'll send it
solemn rain
cool
i have to go do something
so gl but surely some1 will help u if u still have any problems
shy bluff
@solemn rain like this?
solemn rain
cool
silk drift
For some reason I can't figure out how to show that conjugation from a conjugacy class to itself is a bijection
woven delta
Lol
worldly nexus
How are the Fibonacci numbers related to the Golden Ratio?
hot lake
closely
woven delta
Wrong channel @worldly nexus
worldly nexus
which one then?
woven delta
A general question one maybe
worldly nexus
can u not just help me here?
woven delta
#help-8 seems to be open
Ask in there
Anyway @silk drift it is a bijection because there is an inverse
silk drift
Oh wait
I am dumb wow
Okay so since its a bijection then it acts as a permutation on the members of each conjugacy class of finite groups
solemn rain
ye
silk drift
Which means that the sum of the elements of the conjugacy class is in the center of RG which is what I was looking for
Okay I am officially braindead omg
silk drift
) Let K = {k1, . . . , km} be a conjugacy class in the finite gruop G.Let R be
a commutative ring.
(a) Prove that the element K = k1 +ยท ยท ยท+km is in the centre of the group
ring RG.
Man
What am I even doing
silk drift
No I mean
I think I got it
how does this work
with questions btw
Wait
Group conjugation is invertible?
That can't be right
At least not 2 sided since conjugacy classes exist in the first place right
Wait never mind
shy bluff
Um what exactly does this look like?
Does this basically just make a binary string
And the binary string is 1 if x is in S, 0 if x not in S?
woven delta
yes
shy bluff
No shortcuts yea? just show injectivity + surjectivity?
upbeat juniper
you could explicitly show an inverse
shy bluff
I dont' know if explicit inverse is what they want, I'll just do invjcetivity and surjectivity ๐
woven delta
lol
upbeat juniper
as in find a way to 'recover' the set S from the function
shy bluff
Ye
upbeat juniper
but injective+surjective works as well
severe abyss
@shy bluff dm check
half nebula
Hey so I'm having some trouble understanding this.
Like I have a basic understanding of what cosets are from the last problem.
upbeat juniper
basically for each element x in D_4, you multiply {e,a^2} by x from the left and right
so e.g. a{e,a^2} = {e,a^2}a = {a,a^3}
and then you list all the distinct sets obtained this way
scenic sage
the number of left (right) cosets of subgroup H in G is |G|/|H| btw
mild laurel
(in a finite group G)
half nebula
Wait what are the elements in D4 though?
dawn kiln
Symmetries of a square
upbeat juniper
you know the flip and rotate operations in D4 right
upbeat juniper
if f is flip and r is rotate then the elements are 1, r, r^2, r^3, f, fr, fr^2, fr^3
half nebula
Sorry okay now I'm here.
Okay so, am I multiplying both of them by 1, r, r^2, ... fr^3?
Or by a?
scenic sage
well, letters do not matter
scenic sage
i can have ๐ฅ instead of r and ๐ฑ instead of f
scenic sage
well, it depends onwhat you want to have as r and what as s
(s is reflection through the axis of symmetry)
olive mirage
Just make yourself a square out of paper and use that.
oblique river
@half nebula have you figured out the problem?
half nebula
I am a bit confused on this one bit though. So on the first one, I'm multiplying the subgroup by ab. That would make it (ab, aba^2). Can I just make that (ab,a^3b)?
With the subgroup (e,a^2).
oblique river
yeah usually for a group like D4 you want to write everything in its canonical form
so instead of ba you'd write a^3b
half nebula
I know but can I just like take the ba^2 part of aba^2 and make it a^2b (it turns into a^-2b which equals a^2b)
oblique river
yes
ba^2 is equal to a^2b
they're the same
so you can use them interchangably
sry that sounded aggressive
half nebula
I know, I just didn't know if I can make aba^2 into a(ba^2), like split it up like this.
And nah you're good ^^
oblique river
basically this is what associativity is for
abc is equal to (ab)c which is equal to a(bc)
you can do that all you want
shy bluff
Oh so each um element of v?
verbal lodge
yh
shy bluff
i.e, we have k distinct elements of v, each of which occur m_i times, meaning a total of n elements
Ok that makes sense
Ok, then what is G_v? Is G_v = P_sigma(v)?
verbal lodge
i dont think v is a set so idk if it's fair to say element of v btw
elder valley
i would guess G_v is the stabilizer of v
ex: if v=(1,2,3,2), then v is only stabilized by the identity permutation and (2,4), so |G_v|=2 = 1! 2! 1!
solemn rain
does this question make any sense
olive mirage
Yes
solemn rain
okay cool
solemn rain
i am having a very bad hard time
is this a ring : Z/4Z x 4Z
?
i am supposed to find the char of this ring
in a test
and i just cant even find its identity element
(1,1) doesnt exist
smoky cypress
Doesn't seem like a ring to me because multiplicative identity doesn't exist like you said
If you define product and sum to be component-wise
golden pasture
uh isnt (1,1) the identity?
mild laurel
wow I didn't realize 1 was a multiple of 4
golden pasture
Oh wait 4Z oops misread
i kept thinking Z/4Z x Z
rip
then yea 4Z isnt even a ring to begin with
solemn rain
yea cool
fallen bluff
Given an equality of two finitely generated ideals (x1,..,xn) = (y1,...,yn) in an R-algebra, what can we say about the relationship between xi's and yi's ?
