i miss wew's "blunder"
#groups-rings-fields
coral shale
delicate orchid
anyway Q is still definitely torsion free as a Z-module
rustic crown
sebb
all we're saying is that Q doens't have order 2 elements as an abelian group
because 2a = 0
means a = 0
pastel cliff
all we're saying is that Q doens't have order 2 elements as an abelian group
and this implies that any subring won’t either
rustic crown
but that other inifnite direct sum does have order 2 elements
pastel cliff
does treating them as Z modules change anything
no right
bc if underlying group isnt iso then module cant be
delicate orchid
abelian groups are just Z modules anyway
pastel cliff
and also torsion is inherent to module structure ig
delicate orchid
do you have the solutions lol
pastel cliff
i do not
delicate orchid
wtf...
rustic crown
i don't understand half the stuff wew says 🙈
delicate orchid
erm moderators these users are off topic please ban them immediately
pastel cliff
idk how much of an insult wanker actually is
delicate orchid
very mild
rustic crown
what does it even mean 
pastel cliff
he who wanks
chilly ocean
pastel cliff
i dont know either det 
rustic crown
british english sounds hard
like imagine learning it
you write something
but read something else
pastel cliff
extra letter in a bunch of words too
rustic crown
sometimes you just mis-spell things
and sometimes you don't write half the letters
like
gotta catch 'em all
where my "th"
delicate orchid
I think you're confusing british english with me just not being able to operate a keyboard
rustic crown
coral shale
british english sounds hard
silly det, its easy peasy
you just have to avoid interacting with non-british englishers 
and no, dont teach det bad words silly billies
pastel cliff
frick
rustic crown
you just have to avoid interacting with non-british englishers <:sotrue:82697135...
oooh
the word "silly" sounds cute
formal ermine
the word silly is silly
summer path
Siwwy
solar totem
sully
rustic crown
what does sully mean anyway?
delicate orchid
it's the funny blue dude "
"
rustic crown
oh its name?
long geyser
doesn't look funny to me, serious business
chilly ocean
Could someone help me understand this proof here? Especially the part where the left-inverse for phi is implying that the sequence splits. How I understood this was that we should have been required to show that N is isomorphic to M \oplus coker(phi), but it seems that they are showing that it's actually isomorphic to M \oplus ker(phi)?
Ah sorry the pictures came in the wrong order...
rustic crown
that's ker(psi)
which is easily seen to be coker(phi)
because the sequence then is 0 --> M --> M⊕ker(psi) --> ker(psi) --> 0
chilly ocean
Wouldn't the map M \oplus coker(phi) with (m, [c]) \mapsto phi(m) + c work as the isomorphism just as well?
I.e. taking the representative of the equivalence class in coker(phi)
rustic crown
is that well defined
chilly ocean
Not sure haha
rustic crown
hehe :p
chilly ocean
Is it not?
rustic crown
yea it's not :p
because if you pick a different representative then phi(m) + c would definitely change
and other reason is that you never used that phi has a retract
and not every SES splits
chilly ocean
This ker(psi) = coker(phi) equality wasn't quite clear to me so I was a bit confused with this kernel consideration
rustic crown
yea also another thing
maybe you have seen thsi before
if you have a map p : M --> M such that p^2 = p
then you can write M = (im p) ⊕(ker p)
so p projects onto im p
which is why if you do it twice, nothign happens
i think it might be useful to digest this first
chilly ocean
If for psi we have that psi \circ phi = id_M, then do we also get that psi is surjective? This would mean that coker(psi) = 0 which then of course agrees with ker(phi)?
rustic crown
because that phi-psi can be confusing :p
rustic crown
If for psi we have that psi \circ phi = id_M, then do we also get that psi is su...
yep, if the composite is surjective, then the second map is surjective. similarly if a composite is injective the first map would be injective.
chilly ocean
Oh no I'm confusing psi and phi again
chilly ocean
How do you get the equality ker(psi) = coker(phi) then?
rustic crown
oh because you the two exact sequences
0 --> M --> N --> coker phi --> 0
and
0 --> M --> M⊕(ker psi) --> ker psi --> 0
are isomorphic
aluffi defines isomorphisms of exact sequences before doing this lemma right?
(not fully sure)
this follows from that fact that the map phi is identified with the inclusion into the first factor like they say
chilly ocean
He does yeah, but I didn't really see how those two exact sequences are isomorphic?
rustic crown
ig that's left in "all necessary verifications" lol
rustic crown
this follows from that fact that the map phi is identified with the inclusion in...
all i can do is elaborate this if you wish 
the map phi "looks like" i : M --> M⊕(ker psi)
did you understand this?
because after that we're just saying coker phi is isomorphic to coker i
and coker i is clearly ker psi
chilly ocean
I kinda understood this so that 0 --> M --> N --> coker phi --> 0 splits if we can show that N = M ⊕ coker(phi). And yeah I see that phi kinda acts as an "embedding" and the map from N --> coker(phi) as a projection so that it would make sense to say N = M ⊕ coker(phi).
chilly ocean
We don't always have this N = M ⊕ coker(phi) condition even if we have a SES right?
rustic crown
if 0 --> A --> B --> C --> 0 is an SES then A --> B is like an "embedding" and B --> C is like a "projection"
rustic crown
We don't always have this N = M ⊕ coker(phi) condition even if we have a SES rig...
yea that's correct.
it depends on how you think about projection basically
if for you projection means any quotient map, M --> M/N, then that would be true for any SES and not just the split ones
chilly ocean
Yeah this is what I meant. So now the difference is that the given SES splits if it's isomorphic to 0 --> M --> M ⊕ coker(phi) --> coker phi --> 0, but then they seem to show that it's isomorphic to 0 --> M --> M ⊕ ker(psi) --> coker phi --> 0 instead
rustic crown
yea but ker psi is coker phi right
chilly ocean
Right, that's my issue I didn't see this. Should this be some known fact which I have missed or something?
rustic crown
wait so before that
how do you see the third object in the second exact sequence is coker phi?
chilly ocean
Which one of them?
