Just say phi is the projection map from G to G/N
#groups-rings-fields
woven delta
The natural projection
Lol
Ooh here's a fun question which is kind of similar to the last question
Suppose Aut(G) is cyclic. Show G is abelian
fringe nexus
uh
isnt a automorphism a isomorphism from one group onto itself?
then doesn't cyclic imply abelian?
woven delta
the automorphism group is cyclic, not the group itself
fringe nexus
Also I spent 50 minutes on that question on a 60 minute exam so oof
covert vector
woah that sounds like a fun problem!!
oblique river
@woven delta imo that question feels a little cheap
cuz you're not really using anything about the automorphism group other than that G/Z embeds in it
woven delta
Yeah thats true
oblique river
the problem is really about "G/Z cyclic => G abelian"
woven delta
It's still a nice insight
oblique river
maybe im just disappointed cuz the problem looked more interesting at first glance haha
woven delta
yeah I hear
oblique river
I was like "oh this is cool, how could i ever leverage the fact that Aut(G) is cyclic?"
but you dont lol
woven delta
also I guess the intuition it gives you about the automorphism group isn't really correct
oblique river
yeah
brisk granite
Is this a proof that the inverse of a rigid motion is a rigid motion?
Lets say there exists points $P$ and $Q$ and a rigid motion $f$. $f(P,Q) = (P^\prime, Q^\prime$), and, because $f$ is a rigid motion, it must hold true that $d(P,Q) = d(P^\prime, Q^\prime)$. Thus, if an inverse function was defined such that $g(P^\prime, Q^\prime) = (P,Q)$, then it must also hold that $g$ is a rigid motion because, as stated previously, $d(P^\prime, Q^\prime) = d(P,Q)$
cloud walrusBOT
CabbageGuy:
sour plume
If you already know that every rigid motion has an inverse, then the idea is correct, but it's not expressed very mathematically
brisk granite
Also, how do I make it more mathematical?
brisk granite
Also, how do I show that every rigid motion has an inverse
somber bramble
show that they're bijective
injective follows almost immediately from the definition
brisk granite
How do I do that?
somber bramble
show it's injective (almost trivial) and surjective (bit more involved)
brisk granite
Umm... I completely understand how it's injective, but I feel like every time I try to prove it, I end up just saying that it's injective
somber bramble
for surjective i'd personally do sth like take the image of two points, and an arbitrary third point, then take the almost unique point that lies on the similar triangle
chilly ocean
assume some distinct points a,b get mapped to the same point, then the distance will be 0. contradiction since rigid motions preserve distances
chilly ocean
of course
chilly ocean
im not a neet
brisk granite
How do you show that they're injective tho?
chilly ocean
we just did that
somber bramble
as said, surjective is a bit more annoying and actually thr proof I just wrote only works in 2D
with the triangles
somber bramble
I would honestly argue that
- translations are bijective
- a rigid motion is the composition of a full rank linear transformation (bijective) with a translation
oh I guess what you can do is this:
nvm doesn't work
chilly ocean
Does anyone have any intuition for why finite abelian groups are isomorphic to a direct sum of cyclic groups each with prime power order? I know the proof but it doesn't strike me as an "obvious" fact
woven delta
Tbh in order to get that intuition you need to prove the decomposition theorem for finitely generated modules over a PID
But it helps if you work a bunch of examples
wind cypress
Got abstract algebra lectures in 2 hours... groups
plain sequoia
If K is a field with an ideal I, and a in K, then why should a in I imply that I=K
I don’t understand why
I mean if a=1 its obvious for any commutative ring, but thats just a special case
cursive kiln
From what i remember,
A field only has trivial ideals
So if K is a field and I is an ideal of field K,
Then I is either K or the empty set
So since I contains an element a, it can't be the empty set
So I has to be K
plain sequoia
Yes thats true
But we proved that
And there we used what I wrote above
If I remember correctly
R is a field $\iff$ R has only ${0}$ and $R$ as its ideals
cloud walrusBOT
markus:
plain sequoia
So in the process of showing this
We used the fact that I can’t understand why is true
I am probably just stupid
stone fulcrum
I'm no longer sure what you're asking
Are you looking for a proof that fields only have trivial ideals?
plain sequoia
Kind of, we used that if F is a field and I an ideal, then if a in I, then I=F
In the proof
Why is this true?
stone fulcrum
Let's say R is a ring, and u is a unit in that ring. If u is in an ideal, then u¯¹u = 1 is as well.
And if 1 is in the ideal, everything is in the ideal
So units cannot be in ideals, or else the ideal loses all information
In a field, every element is a unit. Fields have no important ideals.
plain sequoia
Of course
Thanks
It was the u^-1u part I didn’t see
Makes a lot of sende
*sense
But every element in a field isn’t a unit? What about 0? Or have I misunderstood the def of a field?
cursive kiln
oo i didnt get that you wanted the proof!
hold on, what do you guys mean by unit? i'm not a native speaker, sorry
plain sequoia
$r\in R$ is a unit if there exists some $s \in R$ such that $rs=1$
fickle brook
Then I is either K or the empty set
either K or {0}
cloud walrusBOT
markus:
fickle brook
unit = invertible
cursive kiln
Clear. Thank you
Then yup, marcus, every element of a field is a unit, except zero
Then {0} is an ideal
cursive kiln
If an ideal of a field contains an invertible element, then it's the entire field
stone fulcrum
A field is a commutative division ring
A field is two abelian groups with distibutivity
cursive kiln
A field is a commutative ring (A, +, ●) where every element of A - {0} has an inverse with respect to ●
And theres distribitivity
plain sequoia
In my native language the word for field is (translated to english) a body
No correlation with the english word haha, makes no sende
*sense
oblique river
its ok, the word "field" also has another meaning in english that is completely unrelated
cursive kiln
what language?
