Reminds me of that one lemma by dedekind
#groups-rings-fields
slim kayak
valid night
I've tried using the theorem that says that for every group homomorphism f there exists a unique g: G/[G,G] -> C^x such that f = g o phi
dusty verge
Reminds me of that one lemma by dedekind
dedekind's lemma?
delicate orchid
I've tried using the theorem that says that for every group homomorphism f there...
G is cyclic, G/[G,G] is G
dusty verge
Wait what's G^X here?
delicate orchid
Try considering maps onto roots of unity
valid night
G is cyclic, G/[G,G] is G
I noticed that but it still didn't lead me any where
dusty verge
Let's start with the easy part
What's the obvious cyclic subgroups of C^times?
Oh dammit wew
valid night
powers of i?
dusty verge
Yes, but moreso complex numbers with a norm of 1 right?
delicate orchid
subgroups generated by nth roots of unity specifically
dusty verge
That are a root of unity
valid night
Oh yeah of course
slim kayak
dedekind's lemma?
the one that states that any set of group homomorphism from G into the multiplicative group of L (L is a field) is linearly independent in the L vector space L^G (homomorphism from G to L)
dusty verge
the one that states that any set of group homomorphism from G into the multiplic...
no I was just suggesting a name for the lemma lol
But also, neat
slim kayak
But also, neat
galois theory really flexing with their lemmas
delicate orchid
the one that states that any set of group homomorphism from G into the multiplic...
This is just a very specific case of the fact that irreducible characters are orthogonal
Didn’t realise it had a particular name
slim kayak
probably need to make up for how ugly the category of fields is
delicate orchid
This is just a very specific case of the fact that irreducible characters are or...
Nvm, if L is of a weird characteristic this won’t necessarily work
dusty verge
dirichlet characters
delicate orchid
Who
slim kayak
ah yes, characters
south patrol
Character theory
dusty verge
Now Grace? She was a character
delicate orchid
Character theory
Character hypothesis
south patrol
What tha5
delicate orchid
There are characters that’s my hypothesis
south patrol
Man character theory is reminding me of the work I just handed in 
slim kayak
So yeah
south patrol
Think about the general description for maps from a cyclic group to any group
dim widget
Man character theory is reminding me of the work I just handed in <:bleak:877093...
I'm sure it was character building!
south patrol
It was K theory but used character theory of compact groups a bit
But im not very knowledgeable about that area so I am scared I got smth wrong but it ok.
dusty verge
Can someone prove that algebra is better than analysis? I've felt like it's true for a long time, but haven't been able to come up with anything rigorous
south patrol
Lmfao
slim kayak
Can someone prove that algebra is better than analysis? I've felt like it's true...
That's logically equivalent to the consistency of mathematics so sadly unprovable but still true
dusty verge
That's logically equivalent to the consistency of mathematics so sadly unprovabl...
So what, it's unprovable in a mathematical sense, but any actual person knows it's true?
I love that, very poetic and mean
slim kayak
thanks
dusty verge
Hey can someone explain langlands to me?
delicate orchid
Good lord no
dim widget
Hey can someone explain langlands to me?
Have you considered asking in #1100505349427036191 ?
teal void
i feel like i'm being dumb. given some ring $R$ and some ideal $M$ with the canonical projection $\pi$, is it true that $\pi(-r) = -\pi(r)$? is $\pi(-r) = \pi(0-r) = \pi(0) - \pi(r) = -\pi(r)$ valid?
cloud walrusBOT
fluff
solar shore
just got done with some more practice exercises for abstract algebra 
feeling good about my final
dusty verge
i feel like i'm being dumb. given some ring $R$ and some ideal $M$ with the cano...
This pi from R to R/M?
teal void
yes
south patrol
Otherwise we wouldn't really care about it as much
dusty verge
Then yes
teal void
for some reason it is not clicking why $\pi(0-r) = \pi(0) - \pi(r)$, i feel like it should be $\pi(0) + \pi(-r)$
cloud walrusBOT
fluff
dim widget
for some reason it is not clicking why $\pi(0-r) = \pi(0) - \pi(r)$, i feel like...
Because it's a homomorphism of abelian groups.
dusty verge
p(-r) = -p(r)
dim widget
And then depending on how you're interpreting some of those minus signs it's also a homomorphism of R-modules
teal void
oh $\pi(-r) = -r + M$ and $-\pi(r) = -(r+M) = -r + M$?
cloud walrusBOT
fluff
teal void
cool thanks!
dusty verge
oh $\pi(-r) = -r + M$ and $-\pi(r) = -(r+M) = -r + M$?
Well that too yeah, but also like pi(r) - pi(r) = 0 = pi(0) = p(r - r)
dim widget
oh $\pi(-r) = -r + M$ and $-\pi(r) = -(r+M) = -r + M$?
You can also see this "by hand" because -(r + M) = -r + (-M) and because M is a subgroup of R: -M = M.
dusty verge
oh $\pi(-r) = -r + M$ and $-\pi(r) = -(r+M) = -r + M$?
Think of the quotient operation as kind of squashing all the elements in a given coset together
It does not relate to the thing I quoted btw
Like when you quotient out Z by (2), you get Z / 2Z, but also you just lump all the even numbers together and all the odd numbers together, and call the even numbers zero and the odd numbers 1
The first isomorphism theorem on rings is just an extension of the idea that even + even = even and even + odd = odd (and so on)
slim kayak
i feel like i'm being dumb. given some ring $R$ and some ideal $M$ with the cano...
keep in mind that any ring homomorphism is still a group homomorphism according to the group structure
static temple
Think of the quotient operation as kind of squashing all the elements in a given...
well, I personnally see it more like an universal property but well
can be both
dusty verge
Hmm, how so?
static temple
Hmm, how so?
well, it's about how a morphism can be factorized on this quotient
dusty verge
don't link the picture
static temple
I won't latex this but it's like you have a morphism A->C and it can be factorized into a commutative diagram A->A/B -> C
delicate orchid
it is way easier to think about it as an equivalence relation/just set operations than a universal property at first
delicate orchid
squash
dusty verge
If topologists get to sniff glue, I can have this
static temple
it is way easier to think about it as an equivalence relation/just set operation...
I think both view have to be understood in parallel, working with morphisms without the universal property-view seems so much a pain
delicate orchid
my dude is in an intro ring theory course
that is not the time to introduce universal properties
slim kayak
that is not the time to introduce universal properties
True
That should've been linear algebra
teal void
back with another related question... for rings $R,S$ (not necessarily a ring with identity) and $f:R\to S$ is a ring homomorphism, is $f(a^{-1})= (f(a))^{-1}$? yea this is like the same thing as before but i don't see why it's true if $f(1_R) \neq 1_S$. i see why it is true in the group homomorphism or if they were rings with identity AND $f(1_R) = 1_S$ as we would have $$1_S = f(1_R) = f(aa^{-1}) = f(a)f(a^{-1})$$ and thus have what we need
cloud walrusBOT
fluff
teal void
but this is reliant on $f(1_R) = 1_S$ which may not be true?
cloud walrusBOT
fluff
delicate orchid
if there's no identity then firstly, how can there be multiplicative inverses
delicate orchid
**fluff**
secondly, of course this isn't true, either 1_R or 1_S don't exist
*doesn't
w/e
static temple
well, usually it's true for "unitary" rings (do we call them like that?)
dusty verge
but this is reliant on $f(1_R) = 1_S$ which may not be true?
f(1_R) = f(1_S) should apply when they both have identity
If they don't, who cares go wild
teal void
not necessarily?
slim kayak
Necessarily
delicate orchid
f(x) = f(1x) = f(1)f(x)
seems pretty necessary to me
:smoke:
nah I kid, but yes necessary
by definition
dusty verge
not necessarily?