rustic crown
0 --> M --> M ⊕ ker(psi) --> coker phi --> 0
chilly ocean
Oh yeah I don't. My only objective was to show that 0 --> M --> N --> coker phi --> 0 and 0 --> M --> M ⊕ coker(phi) --> coker phi --> 0 are isomorphic and for the first and last maps I just took the identities, but couldn't find a map N --> M ⊕ coker(phi) or M ⊕ coker(phi) --> N.
cloud walrusBOT
det
formal ermine
fancy diagram
rustic crown
f(n) = (psi(n), n - phi(psi(n)))
i is the inclusion into the first facto
i(m) = (m, 0)
do you see that the left square commutes?
and f is an iso?
this is what they mean by this
just that i drew f in the opposite direction :p
but that should be fine for you i hope
i'll change it, wait
cloud walrusBOT
det
rustic crown
g(m, k) = phi(m) + k
i(m) = (m, 0) again
see that the rows are exact
and the left square commutes
and that 1 and g are isos
from this you get an induced map ker psi --> coker phi
and that would also be forced to be iso
chilly ocean
Perhaps I'm bad at explaining myself haha. I can see how this plays out, but the biggest issue was that I couldn't figure this diagram out. I was staring at $$[\begin{tikzcd}
0 & M & N & {\mathrm{coker}(\varphi)} & 0 \
0 & M & {M \oplus \mathrm{coker}(\varphi)} & {\mathrm{coker}(\varphi)} & 0
\arrow[from=1-1, to=1-2]
\arrow["\varphi", from=1-2, to=1-3]
\arrow[from=1-3, to=1-4]
\arrow[from=1-4, to=1-5]
\arrow[from=1-2, to=2-2]
\arrow[from=1-3, to=2-3]
\arrow[from=2-3, to=2-4]
\arrow[from=2-4, to=2-5]
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\arrow[from=1-4, to=2-4]
\end{tikzcd}]$$
chilly ocean
So I wasn't reading the proof I was trying to come up with it myself haha
rustic crown
ig maybe now i understand what you want lol
or maybe not
cloud walrusBOT
johannesu
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
rustic crown
the definition here is allowed to have M1' and M2'
it doesn't need to be exactly M1 and M2
chilly ocean
Right so the groups in the below SES could be anything and not neccessarily M and coker(phi)
rustic crown
because they're isomorphic
and this is exactly what we have here
below SES can be anything
rustic crown
from this you get an induced map ker psi --> coker phi
i already said this, so since this is an iso, you can always replace ker psi with coker phi
but lets not confuse ourselves
chilly ocean
Perhaps I'll take another read of the chapter. I feel like I'm wasting your time here haha
rustic crown
and this is exactly what we have here
yee but i think, this diagram should answer most of your confusions
chilly ocean
(i'm happy to help >.<)
Oh good to know 🙂
chilly ocean
and this is exactly what we have here
I'll see if I can find the motivation to come up with this myself. Currently I don't yet see where this SES below is originating from, but perhaps with a bit more thought it'll come clear...
rustic crown
originating in the sense "why is it an SES" or "why should you think about that in this proof"
chilly ocean
The latter
rustic crown
hmm i see
rustic crown
if you have a map p : M --> M such that p^2 = p
give this a try as well... since there is a single map, this is a little less confusing
and this implies your other case lol
you can let p = phi o psi
p^2 = phi o (psi o phi) o psi = phi o 1 o psi = p
chilly ocean
Wouldn't the splitting lemma imply this directly as phi has a left-inverse?
rustic crown
what you asking that for
yea ig both imply each other
p^2 = p can be phrased like p o (1-p) = 0
pastel cliff
why are free modules called that
rustic crown
cause you don't need to pay anything
pastel cliff
:O
rustic crown
anyway, if X is a set and M is an R-module then defining a R-module homomorphism F^R(X) --> M is equivalent to giving a set function X --> M
so it's free in the sense of freedom
no relations whatsoever
pastel cliff
wha
delicate orchid
yeah it's the no relations part
pastel cliff
im not sure i see that
delicate orchid
cause if u had to obey something stupid like a^2 = b then that's just not free enough for these all american modules
rustic crown
wew confuse me again
delicate orchid
it's free in the sense that the generators aren't bound by any rules/relations, sebbb
rustic crown
im not sure i see that
F^R(X) has a basis given by the canonical inclusion X --> F^R(X)
pastel cliff
it's free in the sense that the generators aren't bound by any rules/relations, ...
and is this equivalent to having a basis?
rustic crown
so the (image of) X satisfies no relations
pastel cliff
like that's the defn of a basis right
delicate orchid
yeah a module is free iff it has a basis
formal ermine
embrace just repeating everything
tribal moss
By analogy with free groups, etc: they are the modules in the image of the free functor Set -> RMod.
dreamy chasm
could anyone help me with this question? I need to determine all elements m of Z_42 such that (30,m) is a proper ideal of Z_42
agile burrow
An ideal is proper iff it does not contain 1. Can you think of a criteria to determine when 1 is in the ideal (30, m)?
tribal moss
Perhaps first find a more intuition-friendly way to describe (30).
plain solstice
fuck intuition machine code is all that matters in reality
tribal moss
Good luck writing any of that without intuition,
agile burrow
Here's a fun free group fact: subgroups of free groups are free. More specifically, if F_n is the free group on n generators and G is a subgroup of F_n of index k, then G is free on (n-1)k + 1 generators
plain solstice
isnt there a nice topological proof of that
im trying to learn enough topology to understand it
agile burrow
There is indeed
plain solstice
wait im really curious about that how much topology do i need to learn it
agile burrow
You should probably be comfortable with covering space theory and it's connection with the fundamental group
delicate orchid
I have not seen the topological proof of nielsen-schreier 
delicate orchid
I have not seen the topological proof of nielsen-schreier <:opencry:586078614865...
this was how it was proved in my combinatorial group theory class
graph memes
it looks far neater than it is, lemma 2.5 is a multipage mess of random words
delicate orchid
how does the topological proof go?