plain sequoia
Norwegian
oblique river
also, my understanding is that the original word for field is "korpus" which means body
plain sequoia
I guess that its inspired by german
cursive kiln
I'm italian, and the word we use for body is also used to mean something that's Almost a field
oblique river
yes
cursive kiln
Take out the commutative property, and that's a "corpo"
wind cypress
Learned a lot about abstract algebra today, groups in particular. Was a long day...
cursive kiln
A field is a "commutative body"
plain sequoia
Körper
oblique river
Yeah back in the day there were "fields" (now commonly called division algebras) and "commutative fields" which we now call "fields"
my guess is that for whatever reason, italian didn't switch over
there was a while in the early-to-mid 20th century where italian mathematicians kinda... did their own thing
plain sequoia
But fields
cursive kiln
Ann, thank you for the correction, i saw it now! @fickle brook
oblique river
so whereas everyone else switched notation, italians kept the old notation
plain sequoia
I think in terms of vector fields rhen
oblique river
is my guess @cursive kiln
plain sequoia
Nice @wind cypress 😁
cursive kiln
Oh! So when you say field...?
wind cypress
Felt hopeless when I tried to do the exercises, got some help with those though.
cursive kiln
Is it commutative or not necessarily?
oblique river
so basically whenever someone in english says "field" it always means commutative
always always always
(and probably most other western european languages, like german/french/spanish)
cursive kiln
Italian too!
oblique river
(it's definitely true in french at least)
fickle brook
fields which aren't commutative are called division rings
plain sequoia
Commutative vector field xd
oblique river
what? oh my b I thought you said that in italian the word for field meant not commutative
fickle brook
the quaternions are one
cursive kiln
Okay, thank you ♡♡
oblique river
and I was trying to explain why but maybe I misunderstood
plain sequoia
@wind cypress what kind of abstract algebra?
cursive kiln
Nope!
Campo = field
Corpo = division ring
wind cypress
@plain sequoia basics of groups
Very different kind of thinking I'm used to in previous courses
cursive kiln
(I think. This is the first time i read division ring. )
Quaternions are a corpo but not a campo because they're not commutative.
oblique river
yeah makes sense
plain sequoia
In my country:
Felt = field (like in terms of vector fields, scalar fields, electrical fields etc)
Kropp = field (in terms of a commutative division ring)
In german the same
I think
wind cypress
Groups, subgroups, permutations etc
plain sequoia
Thats a lot to cover in only 2 hours
wind cypress
Naah I meant the course is about those, so far we've covered those areas, and not fully
And the hard part wasn't the lectures, I spent the rest of the day reading material and doing exercises, which is where you really learn
plain sequoia
True
@oblique river Do you know where the name ring is from?
It seems so arbitrary
wind cypress
This one was hard to even understand:
Let X and Y be non-empty sets and let f:X->Y be bijective. Show that permutation groups Perm(X) and Perm(Y) are isomorphic.
stone fulcrum
I would argue for better names. "ring" has always struck me as very weird
wind cypress
If I translated that correctly, it was in Finnish the actual exercise
stone fulcrum
@wind cypress
The permutation group is just Sn, where n is the number of elements in the set.
Since a bijection exists between A and B, they have the same number of elements
wind cypress
Definition of Perm(X) is:
Let X be non-empty set. Perm(X) = {f: X->X : f is bijective}. Group Perm(X) with an operation is a permutation group of set X.
Yeah I think i got that solved, but that was very hard to grasp what that meant
Interesting, but very different from what I've studied previously 😃
plain sequoia
What do you study?
wind cypress
Well, computer science, but mathematics too. But I meant, very different from other math courses I've taken, they have been much more basic
"Show that the set of rational numbers with addition is not a cyclic group"
I got that done too
oblique river
@plain sequoia @stone fulcrum https://math.stackexchange.com/questions/61497/why-are-rings-called-rings
is the best answer I have
plain sequoia
Assume there is a generator $\frac{1}{k}$, then $\frac{n}{k}$ needs to «hit» every rational number when $n\in \mathbb{N}$. Why is this impossible?
cloud walrusBOT
markus:
Compile Error! Click the 
reaction for details. (You may edit your message)
plain sequoia
@wind cypress
wind cypress
Yes, like I said, I think I got that exercise solved, through antithesis or whatever its called in english 😃
wind cypress
Yes contradiction
Still need to write the exercises properly so I can return them tomorrow... and I'm very tired
brisk granite
If I have the arbitrary elements $a_1$ and $a_2$ from an arbitrary group, how would I write that $a_3$ equals those other two elements under the group operation?
cloud walrusBOT
CabbageGuy:
marble wagon
a_3 = a_1 a_2
woven delta
I guess I would say the word times
lyric blade
yeah, you usually refer to the ring operations as multiplication and addition no matter what they really are
this can be confusing when you're talking about homomorphisms...
plain sequoia
Looking back at my elementary number theory course last semester, its fun seeing how abstract algebra makes proving all those main congruences theorems in elementary number theory, just a walk in the park
woven delta
You refer to group operations as multiplication usually unless you know the group is abelian, in which case it's addition
plain sequoia
You may very well use multiplication when you have an abelian group as well, e.g. $(\mathbb{R}-{0},\cdot)$ (R without 0 under multiplication). Point is that it is usually very obvious from the context
cloud walrusBOT
markus:
plain sequoia
Could use some help here: Find all the units in $D[x]$, where $D$ is an integral domain. \newline
Let $f(x),g(x) \in D[x]$ with $\deg(f)=n$ and $\deg(f)=m$. Proving that $D[x]$ is also an integral domain is easy. Thus we must have $\deg((f\cdot g)(x))=n+m$. The only way we can have $f(x)g(x)=1$ is if $m=n=0$. Thus all the units in $D[x]$ are the units in $D$.
Is this correct?
cloud walrusBOT
markus:
plain sequoia
The reason I am asking is because the proposed solutions says that the units are polynomials of degree zero which is not zero. I don't see the equivalence between the statements I arrived at and the proposed solution.
plain sequoia
Yes thats what I tought
For the proposed solution to be true
Then D must be a field?