The reason the professor guy said that it doesn't necessarily apply is that rings don't necessarily have one
I mean he was lying, but still
static temple
f(x) = f(1x) = f(1)f(x)
= f(x)f(1) :smoke:
dusty verge
you're taking a class on rng theory not ring theory
slim kayak
nah I kid, but yes necessary
No, 1_S-f(1_R) is just the zero divisors of f(x) 😎
static temple
but @teal void it depends on the teacher and how he defined rings
delicate orchid
No, 1_S-f(1_R) is just the zero divisors of f(x) 😎
dusty verge
No, 1_S-f(1_R) is just the zero divisors of f(x) 😎
can that work?
slim kayak
Idk, would be funny if it did
teal void
ah uhm i kind of never really thought about this, but is what my professor stated not a homomorphism?
slim kayak
If 1_S-f(1_R) was a zero divisors for all elements in the image of f then... this should work no?
dusty verge
Oh but it is an identity of the subring
delicate orchid
u people are asking about 1_S when we've established it is not unital
dusty verge
u people are asking about 1_S when we've established it is not unital
The map sending R into the top left corner of a 2x2 matrix sends an element from a unital ring to a unital ring, but doesn't send id to id
It's in fact, even an embedding
delicate orchid
" 1_S-f(1_R) was a zero divisors for all elements in the image of f" what does this mean
dusty verge
It means the only one that did not get got was kerr I think
delicate orchid
The map sending R into the top left corner of a 2x2 matrix sends an element from...
so it's a homomorphism of non-unital rings
we have not seen the definition of ring homomorphism in use here
dusty verge
so it's a homomorphism of non-unital rings
I mean are you assuming f(1) = 1 is an axiom for homs?
static temple
I mean are you assuming f(1) = 1 is an axiom for homs?
personnally I would asume so if the context was not given X)
really it depends on how they define the rings and their morph
delicate orchid
if it is a homomorphism of unital rings - which that map clearly is not, the identity is not in the image - then it is part of the definition that f(1) = 1
slim kayak
" 1_S-f(1_R) was a zero divisors for all elements in the image of f" what does t...
Let a=1_S-f(1_R) and af(x)=0 for all x in R, then works out that f has the multiplicative property
So f(1_R) is the identity on the image of f, but not necessarily in the larger ring S
dusty verge
back with another related question... for rings $R,S$ (not necessarily a ring wi...
So to explain, the example you give illustrates that when you define homomorphisms with rings on identity, you have to specify that f(1) = 1. If you don't specify this, then sometimes f(1) doesn't equal 1, even if they have identity. So to answer your question, no unless you force it to be yes, but sometimes it's reasonable to force it.
dusty verge
So f(1_R) is the identity on the image of f, but not necessarily in the larger r...
You win the day, because that's what the example he linked was about I think
slim kayak
It still holds that without it an bijective ring morphism is still an isomorphism of Ring structure, unless I miss sth?
dusty verge
unless structure means something yes
teal void
so i guess this came up because i was trying to prove the fact that "a ring homomorphism is injective iff the kernel is trivial". the forward direction is obvious but i cannot see how the backwards is true (without $f(1) = 1$, which was not assumed as the definition of a ring homo)
cloud walrusBOT
fluff
slim kayak
unless structure means something yes
Well, the image is isomorphic to a ring with unity
dusty verge
so i guess this came up because i was trying to prove the fact that "a ring homo...
cosets my dude
they're all the same size
slim kayak
So when focusing on f(R) it might as well be the identity element for all intends and purposes
dusty verge
ignore me I got my implication backwards fuck
Wait did I? Which direction's the obvious one?
teal void
injective => kernel is trivial is trivial
slim kayak
Idk both are really trivial if you done it often enough
slim kayak
Injective to trivial is just the more general trivial result
dusty verge
Ring kernel is trivial implies group kernel is trivial implies injective
slim kayak
For the other direction use the homomorphism property
teal void
idk i feel like i have no intuition on this ...
dusty verge
idk i feel like i have no intuition on this ...
Okay, so the map's injective if every preimage of everything is unique right?
Like no a and b such that f(a) = f(b)
teal void
yes
dusty verge
Okay, so if the kerel is 0, then there's no a such that f(a) = f(0)
And you can use this to derive that all the other cosets are trivial by like subtraction and shit idk
but that's the basic idea - to show that because f(0) is unique, every thing else must be too
And that just comes from the definition of a hom
teal void
wait bruh can i just
assume $f(a) = f(b)$. then $f(a) - f(b) = f(a-b) = 0$ but the only element that sends to $0$ is $0$ so $a-b = 0$ thus $a=b$ oh why do i care about multiplication at all
cloud walrusBOT
fluff
teal void
for some reason i was really caught up on multiplication because it was a ring homo? but i understand fully
dusty verge
By the way if you wanna piss off your professor, just say that the ring hom is a group hom, and so if it has a trivial kernel it's injective
It's right and also it says nothing (well it actually just says the thing you said, but it provides no actual reasoning)
teal void
i can use it without proof, i just wanted to make sure i understood for myself lol
dusty verge
huh
Professors are really all the same
cool
Sounds exactly like something I've been told a billion times
lament dawn
turns out some things have one line proofs
dusty verge
everything has a one line proof
Especially the proof that arbitrarily long lines exist
lament dawn
godel in shambles
dusty verge
why'd you have to go there?
slim kayak
Okay, so the map's injective if every preimage of everything is unique right?
Aaaah
f(a) = f(b) => a=b and anything else is heresy
dusty verge
who?
dusty verge
That's not abstract algebra
That's probably better suited for number theory
Please try and keep stuff algebra related
I really don't wanna hear this much about how your parents didn't love you
delicate orchid
*you're
ebon wyvern
chmowned
ebon wyvern
but yes please try and keep stuff algebra related don't leak military intel on a math server thanks folks
dusty verge
well that was fun
dusty verge
f(a) = f(b) => a=b and anything else is heresy
I like the preimage way because it's a more natural segue into the inverse being a function
elder wave
but yes please try and keep stuff algebra related don't leak military intel on a...
don't do what now
dusty verge
don't do what now
some guy came here leaked pentagon plans called me a fatass and left
slim kayak
Thug shaker central and it's disastrous consequences
slim kayak
I like the preimage way because it's a more natural segue into the inverse being...
I find it spooky
It means the last axiom of ZFC is nearby, a surjective function being left-invertible
tribal moss
What is "the last axiom of ZFC"? Choice?
south patrol
It depends on your choice of ordering
south patrol
Z (for Zorn's lemma)
slim kayak
FrankelC
static temple
how tf did you get from ring theory to zfc, is the axiom of choice your godwin point?
dusty verge
man I'm trying to remember why the axiom of choice does weird shit with the cantor set
slim kayak
What is "the last axiom of ZFC"? Choice?
If you only care about the logical equivalence class, like I said surjective functions being left invertable
tribal moss
Frænkel.
dim widget
Actually without the umlaut afaik
Afaik in english orthography it's the same for german names?
slim kayak
It is, but the fact is that it is fraenkel in german too
Which is a bit funny since you'd it be expect it to be just the English orthography at first
tribal moss
Just like it isn't Göthe either.
dusty verge
Halliday
dusty verge
what asking a lot of questions?
And also answering still a lot but maybe not as many?
what's that reaction mean?
it's good right?
Okay good
white oxide
how does the preceding sentence show that E is a splitting field for g(x) over F(a)? because we don't even know what the zeros of g(x) are
is it because f(x) = p(x)h(x) = (x-a)g(x) because p(x) has a linear factor since it has a zero?
coral spindle
Since g(x) is a factor of f(x), any splitting field for f(x) is also a splitting field for g(x).
next cypress
does anyone have mathematica experience?
coral spindle
Please ask in #computing-software unless you have an AA-specific question.