agile burrow
The wedge of n circles has fundamental group F_n. A subgroup G of F_n corresponds to a connected cover of the wedge of n circles, and this cover will have fundamental group G. A connected cover of a wedge of circles is a graph, any graph is homotopy equivalent to a wedge of circles, so the fundamental group of the cover is free.
delicate orchid
no I believe it dw
agile burrow
If you quotient out by a contractible subspace then you preserve the homotopy type
delicate orchid
I can "visualise it" which as we all know is completely rigorous
dreamy chasm
An ideal is proper iff it does not contain 1. Can you think of a criteria to det...
if 30 and m are coprime?
agile burrow
yes, that's right
primal tusk
Wait why wouldn’t it be 30,m, and 42 are all coprime?
agile burrow
Oh, perhaps you're right.
dreamy chasm
it seems this is not correct, so maybe it is 30, m, 42
dreamy chasm
well I think gcd only requires one integer to be nonzero
chilly radish
Also would 0 be included?! I don’t know
yes, 30 is not a unit in Z_42, so that (30,0)=(30) is a proper ideal
agile burrow
it seems this is not correct, so maybe it is 30, m, 42
Ok yes, I see the issue. I should've phrased this as an ideal is proper iff it does not contain a unit. We can characterize when (30, m) does not contain a unit of Z/42Z
dreamy chasm
unless I am mistaken, the only units I included are 5 and 25 since 5x17=1 and 25x37=1 but it seems that removing those is also not correct
agile burrow
There is one more
Or rather, there is one other element m such that (30, m) contains a unit
dreamy chasm
do you have another hint? I don't really follow
agile burrow
So (30, m) will contain a unit if gcd(30, m) is a unit in Z/42Z, right?
but k = gcd(30, m) is a unit in Z/42Z iff gcd(k, 42) = 1
tribal moss
That sounds wrong. Won't (30,7) be the whole Z/42Z?
agile burrow
Right, but gcd(30, 7) = 1, which is a unit, no? And gcd(1, 42) = 1
tribal moss
Ah, I somehow misread your last condition as being about gcd(m, 42).
agile burrow
but k = gcd(30, m) is a unit in Z/42Z iff gcd(k, 42) = 1
So now you should be able to use this to determine something about the prime factors of m to ensure that (30, m) does not contain any units
static needle
Guys
Ik this is basic
But im revisiting modular arithmetic and In confused about Z/nZ
Like i get the idea and hoe to construct those sets, but how about negative n? How about negative integers? Why arent they part of the set?
static needle
They are part of the set
Whaa
south patrol
Okay I think the confusion is coming from the fact that there are a couple of ways to construct Z/nZ
static needle
The elements of Z/nZ are **equivalence classes**
Ohhhhhh
barren sierra
so in Z/4Z
south patrol
In a basic course, one often introduces Z/nZ as the set {0,1,...,n-1} with some addition etc, but it's easier / better in the long run to think of them as equivalence classes of integers
barren sierra
-4, 4, and 0 are all in the same equivalence class
south patrol
in which case yes, one can deal with negative n and classes of negative integers &c.
static needle
-4, 4, and 0 are all in the same equivalence class
So are -2,2
static needle
By what equivalence map/ relation tho?
barren sierra
-10, -6, -2, 2, 6, 10, etc are all in one class
In Z / nZ, x = y if and only if n | x - y (as integers)
delicate bloom
might help to see the case of Z/2Z since you are probably already familiar with even+odd=odd and even*odd = even etc (this is literally what we're doing)
static needle
In Z / nZ, x = y if and only if n | x - y (as integers)
This is the equivalence relation?
delicate bloom
even and odd numbers are sets of numbers, which are the equivalence classes for Z/2Z
static needle
I mean right
It just seemed a but weird to me how it was first defined
So every element of Z/nZ is a symbol of an equivalence class under m ≈ g (mod n)
Where n | g-m
barren sierra
yea
It's just that these things are basically numbers so we treat them as such (i.e. we don't use notation for equivalence classes)
and write them as such
so if you don't keep the structure in mind, it can look confusing
static needle
Fax
At first i thought it was a matter of definition
But there is a deeper meaning to it
knotty frigate
tawdry crystal
What does it mean for the Killing form to be positive definite?
chilly ocean
are you asking for the definition of "positive definite"? or for the consequences of that in this case
tawdry crystal
the definition
I've seen positive definite in the context of metric before but I don't know if it's the same thing
chilly ocean
it's what you think it is. a bilinear form B is positive definite if B(v, v) > 0 whenever v is non-zero
tawdry crystal
Oh I thought it's B(u, v) > 0 except for when u=v
livid willow
Does anyone know if there is any hints or solutions for dummit&foote? I am using it as supplement but I find the problems in it often confusing. I am also the only person in my college doing abstract algebra, so I don't have any peer group.
void cosmos
yo
is 2 prime in Z[sqrt(5)]
no right
cuz (3+sqrt(5))(3-sqrt(5)) is in (2) but neither of those is in (2)
?
void cosmos
but 2 is irreudicble tho
so Z[sqrt(5)] cant be a unique factorization domain
true ?
rustic crown
yep
void cosmos
cool af
rustic crown
how do you see 2 is irred btw?
void cosmos
suppose 2 = ab for a,b nonunits in Z[sqrt(5)]
write everything out
and then i just noticed that none of this shit can actually work
unless one side is trivial
rustic crown
lol
void cosmos
or ie a unit
rustic crown
one quick was is to use a norm argument
rustic crown
there is another cool argument
void cosmos
no way
rustic crown
so you know that if D is a UFD, and F is its field of fractions then D is integrally close in F?
rustic crown
void cosmos
1 more bad question
next obsidian
Hi det
rustic crown
Hewwo 
void cosmos
now this is poorly worded right
cuz not all textbooks
have the same definiton for a norm
some just require euclidean algorithm
while other have this N(a)<=N(ab)
i can do it with the second propoerty
but can this be done using only first
question c and d.
rustic crown
but can this be done using only first
i would guess so
because you can "wlog" assume the second property
void cosmos
why
rustic crown
in the sense that if you have a delta satisfying the first
void cosmos
u can imply second?
