ty ill have a think on this topic
#groups-rings-fields
coral shale
prisma ibex
yet another example why people who reject choice are insane
proud bear
nLab 
woven delta
You can do a lot of good math by studying things which are weaker than choice
coral shale
non-free vector spaces 
woven delta
Not quite what I'm talking about
You should be able to get analogous results for subvector spaces of bounded complexity (in the sense of Computability) to be free with a basis of bounded complexity
So you can do math with weaker stuff and you get weaker results, but how weak they are is related to the strength of your replacement for choice
umbral plume
why would you want weaker results, though?
vestal snow
Can someone clarify some category theory stuff for me?
In arbitrary categories, a functor F is left exact if and only if it commutes with finite limits (by definition)
For abelian categories, this is equivalent to showing that F commutes with taking kernels and direct sums
This then implies that given a SEQ 0 -> A -> B -> C -> 0, we have the exact sequence 0 -> F(A) -> F(B) -> F(C)
My question is, lets say F is a functor between categories satisfying the property that 0 -> A -> B -> C -> 0 being exact implies 0 -> F(A) -> F(B) -> F(C) is exact
Does that then imply that F is left exact (it clearly would commute with taking kernels, but I am not so sure about the direct sum)?
I feel like this should be true as (left) exactness is often defined as: F is said the be left exact if 0 -> A -> B -> C -> 0 being exact implies 0 -> F(A) -> F(B) -> F(C) is exact
next obsidian
Does that then imply that F is left exact (it clearly would commute with taking ...
this is the actual definition of left exact
at least in every single treatment of exactness I've seen
I would say that the other definition is what you show is equivalent
vestal snow
I can see why the definition involving limits would be preferred since it generalizes to all categories
But let's try something that might be easier first
Say F preserves SES, then can we conclude that F preserves direct sums?
I don't see how this is the case
Even finite direct sums
We would need some sort of projective condition on all but one of the summands
Ah I think I get why the direct sum property is neccessary
woven delta
If a short exact sequence splits and has a section, can't you push the section to the F short exact sequence
vestal snow
It's probably there to ensure that F takes 0 to 0
Yeah you can
But general SES don't have to split
cloud walrusBOT
Emma
woven delta
Yep
vestal snow
But I think the sum being infinite might mess with that? Same with F being just left exact
woven delta
Yes, definitely
vestal snow
Okay I found this
Since the corollary is not and if and only if, I don't think that the definition that Chmonkey gave is equivalent to mine
It is certainly implied by mine though
vestal snow
You mean stronger?
Something that Chmonkey would call left exact would also need to satisfy the direct sum thing for me to call it left exact
woven delta
Oh I see now
I think you're a little off
One sec
Yeah I was being silly above
About it being stronger, I misread something
I think your definitions are equivalent actually
vestal snow
Oh?
woven delta
One sec
vestal snow
Take your time
woven delta
Okay in the case of a section existing we know that a left exact functor will preserve the entire exact sequence
$0\to A \to A\oplus B \to B\to 0$
cloud walrusBOT
Emma
woven delta
If $i:B\to A\oplus B$ is a section then pushing it forward should still give you a section
cloud walrusBOT
Emma
woven delta
Which should give you a map from F(B) to F(A\oplus B), which shows that the map F(A\oplus B) \to B is surjective
vestal snow
Ok yeah
woven delta
So it's exploiting the definition of products in terms of sections
woven delta
Yeah that's not going to happen
vestal snow
I see
woven delta
When they say direct sum in the definition, they just mean applying it as a binary operation
Not arbitrary direct sums
vestal snow
Are you sure?
They are very particular about specifying finiteness conditions
For example, they use finite limits/colimits
woven delta
They are specifying it by just saying that you preserve the operation A\oplus B
As a binary operation
You are interpreting it as indexed direct sums
woven delta
👍
safe thistle
hello
anyone know how specifically how much set theory will be necessary?
and if it might be in that pdf
@magic owl
vestal snow
That's probably overkill
You should know what all of those symbols mean
And know basic arguments like how to prove two sets are the same etc
safe thistle
ah
agile pine
yeah that pdf is definitely overkill for dummit & foote
vestal snow
ah
I know this wasn't your question, but I have other books for algebra that I think are better
safe thistle
i’m fine with overkill
safe thistle
I know this wasn't your question, but I have other books for algebra that I thin...
shoot
vestal snow
Aluffi's Algebra Chapter 0
That one even has a short intro to set theory and category theory
I can dm you the pdf if you would like
safe thistle
i think it should be on libgen
vestal snow
Yeah it is
safe thistle
“suitable for a first sequence on the subject at the beginning graduate or upper undergraduate level”
what are prerequisites?
vestal snow
I don't think any prerquisites except "mathematical maturity"
I.E. you can write and follow mathematical arguments
safe thistle
I.e. baby rudin?
vestal snow
I.e. baby rudin?
I don't understand what you're trying to say
vestal snow
If you mean that you have done baby rudin and you are asking if that is enough of a prereq, then the answe is yes
Even though I have never touched that book cause its on real analysis
agile pine
https://www.math.toronto.edu/weiss/set_theory.pdf
i mean your uoft pdf here is also supposedly meant for a grad course so if you were thinking about diving into that then you should be alright for aluffi too
it is pretty fun
vestal snow
My problem with Dummit and Foote is that it is really dry and a lot of the arguments in it can be simplified a lot using basic category theory
safe thistle
Aluffi's Algebra Chapter 0
classes wise what does this cover?
vestal snow
This class should prepare you for an algebra qual and then some
quals are exams that phd students take during their first few years
The book is not dense at all
It has like 800 pages
agile pine
I can dm you the pdf if you would like
would you mind sending me a copy - the category lens intrigues me lol
vestal snow
Easy or difficult depends on the individual so i can't really say
I know that I would find baby rudin to be much harder than Aluffi
The content is fairly in depth
The only topic that I feel like is not explored much is infinite galois theory
Post them here?