Also, please do not post your question in multiple channels: one is enough.
white oxide
how do we know that f(x) has a factorization in F(a) tho, because we're only given f(x) in F[x]
also aren't there less zeros for g(x) then that of f(x), so the splitting field should be smaller? where am i going wrong
E is the splitting field for g(x) over F(a), and E contains say n zeros of f(x). but could there exist a splitting field E' for g(x) that contains n - k zeros of g(x) (since g(x) definitely has less zeros than that of f(x)) and since splitting fields are defined to be the smallest fields containing the zeros of a given polynomial and field its over, that would be the field instead?
idk if im making sense
nvm
i got it lol
slim kayak
Bro is giving himself a thumbs up
near star
is this a valid presentation of the quaternion group?
near star
I thought it looked a bit big
south patrol
You can use the like famous things Hamilton wrote down I believe
Also you only rly need 2 generators if you want minimality
next obsidian
I like a presentation where you add a symbol l which is i^3
white oxide
<@&286206848099549185>
coarse kestrel
p(x) is a factor of f(x).
a is a root of p(x)
so p(x) is reducible over F(a)
thus p(x) = (x-a)r(x) where x-a and r(x) are in F(a)[x]
Carry this up to f(x)
f(x) = p(x)q(x) where p(x) and q(x) are in F[x]
now factor out the x-a
f(x) = (x-a)r(x)q(a) where r(x) is in F(a)[x]
Now let r(x)q(x) = g(x) which is still in F(a)[x]
frigid lark
why is g(x) in F(a)[x]?
What book is this, it looks like a high school book
white oxide
What book is this, it looks like a high school book
damn bro I just got flamed
Gallian
tranquil parcel
I hope this is sensible to ask here, and if anyone can point me to literature on it.
If I have a homomorphism of rings $\varphi : R \to S$, one can ask: does that induce a lattice homomorphism on the ideal lattices $\varphi^{*} : I(R) \to I(S)$ by sending an ideal $I\subseteq R$ to the ideal generated by its image $(\varphi (I)) \subseteq S$?
The answer generally is, no. For example, if $R = \mathbb{Z}[x,y]$ and $S = \mathbb{Z}[z]$, define $\varphi$ so that $x,y \mapsto z$ (obvious map on the integers themselves). Then the meet of the ideals $(x),(y)$ is their intersection $(xy)$ in $R$, but their meet in the image is (z). This happens because, generally, the image of an intersection can be properly contained in the intersection of the images. But I'm certain joins are preserved under these induced homomorphisms, so my question is, what conditions are sufficient/necessary for a meet homomorphism to be induced as well?
cloud walrusBOT
Michael Harp
tulip hawk
Why we can assert that? (dummit page369)
chilly ocean
did you try applying the map to 1 ⊗ 1?
tulip hawk
you are right
frigid lark
Can I go any further for this question?
Wait I can make the summation be formal and I think it works then
wooden ember
alright im making a dumb mistake somewhere but I can't figure out where. I've somehow shown that $\mathbb{Q}(\zeta_n,\zeta_m)\subsetneq \mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)$ in a few cases, which is of course false. I argued as follows:
$$\begin{align}[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)]&=[\mathbb{Q}(\zeta_n,\zeta_m):\mathbb{Q}(\zeta_n)]\nonumber\
&=[\mathbb{Q}(\zeta_{[n,m]}):\mathbb{Q}]/[\mathbb{Q}(\zeta_n):\mathbb{Q}]\nonumber \
&= \phi(nm/(n,m))/\phi(n) \nonumber \ &=\phi(m/(n,m))\end{align}$$
where the last line follows from $n$ being coprime with $m/(n,m)$. However I also have that $$[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)]=[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_{(n,m)})]=\phi(m)/\phi((n,m))$$ and in general these two quantities arent equal
cloud walrusBOT
𝓛ittle ℕarwhal ✓
alright im making a dumb mistake somewhere but I can't figure out where. I've somehow shown that $\mathbb{Q}(\zeta_n,\zeta_m)\subsetneq \mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)$ in a few cases, which is of course false. I argued as follows:
$$\begin{align}[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)]&=[\mathbb{Q}(\zeta_n,\zeta_m):\mathbb{Q}(\zeta_n)]\nonumber\\
&=[\mathbb{Q}(\zeta_{[n,m]}):\mathbb{Q}]/[\mathbb{Q}(\zeta_n):\mathbb{Q}]\nonumber \\
&= \phi(nm/(n,m))/\phi(n) \nonumber \\ &=\phi(m/(n,m))\end{align}$$
where the last line follows from $n$ being coprime with $m/(n,m)$. However I also have that $$[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)]=[\mathbb{Q}(\zeta_m):\mathbb{Q}(\zeta_{(n,m)})]=\phi(m)/\phi((n,m))$$ and in general these two quantities arent equal
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(amsmath) trying to recover with `aligned'.
See the amsmath package documentation for explanation.
Type H <return> for immediate help.
...
l.60 ...n) \nonumber \\ &=\phi(m/(n,m))\end{align}
$$
Try typing <return> to proceed.
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wooden ember
(say take m=12, n=2)
slim kayak
I don't understand the first line under the hypothesis
If the field extension by those two roots of unities is contained in their intersection it's also contained in of either the field extensions by just one of the roots.
So the RHS should just be 1
wooden ember
oh oop i miswrote
i meant Q(zeta_{[n,m]}) strictly contained
not Q(zeta_n,zeta_m)
lmao
slim kayak
What is zeta_n,m ?
dim widget
By the principle of explosion either mistake is equally bad!
wooden ember
What is zeta_n,m ?
zeta_[n,m] no zeta_n,m
slim kayak
Sure, still don't know what you mean by that
wooden ember
Sure, still don't know what you mean by that
[n,m]=lcm(n,m)
dim widget
Sure, still don't know what you mean by that
In this notation [n, m] is the lcm and (n, m) is the gcd.
slim kayak
Oh, yeah that's more coherent
low void
Why does A=R[X]/(X²+1) as an A-Module only have 0 and A as submodules?
dim widget
A-submodules of A are ideals of A. Can you think of why a ring might only have trivial ideals?
slim kayak
Or from the other direction, do you know what A as a ring is isomorphic to?
delicate orchid
I swear you asked this like 14 hours ago lol
wet zodiac
yeah i didnt get a response
delicate orchid
and I'll repeat what I said before, there is a 1:1 correspondence between equivalence relations on a finite set and the partitions of that set, so check how many partitions of 5 there are
wet zodiac
yeah then its 7
delicate orchid
yus
wet zodiac
i was thinking like
elements that make up a partition have to be unique
so i was thinking it had to be more
delicate orchid
no u just draw le funny tetris
although if you drop the "up to "isomorphism" (whatever that means)" condition it becomes much more fun
wet zodiac
yeah i was thinking without isomorphism
dim widget
Yes it seems like the question is not about "up to isomorphism"
wet zodiac
delicate orchid
well the case of the partition 5 only has one possible equivalence relation on it, as {1,2,3,4,5} is the only option
wet zodiac
case of partition 41 has 5
delicate orchid
indeed
3+1+1 is probably the next easiest one
and you may start spotting a pattern in your answers, or maybe not
wet zodiac
311 has 5 pick 3
delicate orchid
omg using "pick" for the choose function is incredible
wet zodiac
and then multiply that by 2 to accommodate for 32
delicate orchid
I am stealing that
dim widget
I am stealing that
For notation you could even write it like nPk for n pick k! Oh wait...