rustic crown
then you can find a delta' satisfying both
void cosmos
Ohhh
rustic crown
delta'(a) = min_{a | b, b non-zero} delta(b)
void cosmos
while other have this N(a)<=N(ab)
okay so basically i ca just assume this in my answering?
void cosmos
i still dont have any reference for these exams
rustic crown
you need the second property
void cosmos
like the prof did not give me any textbooks to follow
she just told me okay i am going to examin you in this here is a sample
so i picked up df and in df this second property does not exist
rustic crown
ah
rustic crown
N(0) = 0?
rustic crown
norm where
void cosmos
what i thought is we can just spam euclidean algorithm given a is a unit
to show that the second propeprty holds but not form assumption
rustic crown
not a nice def
void cosmos
by just applying first property
void cosmos
not a nice def
trash ass def
like u see these inequalities
from the algorithm
i thought maybe we can just do some random ass magic
but turns out no cuz this is strict while the other is not
so no way
ig
rustic crown
but turns out no cuz this is strict while the other is not
yea but that's not a proof
void cosmos
so now for c) : suppose u is a unit --> exists v ssuch that u.v = 1 ---> N(u) <= N(uv) = N(1) but from second property this norm must be increasing (?? xd) so N(u) is min{min(x) | .. }
is that correct
rustic crown
yea just directly say that N(u) <= N(u(vx)) = N(x)
rustic crown
yea i don't hate it, but it doesn't tell the full thing :p
void cosmos
okay
rustic crown
you still have to show that N(1) <= N(x)
but like you say that's because it's increasing wrt |
rustic crown
trash ass def
i would really want deg : k[x] --> N to be a nice valuation.
and deg 0 is not 0
void cosmos
fuck df
rustic crown
to see why you need the second property
consider any function d : k* --> N
this is a euclidean valuation on the field k
because remainder is always 0
so never need to know what the actual function is
but ofc the second property can fail here
by that we know all units would have same d-value
but i can easily find functions for which that is not true
void cosmos
yea
yea makes sense
cool
i thought the exam was super hard cuz of this
yo for d)
suppose a = ub for some unit u --> N(a) = N(ub) <= N(u * v * b) = N(b)
uv=1
and then N(b) = N(av) <=N(avu) = N(a)
so done?
rustic crown
yup
void cosmos
col
rustic crown
void cosmos
now for e) ig in Z[i] we have 5+2i and 5-2i are of same norm
but cant be associates ?
under a^2+b^2
void cosmos
cool af
rustic crown
how about square summable sequences?
l^2 is a hilbert space right
and you have the right-shift operator
chilly ocean
Some more hints for this problem? I guess we want to find g(x) such that g(x) = f(x)h(x) and that all the monomials of g have prime exponents, but they are hinting to consider the quotient, but g(x) = 0 in k[x]/(f(x)) so I think we are not supposed to consider g in the quotient space?
rustic crown
being a multiple of f is captured by being 0 in the quotient
so you wanna show a linear combination of certain monomials is 0
chilly ocean
How does that give information about the exponents being prime?
north sand
start with the exponents being prime
chilly ocean
So like g(x) = x^{p_1} + ... + x^{p_n}?
north sand
well not immediately in a sum
you want to show that for the monomials x^2, x^3, x^5, ... eventually there is a linear combination of them which equals 0
chilly ocean
I think it would be essential to have some notion about the degree of f here?
north sand
the degree isn't that important. I would focus more on the linear algebra point of view
chilly ocean
I somehow tried to suppose that if f has degree n and I take n+1 primes, then I would get some condition on linear dependence, but are you suggesting there is another way to look at this?
north sand
yes that's also what I would do but I wouldn't be that specific. clearly the degree is finite, there are infinitely many primes etc
pastel cliff
are free modules just a flavor of direct sum of R-modules
in my defn it has an index set X but im not sure what it does
is it just there to say how many elts are in a basis
coral spindle
Yes
cloud walrusBOT
hausdorff
median pawn
a is in R, and so is b, so b/a in K is obviously clear. if b/a is in R, let b/a = r, so b = ra. what goes wrong?
vagrant zinc
Guys how can I find the order of 5 belonging to Z_12
I have this theorem then it would be 12
hot lake
you could try using the definition ?
rustic crown
a is in R, and so is b, so b/a in K is obviously clear. if b/a is in R, let b/a ...
because b doesn't lie in (a)
hot lake
yeah the theorem works to show it's 12
rustic crown
**hausdorff**
for the second thing, use that A is contained in P1, so bA is contained in P1P2...Pr which means bA is contained in (a) = aR.
vagrant zinc
If of course n is the modulo 12 then n=12
12 and 5 are coprime so 12/1 =12 then the order of the 5 belonging to Z_12 is 12
vagrant zinc
yeah the theorem works to show it's 12
thanks
vagrant zinc
5
hot lake
a=5 ?
the theorem tells you how to find out the order of a^k if you know the order of a
vagrant zinc
yeah 5 is generator of Z_12
hot lake
but you don't know the order of 5
vagrant zinc
I think I'm a little confused
hot lake
if you want to apply the theorem to find out the order of 5
then you need to find some a and k such that 5 = a^k (I'm using the multiplicative notation because that is what is used in the statement of the theorem) and such that you know the order of a
but if you know the order of 5 already then you don't need to apply the theorem
or maybe you can go back to all the definitions
vagrant zinc
then what would be the order of 5
because I am told to find the order of 5 belonging to Z_12
hot lake
well to get the order of 5 do you want to apply the theorem somehow or do you want to use the definitions
vagrant zinc
I'm just a little confused
one is the order of an element and the other is the order of a group.
hot lake
5 is an element of the group Z/12Z yeah
vagrant zinc
This definition tells me the order of an element I suppose and the theorem above is the order of a group, right?
median pawn
for the second thing, use that A is contained in P1, so bA is contained in P1P2....
thank you!
rustic crown
median pawn
This definition tells me the order of an element I suppose and the theorem above...
both tell you about the order of elements
median pawn
Guys how can I find the order of 5 belonging to Z_12
I have this theorem then i...
here it says that the element a in G has order n
vagrant zinc
here it says that the element a in G has order n
then 5 which belongs to Z_12, if a=5 and belongs to group Z_12 then it has order 12
This is what I have just interpreted
vagrant zinc
This definition tells me the order of an element I suppose and the theorem above...