Look on MSE
safe thistle
ah
safe thistle
quals are exams that phd students take during their first few years
this would be rather good to know
vestal snow
Yeah. At most universities, you can take the qual as an undergrad too and if you get a high pass that looks really good on your grad school application
safe thistle
are you getting a phd?
vestal snow
It would be pretty cool if qual results transferred, but I would be very surprised if they did
Eventually
right now I'm a junior
safe thistle
i assume a college junior?
agile pine
Not at UW
which UW are you referring to, I go to a UW but its probably a different one lol
vestal snow
yeah
agile pine
vestal snow
Can*da
chilly ocean
Why is it that if I have a center of index n, I cannot have a conjugacy class with more than n elements?
Does it have to do with the cosets of the center of the group?
woven delta
Let G act on the cosets of Z(G) by congugation. Then the stabilizer of each x is going to contain the center, so the stabilizer has index dividing n. By the orbit stabilizer theorem each orbit will be of size <=n. You can then show that conjugacy classes in G get mapped bijectively to orbits in G/Z(G) under the quotient map
chilly ocean
Thank you
woven delta
Or maybe a nicer way of doing this would be
Since Z(G) is normal in G, every element in G can be written as mg_i where g_i is a representative of the ith coset and m is in Z(G). Then if you take an element x of G and look at all the possible congugations, you have (mg_i)^{-1}x(mg_i)=g_i^{-1}xg_i, so any congugacy class can be determined by choosing one representative of each of the n cosets and just congugating by that @chilly ocean
chilly ocean
Yeah, I need to spend more time learning to use group actions and getting comfortable with cosets, but thank you!
wanton moat
lethal dune
What are the [x, y]? Like what are those objects
the equivalent class containing (x, y)
terse crystal
G is the subgroup generated by all elements of the form 1) or 2) or 3)
Subgroup of abelian group is normal so you can take quotient
barren sierra
Right but [x, y] is an equivalence class of Z[M, N]/G where G has elements of these 3 forms
But those 3 forms are defined by equivalence classes themselves which seems circular
next obsidian
[x,y] is not equialence classes
Z[M x N] is a free Z-module
with basis element of the form [x,y] with x in M, y in N
Z[M x N] are formal sums of the elements [x,y]
heavy dagger
Let R be a commutative ring. Prove that N = {all x in R | x^k=0 for some natural number k} is a subring of R.
The above statement uses the defn of a ring without 1 right?
And it is convention to define a^0 as 1 isn't it?
Where a is an element in the ring
delicate orchid
yeah I think this is a ring without 1
units can't be nilpotent and 1 is clearly a unit (unless the ring is trivial)
south patrol
It is an ideal though, the nilradical, if ur interested tho
heavy dagger
Yeah I can see that
coral shale
units can't be nilpotent and 1 is clearly a unit (unless the ring is trivial)
trivial ring = {0} right? 0=1 is a unit in this case too, no?
iron vessel
Determine the degrees of the irreducible factors of $f = X^{11} -1$ in $\mathbb{F}5$ and $\mathbb{F}{25}$. How do I solve this?
cloud walrusBOT
Évariste Galois
iron vessel
Is there a program I can use to find the answer?
coral shale
🤷♂️
iron vessel
I tried using sage but for some reason I cant manage to factor it over a finite field
it just factors in Q
rustic crown
So you find like smallest n sich that 5^n = 1 mod 11
coral shale
iron vessel
hmm ok
coral shale
im confused
delicate bloom
I tried using sage but for some reason I cant manage to factor it over a finite ...
you have to make a kind of finite field object and make your polynomial an element of that I believe
coral shale
So you find like smallest n sich that 5^n = 1 mod 11
why is this helpful???
iron vessel
you have to make a kind of finite field object and make your polynomial an eleme...
I tried to, but when i do f in R where R is my finite field it says false
rustic crown
sowwy my internet wasn't being nice to me >.<
iron vessel
im using the same variable too
delicate bloom
idk sounds like a programming problem at this point to search through to work it out
if I had time I'd look into it more
rustic crown
So you find like smallest n sich that 5^n = 1 mod 11
if this happens then you know that x^11 - 1 divides x^(5^n) - x, and we understand how the latter thing factors nicely
rustic crown
the factorization of x^(q^n) - x over F_q is the product of all monic irreducibles of degree d dividing n
delicate bloom
interesting, the program I used only factored it into 3 things of deg 1, 5, 5
delicate bloom
I use pari gp
iron vessel
ah ok let me check it out
coral shale
I... have yet to see this in my course I suppose. Next section I think
iron vessel
damn i was really hoping to get that right, in all fairness it was so long i had like 3 minutes to do it
rustic crown
the factorization of x^(q^n) - x over F_q is the product of all monic irreducibl...
so if you assume this, then
you see that x^11 - 1 would factor into irreducibles with degree dividing n = 5
iron vessel
but the rest went very good
iron vessel
so if you assume this, then
you see that x^11 - 1 would factor into irreducible...