Because every constant polynomial in $D[x]$ isn’t nesecarry a unit if $D$ is just an integral domain. For that to be true, then $D$ must be a field, yes?
cloud walrusBOT
markus:
woven delta
Yeah of course
There is an isomorphism between D and the subring of 0 degree polynomials in D[x]
@plain sequoia
woven delta
Lol
plain sequoia
And polynomialaddition and polynomialmultiplication on constant polynomials just falls down to addition and multiplication on the coefficients
woven delta
Yep
plain sequoia
Is the isomorphism just the identity mapping here?
woven delta
Yeah, if you view it as a subset
But it's kind of the same as viewing Q as a subfield of R
plain sequoia
True but Q and R aren’t isomorphic?
Ofc not
They are homomorphic tho
I see what you mesn
*mean
This reminds me that I need to learn the aleph thingy. Perhaps some clasical set theory first lol. Have you read Goldrei, Classical Set Theory? I got it recommended, but I don’t want a bad book on that topic
Like, what I am trying to say is that if you have read any other good book on set theory I would like to hear 😊
uncut girder
Q and R as groups?
What does homomorphic mean?
Because any two groups are homomorphic
Just take the 0 homomorphism between them
So I don't see what you mean
@plain sequoia
plain sequoia
Yes as groups
wind cypress
Homomorphic, means that in one group say f(ab) = f(a)*f(b) but the * is not necessarily same operation
that is one of the two things ijn it
(G, *) -> (H, ***) where * and *** are not necessarily same operation
ab doesn't necessarily mean a times b as in real numbers for example
hollow comet
Yes, but what @uncut girder meant is that there exists a group homomorphism between any two groups, namely f: G -> H: x |-> e, where e denotes the identity element in H
wind cypress
hmm that got messed up a little that text
hollow comet
In that sense, any two groups are homomorphic
wind cypress
I'm actually just studying groups, third week :). Or is it fourth
woven delta
Yeah, homomorphic is not really a thing
marble wagon
he probably wanted to say that Q embeds in R
plain sequoia
What Cloud said
oblique river
the only time people use the word homomorphic is to say something like
"the subgroup K of G is a homorphic image of H"
which just means that there is a hom H -> G with image K
bleak abyss
That makes my skin crawl
timber bay
hey guys
galois theory time
looking at the subgroups of Q(rt(2),rt(3),rt(5))
the first subgroups of this will be like Q(rt(2),rt(3)), Q(rt(2),rt(5)), Q(rt(5),rt(3))
is this right? there will be no conflictions since any of the elements can't be generated by the other elements?
this doesn't look right
and the gal group diagram will be this switched, with the top fixing Q, second layer fixing Q and a single element,
third layer fixing 2, and bottom layer fixing Q(rt(2),rt(3),rt(5))?
is that right?
oblique river
Do you mean "subfields of Q(rt(2), rt(3), rt(5))" instead of "subgroups"?
timber bay
yeah i do
oblique river
you are missing a lot of fields
let's start with a simpler example, Q(rt(2), rt(3)). What are the subfields of that?
timber bay
oops
oblique river
so there are a couple ways to think about this, and I think the easiest is using galois theory
(you're familiar with the fundamental theorem of galois theory, yes?)
timber bay
yeah
oblique river
ok
timber bay
thats what this question is about
oblique river
what is the galois group of Q(rt(2), rt(3))/Q
timber bay
its the set of automorphisms from those elements to their conjugates
oblique river
yes but what is it
like, it's some actual finite group
it's abelian in fact
so, which one is it?
timber bay
like what is it isomorphic to?
oblique river
yes, like what is it's group structure
when people ask "what group is this" they mean they want some kind of description in terms of groups that we already know
so a valid answer would be like "trivial group" or "S_3" or "D_10" or "A_4 x Z/3Z"
timber bay
dammit
oblique river
(fyi, nobody ever calls that group S_2. Everyone calls it Z/2Z)
oblique river
new question: how many groups of order 2 are there?
like, isomorphism classes of groups
timber bay
1 right?
oblique river
(if you're not confident about that, you might need to spend some time reviewing some group theory.)
yes
now, what are the orders of Z/2Z and of S_2?
oblique river
yes, so then they have to be the same
timber bay
yeah
oblique river
so back to Q(rt(2)) / Q
you seemed to think at first that the galois group was Z/2Z but then you questioned it
oblique river
well, what is the relationship between the degree of an extension and the size of its galois group
timber bay
equal for finite separable extensions?
oblique river
you should probably be reviewing that kind of stuff...
you're missing a hypothesis
you also need the extensions to be normal
also, you should know that extensions of Q are always separable
and I'm not asking you about the general theory
well actually nvm I guess I did ask you about the general theory :P
timber bay
lol
oblique river
what I want to ask you now is: what about for Q(rt(2))?
timber bay
yeah i meant to say normal
oblique river
it's great if you can say the theorems
timber bay
actually lol
oblique river
but you need to actually know how to use them with examples
timber bay
so Qrt2 has a gal group of size 2?
timber bay
yeah
oblique river
(can you give me a polynomial whose splitting field is Q(rt(2))?)
timber bay
so wouldnt Q(rt2,rt,rt5) have size 8?
oblique river
the identity element corresponds to the identity automorphism of Q(rt(2))
timber bay
order 8
oblique river
we're not there yet!!!
oblique river
what automorphism of Q(rt(2)) does the non-identity element of that galois group correspond to
oblique river
to say what it does on all of Q(rt(2)), you only need to specify what it does on 1 and sqrt(2)
explicitly, where does it take 1 and where does it take rt(2)?
timber bay
fixes 1, sends rt(2) to -rt(2)
timber bay
alright
oblique river
what is the galois group?