wet zodiac
2111 has 10 and then we can generate all 221 partitions by multiplying it by 3
delicate orchid
throwback to that time I tried to defined the choice function over an arbitrary field
hold on lemme think about 2111
yeah it's 10
dim widget
2111 has 10 and then we can generate all 221 partitions by multiplying it by 3
Waiiiiiiiiiit what is your answer for 221?
wet zodiac
30 partitions of 221
dim widget
Are you sure?
wet zodiac
because with 2111 we can have the first 2 1s grouped together to generate a new one
last 2 ones to generate another
dim widget
Yes but you are over counting some things
dim widget
{a, b}{c, d}{e} is the same as {c, d}{a, b}{e}
wet zodiac
delicate orchid
{a,b}, {c,d}, {e}
5 choices for a, 4 choices for b invariant under permutation
3 choices of c, 2 choices of d invariant under permutation
dim widget
So you have to divide by 2
Yes and then you have to divide by the permutation switching the first two pairs
wet zodiac
ah i see why dividing by 2 works now
delicate orchid
this reminds me of conjugacy class sizes of S_n but we don't care about order
in fact
delicate orchid
yur way more fun than up to homeomorphism
dim widget
so then the total possible different equivalence relations is 52
I actually have no idea what the right answer is so double check your arithmetic if this is being graded
delicate orchid
SURPRISE they're open sets in topologies on {1,2,3,4,5} now
wet zodiac
not graded
delicate orchid
darn...
dim widget
What is U^\circ?
But anyway, absent other context, probably not: if U \subset V then for most defintions U^* would have the same dimension of U, so it wouldn't make much sense for it to contain V^*
silent oxide
What is U^\circ?
annihilator of U
silent oxide
every linear algebra definition ever
silent oxide
U^\circ sits in V*
dim widget
but annihilators generally flip things don't they
Yes but you're not taking the annihilator of V
V^\circ is just 0
frigid lark
They sound so cool and then are actually so boring
Just like Radical Groups
dim widget
So yes P(V^\circ) \subset P(U^\circ) if you decide the first is the empty set.
dusty verge
Just like Radical Groups
They just sounds like they wanna overthrow the government
vague granite
U^\circ sits in V*
oxon?
vague granite
fairs
dim widget
NT, groups, calculus of variations, integral transforms
who is teaching NT
vague granite
who is teaching NT
you would think James Martin, but he's actually teaching Calculus of Variations
silent oxide
think he's lecturing calculus of variations?
vague granite
yep
silent oxide
you would think James Martin, but he's actually teaching Calculus of Variations
maynard*
vague granite
I was stunned when I saw him walk in and write down calc of var
dim widget
probably not james
What about James Newton?
silent oxide
What about James Newton?
right forgot there are a few james
oh it's kobi, just checked
vague granite
he's a cool guy, a bit quiet
NT seems pretty chill. Just basic theorems...Euler-Fermat, Wilson, Gauss' Reciprocity and some appendix on cryptography
silent oxide
NT seems pretty chill. Just basic theorems...Euler-Fermat, Wilson, Gauss' Recipr...
ye it's just elementary/olympiad NT
silent oxide
3rd year has algebraic NT and 4th yr has analytic NT, and other algebraic NT ish things
vague granite
yessiree
dim widget
Everyone went to a better school than me
england is cringe dw
silent oxide
going to school in england is cringe dw*
vague granite
I think it's a shame anal NT is so late in the degree cause it seems pretty cool tbh
frigid lark
Everyone went to a better school than me
Don't worry, I probably go to a worse school
silent oxide
I think it's a shame anal NT is so late in the degree cause it seems pretty cool...
yeah it doesn't require that much machinery to be a 4th year course imo
vague granite
yeah it doesn't require that much machinery to be a 4th year course imo
Absolutely, some of the elementary proofs use analysis I techniques
dim widget
Better than France, at least
They deserved each other in the 100 years war fr fr
dusty verge
Don't worry, I probably go to a worse school
University of Maine
frigid lark
Monash university
silent oxide
Absolutely, some of the elementary proofs use analysis I techniques
at most uses residue theorem or some fourier transform things (but for Fourier the course will teach the relevant results)
dusty verge
Do you guys have a math department with more than five people?
silent oxide
so it couldve been a 3rd year course imo
vague granite
FT based
dusty verge
We regularly got confused for a burger chain
frigid lark
Do you guys have a math department with more than five people?
Like all of math, or a specific area
vague granite
I'm actually doing research in CMS in cam using FT to study orthogonal series
dusty verge
All of
frigid lark
My condolences
dusty verge
Also it's more like 15 people I was exaggerating
The reason I don't have a PhD right now is we didnt have a PhD program
frigid lark
I'm an undergrad who only cares about algebra, I don't know how big the math faculty is
dusty verge
And it's hard to get in to math PhD programs if you have shit grades
frigid lark
And it's hard to get in to math PhD programs if you have shit grades
Isn't math a Wam booster
dusty verge
Wam?
silent oxide
I'm an undergrad who only cares about algebra, I don't know how big the math fac...
it's better to keep an open mind unless you've already looked at other topics
since you're only an undergrad
frigid lark
Weighted average mark
dusty verge
it's better to keep an open mind unless you've already looked at other topics
Bullshit
frigid lark
it's better to keep an open mind unless you've already looked at other topics
I'll do other stuff once I'm done with Lang
dusty verge
That's just what we tell analysis guys
silent oxide
(I'm an analysis guy)
frigid lark
Also I'll osmosis some analysis through Lang
dusty verge
Don't tell someone to try new things if they're already right
dusty verge
(I'm an analysis guy)
I just threw up a little
frigid lark
i threw up more studying category theory
Fair
silent oxide
It's all categories my friends
true if you allow your statement to be meaningless
dusty verge
You say that like Hilbert didn't invent half your shit
frigid lark
Arrow theoretical means we can shoot arrows at people
dusty verge
Only theoretically
frigid lark
So it is abstract nonsense
dusty verge
Nah it's really not
It's just defining math shit by their structure instead of by the things they are
Also yes yes, you can do abstract bullshit with it too, but who cares
formal ermine
dusty verge
Because they're something inherently cool about the fact that you can uniquely identify a group by only it's homs to other groups
And do that with everything else too
frigid lark
Ok, let me try, the following sequence is exact, where the last arrow sends all the not cool stuff to 0. $$ \text{category theory} \to \text{math} \to A$$
cloud walrusBOT
Parrot Tea
formal ermine
oblique river
Gottem
delicate orchid
the following sequence is exact:
0 -> Me -> Cool -> 0
dim widget
the following sequence is exact:
0 -> Me -> Cool -> 0
*0 -> Wew -> Cool -> Cool -> 0
delicate orchid
now hold on just a moment
frigid lark
It's fine, Cool is projective
frigid lark
My brain is too small for this, does noetherian + projective give you something
Or is it just cause Noether is cool
dim widget
Noetherian means that it cannot be true that Cool/Wew \cong Cool unless Wew = 0
frigid lark
Ahh, ok
dim widget
But the fact that Emmy Noether was cool is a proof that Cool is Noetherian, thus we can be sure that that exact sequence implies that Wew = 0.
frigid lark
Yeah otherwise you get an infinite filtration
frigid lark
But the fact that Emmy Noether was cool is a proof that Cool is Noetherian, thus...