I see, I would not have come up with it on the spot
delicate bloom
here's how you can do it for the F_5 case in pari gp, the second thing is just lifting it to clean it up
iron vessel
Thanks!
delicate bloom
for the higher finite fields k>1 you gotta construct it a little differently
cause it's not mod 25 obviously
iron vessel
ok ill check it out
wait it isnt working, It open terminal but I cant type anything lol
delicate bloom
idk can't help you there
also I should mention the pari library is part of sagemath btw
iron vessel
ah
delicate bloom
so anything you can do in gp you can do in sagemath
but I kind of hate fumbling around with sagemath sometimes
iron vessel
alright, thanks 🙂
jade kite
Anyone know the formulas for quadratic/exponential/logistic regressions to be done by hand?
dull root
Question: If G is a group (can be infinite) with an abelian subgroup A such that |G:A| = 5, how to see that G is not simple. The 5 sticks out like a sore thumb and immediately suggests that we look at the action of G on the cosets of A, so we see |G:ker| divides 5!, but what can we deduce from this?
It seems if |G| > 5!, we already see this kernel is nontrivial, so G is not simple, but the issue is if G is small
delicate orchid
how would it generate N if it wasn't in it
rapid bramble
Why g(x) belongs to N
you need g(x) to be in N, for the ideal generated by it to be N
chilly ocean
Why..
delicate orchid
ok that kinda makes me think you don't really get what "generated by" means, no offense
can you recall what it means for an ideal to be generated by some elements?
chilly ocean
I know what generate by means..
delicate orchid
ok then it should be trivial
chilly ocean
<a> includes a1=a
worthy haven
$\langle g(x) \rangle$ is by definition the smallest ideal contaning $g(x)$, so if $N = \langle g(x) \rangle$ then $N$ contains $g(x)$ by definition
cloud walrusBOT
Nick
woven delta
Question: If G is a group (can be infinite) with an abelian subgroup A such that...
Being simple means that any kernel must be G or 0. This homomorphism from G to S_5 will have kernel G only if A is normal
worthy haven
<a> includes a1=a
so then what's your confusion?
woven delta
Oh one sec, maybe what I'm saying is way too strong
Yeah it is
Because you can't act by congugation in general
dull root
I have a related question to this. If I know A is an abelian subgroup of G with order A | 4!, and 5 divides |G|, with |G:A| =5, is it true that G = C x A where C is a cyclic group of order 5?
woven delta
Question: If G is a group (can be infinite) with an abelian subgroup A such that...
Why can we see that the kernel is nontrivial?
What action are you using? You can't do action by congugation unless your subgroup is normal
dull root
What action are you using? You can't do action by congugation unless your subgro...
By right action
woven delta
And where does the abelian come in?
dull root
We do not need abelian when |G|>5!. We consider the action of G on the cosets of A
Let phi: G \to S_5 be right action of G on the (five) cosets of A since G:A = 5
dull root
We can consider them as right cosets. Well if we want to work with left cosets, lets just use the left action
but the idea is the same
dull root
So G/ker(phi) is isomorphic to the image which is a subgroup of S_5
if |G| >5!, this kernel >1
woven delta
But why can't the kernel be G
dull root
because the kernel left action on a proper subgroup can not be all of G
dull root
np, so the only case that needs to be dealt with is if G is small, that is |G| < 5!, and I think that A abelian must play a part in this case
since we did not use it before
Certainly, we know 5 divides G, so A is an abelian subgroup of G where A | 4!
and G | 5!
Seems like some Sylow trickery is going on here
if we had unique sylow groups we are good
woven delta
Okay let's do some Sylow then
dull root
Hmm, what if we loosened the statement, to G is not necessaily nonsimple, but there exists a normal subgroup (\neq 1) in G
woven delta
You can just say that A is not 0 or G
To avoid that case
Which is a very reasonable assumption
dull root
Yea, lets go with that
woven delta
Hmm, what if we loosened the statement, to G is not necessaily nonsimple, but th...
Also this is the definition of nonsimple
dull root
Simple groups have no nontrivial normal subgroups, where nontrivial means not 1 or the whole group right?
here we allow the whole group
woven delta
Well you wouldn't want that
woven delta
Yes
Okay so a counting argument should work
Let's look at the factors of 4! That are 1 mod 5
dull root
Wait, just knowing a sylow 5 group is cyclic (thus abelian) is not enough to give normality right?
dull root
So you're looking at the factors of 4! mod 5 to see if this sylow 5 group is unique?
woven delta
The only one that's 1 mod 5 is 6
Yeah
So if there isn't 1 Sylow 5 group there are 6 of them
dull root
yes
dull root
wait i don't following that? I know the sylow 5 groups are all isomorphic
woven delta
So they are all prime cyclic
dull root
so we have 6 sylow 5 groups
woven delta
So every non identity element generates
Therefore there can't be any nontrivial intersection between the Sylow 5 groups
dull root
So that means if two sylow 5 groups share a nonidentity element, those two groups must be the same
woven delta
Yes
woven delta
This is a pretty standard technique for Sylow arguments
So it's good to see it
Okay so we have exactly 24 elements of order 5
dull root
So none of those eleemnts can be in A, since A|4!
woven delta
Yes
Okay I think A being abelian is pretty interesting actually
Let's think about that
Okay so fix an element g of order 5. Then every element in G is of the form (g^i a) for some a in A
Then let's say you have a Sylow 3 subgroup in A. Then congugating by (g^i a) will be the same as congugating by g^i
Because A is abelian
dull root
Because G:A = 5, so A, gA, g^2 A , g^3 A ..., g^5 A gives you the parition of G?
woven delta
Yes
Okay so we have only 5 possible congugations of the Sylow 3 subgroup
So that's pretty interesting
I think that should give you n_3 is 1
Because 5 is not 1 mod 3
Yeah that does it
You don't need to do the counting argument
dull root
What it can happen that A does not have a sylow 3 group
dull root
What if |A| = 2* 4?