(maybe first: is it a normal extension? what is its degree?)
timber bay
it is finite normal, degree 4? so S_4?
oblique river
what is the size of S_4?
timber bay
24
oblique river
which is not equal to 4
timber bay
yeah
oblique river
so how could that be the galois group of an extension of degree 4?
oblique river
Z/4Z is a group of order 4, are there others?
timber bay
is the dihedral group order 4?
oblique river
(as I said before, this kind of stuff you really really really need to know just off the top of your head)
I... guess so? Except just like S2 vs Z/2Z, literally nobody calls it that
for one, it's abelian, and every larger dihedral group is nonabelian
timber bay
yeah. Ill need to improve on that. We havent had much applications like that in my algebra class.
oblique river
and for two, it has a much simpler description
timber bay
true
oblique river
but like do you see what I mean? if you want to answer the original question using explicit galois theory, you're going to need to use explicit group theory
including knowing how many groups of order 4 there are
timber bay
yeah i get what youre saying
oblique river
ok but let's keep going. what explicitly are the automorphisms of Q(rt(2), rt(3))
there should be 4 of them
so list them for me
timber bay
fixes it
oblique river
great
so we know that there have to be 4 autos, and we just found 4, so that must be all of them
so now we can figure out the group structure
what happens if you compose these?
timber bay
would the combination one be a problem? like what it does to rt6?
oblique river
also let's give them names. let's call the one that swaps rt(2) A, the one that swaps rt(3) B, and then their product is AB
so our group is {e, A, B, AB}
timber bay
sure
oblique river
we can see that it's abelian
and that everything has order 2
A^2 = B^2 = (AB)^2 = e
which means our group is Z/2 x Z/2
(which is not Z/4, because our group has no elements of order 4.)
timber bay
yeah that notation is something im not too familiar with
oblique river
direct products?
timber bay
weve never really used it in class
oblique river
oof
timber bay
i can learn that on my own
oblique river
I guess it's fine, we can think of {e, A, B, AB} for now
yeah I mean you probably should. just as a general fact, suppose you have galois extensions K/Q and L/Q such that the intersectino of K and L is Q
then the extension KL / Q is also galois and its galois group is the product of Gal(K/Q) and Gal(L/Q)
which is exactly what's happening now, actually. K = Q(rt(2)) and L = Q(rt(3))
this is just to say that you should know about direct products cuz they come up a lot in these kinds of field constructions.
in any case, let's go back to our specific example
we want to calculate the subfields of Q(rt(2), rt(3))
by the fundamental theorem of galois theory, we know that they are in bijection with subgroups of the galois group
so, how many subgroups of the galois group are there?
(this is often much easier to count than trying to directly count subfields.)
timber bay
4?
timber bay
oh wait let me just write them
i think 4 isnt right
<id, A>, <id, B>,<id, AB>,<id>
and whole group
oblique river
uh, do you mean the whole group?
oblique river
so there are 5, that is correct
let's go down the line and go from subgroups to subfields
i'll give you the subgroup and you give me which subfield it corresponds to
{e}
oblique river
hrgh
timber bay
whole field
oblique river
yes
timber bay
i have to use my brain before I type
oblique river
remember the correspondence. a subgroup goes to the subfield that is fixed by the subgroup
timber bay
yeah
oblique river
and the entire field is fixed by {e}
timber bay
yep
oblique river
ok now the other end: {e, A, B, AB}
timber bay
only fixes Q
timber bay
Q(rt3)
oblique river
yay :)
timber bay
haha
oblique river
so then {e, B} will fix Q(rt(2))
timber bay
i feel like a child
timber bay
Q(rt6)?
timber bay
Q(rt6)
oblique river
great! now why?
timber bay
it fixes rt6 and multiples of it, but not rt3 or rt2
oblique river
perfect
a "only thought" explanation could be something like: we know it fixes rt(6), so it fixes all of Q(rt(6)). We also know that the degree of the fixed field of H is equal to the index of H in the galois group. Since H has index 2, the degree must be 2, Q(rt(6)) must be all of it"
(basically the idea is just to justify how you know that there can't be some other random element that somehow snuck into the fixed field of {e, AB}
(and the idea is that by counting degrees, you know that Q(rt(6)) has to be all of it.)
but in any case, we just found 5 subfields and by the fundamental theorem of galois theory, they must be all of them
so NOW we can return to the original example
you are correct that the galois group has size 8
but there are a lot more than just 8 subgroups.
timber bay
what do you mean by has to be all of it?
oblique river
like
the fixed field of {1, AB}
that's the correspondence
you go from subgroups of the galois group G
to the fields that are fixed by those subgroups
you did correctly identify that Q(rt(6)) is fixed by {1, AB}
but what if there was something else that was also fixed?
like what if some other weird combination of rt(2) and rt(3) happened to be fixed by AB? how could we ever tell?
timber bay
ah okay
oblique river
so then I gave some explanation to show that indeed, Q(rt(6)) is the entire fixed field of {1, AB}, i.e. there is nothing else
timber bay
so because qrt6 has degree/index 2,we know there are 2 elements that fix it, and not more?
ah ok
oblique river
it's unfortunate that 2 is equal to 4/2
the correspondence is the degree of the fixed field is the index of the subgroup
in this case, our subgroup has index 2 (it also has size 2)
so we are looking for a field of degree 2
Q(rt6) has degree 2
so that's it
i have to go now
but you should think about the Q(rt2, rt3, rt5) case
here's what you should focus on in particular:
write down all of the automorphisms explicitly
(we already said that there are 8) and determine the group structure
you dont have to write it out like Z/2 x Z/2, just know how elements multiply
e.g. is it abelian, what orders do they have, etc
then you have to figure otu all the subgroups (and there are quite a few)
and then go to subfields
and you will find the ones you originally missed
for example, in your very first picture you posted, you didn't even have Q(rt6)! but since that's a subfield of Q(rt2, rt3), it's certainly also one of the bigger field Q(rt2, rt3, rt5)
ok gotta go, gl!