Which is intuitive as wew is just one box
delicate orchid
one of the conversations I've read
delicate orchid
You got smoked
simmer down in the back
lethal dune
But I think that Cool is also Noetherian.
aren't all rings?
south patrol
lethal dune
south patrol
Ring of continuous functions on [0,1] my beloved
south patrol
What lol
lethal dune
no 'N'
south patrol
if f is a cts function so is -f
This is [0,1] -> R not [0,1] -> [0,1]
Okay yeah fair I guess
saying on [0,1] implied that ig
Continuosu real-valued functions on [0,1]
lethal dune
fallen haven
what on earth is intro to complex numbers?
do you introduce euler's formula mid semester?
celest cairn
Hi
I asked the new Snapchat AI bot on how to find the minimal polynomial of sqrt(3) over Q_2, but I feel like it’s not correct. I don’t know much at all on p-addic numbers, so I don’t know for sure if it’s incorrect. Here’s what it gave me:
“To find the minimal polynomial of sqrt(3) over Q_2, we need to find the smallest degree polynomial with coefficients in Q_2 that has sqrt(3) as a root. We can start by writing down the general form of a quadratic polynomial with coefficients in Q_2:
ax^2 + bx + c
We want this polynomial to have sqrt(3) as a root, so we set x = sqrt(3) and simplify:
a(sqrt(3))^2 + b(sqrt(3)) + c = 0
3a + bsqrt(3) + c = 0
Since a, b, and c are in Q_2, we know that 3a, bsqrt(3), and c are all in Q_2 as well. Therefore, we can write:
3a = 2^n * m, where m is an odd integer and n is an integer >= 0
bsqrt(3) = 2^k * l, where l is an integer
c = 2^p * q, where q is an odd integer and p is an integer >= 0
Now we can substitute these expressions into the equation 3a + bsqrt(3) + c = 0 and rearrange:
2^n * m + 2^k * l + 2^p * q * sqrt(3) = 0
If we assume that m and q are both equal to 1, we can solve for sqrt(3) and get:
sqrt(3) = -(2^n * l) / (2^k + 2^p * sqrt(3))
This expression shows that sqrt(3) is in fact an element of Q_2(sqrt(3)), so it is a root of some polynomial with coefficients in Q_2. To find the minimal polynomial, we can try plugging in a few different values of n, k, and p and see what works. For example, we might try n = 0, k = 1, and p = 1, which gives us:
sqrt(3) = -l / (2 + sqrt(3))
Multiplying both sides by 2 + sqrt(3) gives:
sqrt(3)(2 + sqrt(3)) = -l
Simplifying gives:
5sqrt(3) = -l
This means that the minimal polynomial of sqrt(3) over Q_2 is x^2 + 5.”
delicate orchid
pfftttt reddit spacing
rustic crown
never ask ai some math >.<
south patrol
Why are we having to correct AI lol
rustic crown
x^2-3 is a poly that has sqrt3 as a root
dim widget
south patrol
If you don't know about p adscs anyway
rustic crown
and to show its minimal, go mod 4 and say 3 is not a perfect square there.
sharp sonnet
If you don't know about p adscs anyway
or apparently field extensions
delicate orchid
gah if it wasn't Q_2 I might've been able to actually think about this
sharp sonnet
classifying squares in Q_2 is a bit annoying actually iirc
delicate orchid
yeah it's not like the other p-adics
sharp sonnet
its similar
delicate orchid
I'm not like the other girls
delicate orchid
idk ur the number theorist
sharp sonnet
you have like Q_p^x = <p> x roots of unity x (1+ (pi))
and then something p-adic logarithm
but if p=2 some cases are left
sad!
dim widget
The complicated thing is that convergence for square root in Q_2 is worse
sharp sonnet
anyways, this server is more for helping humans, not debugging AI
dim widget
The complicated thing is that convergence for square root in Q_2 is worse
A principal unit needs to be congruent to 1 mod 8 for the square root to converge.
dim widget
Yes and in fact they don't have square roots.
delicate orchid
false. The square root of 1 is one. Libreal destroyed
sharp sonnet
ah hmm
dim widget
false. The square root of 1 is one. Libreal destroyed
1 is congruent to 1 mod 8, I just checked
delicate orchid
1 is congruent to 1 mod 8, I just checked
t-t-t-t-there must have been... some some... SOME sort of m-mistake!!! No... NO!!!
dim widget
3, 5, 7 have to be checked by hand
I think actually you can make the argument that if the formal power series for (1+x)^{1/2} doesn't converge that it's impossible for 1 + x to have a square root.
delicate orchid
actually wew was destroyed
ngl I just saw "no square roots"
that's the only part i read
dim widget
Just by writing $y = \sum_i a_i 2^i$ and noting that the $a_i$ have to satisfy the same equations as the square root of $(1 + x)$ in the ring of formal power series
cloud walrusBOT
Topos_Theory_E-Girl
delicate orchid
that's nifty
formal ermine
false. The square root of 1 is one. Libreal destroyed
this reminds me of math class
delicate orchid
*maths
formal ermine
my teacher asked "what's the square root of 0?"
then picked this one guy and added "what number multiplied with itself gives 0?"
delicate orchid
the number of undergrads in my analysis class who were convinced sqrt(0) was undefeind
formal ermine
the dude deadass said "every number"
south patrol
the number of undergrads in my analysis class who were convinced sqrt(0) was und...
Bruh
Yikes
rotund aurora
what about -0 tho
delicate orchid
doesn't mean there can't be be others :smokeD:
but lets just say if u ask me what THE sqrt{0} is I'm saying 0 irregardless of if we're in C or Z/4Z
south patrol
If someone asked for the sqrt of 0 in Z/4Z I'd be like huh
delicate orchid
lol
lethal dune
anyway x²=0 supposed to have 2 solutions
solar inlet
because THE square root of zero is zero
"erm sorry I don't think the definite determiner is properly licensed here, there is a failure of presupposition meaning your claim is meaningless and uninterpretable"
🤓🚬
delicate orchid
"erm sorry I don't think the definite determiner is properly licensed here, ther...
SHUT UP
dusty verge
because THE square root of zero is zero
What about the other square root though?
You know, reading the above I feel unoriginal in my harassment of wew here
rustic crown
because THE square root of zero is zero
casual reduced ring enjoyer :p
dusty verge
I'll do better
Clearly the square root of zero is an infinitesimal
In the auto differentiation ring
dusty verge
Ah yes my favorite ring
Z/1Z
novel parrot
Q/Q is my favourite ring
rustic crown
OwO
novel parrot
QwQ
dusty verge
Do you mean Q, the field extension of Q?
I guess it's technically a ring
Good ol Q(6)
formal ermine
Q bar/Z is my favorite ring
novel parrot
whats Q bar
rustic crown
(why tho?)
rustic crown
whats Q bar
alg nums
dusty verge
The complex conjugate of Q
dusty verge
big boi stuff
Roots of polynomials with integer coefficients
formal ermine
Roots of polynomials with integer coefficients
huh
novel parrot
so Q bar = Q ?
dusty verge
No
delicate orchid
x^2-2 moment
dusty verge
Sqrt 2 is root of x^2-2
novel parrot
x^2-2 moment
rip
rustic crown
i is not an algebraic
it is
delicate orchid
yes, it is
smoky ivy
formal ermine
algechill moment
dusty verge
This is number theory but who cares
lethal dune
ok wtf? why did the channel got purged?
dusty verge
Define $p_n$ to be the sequence of prime numbers. For a given number $k$ define $\tau(k ) = {(p_m/p_n)k: p_n|k, m\leq n}$. Let $t_0(k)= {k}$, and let $t_m(k) = \bigcup_{j \in t_{m-1}(k)} \tau(j)$. Finally, define $t_{\infty}(k)=\bigcup_{m=0}^{\infty}t_m(k)$. Find / approximate $|t_{\infty}(k)|$.