woven delta
Sorry I already discounted that case
Because we know n_5 has to be 6
Because no other option is 1 mod 5
I did that in my head but didn't say it out loud lol
dull root
yes, so the abelianness of A is only used here: Then let's say you have a Sylow 3 subgroup in A. Then congugating by (g^i a) will be the same as congugating by g^i
Because A is abelian
woven delta
Yes
dull root
So I see that, buthow does this imply that the Sylow 3 group in A has exactly 5 congjugatons
But if we already know that 3|A in our case, and A|4!, then the only things =1 mod 3 are 1, right, so isn't this sylow 3 group unique
woven delta
So this definitely shows that there are at most 5 congugations
dull root
without any call to the abelianess of A
dull root
nvm 4 = 1 mod 3
woven delta
You also need the 5
You have to look for 1 mod 3 in 5!
Because obviously there's only 1 Sylow 3 subgroup in A
But there can be other copies of A that you find by congugating A
But yeah also 4
Okay anyway so we know there are at most 5 possible congugations, and if g^i (for i not 0) fixes your Sylow 3 subgroup under congugation, then every g^i does
So therefore either every g^i fixes your Sylow 3 subgroup (ie it is normal) or there are 5 distinct ones
dull root
But that is only true if the sylow 3 subgroup of A is unique right?
woven delta
Yeah but that's true because abelian
Hmm
Maybe you can do this easier
Yeah I think you can state this a lot easier than I just did
Do you get my argument though?
dull root
Oh yes, because A is abelian, every subgroup is normal in A, so sylow 3 group is normal in A, it is unique
because all sylow groups are obtained under conjugation
dull root
Why does it not immediately imply this sylow 3 group is also normal in G?
We know it is normal in A
woven delta
Well my argument uses that to show that it's also normal in G
But you need to use something more than just being normal in an abelian subgroup
dull root
So you argue that g^i either fixes this sylow 3 group. then this sylow 3 is normal in G, I see this. However I still am having trouble with the 5 disticnt sylow 3 group case
woven delta
I think maybe there's a theorem saying that if A a subgroup of G of index k, then n_p for G divides k times n_p for A
That would be nice
dull root
hmm perhaps. I think these ought to be a simpler elementary reason tho xD
woven delta
Sure
Okay let Z/5 act on the Sylow 3 subgroups by congugation
Then you are done
Because there can't be 5 of them
And otherwise Z/5 must be the trivial action
I think this is a nicer way of saying it
dull root
Hmm, why can't there be 5 conjugates?
woven delta
Because n_3 is 1 mod 3
dull root
yes this is good, i see it now
woven delta
Okay cool
dull root
ty
lethal cipher
Okay, I am trying to work on part (c), and I have no idea where to go with this. My prof says we can easily find the rest of the roots using a^3=t, but I've been trying and I am extremely stuck
...nevermind. I was trying to dig for another root, but alpha was the only root -_-
pastel cliff
There is a homomorphism $\mathbb{I}_m \to \mathbb{Z}$ defined by $[a] \mapsto a$ - does this statement look true? i think it is but idk
delicate bloom
since |A| = sup |Ax|/|x| then it's bigger than all x, so you can write that as an inequality |A| >= |Ax|/|x|
pastel cliff
ignore the stupid notation I_m is just integers mod m my textbook is weird
delicate bloom
so if you rewrite it like |Ax| <= |A| |x| that's how they apply it, twice
cloud walrusBOT
μ₂ (46/47 🪲)
terse crystal
Question: If G is a group (can be infinite) with an abelian subgroup A such that...
A is an abelian subgroup , abelian subgroup is always normal
pastel cliff
is it normal to say "cosets of a subgroup"
or should is the of a subgrp implied
It follows naturally that the cosets of any subgroup of an abelian group are also abelian.this is my sentence
damn chillgebra dead today
next obsidian
You should say subrgroup
But also like wdym?
A coset unless the trivial coset isn’t a group
So they can’t be abelian
coral shale
All subgroups of an abelian group are normal
You can quotient by any subgroup
The resulting quotient group may or may not be abelian, prove it 
next obsidian
If you mean like the set of cosets (aka the quotient)
That’s different than what you said
next obsidian
Yeah I don’t know what a coset being abelian means
coral shale
$\trianglelefteq$
cloud walrusBOT
coral shale
Use this symbol to denote normal subgroup
pastel cliff
oop
pastel cliff
yeah it's mistaken i forget cosets aren't necessarily groups, it's the set of cosets i should be thinking about
i'll rephrase one sec
coral shale
the set of cosets are precisely what make up the quotient group if these are cosets of a normal subgroup
Just say G/H is abelian.
pastel cliff
isn't that kinda just restating the claim tho
coral shale
huh?
Well yeah, you write that at the end of your proof
I don't get it actually
"It follows......." I do not see how it follows
pastel cliff
it just does
coral shale
or I don't know what you want to follow
umbral plume
it just does<:monkagigagun:651816940754436107>
proof by obviousness 👍
pastel cliff
okok one sec
actually this should be enough right
the first equality is true bc H is a normal subgroup
coral shale
ok.
coral shale
If you're going to invest words in this proof, I would invest them in the reasoning behind each equality
not in that introductory paragraph.
Let G abelian. Let H =< G.
Then H is abelian, so H =<| G.Since G abelian, H =<| G. Then forall x, y in G ...