Now the above definition as it would be applicable
pastel cliff
why do you think that's true
pastel cliff
then 5 which belongs to Z_12, if a=5 and belongs to group Z_12 then it has order...
this is very easy to verify
5 * 5 = 25 = 1 mod 12
vagrant zinc
12 and 5 are coprime so 12/1 =12 then the order of the 5 belonging to Z_12 is 12
hot lake
have you looked at what is the least positive integer n such that n*5 = 0 in (Z_12,+)
you should compute 1*5, 2*5, 3*5, and so on until you get 0
(I'm using the additive notation here because it makes more sense)
so you should compute 5, 5+5, 5+5+5, and so on
formal ermine
n = 0
rotund aurora
Parameterize the following affine plane curves, projecting from the point (0,0) onto the line Y=1 and onto the line X=1
What does this mean?
One of those curves is X^2+Y^2-2X
hot lake
it means you parametrize those lines
then you use that to parametrize the curve by sending a point on the curve to a point on the line
or rather from a point of the line send it to a point of the curve
so the line y=1 for example
is parametrized by the x coordinate
so then you have to find a parametrization (X(x),Y(x)) of the curve
so that (x,1) is the projection of (X(x),Y(x)) from (0,0) onto the line y=1
if you choose functions X and Y correctly this will give you almost a bijection
and an algebraic one at that
rotund aurora
so that (x,1) is the projection of (X(x),Y(x)) from (0,0) onto the line y=1
Im not exactly sure what this means
hot lake
(x,1) is the intersection of (the line going through (X(x),Y(x)) and (0,0)) and (the line y=1)
rotund aurora
ah
hot lake
you should get that x is a rational function of X and Y, and also that X and Y are rational functions of x
rotund aurora
you mean in the example X^2+Y^2-2X ?
hot lake
yeah
rotund aurora
oh so
you just consider the line going through (0,0) and (x,1) and calculate the intersection with the curve?
hot lake
yeah
hot lake
yeah
rotund aurora
Some of the curves involve cubic terms, I guess stuff will factor lol
hot lake
hopefully (0,0) is a singular point of the cubics
hot lake
you should do the computation inside Z_12
and check for the first one that results in 0 in that group
rotund aurora
Now, they ask if we would obtain a parameterization projecting from other points @hot lake
So the exercise only makes sense because the equations you obtain only have two solutions in terms of x (in the case where you project from (0,0) onto Y=1), one of them being, of course, x=0
hot lake
well for the degree 2 curves, any point on the curve would do
rotund aurora
if its not on the curve, then no because some lines will intersect the curve at more than one point, right?
hot lake
yeah
primal tusk
can somebody help me prove this
where R is a ring and a is an element of r
also there should be parenthesis around the x in the top right of the fraction
coral spindle
Find an appropriate map and use the first isomorphism theorem
Hint: the map should be from R[x] into some other ring
primal tusk
Find an appropriate map and use the first isomorphism theorem
i guess im struggling on all of this notation in general
like what even is R[x]/(x,a)
is that just all polynomials in R[x] divided by x+a ? and then you take the remainder?
formal ermine
R[x] quotiented out by the ideal (x, a). I like to think of it using modular arithmetic: y is congruent to 0 iff y is in (x, a)
you "set everything that is in (x, a) to 0"
e.g. R[x]/(x^2 + 1) would be the complex numbers, C, because we set x^2 + 1 = 0, so "x = i"
primal tusk
okay so i can see how everything in (5) and everything in (x) is also in (5,x)
coral spindle
Maybe you should go back and look at the definition of a quotient ring
primal tusk
the set of all congruence classes of R modulo I for some ring R and ideal I
along with some operation definitions
oh okay i think i am getting it
uh so is this "isopmorphism" in any way close to what i want?
formal ermine
the easiest way is to use the first isomorphism theorem, like boytjie already said
coral spindle
I can see above that you wrote Z[x]/Z = Z
this isn't true
Z isn't even an ideal of Z[x]
coral spindle
If by '=' you mean that they are isomorphic, yes
pastel cliff
can someone verify this rq pls
white oxide
so here i'm asked to express this permutation as a product of disjoint cycles. but would it only be possible to express this as only one cycle?
coral spindle
Yup
white oxide
oh so product can be just one cycle
coral spindle
Yup
pastel cliff
can someone verify this rq pls
bump
coral spindle
You say "order doesn't matter since R is commutative" but I don't see why that's relevant.
The proof is fine
pastel cliff
yup
i only said so bc i put the alpha before/after sometimese
but ig that's obv
is the bijection obv enough to omit? i have a hard time gauging that kinda stuff
coral spindle
No, that's kinda the main part of the proof
pastel cliff
oh 
ok yeah turns out the bijection might be a bit annoying to show 
is that last equality true?
coral spindle
You have no guarantee that IM contains N, hence IM/N is meaningless
pastel cliff
ahhh ok that did feel too easy
south patrol
Lol I saw I(M/N) and just assumed it was nakayama
pastel cliff
who
pastel cliff
Nakayama's lemming
next obsidian
Yo mama
pastel cliff
south patrol
Lmao
rustic crown
south patrol
Yo mama so finitely generated
pastel cliff
wouldnt if be worse is she had an infinite set of generators
south patrol
Lol
pastel cliff
shitposting aside is there anything i can use on IM/N
rustic crown
that no make sense like boytjie said
south patrol
I mean these are always a first iso thm thing
rustic crown
since it no contain N
pastel cliff
ohhhh the earlier equality was the wrong one
chilly radish
Dw just a theorem in comm alg that deals w IM
The fundamental theorem of commutative algebra
pastel cliff
i(m + n) and (x + n), where x = im + n...?