Might need to define p_0 to be 1 I forget
Also it should be |t_{\infty}(k)|
cloud walrusBOT
Halliday
dusty verge
Then use it to prove the prime number theorem or some shit idk
(hint: $|t_{\infty} (p)| = \pi(p) + O(1)$ for $p$ prime)
cloud walrusBOT
Halliday
dim widget
algechill moment
dusty verge
Solve my problem you won't
static temple
Q bar/Z is my favorite ring
it doesn't seem so hard, like not harder than bar(Q)
dusty verge
it doesn't seem so hard, like not harder than bar(Q)
My god man what's your definition of not so hard?
Is a train not that heavy?
delicate orchid
Q bar isn’t that bad until you look at it as a field extension
dusty verge
Oh neat
dim widget
What is "the best" proof that $-\frac{B_{n + 1}}{(n + 1)} = \zeta(-n)$?
cloud walrusBOT
Topos_Theory_E-Girl
dim widget
discuss.
dusty verge
What's B?
dusty verge
oh gross
dusty verge
What is "the best" proof that $-\frac{B_{n + 1}}{(n + 1)} = \zeta(-n)$?
the one I don't have to read through
lethal dune
**Topos\_Theory\_E-Girl**
by definition
south patrol
So cringe
lethal dune
dusty verge
By definition... of zeta
dusty verge
No, define zeta to be the function such that $\zeta(-n) =$ the sum of this sequence I just pulled out of my ass called the bernouli numbers
cloud walrusBOT
Halliday
dusty verge
Oh hey look it can do all this other stuff too, neat
lethal dune
formal ermine
delicate orchid
u don't know the half of it
summer path
abstract chillgebra moment
alpine remnant
Yo, is there any way to prove that you can't have an affine space where the set of points A is a field K and the vector space V is a two dimensional vector space over K?
I asked in questions and they told me to ask here.
It's easy to prove for finite fields, but I don't see how I can do it for infinite ones
south patrol
How are you distinguishing the points from the vector space?
willow cargo
what would be prerequisite materials and good entry-level books for AA?
alpine remnant
How are you distinguishing the points from the vector space?
the points are those in the field K, the vector space is of dimension two so I'm guessing you identify them as the addition of two points times vectors in the base of V over K.
That's how you prove it can't be affine over finite fields
because K can't be bijective with KxK for finite fields
but for infinite ones it can
Intuition still tells me it should be impossible to find a function f: AxA-> V such that both:
f(a,b)+f(b,c)=f(a,c), dor a, b and c are in A.
And if you any pick p in A, f(p,a) is bijective between A and V.
But it's hypothetically possible from all I know
lethal dune
what would be prerequisite materials and good entry-level books for AA?
good entry level book does the prereq anyway
but a bit of sets functions relations etc
lethal dune
Yo, is there any way to prove that you can't have an affine space where the set ...
are you asking if you can have an isomorphism from from k → k²? k being a field? I don't get the question
alpine remnant
are you asking if you can have an isomorphism from from k → k²? k being a field?...
My question is: For which choices of fields K does a K-vector space V of dimension two accept an application f: KxK -> V such that (K,V,f) forms an afine space, that is, such that there exists an f: KxK -> V which satisfies:
1.- For all a, b, c in K, f(a,b)+f(b,c)=f(a,c).
2.- For all p in K, by fixing p and studying the function f(p,x), this function is bijective between K and V.
And I said+
Condition two cannot be satisfied for finite choices of K
because if K is finite field, then K can't be bijective with V, since V has dimension two over K, so it is isomorphic to KxK
So I know that for finite choices, this is not the case
for infinite choices, I believe it is not the case either
But i can't prove it
dusty verge
Yo, is there any way to prove that you can't have an affine space where the set ...
I've heard affine spaces are like euclidean space, but what exactly are they?
formal ermine
I've heard affine spaces are like euclidean space, but what exactly are they?
affine n-space over an algebraically closed field k are all n-tuples of elements of k
dusty verge
Oh neat
formal ermine
denoted A^n_k or simply A^n
elder wave
you just forget the extra structure
elder wave
algebraically closed is not necessary
dusty verge
So what's the vector space, and set of points here?
lethal dune
does not sound the same from the op's definition
elder wave
there is no vector space
dusty verge
<@143148906287464448> have you heard of projective space?
Probably not. I've heard of the projective plane, but I forget it
formal ermine
algebraically closed is not necessary
alpine remnant
I've heard affine spaces are like euclidean space, but what exactly are they?
The way i define them here is, a tuple (A,V,f) where A is a set of points, V is a vector space and f is a function that returns vectors between points in said set, that satisfy that f(a,b)+f(b,c)=f(a,c) and that f(p,x) is bijective for any fixed p I choose.
south patrol
there is no vector space
There is lol
formal ermine
Probably not. I've heard of the projective plane, but I forget it
it's like affine space but you add points at infinity
so all lines interesect
even parallel ones
south patrol
It's different from just affine n space
elder wave
valid night
Let $G = GL(2, \mathbb{R})$, prove $[G,G] = SL(2, \mathbb{R})$
cloud walrusBOT
Kroros
valid night
I've already proven [G,G] is a subset of SL(2,R)
But I'm stuck on the opposite inclusion
I've also proven that SL(2, R) is generated by matrices of the form [[1,0],[x,1]] and [[1,x],[0,1]] in a previous exercise and I assume I have to use that
valid night
I edited the message, my apologies
lethal dune
if you can show the generators are jn[G,G] you're done right
lethal dune
lol
warm wyvern
dusty verge
it's like affine space but you add points at infinity
Oh neat I have heard of that
dusty verge
The way i define them here is, a tuple (A,V,f) where A is a set of points, V is ...
What's a set of points? Just some set?
dusty verge
Okay cool
alpine remnant
basically, intuitively, an affine space is a bunch of points and a bunch of arrows that connect them all
dusty verge
Yeah I gotcha
Almost like an metric but different too
No not really hmm
Hmm if you can do it, you can have the vestor space be 2^k dimensional I think, but who cares about that
dusty verge
Yo, is there any way to prove that you can't have an affine space where the set ...
Fix a basis on $V$ and define a map $f^: V \to V$ by $f^(x, y) \mapsto f(x, y)$. This gets finite fields by cardinality and shit I think, but idk if it breaks infinite fields
cloud walrusBOT
Halliday
dusty verge
Hmm, I don't really wanna get out a pen and paper
Wait do vector spaces play as nicely over finite fields as I want them to?
Like is F^2 just FXF?
I'm just gonna assume so
delicate orchid
I've also proven that SL(2, R) is generated by matrices of the form [[1,0],[x,1]...
||SL(2, R) is the kernel of the determinant, so by first iso GL(2, R)/SL(2, R) = R^\times, which is abelian. And since the quotient is abelian [G,G] <= SL(2, R)||
this is quicker imo
shit that's the wrong direction 
zenith crest
i saw ppl ve been asking question here to so i guess its not forbidden, and im desperate, does anyone onow the answer? i need it like for right now :/
lament dawn
Why do you need it right now
zenith crest
Why do you need it right now
i have an exam tommorow and when i wake up i wont have any time to practise
im trying to find answer for almost 3 hours and cant it, like my examples have supremum and infimum so i would say they are sublattice, but that note from lecture is making me crazy and i spend so much time to learn for this exam i cant mess up on sth like that
white oxide
p(x) is a factor of f(x).
a is a root of p(x)
so p(x) is reducible over F(a)
th...
thanks that really helps!
but why is r(x)q(x) = g(x) still in F(a)[x]?
dim widget
My question is: For which choices of fields K does a K-vector space V of dimensi...