You can write proper sentences unlike me, but you can cut down on a lot of words there
coral shale
Technically, for that exact claim
You don't need to start with an arbritrary subgroup H
You should instead start with G abelian, G/H a quotient group with H normal.
next obsidian
> Let G abelian. Let H =< G. ~~Then H is abelian, so H =<| G.~~ Since G abelian,...
You shouldn’t say “so” here
Because H being normal doesn’t follow from H being abelian, it follows from G being abelian
umbral plume
there's also $\forall$ which actually looks so cool
cloud walrusBOT
valley
umbral plume
the symbol deserves to be used tho
umbral plume
noo it flows so much better than "forall"
next obsidian
It is universally considered bad form to use the symbolic for all and there exists
Unless you’re writing a first order sentence
Or like doing logic
coral shale
kek
coral shale
I know.
But in my opinion, bad use of english leads to ambiguity
It's slightly harder to use those symbols incorrectly (but it can happen, ig)
next obsidian
Replacing the word “for all” in a sentence with a symbol doesn’t change that at all
delicate bloom
fetishizing mathematical symbols is cringe don't do it
coral shale
That is true.
next obsidian
If you write an entire statement symbolically for a logic course
Then it is appropriate
But if your sentence was ambiguous with the word “for all” it’s ambiguous with \forall
coral shale
But I see people writing statement forall ....
ie. they put the quantifier at the end
then somewhere along the lines they write
quantifier statement quantifier
sturdy marsh
fetishizing mathematical symbols is cringe don't do it
$\varpi$
cloud walrusBOT
Brofibration
coral shale
(probably referring to epsilon delta statements)
and I go wtf does this even mean
How do I put it - it isn't clear in words this is problematic. Gramatically it might make sense.
In symbols, at least for me, I was taught to put the quantifiers at the start . . .
Yh like I have seen this appear in my notes and I think it is terrible. How to phrase statements unambiguously... or how to use symbols properly... either of these things need to be enforced, and to me, the 2nd seems more natural to teach.
chilly ocean
given f*:Spec B_q -> Spec A_p is f*(Spec(B_q))=V(B-q) = B-V(q)?
im trying to do atiyah macdonald chap 5 exercise 10
languid walrus
did you mean to write f*(Spec(B_q)) = A-V(p)?
tawny pine
and I go wtf does this even mean
thats more an issue of reading comprehension than an issue of whether we should use symbols over english
many times for ease of communication we state a property before quantifying the variables involved
insisting on always using symbols puts needless rigidity on communication, and is harsh on many eyes
novel loom
Hmm...
Since the rank is 2, there are two linearly independent rows.
However, note that it doesn't say which two rows.
So, consider the 3x3 identity matrix I.
Let I' denote the matrix obtained by deleting the top-most 1 in I.
Let I'' denote the matrix obtained by deleting the bottom-most 1 in I.
Both I' and I'' have rank 2.
However, is the rank of a linear combination of I' and I'' necessarily 2?
Heck, there's an even simpler example:
Let M have rank 2.
If the set of all such matrices forms a vector space, it should contain a zero element.
That is, multiplying any element of the space by 0 should give the zero elements of the space.
However, 0 x M is the zero matrix.
And that has rank 0.
So, the answer is (d), none, because the set of all rank 2 matrices of dimension 3 x 3 fails to be a vector space.
sleek wagon
However, is the rank of a linear combination of **I**' and **I**'' necessarily 2...
yes? There are two linearly independent vectors remaining so the rank is supposed to be 2?
novel loom
yes? There are two linearly independent vectors remaining so the rank is suppose...
No.
Add them together.
But that's a non-issue.
sleek wagon
ok questions
novel loom
The simplest explanation is that the zero matrix does not have rank 2.
Thus, we don't have a vector space.
sleek wagon
I've encountered this adding of matrices to check if its still in the right form...for instance
the solution says if we add two matrices of det 0 then it need not be a matrix with det 0
so its not a subspace
sleek wagon
but the definition of subspace is
- w1-w2 ∈ F
- ⍺w ∈ W
or simply
⍺w1+ beta w2 ∈ W
so why are we adding to check if its still the correct one? shouldnt we subtract?
novel loom
so why are we adding to check if its still the correct one? shouldnt we subtract...
Subtraction is just addition when beta is -1
sleek wagon
true that
sleek wagon
W has to contain 0
it was not mentioned in my reference book :/
novel loom
Well, it is technically implies by (2).
But it is usually important enough to be worth stating on its own. xD
Anyhow, that's how you solve the problem: the set of rank 2 matrices isn't closed under scalar multiplication.
Thus, it cannot be a vector space, and so it has no dimension.
Does that make sense?
sleek wagon
it does
novel loom
Excellent.
sleek wagon
However, is the rank of a linear combination of **I**' and **I**'' necessarily 2...
so if I add these two matrices I may get a matrix with rank 3
novel loom
Yes.
sleek wagon
interesting
novel loom
A simpler example: break the identity matrix into 3 matrices, each with a single 1.
The three individual matrices all have rank 1.