rustic crown
finite sums of those, but sure
pastel cliff
so it's (i1m1+...+ikmk + n) + n
pastel cliff
yeah i just mean like
coseets of N
so better said it's i(m + N) and (i1m1+...+ikmk + n) + N
pastel cliff
maybe im being silly but im not sure i see why
rustic crown
oh so I is an ideal, M/N is an R-module
I(M/N) would be a submodule of M/N
which is generated by products i * (m+N)
that's the definition right
pastel cliff
ok in general, how do ideals "interact" w modules
i know an ideal of R is itself an R-module
(any subring of R should be i think)
(it's the other way around whoops)
next obsidian
Look at your last chain of equalities
You rewrote a guy into something you can use 2nd iso on
Well, sort of one step is dicey
chilly radish
A guy
pastel cliff
You rewrote a guy into something you can use 2nd iso on
not necessarily contained tho
chilly radish
Distinguished gentleman
pastel cliff
a man
next obsidian
But I(M/N) = IM+N/N is true
pastel cliff
look at this man
next obsidian
look at this man
Evil
Anyway I did evil and probably derailed det help
Sorry det
pastel cliff
rustic crown
issokie
next obsidian
Det is uwu
And they owe me for helping with étaleness so they can never be mad at me
Kekw
chilly radish
I'll be mad at you on his behalf
rustic crown
yee so sebb, the only input in proving I(M/N) = IM+N/N is the definition of multiplication of ideal with a module
another way sort of to think about that could be using the map M --> M/N. image of IM is exactly I(M/N)... if you can verify this by hand, then you're done. but it's not any "better" than doing the whole thing using elements.
summer path
Don't 2nd and 3rd iso theorems pretty much follow immediately from first iso anyway?
formal ermine
everything follows from the definitions...
pastel cliff
"we start by consider the peano axioms..."
sonic coral
I’m trying to show that if K is normal in G then HK = KH, I should just show that HK and KH are contained in each other right?
chilly ocean
yeah, thats what it means to be equal for sets
I'd write HK as a union hK for h in H
and then, for singleton, show that hK = Kh
and now this sum is exactly KH
sonic coral
id have to try it. I think the idea is that since there exists an element k’ in K such that hk=k’h, we can show that the sets are equal.
i know that’s essentially rewording the proof with the definition of normal but the idea makes sense intuitively to me at least
chilly ocean
reword it as you wish if it makes it easier to understand
as long as it makes sense of course
sonic coral
also, to show that HK is normal in G if H and K are both normal in G, i could just show that (gHg’)(gKg’)=gHKg’ and then i’m done, right?
chilly ocean
also, to show that HK is normal in G if H and K are both normal in G, i could ju...
HK = (gHg')(gKg') = gHKg' you mean?
yes
chilly radish
Please provide adequate context so we can help
lament dawn
send the entire page
chilly radish
Also, I have a nagging sensation this might not fit in #groups-rings-fields
What subject is this
south patrol
I was gonna say like the fact I don't even know whether this is algebra kinda stops any ability to help lol
long geyser
can someone help me understand why the left hand side of (**) is symmetric in the indices? it switches the order of the product of the two groups in the quotient
probably missing some obvious deduction that they lead to the same group but not getting there rn
chilly ocean
can someone help me understand why the left hand side of (**) is symmetric in th...
the product of the two groups in the quotient commutes
long geyser
I know that's what I'm supposed to be deducing but I still don't understand why that is
is it obvious?
chilly ocean
you've essentially shown that those subgroups are normal in G_1 cap G_2 at the start of the proof of the lemma
normal subgroups commute with each other
they don't have to be normal in whole of G but that isn't important
long geyser
oh normal subgroups commute? okay that makes a lot of sense now that I think about it
thanks
chilly ocean
I’m trying to show that if K is normal in G then HK = KH, I should just show tha...
yeah you can look here
H is any set whatsoever
long geyser
oh lol that's quite a coincidence
chilly ocean
iirc Zassenhaus lemma contains a lot of isomorphism theorems as a corollary
it's also called butterfly lemma sometimes
because when you draw it, it should resemble a butterfly
long geyser
yeah I saw the butterfly on wikipedia
wanna learn about lattices of groups (and lattices in general) one day they seem powerful but I suck at imagining them
chilly ocean
I don't know if lattices, on their own, are "powerful" as you say. A lot of theorems only begin to hold with some assumptions like distributivity
There's also a version of Zassenhaus lemma for semigroups, and analogue of its corollary
though in spirit its a little different I suppose
long geyser
I don't know if lattices, on their own, are "powerful" as you say. A lot of theo...
well, speaking from experience, I've seen them pop up here and there all over the math spectrum so I'd at least guess they are powerful in a sense that they can provide intuition and unite common properties of different fields together (sort of like categories do)
for my gain I don't think they need to be good at proving nontrivial theorems to be of use really
chilly ocean
they are surely significant enough to be recognized by pioneers of universal algebra
like Birkhoff
long geyser
chilly ocean
sigma is signature right
and F is the set of function symbols
As and Bs are a family of algebras of some kind?
sigma algebras? 
I think he means it as in (big) Sigma algebras
to refer to the signature
I'll be honest, I've never seen algebras defined this way. A Sigma-algebra here seems to be a family of of algebras A_s for each s in S, where A_s has signature (F, ar)
correct?
I've never seen sort before
wikipedia seems to call the definition I've seen as single-sorted signatures, with single-sorted in brackets
yeah
({IN}, {+,*}, ar) where ar(+) = ar(*) = 2 I think you mean
um. The operation + and * are binary operations
meaning their arity is 2
associativity doesn't play a role here
okay so from what I see, what you're dealing with is that we can have A_1 x A_2 map to A_3 for example
and here S can have 1 x 2 -> 3 as an element for example, meaning that there will be an operation like that (I think?)
ar is a map from F to S
so it tells us how each function will behave for a Sigma-algebra
here h = (h_s)_{s in S} is a homomorphism of Sigma-algebras?