I think this is a very interesting question but I don't know where you got it from
coarse kestrel
but why is r(x)q(x) = g(x) still in F(a)[x]?
r(x) is in F(a)[x] and q(x) is in F[x] which is a subset of F(a)[x]
so closure of the ring F(a)[x]
dim widget
Such an f(x, y) is (I guess) best understood as a 2-cocycle of the abelian group K with values in the (trivial) K-module V.
white oxide
r(x) is in F(a)[x] and q(x) is in F[x] which is a subset of F(a)[x]
so closure o...
ah that makes sense thank u
dim widget
If we could show that all of these vanish, for instance if H^2(Q, Q) = 0, then we would have our answer, because then all such f are of the form f(a, b) = a - b
I suspect this is untrue though.
dusty verge
H^2(Q,Q)?
wraith cargo
Yeah you can define cohomology for groups
dusty verge
H^2(Q, Q) is a cohomology then?
Okay so f is bijective if you fix the second coordinate too.
dusty verge
My question is: For which choices of fields K does a K-vector space V of dimensi...
Point 2 will imply that it's bijective if you fix the second variable too
alpine remnant
Point 2 will imply that it's bijective if you fix the second variable too
no?
like
let's say you fix p=1
dusty verge
Oh right I'm dumb
alpine remnant
Then every vector in V will have a correspondence to a unique vector (1,x) since it is bijective
and same for 2
etc.
dusty verge
However, if its not bijective in the first, then you do have to have a bijection between K and a subset of K
Wait no, I was right
Uh one sec
Yeah I was wrong
dim widget
Then every vector in V will have a correspondence to a unique vector (1,x) since...
Can I ask who asked you this question?
Or did you come up with it?
dusty verge
Feels like you might be able to finagle a fixed point out of this?
alpine remnant
Can I ask who asked you this question?
algebra teach
dim widget
How advanced are you?
I'm very advanced 
alpine remnant
Such an f(x, y) is (I guess) best understood as a 2-cocycle of the abelian group...
yeah, whatever this means!
alpine remnant
Or did you come up with it?
basically our algebra teacher said that we're equipped to solve this using basic linear algebra
like
were not that far into it
dim widget
Hmm are you sure there's not a hypothesis missing? It seems kind of hard as stated.
rapid junco
Is the set of all functions such that f(0) = 0 an ideal for the set of all functions from R to R?
chilly ocean
did you try proving it
chilly ocean
you are correct
rapid junco
What would the dimension of it be as a subspace of R^R?
Sorry
The quotient of it over R^R
R^R was infinite dimensional but I said that that the quotient had dimension 1.
dim widget
You mean the quotient of all functions by that ideal?
rapid junco
Yes
chilly ocean
R^R was infinite dimensional but I said that that the quotient had dimension 1.
you are again correct
a very quick way to see it is to apply the first isomorphism theorem to the map f -> f(0)
rapid junco
Yeah I didnt get that far unfortunately but I guess my intuition was correct
I just said that the equivalence class of 1 was a basis
dim widget
That is at least an argument that it is at least one dimensional.
rapid junco
Actually I guess every equivalence class is a basis except for the set itself
rapid junco
Yeah
dim widget
Modulo that ideal everything is equivalent to a constant function.
Because if you subtract a constant from any function you can make it vanish at 0
rapid junco
yes
dim widget
sounds good then! good job
rapid junco
Thanks
next obsidian
No
next obsidian
Construct an example
void cosmos
probably can do one with chinese
karmic moat
yo im chinese……… u can’t say that…….
next obsidian
How is R isomorphic to R^2?
void cosmos
arent they as additive abelian groups?
rapid junco
How is R isomorphic to R^2?
Just because it is
next obsidian
Can you write down an isomorphism?
void cosmos
u cant write it down explicitly
next obsidian
If you want to appeal to the axiom of choice you’d probably have to show R is free as a an abelian group which isn’t true
rapid junco
Im joking
void cosmos
ig R is free as a Q-module
void cosmos
Can you write down an isomorphism?
but its like its there by AoC
just like u know there is a basis to ur vec spaces
so why wouldnt that be neough
enough*
next obsidian
Okay, so then you need to argue how you do it
I mean if you want to appeal to vector spaces then sure, but you need to use the fact that R has infinite rank as a Q-VS
And sure it’s true that this also implies the isomorphism over Z, but it’s hard to see why this holds when you only consider them as a modules over Z because you don’t have a basis over Z
void cosmos
modules over Z just came to my mind to ease it for me to find examples thats it really
but using AoC u can define some decimal map from R^2 to R cant really remember how tbh but its there
next obsidian
You don’t need to do that
next obsidian
Yes over Q it’s a cardinality thing
void cosmos
or ig as Q modules
next obsidian
R and R^2 are bijective by some cardinal arithmetic implying they both have a continuum-sized basis over Q
Then you take a bijection of their bases and that defines an isomorphism
void cosmos
so its there ig 😄
idk i just remember it as a stated fact
but yea it doesnt work without AoC ( no clue why )
merry plaza
I tried following this convo but I'm so confused
my abstract 2 final is next week and I'm p confident but this still feels beyond me
void cosmos
here is what you can get out of it
A x B iso to A x C does not imply B iso to C
as groups
😄
i figured RxR iso to Rx{0} would work and it does but i took the fact R^2 is iso to R for granted when in reality its not that easy
the map (x,y) --> (x,0) is not even surjective
if thats what ur thinking
merry plaza
yeah R^2 to R feels crazy to me but i guess it's possible
void cosmos
yeah R^2 to R feels crazy to me but i guess it's possible
it is but with AoC
but ig AoC is always there like with ideals and/or with basis
so meh
merry plaza
I'm still in undergrad so I need to find a good resource for axiomatic set theory i think (AoC is set theory right?)
rapid junco
Axiom of choice seems illegal
silk bramble
none of this conversation uses anything beyond basic set theory for general mathematicians
void cosmos
I'm still in undergrad so I need to find a good resource for axiomatic set theor...
not really
as delerik said u WILL see AoC alot
silk bramble
like you could just read chapter 0 of Folland's Real Analysis
south patrol
And also implicitly often lol
merry plaza
ah ok, im in real analysis in fall
silk bramble
it will have all the set theory you will ever use for a long while
void cosmos
basically idk how would u do algebra without AoC like u wouldnt even have maximal ideals
so maybe thats why R^2 is R as groups is more well-known than "not without AoC"
and alot of the proofs use AoC aswell
silk bramble
choice is an obvious thing students will just assume
you mean the equivalent of choice, Zorn's lemma
void cosmos
Baer's criterion aswell
void cosmos
choice is an obvious thing students will just assume
you mean the equivalent of...
yes
void cosmos
Baer's criterion aswell
which proves alot of things for injective modules later on :d
silk bramble
chapter 0 of Folland has it all
void cosmos
folland has it all
great textbook
so now does the counterexample i wrote work with AoC?
@next obsidian but yo what if this product turns into a direct sum
or nvm
dusty verge
I'm very advanced <:pandaScreams:513070256550445083>
I was asking dansam
It's probably just some shit that pops up when you feed it to itself
charred crescent
I have a problem that I have been stuck on for a long time. it seems very simple but i can't seem to prove it:
Let R be a commutative ring with identity. Let a in R be nilpotent, (a^n = 0 for some positive integer n), and let u in R be a unit. Show that (u-a) is also a unit in R.
it also gives a hint, namely "first consider when u = 1"
but the hint hasn't helped me generate anything substantial
upper pivot
right so you know a^n =0, do you see a relation between 1-a and a^n?
charred crescent
let me think for a moment
nothing is jumping out at me
the only thing i can think of would be to condsider (1-a)^n
upper pivot
it is often useful to like, look at simpler cases. what about when n=2
upper pivot
yep
upper pivot
as i said, I thought of n=2 first and it was easy to figure out what to do from there :)
its a good idea to look at simpler cases usually
and you did!
charred crescent
hmm okay
upper pivot
i just pointed you to look at the simpler case
charred crescent
so in the most general case, we consider u^n - a^n = (u-a)(.....) = u^n, which is a unit?
i just want to make sure i understand
upper pivot
yeah that works
charred crescent
that is a lot simpler than i made it out to be
thank you for your help my friend
upper pivot
mhm
dusty verge
Let R be a commutative ring with identity. Let a in R be *nilpotent*, (a^n = 0 f...