But their sum has rank 3.
sleek wagon
yes
sleek wagon
but the definition of subspace is
1. w1-w2 ∈ F
2. ⍺w ∈ W
or simply
⍺w1+ beta w2...
about this
just confirming that I can use these conditions to check if the given something is a vector space or not
cuz the definition is for subspace...a subset has to be a vector space itself to be a subspace so I suppose it can?
tawny pine
just confirming that I can use these conditions to check if the given something ...
yes as long as we're given a parent space (subsets inherit its operations)
sleek wagon
got it...thanks guys
tawny pine
since vector spaces must be nonempty (contains in particular the 0 vector), there should be a 3rd condition there, W is nonempty/contains 0
sleek wagon
okay ill add
I know that dimension of symmetric matrices is n(n+1)/2 and that of skew-symmetric matrices is n(n-1)/2
<@&286206848099549185>
woven delta
if A^T=-A and the diagonal is fixed under transpose then what does that mean about the diagonal?
sleek wagon
if A^T=-A and the diagonal is fixed under transpose then what does that mean abo...
not sure
it should be A^T=A since its symmetric?
woven delta
if A^T=-A and the diagonal is fixed under transpose then what does that mean abo...
think about this
kind temple
?
sleek wagon
think about this
trace is always same?
lethal dune
?
kind temple
,w sum (n-1-k) from k = 0 to n-1
cloud walrusBOT
kind temple
then add one
lethal dune
kind temple
noice
sleek wagon
,w sum (n-1-k) from k = 0 to n-1
can you explain?
kind temple
take an n by n matrix in your vector space
sleek wagon
ok..
kind temple
i’m going to construct a basis for this space.
let E(ij) be the matrix with a 1 in row i column j and a 1 in row h column i, zero everywhere else
a basis for this space is the set
{E(ij) : 1 <= i < j <= n } U {I_n}
show that this is a basis. try and understand why. then count how many basis elements there are
chilly ocean
did you mean to write f*(Spec(B_q)) = A-V(p)?
yeah
latent anvil
~~Say you have an integral domain A, a finitely generated ideal I = (f_1,...,f_n) of A[x], and a polynomial f of degree <= d. Then f in (f_1,...,f_n) iff f = g_1 f_1 +... + g_n f_n for polynomials g_i of degree <= d. Let P_k be the free A-module of polynomials of degree <= d and N the maximum degree of any f_i. Then the map T : P^n -> P_{dN} defined by T(g1,...,gn) = f1 g1 + ... + fn gn is linear, and f in (f1,...,fn) iff f in im T.
Is there a result in basic linear algebra that gives me a polynomial p in the coefficients of f (the "output vector") and the f_i (the coefficients of the matrix T) such that p = 0 iff f in im T?~~ just read my next post lol
So okay my question is actually: let A be an integral domain and say you have free modules V = A^n, W = A^m over A, a linear map T : V -> W, and a vector w in W. Is there a polynomial p in the entries of the matrix T and the coefficients of w such that p = 0 iff w in im T?
(p should be an integer coefficient polynomial depending only on the numbers n, m)
We can pass to the field of fractions of A (tensor everything with the field of fractions, im T is preserved bc tensor product right exact)
So now this is honest linear algebra
So now I'm wondering if you have vector spaces V = k^n and W = k^m, a matrix A : V -> W, and a vector w in W, is there a polynomial in the coefficients of w and A which vanishes iff w is in the image of A
Oh wait umm should it be nonvanishing? Either one is probably good
Oh wait yeah
Check if in column space
Is asking about linear dependence
Like let B be A but we adjoin w as a column
woven delta
I'm trying to figure out how to use determinant here
latent anvil
I got it one sec
Now w in im A = columnspace of A iff B has linearly dependent columns
Er
Wait no...
That assumes the columns of B are already linearly dependent
woven delta
what is B?
latent anvil
Oh shoot I reused the letter A
For the matrix and the ring
Oh but the ring is out of scope nvm
latent anvil
what is B?
B is A but we add in w as a column
latent anvil
if A starts out with column rank r
Then w in im A iff B still has column rank r
right?
latent anvil
Otherwise im A + span w would increase in dimension
So
you can test a bound on the rank by polynomials
But it's only an upper bound
The rank of B is the largest s such that B has an invertible s by s submatrix
So rank B <= r iff all minors of size r vanish?
(minor = determinant of submatrix to be clear)
The issue is like umm
How exactly does this work in A?
Like r sort of depends on A once we passed to the field of fractions
and I wanted a polynomial independent on the underlying ring
like the image of a matrix might not be a free module to start out
Sorry @woven delta lmk if I'm being confusing I've been thinking about this for a whild and I'm starting to get twisted up about things
dreamy geyser
Does somebody know the proof for 8 alpha?
woven delta
no this is fine
latent anvil
Does somebody know the proof for 8 alpha?
I think the associativity is kind of just a garbage computation
From what I remember
woven delta
I'm just processing the situation
latent anvil
Like it comes down to associativity of xor of boolean a
Which you can prove by a truth table
Does what I'm saying make sense?
dreamy geyser
I think the associativity is kind of just a garbage computation
Every time I do it, I get awfully close to proving it.
latent anvil
So think of this
dreamy geyser
Wanna see what I did?
latent anvil
Sure
kind temple
Does somebody know the proof for 8 alpha?
use a truth table, not symbols
latent anvil
,rotate
cloud walrusBOT
latent anvil
So @dreamy geyser I have a suggestion
Show this by elements
Not by set theory algebra
dreamy geyser
use a truth table, not symbols
but theres got to be a way to do it symbolically right? 😔
latent anvil
There definitely is
kind temple
yes but it’s gross, as you can see
latent anvil
Express everything in conjunctive normal form
kind temple
breh
latent anvil
I'm just saying there's a mechanical way to check this
Bc there's a normal form for propositions
So you can just reduce both sides
I'm not saying you should do this
Truth table is better
dreamy geyser
Express everything in conjunctive normal form
what does conjunctive normal form mean? (im new to abstract algebra and im self learning)
latent anvil
Yeah no worries
If you have propositions p, q, r
Like p = "x is in A"
and you combine them using the boolean operators ¬, and, or
You can write it as a bunch of "and"s of "or"s of the propositions p, q, r, ¬p, ¬q, ¬r
So x in A * (B * C) is equivalent to (x in A or x in B*C) and not (x in A and x in B*C)
Right?