timber bay
thanks for all the help!
hollow comet
Are you done? Can I ask something now? (Reading this answer was very interesting btw)
timber bay
im done talking about my question for now
hollow comet
Cool
I wanted to ask whether anyone knows what could be mean by a "morphism of G^D-representations", where G^D denotes the Cartier dual of the constant group scheme associated to a finite group G
In particular I'm reading about graded algebras, and the relationship between gradings and actions
The statement is that if A has a G-grading, then the multiplication map of A is a morphism of G^D-representations
timber bay
would it be true to say that there does not exist an isomorphism from Q(rt(2), rt(6)) to Q(rt(2),rt(3)) but there exists an iso from each of the gal groups of the two?
marble wagon
there is an isomorphism for sure
rt(6) = rt(2)rt(3)
if you embed them in C in the usual way
timber bay
hmm
im trying to find two extensions of q that have no isomorphisms between them, but the gal groups have isos
i was thinking the rt(6) thing interfered with the iso properties
marble wagon
Q(rt(2)) and Q(rt(3))
timber bay
why is there no iso between those two?
marble wagon
yeah
I mean this isnt exactly the reason
you could concievably think there is a root of 2 in Q(rt(3))
but you can check there isn't
because the polynomial x^2 - 2 is irreducible there
timber bay
yeah
whole basalt
So I'm stumped on a minimal polynomial sort of question
Or really any annihilating polynomial
Trying to find a polynomial for $\sqrt{\sqrt[3]{2}-i}$
cloud walrusBOT
DMAshura:
whole basalt
I tried my usual game of playing with the numbers by addition, subtraction, multiplication, by integers as well as raising to powers, but I couldn't get anywhere past squaring
I thought maybe raising to $\sqrt[3]{2}-i$ to the sixth power might help but it didn't at all XD
cloud walrusBOT
DMAshura:
hollow comet
According to Wolfram the minimal polynomial is x^12 + 3 x^8 - 4 x^6 + 3 x^4 + 12 x^2 + 5
Now let's think how we get there
hollow comet
Yes
whole basalt
I tried to use WA to compute it but it didn't seem to know what to do
I must not know the commend
marble wagon
it's not as hard as it looks
hollow comet
I used no command, just write the number and it gives you the poly
marble wagon
$\sqrt{\sqrt[3]{2}-i}$
cloud walrusBOT
Cloud:
whole basalt
Weird.
whole basalt
Yeah I get that
marble wagon
and you reduce to finding minpoly of $\sqrt[3]{2}-i$
whole basalt
But I'm still playing with it, apparently squaring again may help
cloud walrusBOT
Cloud:
hollow comet
And then cyclotomic poly?
marble wagon
this one is tricky, by inspection you should immediately see that it's degree 6
whole basalt
Yeah I see it "should be" 6
marble wagon
so you just take the sixth power
whole basalt
But I guess it requires more clever steps
whole basalt
Yeah I tried the 6th power and it was hairy af
whole basalt
Lemme play with it some more, I think I found something
marble wagon
but you know the basis
so this is linear algebra
1, i, \sqrt[3]{2}, \sqrt[3]{2} i, \sqrt[3]{2}^2, \sqrt[3]{2}^2 i
is the basis
so you just write each power in that basis
and express the sixth in terms of the lower ones
whole basalt
...nope that didn't help 😛
I guess it just takes more "clever" steps than some of the others I've done
whole basalt
I mean the thing I tried
marble wagon
oh ok
I mean, this method takes no cleverness
it uses bruteforce to reduce it to linear algebra
whole basalt
I noticed that $(\sqrt[3]{2}-i)^2=-1-2i\sqrt[3]{2}+2^{2/3}$
cloud walrusBOT
DMAshura:
whole basalt
So I figured I could add 1 and then cube or something
Didn't do me much good.
I do know what you mean about finding a basis, but I was trying to build it based on a step-by-step manipulation of the number.
marble wagon
it's a vector space over Q
whenever 1,a^1, a^2, ..., a^k are linearly dependent, you're done
and the linear dependence gives you the poly
whole basalt
I just said I get you could do it like that
I'm trying to do it as "start with your number, do this, now do that, keep doing things until you get to zero, that's how you build your polynomial"
marble wagon
you can do it if you can find intermediate extensions
for example find the minpoly of \sqrt[3]{2} - i over Q(i)
so you just wanna eliminate the left term
etc
marble wagon
yeah
and then that poly divides the other minpoly
the other part comes from the automorphism of Q(i) over Q
so it's doing the same thing for \sqrt[3]{2} + i
multiply both, you get the minpoly over Q
whole basalt
Cool. Thanks. 😃
timber bay
oops wrong q
so (123) sends the elements s_1 s_2 s_3 to each other, right?
im having trouble viewing what L would look like
I know that L fixes the sum of the s's
is that the answer? is L = E(s_1 + s_2 + s_3)?
it can't be that simple, right?
marble wagon
s1 + s2 + s3 is already in E
you need a degree 2 extension of E
a function of the y that is fixed by (123) but not by (12)
will be a generator
timber bay
so does (123) send s1 to s2, s2 to s3 and s3 to s1 or does it send each of the y's? @marble wagon
must be the y
marble wagon
the y
timber bay
hmm
man, i can't think of anything that would do this. do you have any pointers @marble wagon
timber bay
im not sure what the minpoly is
marble wagon
yeah that's what I'm saying
I would compute it
it has degree at most 6
probably 6
timber bay
in terms of s?]
timber bay
oh okay
marble wagon
if you find a degree 2 irreducible poly, you win
and it's an intermediate extension of E(y1,y2,y3)
so maybe you can find it by factoring the minpoly of the others
timber bay
yeah i should be able to do that
marble wagon
oh
the minpoly is clear
it's (x-y1)(x-y2)(x-y3)
you want the discriminant
that's the degree 2 element
have you seen discriminants?