That's what, from dummit and foote ring theory first set of questions? Edit: That's not quite the problem from D&F, but it's close - I was thinking of sectoin 7.1 problem 14 part C, which wants you to say that 1 + X is a unit for nilpotent X
dusty verge
so in the most general case, we consider u^n - a^n = (u-a)(.....) = u^n, which i...
What's those dots for? Edit nvm
solar shore
the number of endomorphisms of Z_n is n right...
tiny jolt
Suppose ord(g) does not divide |G|. Then g^{ord(g)} = g^{|G|} = 1, and try to derive a contradiction
cloud walrusBOT
Kevin Yang
tiny jolt
Oh, I wasn't sure what version of Lagrange's theorem you were asking
warm ember
oops sorry its this
warm ember
ye
lapis trail
If H is a subgroup of G you first need the fact you can partition the group into cosets of equal size
lapis trail
Oh
tiny jolt
Do you know subgroups have order dividing the main group?
lapis trail
Do you know about Cyclic subgroups?
warm ember
no oops
ill just skip this
it suffices to prove that a primitive root exists
right
does proof for FLT generalize here
tiny jolt
I think the proof of FLT results from this
warm ember
theres an easy proof of FLT
using basic nt
wait this is literally just the same as FLT
im dumb
lusty marlin
How would you prove Fermat's Last Theorem using this?
warm ember
fermat little theorem
lusty marlin
Lol I was just messing around, sorry to bother.
warm ember
no worries lol
lapis trail
@warm ember try first to prove, if G abelian and has order n, that (a_1a_2...a_(n-1))(a_1a_2...a_(n-1))=1
Then use the fact that a_j*a_k=a_1 for some unique pair j,k
alpine remnant
Jeez
I'm still stuck on a question
anybody know advanced abstract algebra?
Like
I'm pretty sure there's no way to make an affine space where the vector space has a higher dimension than the set of points over field of coefficients
Like
if we're dealing with 2D points, 2D vectors should be joining them
etc.
But exercise is asking me if I can have an affine space where I have a field K as the set of points, and. the vector space is of dimension two over K
I know I can't for finite ones
but idk for infinite ones
dire siren
Wasn't this supposed to be the identity on A_i? It would make so much more sense considering the particular case i=j in the first condition.
And pretty much the same issue on Wikipedia. Maybe they meant "the identity on A_i"?
dim widget
I'm pretty sure there's no way to make an affine space where the vector space ha...
Depending on what answer you want: here's a dumb example. If you scroll up you'll see someone giving a proof that $R \cong R^2$ as abelian groups (you can check that they're even $\cong$ as rational vector spaces). Let $\alpha$ be this isomorphism, then $f(x, y) = (\alpha(x) - \alpha(y))$ gives an "affine space" which is bijective in $x$ for any fixed $y$.
formal ermine
why should it not be the identity on A_i
formal ermine
why should it not be the identity on A_i
@dire siren
dire siren
@formal ermine I was saying that I think it should be the identity on A_i instead of what is written in the photos.
the identity function rather than the identity element
formal ermine
1 is the identity function here
dire siren
1 is the identity function here
then it makes sense; but I was confused when I saw "identity of A_i" on Wikipedia
maybe it's a typo
formal ermine
yeah
dire siren
plus I haven't seen before the identity function to be denoted by 1
thanks for confirmation
formal ermine
plus I haven't seen before the identity function to be denoted by 1
maybe you've seen the identity matrix denoted by 1
dire siren
yes
pliant forge
Let $\Lambda$ be a $R$-algebra with $R$ commutative artinian ring and $\Lambda$ finitely generated $R$-module. We know that for any $A \in \text{mod} \Lambda$ that $A$ is a left $\text{End}{\Lambda}(A)$-module by definining $f \cdot a = f(a)$ for $f \in \text{End}{\Lambda}(A)$ and $a \in A$. We also have that for $\Gamma := \text{End}{\Lambda}(A)^{op}$, which is also a $R$-module with the same properties as $\Lambda$, we have $A$ a right $\Gamma$-module. Now we want to show that $A$ is a bi-module, that is we want to show $(\lambda a) \gamma = \lambda (a \gamma)$ for $a \in A, \lambda \in \text{End}{\Lambda}(A)$ and $\gamma \in \Gamma$. I end up with $\gamma(\lambda (a)) = \lambda( \gamma (a))$ if i havent made any mistake.
summer path
Wrong channel and also wrong server
severe willow
sorry about that im not finding any servers related to this question
lethal dune
try to find the homotopy equivalent circuit of this then use van kampen
dim widget
Let $\Lambda$ be a $R$-algebra with $R$ commutative artinian ring and $\Lambda$ ...
I think you're trying to prove something that is not true, also seemingly this has nothing to do with R being Artinian. In general if R is a commutative ring and M is a module then M is either an End_R(M)-R bimodule or an R-End_R(M)^{op} bimodule. It's pretty unlikely that it will end up being an End_R(M)-End_R(M)^{op} bimodule in a natural way (think of R^2 as a module over R, it is not a bimodule for M_2(R) and its opposite because the natural "left and right actions" don't commute). On the other hand End_R(M) is an End_R(M)-End_R(M)^{op} bimodule because matrix multiplication is associative. Maybe you should explain what you think the End_R(M)^{op} action is in your setting?
pliant forge
I think you're trying to prove something that is not true, also seemingly this h...
this is all the information ive been given
dim widget
this is all the information ive been given
Yes this is different from what you wrote.
pliant forge
i wouldve just assumed the op meant f\cdot_{op}g = gf
dim widget
i wouldve just assumed the op meant f\cdot_{op}g = gf
I don't know why you would write it that way.
But anyway they want you to show that if A is an S module for S a ring, then an S-module M is an S-End_S(M)^{op} bimodule.
It seems to have nothing to do with R really or any Artinian conditions for the moment.
pliant forge
yeah im not saying the artinian fact because i found it useful. it was just given
because thats what an artin R algebra is
pliant forge
I don't know why you would write it that way.
emphasize that the operation is different, switching the order for the original one
dim widget
Yes but you confused yourself about what they were trying to prove. The left action is just by scalars in \Lambda (= S in my notation), then the right action is by those matrices which commute with the scalars in \Lambda
pliant forge
Yes but you confused yourself about what they were trying to prove. The left act...
anyways ye ic
my bad
alpine remnant
Depending on what answer you want: here's a dumb example. If you scroll up you'l...
This was very helpful. Thank you.
dim widget
This was very helpful. Thank you.
No worries! I'm not sure about answering this for other fields in general, but this shows that "most of the time" it is possible
alpine remnant
No worries! I'm not sure about answering this for other fields in general, but t...
Well, I know K can't be isomorphic to KxK for finite K, because cardinalities don't check out, but if we keep axiom of choice, K and KxK are isomorphic for all infinite cardinals
so that settles it!
dim widget
Well, I know K can't be isomorphic to KxK for finite K, because cardinalities do...
That's not true! They're just in bijection. For Q and Q^2 for instance it is an easy exercise to see that they are not isomorphic.
(as abelian groups)
Having that isomorphism of abelian groups is an important part of my example.