kind temple
you could also convince yourself that x is in A * (B*C) iff it’s in an odd number of the sets A,B, or C (sry sham for interrupting)
latent anvil
I'm saying $A\ast (B\ast C) = (A \cup (B\ast C))\cap (A \cap (B\ast C)^c)$
cloud walrusBOT
Hom(-, Shamrock)
latent anvil
And then you can expand B * C the same way
You get some "boolean combination" of the sets A, B, C
using complements, intersection, and union
And then you write this as an intersection of unions of A, B, C or their complements
You do this to both A * (B * C) and (A * B) * C and you'll get the same thing up to commutativity of and and or
latent anvil
you could also convince yourself that x is in A \* (B\*C) iff it’s in an odd num...
Np. Also I'm not endorsing the solution I'm giving, a truth table or what you said here is better
(also a truth table is implicitly giving you a disjunctive normal form)
kind temple
was just about to say that lol
woven delta
pick a spanning set for the image of A as a function of the coefficients of A (just the columns). for a fixed w you take rank(A)-1 many of the column vectors, throw them into a matrix with w, and take the determinant. add up all the determinants squared
for each subset of the spanning set of size r-1
latent anvil
What would squaring the determinants do?
latent anvil
Okay so it might be good for me to explain why I was this lol
woven delta
What would squaring the determinants do?
to avoid negative entries in the sum
latent anvil
to avoid negative entries in the sum
We're over an arbitrary field (or integral domain) though
woven delta
oh oof
yeah that's silly
there are ways to do a similar thing with irreducible polynomials
but idk how general that is
latent anvil
So I have an integral domain A and a finitely generated ideal I = (f1,...,fn) and an element f of A[x] not in I. I want to find an element a in A such that for any map φ : A -> K into an algebraically close field, φ(a) ≠ 0 implies φ(f) isn't in the ideal (φ(f1),...,φ(fn))
latent anvil
~~Say you have an integral domain `A`, a finitely generated ideal `I = (f_1,...,...
I thought you could take a to be a polynomial dependent on that linear map I described here
And then since it's a polynomial, φ(that polynomial) = that polynomial applied to φ(f) and φ(fi)
so if I can get a polynomial that measures whether f in (f1,...,fn) I'm golden
but it needs to work both in A and K without knowing anything about K
So I can't really compute the rank
Which is awkward
Although I guess the map φ can only drop the rank the matrix whose columns are the coefficients of the fi?
That's not even true lol
I think I figured out how to state this? For fixed n, m, is there an integer polynomial p in variables x_{i, j} for 1 <= i <= n and 1 <= j <= m and y_j for 1 <= j <= m such that for every integral domain A, matrix T : A^n -> A^m, and vector w in A^m we have w in im T implies p(x_{i, j} = ijth coeff of T, y_j = jth coeff of w) = 0
I think that's a clearer statement of what I want
Now if p works for all fields A it also works for all integral domains
so it's okay to assume A is a field
but p can't depend on the rank of T
By how quantifiers work
woven delta
okay if we have a irreducible polynomial $p(x)=a_nxx^n+a_{n-1}x^{n-1}+...+a_0$ and you homogenize via $F(x,y)=y^np(x/y)$ and plug in two polynomials f and g for x and y, then you get $F(f,g)$ is 0 iff both f and g are 0
cloud walrusBOT
Emma
woven delta
that's something that may help in terms of what I was thinking of before
latent anvil
This is true, but the field might be algebraically closed, in which case F(x, y) = x-y
woven delta
yes
woven delta
yeah but it works in every other case
latent anvil
I'm confused about where F comes in
woven delta
it's homogenizing p
latent anvil
Where does it come in for this problem?
latent anvil
Oh you want something which vanishes iff two given polynomials vanish?
woven delta
yeah
latent anvil
Yeah this is possible iff the base field is non algebraically closed
Ultra was giving this problem out earlier
Like a couple weeks ago
woven delta
yeah I know
latent anvil
Ah sorry
woven delta
no worries
woven delta
okay so sum over all minors and compute the determinant of each minor. If that determinant is nonzero and all the minors of higher dimension have determinant 0 then we have found a basis for the image. (think about how you would do this) then we take such a basis, add in w, and look at the determinants of all the minors of the size of the basis
the thing I'm designing here is very unwieldy and I'm relying on some things we may not be able to do. However hopefully I think you can find the rank of the matrix inside the polynomial to some extent
I guess if I could build logical symbols inside polynomials that would help with this approach
woven delta
yeah
woven delta
right
woven delta
yeah I think so too
latent anvil
And we should allow ourselves to use the rank of the matrix in the original integral domain
(ie its rank after tensoring with the field of fractions)
Okay so
We have an integral domain A
And a linear map T : A^n -> A^m
And a vector w in A^m
woven delta
okay so I think in this case my plan goes through
latent anvil
I want a polynomial p in A[xij, yj] such that for any field K with φ : A -> K, we have φ(f) (×) 1_K in im(T (×) id_K) iff p vanishes when we evaluate it at the coefficients of the matrix T (×) id_K : K^n -> K^m and w (×) 1_K in K^m
I think this is the appropriate level of generality
What plan?