timber bay
not in terms of this class
marble wagon
the idea in the case of the cubic is this
a root of the cubic is a degree 3 extension
and you possibly have another degree 2 extension
in the fraction field of the cubic
this is captured by the discriminant (y1-y2)(y2-y3)(y3-y1)
which is clearly an element in the extension
and its square is always in the base field
so it gives a degree 2 intermediate extension
(or possibly a trivial extension)
timber bay
ok sorry. was kind of busy for a second
marble wagon
(y1-y2)(y2-y3)(y3-y1) is what you want
timber bay
that's the minpoly of y1?
marble wagon
into what
marble wagon
F(y1,y2,y3) is the field of fractions of (x-y1)(x-y2)(x-y3) in F(s1,s2,s3)
so we can use the polynomial to find the intermediate extensions
timber bay
wow the cubic discriminant is not fun
marble wagon
what do you mean
it's just (y1-y2)(y2-y3)(y3-y1)
if you know the roots it's very easy to calculate
timber bay
sorry, i got a call, and got lost somewhere i think
but that makes sense
i actually remember talking about the descriminant for like 3 minutes in one of my lectures
daring wolf
Hey guys! Anyone know anything about coding theory regarding to Linear Cyclic code? I was given a generator matrix of a binary linear cyclic code and need to find out the smallest length of such code ><
If I am asking in the wrong area feel free to re-direct me
daring wolf
If anyone else have the same question I found out that we just need to find a polynomial that is divisible by the generator polynomial then it should tell you the required smallest length
somber bramble
tf when your galois theory class is suddenly about constructions with straightedge and compass
stone fulcrum
I think everybody has that tf at that part
somber bramble
is it a standard thing to cover there?
stone fulcrum
For galois on constructions? It's in the books I've read. The standard theorems prove a few big things about constructions, and some extra work blows them wide open
You can tell exactly which polygons are constructable using it
somber bramble
well it was a fun lecture at least
timber bay
anybody got any examples or something out the main theorem of galois theory?
im kind of iffy on it
wind cypress
Let G be a commutative group. Let's define subgroup Tor G = {g belongs to G: ord g < infinity } . Show that Tor G is a subgroup of commutative group G.
Right....
fickle brook
what have you tried and what's holding you up
woven delta
Are you doing the fundamental theorem of finitely generated abelian groups @wind cypress
If you multiply 2 numbers a, b with finite order together, and the elements commute, what happens if we raise (ab) to the |a| or the |b|?
Does 1 have finite order?
fickle brook
somber bramble
either use 1 as the neutral element or don't use "to the n-th power"
woven delta
Fair enough
wind cypress
Oh, about the earlier thing, it is just an exercise, I'm studying abstract algebra and that is one of the assignments for next week. I'm really a beginner in abstract algebra, groups is the topic.
I am fascinated about it though. But maybe Friday evening is not the best time to think about that 😃
@woven delta So, I don't really know what you mean about fundamental theorem on finitely generated abelian groups 😃
marble wagon
If you multiply 2 numbers a, b with finite order together, and the elements commute, what happens if we raise (ab) to the |a| or the |b|?
wind cypress
Oh yeah, commutative groups are abelian groups, right?
marble wagon
this is the key observation
woven delta
The first step in the proof of that theorem is to show that the torsion subgroup is a subgroup
wind cypress
I was just interested, you said fundamental theorem....
woven delta
Look it up, it's a very useful theorem
wind cypress
Oh okay 😃
woven delta
It tells you that you can decompose a finitely generated abelian group as a product of Z_p^k's and Z's
wind cypress
Uh...
woven delta
My statement isn't very nice, but just look it up
wind cypress
I will. Not sure that i will do it tonight though, it's friday evening 😃
But I got all weekend
marble wagon
literally takes 3 minutes
chilly ocean
cayley table for Z_{2} x Z_{2} where x is defined as the direct product
just wondering what is will look like I know what the generators are from solving that part
stone fulcrum
What the Cayley Table will look like?
We normally call Z2xZ2 the "Klein Four group". You can see it's kind of like four Z2 cayley tables put together
plain sequoia
Maybe a stupid question, but a polynomial ring $F[x]$, always needs $F$ to be a field to actually be a ring yes?
cloud walrusBOT
markus:
plain sequoia
If thats true, why does my book keep talking about $\mathbb{Z}[x]$, when thats not even a polynomial ring because $\mathbb{Z}$ is not a field..?
cloud walrusBOT
markus:
fickle brook
@plain sequoia no, F does not need to be a field for F[x] to be a ring
R[x] is a ring for any ring R
plain sequoia
Ah, that explains a lot
I just looked at wikipedia which said something like for a field $K$, $K[x]$ is blabla
cloud walrusBOT
markus:
fickle brook
well
plain sequoia
Thanks anyway
fickle brook
polynomial rings over fields have nicer structure
plain sequoia
Yeah, and stuff like the division algorithm only works if the polynomial rings is over a field right?
fickle brook
yeah i think
oblique river
@plain sequoia correct
marble wagon
in general F[X] is a PID if and only if F is a field
(cuz it must be a domain and have dimension 1 - 1 = 0 since X adds 1 dimension)
plain sequoia
Sorry but what is a PID
Never heard of it before
Guess ID stands for integral domain or something
wind parrot
principal ideal domain, it means the ring is an integral domain in which every ideal is actually a principal ideal, i.e. generated by a single element
somber bramble
as an example, Z is a PID because if you take any ideal, it's generated by the least common divisor of the elements in it
is there a name for rings where every ideal has a finite set of generators?
stone fulcrum
Something something finitely generated
This term comes to mind but only from modules. I'm not sure if there's a term for rings?
fickle brook
@somber bramble noetherian
somber bramble
wouldn't it be descending, rather?
we said that there are only finite chains of subideals
we only used noetherian in the proof that PID are UFD tho
and he just mentioned by the side that what we just proved was PID implies noetherian and then noetherian implies UFD
fickle brook
not descending no
(2) ⊃ (4) ⊃ (8) ⊃ (16) ⊃ ...
that's an infinite descending chain in Z, a noetherian ring
bleak abyss
Noetherian = ascending, and Artinian = descending
Artinian integral domains are fields so yeah
granite solar
Hey guys, I somehow ended up in a course on algebraic curves without having much of an algebra background and I'm wondering how one would go about showing that symmetric group elements share a cycle type iff they also share a conjugacy class. Could anyone walk me through the logic on this?
marble wagon
notice that conjugation by x is relabeling every number i to x(i) @granite solar
the result follows immediately
plain sequoia
Can I get a hint on this one?