next obsidian
I donโt think you can say much in general. If n = 1 then x1 and y1 are the same up to a unit, but you canโt say anything in general. Like what if theyโre both the unit ideal, but none of the xi are 1, but y1 is 1 and the rest of the yi are just randomly selected, totally unrelated to each other or the xi (take R as an algebra over itself for this example)
Just by definition you know the yi can be written as linear combinations of the xj, and vice versa, but past that I donโt think you can really extract much in general
chilly ocean
Let V and W be vector spaces over a field F. Prove that if a linear transformation
f : V โโ W is one-to-one then, {f(v1), f (v2), . . . , f (vn)} is a linearly independent
subset of W whenever {v1, v2, . . . , vn} is a linearly independent subset of V
Any help please
olive mirage
This is really #linear-algebra , but you want to use the fact that if a linear transformation is one to one then it has trivial kernel.
somber rivet
@chilly ocean Don't post the same question in multiple channels. I've already given you an outline of the proof.
chilly ocean
@somber rivet oh okay
shy bluff
Does this basically just say that you can gog and "factorize" a finitely abelian group the same way you can prime factorize integers?
knotty mason
yeah
shy bluff
i.e, if |G| = x, where x is some integer, then we can write it as being the direct product of ... integer groups of the size of each prime factor?
Oh ok
thank
knotty mason
it's not exactly the same
shy bluff
What is different? I dont' really understand b)
shy bluff
Yea but aren't those all isomorphic?
knotty mason
they are distinct groups, C_8 has an order 8 element and none of the others do
shy bluff
I see
But we can write G as being any of those right? And that they're all isomorphic even thouggh they're distinct groups? Like C_2xC_2xC_2 won't have an element of order 8 but there is an element that the element of order 8 will map to/from in it right?
knotty mason
sorry when I said distinct groups I meant the stronger statement that they are non-isomorphic
they are completely different groups
there are 3 different abelian groups of order 8
shy bluff
Oh, they're not isomorphic?
knotty mason
as permutations you can see C_8 as generated by (1 2 3 4 5 6 7 8) and C_2 x C_2 x C_2 generated by 3 permutations (1 2),(3 4),(5 6)
knotty mason
If G is an abelian group of order 8 it is isomorphic to one of those 3 groups
knotty mason
yeah
knotty mason
you could search for an order 8 element, if it has one it's C_8, otherwise search for an order 4 element, if it has one it's C_2 x C_4, otherwise it must be C_2 x C_2 x C_2
shy bluff
Oh I see
knotty mason
you can do the same process for any abelian group of order p^r
shy bluff
So basically it lets you go and classify groups? Like "Groups of order 8 are isomorphic to one of these, groups of order 10 are isomorphic to one of these, etc etc"?
knotty mason
yes it completely classifies the finite abelian groups
olive mirage
Be warned that C_3 x C_4 is isomorphic to C_12
shy bluff
?
olive mirage
So, if m and n are relatively prime, C_m x C_n is isomorphic to C_mn. That's called the Chinese Remainder Theorem.
olive mirage
well, C_4 x C_6 isn't isomorphic to C_24, for instance
shy bluff
I see
For this question, part b, I don't really understand how to prove it? Like I dont' really see why that's true either
The uh action that they're referring to is this
Like um so say for example thatwe have X = {1, 2, 3, 4}, and that we have G = S_4
And say that we have S = {1, 2}
And that we have g = (1 2 3)
Then we have that $\chi_S(1) = 1, \chi_S(2) = 2, \chi_S(3) = 0, \chi_S(4) = 0$ right?
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
Then we see that $(g \cdot \chi_S)(x) = (1 2 3) \cdot \chi_S((3 2 1)(x))$ right?
hot lake
chi_S(2) can't be 2
cloud walrusBOT
Liria ^(;,;)^:
hot lake
yes
shy bluff
But how do I show that this is equal to $\chi_G \cdot S$?
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
Like what is $\chi_g$?
cloud walrusBOT
Liria ^(;,;)^:
hot lake
this question is just a matter of using all the definitions
chi_G ?
who ever talked about a chi_G ??
shy bluff
cloud walrusBOT
Liria ^(;,;)^:
hot lake
probably
shy bluff
Wait that doesnt' make sense though? S is a set, not a function right?
But also like if we have that $g \cdot S$ just means "application of g onto each element of S, then $\chi_{g \cdot S}$ is 1 if $gs \in X$ and 0 if $gs \not \in X$ for all $s \in S$
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
Whereas like $g \cdot \chi_S$, using the left action from the previous question is $g \cdot \chi_S(g^{-1} \cdot x)$, which would mean....applying the left action of $g$ on either 0 or 1?
cloud walrusBOT
Liria ^(;,;)^:
hot lake
S is a set yeah
g.S is {g.s ; s in S}
usually
it's not in one of your copy pastes so I could be wrong about that but the chance is low
no ; chi_(g.S) (x) is 1 if x is in g.S and 0 if it's not
you are not applying the definitions correctly
shy bluff
What am I doing wrong?
hot lake
chi_(g.S) is a function from X to {0;1}
you can't say it is 1 if gs is in X
that makes no sense at all
instead
you copy paste this
but with replacing S with (g.S)
so chi_(g.S) (x) = 1 if x is in g.S and chi_(g.S) (x) = 0 if x not in g.S
and that's for all x in X
cloud walrusBOT
Liria ^(;,;)^:
hot lake
you also copy paste the relevant definitions to simplify (g. chi_S) (x)
shy bluff
Like we have that $(g \cdot \chi_S)(x) = g \cdot \chi_S(g^{-1} \cdot x)$ right?