h_s are just functions I think
and we define what relations does those functions have to satisfy for h to be consider a homomorphism
you want to think as h to be a morphism from A = (A_s) to B = (B_s)
and h is this family of maps
Isn't f_A just operation on A corresponding to the function symbol f
so, presumably, h has to preserve all operations of A
pastel cliff
blitz
chilly ocean
Which means that if $f_A:A_{s_1}\times ...\times A_{s_n}\to A_s$ then $$h_s(f_A(a_{s_1}, ..., a_{s_n})) = f_B(h_{s_1}(a_{s_1}), ..., h_{s_n}(a_{s_n}))$$ for $a_{n_i}\in A_{n_i}$
pastel cliff
what's the new pfp
cloud walrusBOT
Blitz
chilly ocean
well yes. I was being neglectful and denoted both just by f
h_s is not an operation
h_s is a function from A_s to B_s
yes thats it
we don't require anything else from h_s
but we do require things from h = (h_s) which means that the maps h_s have to satisfy some relations between themselves
and h here is the homomorphism
the homomorphisms will be families of functions between those underlying sets in those algebras
oh wait
I think I'm wrong
S is not just a set of indices but it contains much more
okay no I'm correct
just ar isn't a function from F to S
but from F to symbols made from S and x, -> called relational and functional symbols
what do you mean
does ar(f) only output you function symbols?
so there's no relations?
in your definition
since we're talking about Sigma-algebras that makes complete sense that no relations are included
okay
so ar(f) = (s_1, ..., s_n, s) means that f_A should be a function from A_s_1 x ... A_s_n to A_s
nothing else to it
because F is just a set of symbols
you need to know what each element f of F will do to your Sigma-algebra A = (A_s)
no
they're not functions, just symbols
sorts are your indices
they signify the amount of stuff you have in your Sigma-algebra
we have a set A_s for every sort s in S
lower index
something you denote it by
A_1, A_2, A_3, A_4, ... could be like this
or could be A_a, A_b, A_c, A_d
for S = {a, b, c, d}
no, S is the set of sorts
so a is a sort
element of S
it's not important what S is
we could have two sorts for example right
and we would probably just go S = {1, 2}
pastel cliff
look at this man
ignore this, linking myself to convo
chilly ocean
and we would probably just go S = {1, 2}
and Sigma-algebra here would consist, among other things, of sets A_1 and A_2
sure why not
pastel cliff
quick interruption: are ideals of a ring in direct correspondence with the submodules of an R-module
let R be the R-module
chilly ocean
you don't see it with multi-sorted algebras
but if S has just one element, we call the algebra one-sorted
and here we could for example model the natural numbers as a set N with addition and multiplication
- would be a function symbol
why function symbols and not functions?
because you will consider many different Sigma-algebras with the same function symbols
the function itself is not as important
idk what you mean by the same thing
S is the number of your sorts which tells you how many of those A_1, A_2, A_3, ... have have basically
and F is the function symbols which tell you what sort of operations can you expect on your Sigma-algebra
S is a set yes
well sure. But it'd be pretty bad notation
If someone wanted to consider algebras with 3 sorts, they'd most likely just say S = {1, 2, 3}
or, if this is in some sort of context, they would name elements of S according to that context
I kind of hate this but in principle yes
in normal kind of universal algebra you consider just one set and operations on them
here S just means you have multiple sets and its more complicated
you have a whole structure of sets
and they can act between each other
according to your functions corresponding to your function symbols
I think it's the easier to understand the rest by looking just at the single-sorted algebras
because we don't really lose anything important to the intuition
an algebra must consists of family of sets A = (A_s) and a function f_A for every function symbol f
and this function f_A has a specified domain and codomain
and this information is contained in ar(f)
you can't write ar(f_A)
that doesn't make sense
ar is only defined on function symbols
ar(f) tells us what the domain and codomain of f_A or f_B or f_C should be
no matter if its algebra A, algebra B, algebra C, it should work the same for all of them, as long as they all have the same signature
A_+?
+_A you mean
+_A will be a function from A_1 x A_1 to A_2 then
and +_B will be a function from B_1 x B_1 to B_2
I don't see why not
those can be any sets
we need h_s because the structure of an algebra doesn't consist of just one set
it consists of a set A_s for each sort s
h_s is not a homomorphism
h_s is just a function
I need to draw you this
you can think of it as A consisting of those two parts
A_1 and A_2
1
and B also consists of two parts B_1 and B_2
the function h is a function from A to B
and it needs to map each part to its corresponding part in B
what is
nowhere
we take two Sigma-algebras
you take a family of functions
h
and now you define what it means for this family of functions h to be a homomorphism from one Sigma-algebra to another Sigma-algebra
using the sort
A_1 has sort 1
A_2 has sort 2
but actually like
A = (A_s) as a tuple
so its like, even though A_1 is not related to its sort 1
even it can happen that A_1 = A_2
but A = (A_s) is a tuple so that it tells you, the set in algebra A with sort 1 is A_1
and the set in algebra A with sort 2 is A_2
I just denoted it like this
Formally we would probably distinguish between bold A and A
bold A containing all the structure, that is function for each function symbol and A
and A = (A_s)
but its a simplification I can live with
just how group is not just a set but we treat it like one
?
give me a different question
A = (A_s)
it means that A is a tuple consisting of set A_s for each sort s in S
$$\mathbb{A} = (A, \mathcal{F}\mathbb{A}),\ A = (A_s){s\in S}$$ where each $A_s$ is a set and $\mathcal{F}_\mathbb{A}$ is the set of operations on $\mathbb{A}$
cloud walrusBOT
Blitz
chilly ocean
If $\Sigma = (S, \mathcal{F}, \text{ar})$ is a signature, a $\Sigma$-algebra is a tuple $$\mathbb{A} = (A, \mathcal{F}\mathbb{A}),\ A = (A_s){s\in S}$$ where each $A_s$ is a set and $\mathcal{F}\mathbb{A} = {f\mathbb{A} : f\in \mathcal{F}}$ is the set of operations on $\mathbb{A}$, that is if $\text{ar}(f) = (s_1, s_2, ..., s_n, s)$ then $$f_\mathbb{A}:A_{s_1}\times A_{s_2}\times ... \times A_{s_n}\to A_s$$
cloud walrusBOT
Blitz
chilly ocean
everything is a set in mathematics. Unless its a proper class
because I say so
well I'm using a tuple in the more general sense of the word
it doesn't have to be finite
Just treat A as a function which after inputting s outputs A_s
I'm tired
you keep on repeating the same questions
Ass
thats not how you're going to understand this
you can only understand this by analyzing every part it consists of
and understanding what works where
how to interpret what
tell me what you don't understand about this definition and I'll answer you
you should know what a signature is already
because the latter doesn't contain the information about which set corresponds to which sort
heck, what if A_1 = A_2 = ... then the latter set becomes very degenerate
by writing A_2 for example
you already know that it means the set corresponding to sort 2 but thats only because you put 2 in the lower index
so...