Let G be a finite group of order p^2q, where p and q are distinct primes. Show that if p>q, then G has a normal subgroup of order p^2
fickle brook
plain sequoia
Yeah I figured it out 😃
bleak finch
Quick question. Let x be the element of a ring. What does x + x equal?
bleak finch
I made a grammatical error
somber bramble
well, then what x+x is will depend on the ring and the element
but it’ll be the same as 2*x, where 2 is 1+1, where 1 is the multiplicative identity
bleak finch
What happens when x doesn't divide y and y doesn't divide x?
somber bramble
what are these super broad questions?
like, 5 doesn’t divide 7 and 7 doesn’t divide 5
what happens?
i dunno
they exist, you can’t divide one by the other
bleak finch
If x doesn't divide y and y doesn't divide x are they coprime?
somber bramble
that’s the definition of coprime
bleak finch
Oh
somber bramble
but that’s not a thing that happens
fickle brook
no
fickle brook
that isn't the definition of coprime
somber bramble
fuck
somber bramble
yea you’re right I misremembered what coprime means :P
does coprime even make sense in all rings though
I think only in UFDs right?
bleak finch
SaschaBaer it holds in any GCD domain
bleak finch
No it is not
A euclidean domain is more specific than a UFD
I need to know what can I conclude if x ∤ y and y ∤ x
somber bramble
I had to look it up and promptly read the inclusion the wrong way round
bleak finch
What conclusion can I draw if two numbers are mutually indivisible?
somber bramble
these questions are just too broad
fickle brook
is there a problem you're doing that made you ask this
bleak finch
I am proving Bezout's identity without using the euclidean algorithm
bleak finch
How do I prove Bezout's identity in a GCD domain (which may not have a total ordering?)
bleak finch
I'm sorry for repeating, but this is a priority. How do I prove Bezout's identity in a GCD domain (which may not have a total ordering?)
plain sequoia
Doesn’t well ordering imply Bezout in the first place?
I am only known with Bezout over the ring of integers tho, so I may be wrong. However I am pretty sure you’ll have a very hard time proving it over Z without using well-ordering, which makes me think if that should also be a requirement in other rings
plain sequoia
I think there was a proof on proofwiki which only used the division algorithm
mild laurel
What definition are you using for coprime?
If you use the definition that a,b are coprime if and only if their only common divisors are units
@bleak finch
Then you're basically done since if you have coprime a,b then, there is a unique minimal principal ideal containing the ideal (a,b) and the generator of that ideal must be something that divides both a and b
So it must be a unit which implies that 1 is in (a,b) which implies that xa + yb = 1
marble wagon
you are using what you want to prove
that Z is a PID
or are you?
how do you get that the generated of the minimal principal ideal is in (a,b)?
seems like you are using that (a,b) is principal
mild laurel
definition of a GCD domain is that any two elements have a gcd
which implies that there is a unique maximal principal ideal that contains the ideal generated by (a,b)
marble wagon
yeah
but you are using the other direction
that the gcd is in the ideal (a,b)
which is what you want to prove
mild laurel
Yeah you're correct
I actually don't think this statement is correct, there exist gcd domains where bezout's lemma doesn't hold
Let K be a field and consider K[x,y]. Then bezout's lemma doesn't hold for the polynomials x and y, but I think this is still a gcd domain
brisk granite
How do we know that all rigid motions are translations, reflections, rotations, or glide reflections?
Is there a proof
somber bramble
there's one hidden among the first chapter of Visual Complex Analysis
it's a bit all over the chapter though, iirc
because he first proves that any rigid motion can be decomposed into one, two or three reflections, and it follows from that (two reflections make up a rotation or translation, three makes a glide reflection)
oh actually, can't there also be a rotation-reflection, which isn't on your list
somber bramble
But the basic idea is essentially this:
-a point is uniquely charcterized by its distances to the corners of an arbitrary triangle. Therefore you only have to consider what happens with a triangle.
-first consider what happens with one of its sides. a rigid motion leaves distances invariant and straight lines as straight lines (to prove that simply consider a third point on the line segment, the only place it can land is directly between the images of the two points again, preserving the collinearity), so the image of the line segment must be a line segment of same size somewhere in the plane
-then you can get the line segment there by a translation (moving A onto A') followed if needed by a rotation about A' (moving B to B')
-now either C is already on C' or opposite of it, in which case you have to apply a reflection about A'B'
Thus, the rigid motion can be written as a composition of at most one translation, rotation and reflection, now you can classify what sorts of motions you can build with these
@brisk granite
lusty swallow
if we've been given an R-Module, over a monoid M, do we implicitly get a homomorphism from R to M by looking at the scalar multiplication?
fickle brook
an R-module over a monoid? 
lusty swallow
is that sensible?
suppose we have a monoid (M,+)
then we equip it with scalar multiplication over some commutative ring R
is that a definition that makes sense? sorry, the terminology isnt something im comfortable with
mild laurel
usually when people say over something, they're talking about the base ring
But your definition does make sense
lusty swallow
i guess what im trying to ask is that does this scalar multiplication allow us to derive a homomorphism from R to M
mild laurel
What are you saying is a homomorphism though? A monoid homomorphism?