Maybe the notation A = (A_s) is just confusing
but its standard
I didn't define S as a set of numbers
that was an example to help you see what S could be
when I wrote this
I meant it regardless of what I said before
we spend at least an hour on this definition can we just move on
it felt like an hour for me
so a homomorphism of $\Sigma$-algebras $h:\mathbb{A}\to\mathbb{B}$ is a tuple $h = (h_s){s\in S}$ of functions $h_s:A_s\to B_s$ such that for each function symbol $f\in\mathcal{F}$ with $\text{ar}(f) = (s_1, ..., s_n, s)$ and for each $(a_1, ..., a_n)\in A{s_1}\times ...\times A_{s_n}$ we have $$h_s(f_\mathbb{A}(a_1, ..., a_n) = f_\mathbb{B}(h_{s_1}(a_1), ..., h_{s_n}(a_n))$$
cloud walrusBOT
Blitz
chilly ocean
that is, h commutes with every operation f
ommiting the lower indexes we could write this as
$$h_s(f(a_1, ..., a_n) = f(h_{s_1}(a_1), ..., h_{s_n}(a_n))$$
cloud walrusBOT
Blitz
chilly ocean
and now its clear that it just sorts of commutes with f
?
arity is for function symbols
it means a function which commutes with all the operations
and here we have multiple sorts so we need a function for each sort
be precise in your question
I never wrote n
oh I wrote n
n depends on f and is part of the data of arity of f
n is not fixed, no
each f can have different n
okay okay
formally you should probably write
cloud walrusBOT
Blitz
chilly ocean
**Blitz**
I mean this abuse of the language is also fine imo
this is just like, a group homomorphism
same thing
sonic coral
did i do these right?
tender wharf
yeah looks fine although you should probably clean up how the argument is presented
sonic coral
probably in the <- direction right
something like since gag’ = a is the definition of the centralizer in G of a, we’re done maybe
i’m bad with wording
tender wharf
your argument is fine
and I can follow your wording, but it can be confusing to say your prof
so for <= what I'll do is
let a in Z(G). let g in G
then by definition of Z(G) ga=ag so gag' = a
chilly ocean
quick interruption: are ideals of a ring in direct correspondence with the submo...
Wdym? Not all submodules of R are ideals
tender wharf
then by definition of Z(G) ga=ag so gag' = a
and then blah blah blah qed
also why was that homework not typeset in LaTeX
chilly ocean
Submodule of what
chilly ocean
R is its own R-submodule sure
pastel cliff
but then an ideal I of R is also and R module
idk there was a remark abt it in my notes
studying for exam tm
chilly ocean
Yeah its an R-module with r•x = rx
I see what you mean. I was wrong I think
yeah if M is a submodule of R then that means M is closed under subtraction and r•x is in M for r in R and x in M
chilly ocean
Since here r•x = rx is multiplication in R this is precisely the definition of an ideal
Assuming R is commutative of course
prime sundial
if i have a sequence of normal subgroups $$G_0 \trianglerighteq G_1 \trianglerighteq G_2 \trianglerighteq \dots$$ where $G_i \trianglelefteq G_0$, does this mean $G_i \trianglelefteq G_j, i < j$?
cloud walrusBOT
maximo
prime sundial
we defined normal series as the sequence above where every subgp is normal to the initial group, does that imply every subgp is normal to every subgp before it?
ok now i see that it is trivial
pastel cliff
when is the cokernel and its universal property relevant
lethal dune
when is the cokernel and its universal property relevant
what do you mean when
one example is in homological algebra
pastel cliff
i guess that is vague
there's a brief section on it in my notes but it doesnt come up again i dont think
was just wondering
lethal dune
I can't think of any module theory specific example atm
pastel cliff
all goood
lethal dune
apart from proving some random exercises
next obsidian
Do you like quotients
pastel cliff
should i
lethal dune
next obsidian
If so, you have used the univ property of a cokernel
pastel cliff
:O
prime sundial
what is meant by "every integral domain can be embedded in a field"? that every integral domain is a subset of some field?
pastel cliff
something something injection map i believe
maximo are you also studying for something
prime sundial
i have an exam tuesday
pastel cliff
,ti
cloud walrusBOT
The current time for stμ₂dying is 12:20 AM (EST) on Mon, 20/02/2023.
pastel cliff
exam in 10 hours
prime sundial
oh man. good luck
pastel cliff
having a hard time understanding this question
$cR/aR$ is cosets of $\ang{a}$ by $\ang{c}$...?
cloud walrusBOT
sebbb
agile burrow
what is meant by "every integral domain can be embedded in a field"? that every ...
every integral domain is a subring of a field, namely its field of fraction
pastel cliff
walter
agile burrow
$cR/aR$ is cosets of $\ang{a}$ by $\ang{c}$...?
for this, you know what R/aR is. Now you're just looking at the elements of the form cr + aR where r is in R
agile burrow
right
pastel cliff
ok gonna try and write something up then
ok something something first iso
i just proved this actually
so im gonna assume $\ang{\frac{c}{a}}$ is the annihilator of $cR/aR$
cloud walrusBOT
sebbb
agile burrow
better than assuming it, can you prove it?
pastel cliff
i'll try rn just talking out thots and ideas
oh wait
so first of all since c | a, there exists an element b such that cb = a, b = a/c (not sure if this is necessarily well-defined?)
tough raven
b is well-defined because multiplication by c is cancellative, because R is an integral domain.