lusty swallow
yes, you're right! a monoid homomorphism from (R,*) to M
sorry again. i'm still grappling with the terminology.
lusty swallow
i think i know that i can go the other way.
like, if i had a monoid homom from R to M
i could then use that to define scalar multiplication
mild laurel
Ah I see what you're saying
lusty swallow
by defining the left-scalar mult by r*m in the module to be \phi (r) * (m)
i think the more i talk about it im concluding that i cant go the other way.
mild laurel
I don't think it works out in general since
Don't think it would satisfy the distributive law
\phi(r) * (m x n) = \phi(r) *m x \phi(r) * n
where I write * for the scalar multiplication and x for the monoid operation
lusty swallow
got you. i see
mild laurel
Or that's what the distributive law would have to be, but you don't have that doing what you said
lusty swallow
you're right, thanks for clarifying that
back to the drawing board i suppose
thanks!
mild laurel
Yeah I guess the way to think about it
Is that scalar multiplication is a ring homomorphism from the ring to the ring of endomorphisms on the group (when talking about normal modules)
And when you replace the group with a monoid, elements aren't invertible so multiplying by an element isn't automatically a monoid automorphism
lusty swallow
i see! that helps thinking about it.
i think perhaps i should be thinking of a group module instead.
if you're not too busy, could you explain your comment about scalar mult being a homom from ring -> endomorphisms on the group?..
im not very quick at algebra
mild laurel
Algebra definitely takes a lot of thinking time
lusty swallow
how am i supposed to think about the endomorphisms of the group? its permutations?
i mean, if its finite.
mild laurel
They are permutations, but they're also group automorphisms
As in they're isomorphisms from the group to itself
And each element of your ring gets mapped to a specific group automorphism, which is how scalar multiplication happens
mild laurel
yeah no problem
I guess I should correct myself a little bit
endomorphisms are group homomorphisms from a group to itself, not necessarily isomorphisms
Automorphisms are for when it is an isomorphism
lusty swallow
oh yes. that's right. i see
marble wagon
a monoid map into a group factors through a group
lusty swallow
ah, im still struggling with this a little. im thinking that such a scalar mult within an R-algebra (formed from a group G and a ring R), from R to a group G, produces a map from R to an endomorphism of G
marble wagon
which is universal in that sense
lusty swallow
R-Module* sorry
marble wagon
so monoid maps into modules are actually just group maps
and then again this factors through the free module R<G>
lusty swallow
sorry, does that monoid map mean a map from a monoid to the underlying group of a module?
marble wagon
yes
lusty swallow
sorry, when you say free module
do you mean the linear combinations of G
with coefficients from R
marble wagon
hmm what do I mean
lusty swallow
thats the only way ive seen the term free module used, sorry
marble wagon
I mean that's what I said but I'm not sure that's what I mean
lusty swallow
I see
marble wagon
ok yeah it is
that's the universal R-module structure to put on it
so it ends up just being a map of R-modules
lusty swallow
sorry, sorry, that's a little fast. what are you referring to when you say its a map of R-Modules?
marble wagon
so say you have a group map G -> H where H is also an R module
we can put an R-module structure on G which is gonna be universal
in the sense that all such maps become maps of R-modules
it's not exactly the free one
lusty swallow
oh, i see.
oof, maybe this is a little too advanced for me at the moment.
thanks for the help
marble wagon
are these abelian?
like is G abelian?
that's what I'm missing
if G is not abelian first you need to take Ab G
lusty swallow
i hadnt thought of that, but in generally i consider modules formed from some ring and some group to have commutativity?
is Ab G the center of G?
marble wagon
since any map G -> H for abelian H factors through Ab G
abelianization
now we're in the category of Z-modules
so the universal R-module structure on Ab G that I mentioned has an easy description, it's just tensoring with R
over Z
lusty swallow
i havent seen that before, sorry.. this is a little difficult for me to keep up. i was planning to start reading about tensor algebras this week but i havent got around to it yet
marble wagon
where did the motivation for the question come from?
lusty swallow
thanks a ton still
marble wagon
its a strange one
lusty swallow
oh, i suppose it's stupid to now admit what im actually looking for.. might be in the complete wrong direction.
the actual question is
prove that a representation of G on V consists of the same data as the structure of a kG-module on V, where G is a group, V a vector space, kG a k-algebra and the representation a group action of G on V via k-linear maps..
i think my lecturer also means that k is a field.
i might have been barking up the entire wrong tree.
marble wagon
yeah so if you notice a lot of what I did was say that a map from A to B where B has more structure is the same as putting more structure on A
I did this a bunch of times for different objects
but here you have another one
and you kinda have to chase the definitions
to see it work
k is a field
yes
basically k[G] is the universal vector space taking maps from G
so you have to show every map G -> V factors through k[G]
both the maps G -> k[G] and k[G] -> V have very explicit descriptions
so you should figure out what they should be
write them down
and check it works
lusty swallow
okay. i'll try to approach it like how you said
thanks heaps
really appreciate the patience
timber bay
not sure how to deal with this one. We talked a bit about isomorphic refinements, but only looked at the sizes of the subfields. And im not sure exactly how they can be isomphic. I'm also thinking that something like Z has infinite refinements
hollow comet
Quotients of Z (as a group) are completely determined by their number of elements
There is a standard way to get isomorphic refinements
Do you know it?
chilly ocean
that's a beautiful diagram
timber bay
damn butterfly theorem
hollow comet
The idea is that at each step in one of the series, you put in all of the steps of the other series
And vice versa
Say that you have two series, which I denote by V_i and W_j (so that V_0 = W_0 = 0 and V_n = W_m = G)
Then you define the following
$V_{i, j} := V_i (W_j \cap V_{i+1} )$
(I don't know why it does not compile)
Notice that this is between $V_i$ and $V_{i+1}$
You do the same with W_{j, i}
And arrive at two isomorphic series
@timber bay
Is that clear?
timber bay
what is \cap?
hollow comet
Intersection