woven delta
oh lol
latent anvil
Sorry haha
This is a really technical commutative algebra thing for chevalley's theorem (the version for finite type maps of noetherian schemes, so you can't prove it using quantifier elimination of ACF)
So I'm still figuring out what the appropriate lemma is
woven delta
yeah nevermind I was using the integral domain being fixed
and also that it isn't algebraically closed
latent anvil
Yeah sorry
woven delta
but in that case I'll outline the plan
latent anvil
So like, we have a rank r of A (ie the rank after tensoring with the field of fractions of A)
Sure
woven delta
so you take all the minors of the appropriate rank and take the determinant of them, which you use to indicate when you have a basis or not. Then you append w to the full column vectors from the minor, look at all the rank+1 minors and check to see if that determinant is 0
latent anvil
Ah yeah that makes sense
woven delta
and then you use the existence of a polynomial in 2 variables that is 0 iff all of it's inputs are 0
latent anvil
Yeah I'm starting to think maybe it's okay to have multiple polynomials
And AND them
But I'm still a little fuzzy
Yeah it's totally fine
woven delta
okay well great
latent anvil
So the issue with extending this to a field over A
Is whether the rank r of A changes
If it doesn't change we're fine, but I find that highly suspicious
Like 2 vanishes if we kill off 2
woven delta
right so before I was trying to encode that into the polynomial
woven delta
because I wanted to say that if you aren't of the highest rank so that every minor of higher rank has determinant 0 you should be 0
but maybe I don't need to do that
we in lattice mode
(because we have and and or but not negation)
lol
chilly ocean
could someone help me parse this problem? isnt <(1,1)> ={(1, 1), (2, 0), (3, 1), (4, 0)...}?
latent anvil
Right so take A = Z and T the matrix with columns [1 2] and [1 0]. Let K = F2. Then the rank of T drops from 2 to 1 as we push it under φ
latent anvil
could someone help me parse this problem? isnt <(1,1)> ={(1, 1), (2, 0), (3, 1),...
Don't forget (-n, n mod 2) also
Do you know what a coset is?
chilly ocean
I do but I find them really hard lol
havent really grasped the concept fully if I am being honest
woven delta
okay let's work some more on encoding the rank
is there a polynomial that can never be 1 and is 0 at 0?
oh right
fundamental theorem of algebra
latent anvil
havent really grasped the concept fully if I am being honest
So the question is saying we have a subgroup H = <(1,1)>, let's look at all the sets x + H for Z in Z×Z/2Z
Sometimes we'll have x + H = y + H but x ≠ y, eg x = (1,1) and y = (2,0)
latent anvil
so find a set of x-es such that x + H is never equal to any other x' + H (for x' in your chosen set not equal to x) but for any y in Z×Z/2Z there's an x with x + H = y + H
Does that make sense @chilly ocean ?
woven delta
okay no I got it
latent anvil
hype
woven delta
take x^2+x and x. Then x^2+x=x only when x=0 no matter what field you live in
chilly ocean
x is in Z I assume?
latent anvil
x is in Z I assume?
No, x is in Z×Z/2Z
woven delta
so you can use this to detect when something is nonzero
latent anvil
so you can use this to detect when something is nonzero
I'm not sure what you mean
woven delta
same
chilly ocean
omg I am an idiot lol, I forgot that ZxZ/2Z also has (1,1) and (1,0) so it makes sense
latent anvil
Okay so
Here's my original hint from Hartshorne:
Show that for any (noetherian) integral domain A and ideal I = (f1,...,fn) of A[x], for any polynomial f(x) not in I there's an element a in A such that for each map φ : A -> K into an algebraically closed field K with φ(a) ≠ 0 there's an element c of K such that φ(fi)(c) = 0 for all i but φ(f)(c) ≠ 0
So we can actually have a finite number of a's
Because we can take their product
a1...an ≠ 0 if ai ≠ 0 for all i
So start with all the nonzero minors of the matrix T
Know we can assume T still has rank r over K!
So we can a to be the product of all the nonzero minors of the matrix you get by adjoining the coefficients of f as a column vector to the matrix of the map T(g1,..,gn) = f1 g1 +... + fn gn from (polynomials of degree <= deg f) to (polynomials of degree <= (deg f)*(max_x deg fi))
Then if φ(a) is nonzero
We know the map T will still not have f in its image
Because if it did then the rank of the matrix where you adjoin f as a column vector would have to drop in rank after applying φ
woven delta
so I guess going back to the original setting is good
latent anvil
Yeah sorry
woven delta
because that let you disregard the rank issue
latent anvil
Yeah
woven delta
and we weren't doing anything special to fields
latent anvil
Right
Well I kind of am
I'm thinking of things happening in the field of fractions of A to justify the rank -> nonvanishing minor thing
And the codomain is a field
woven delta
So we can a to be the product of all the nonzero minors of the matrix you get by...
from (polynomials of degree <= deg f)
why is that the upper bound?
oh I think I see
latent anvil
You mean why does f in (f1,...,fn) imply f = g1 f1 +... + gn fn for polynomials gi of degree <= deg f?
woven delta
no actually I don't
latent anvil
Thinking
latent anvil
Yeah I thought I had an argument but it relied on all the gi fi having different degrees or something
hmm
woven delta
okay I'm going to sleep
woven delta
okay cool
latent anvil
I just need a bound on the deg gi
woven delta
have a good night
chilly ocean
@latent anvil would x = (0, 1)?
latent anvil
You're going to need more than one x
you want a set S such that for every y in Z×Z/2Z, there's a unique(!) x in S with y + H = x + H
Where H = <(1,1)>
So if you took S to be just {(0,1)} we would be able to find an x for y = (1,1)
Does that make sense?
chilly ocean
but doesnt <(1,1)> already give us half of ZxZ/2Z?
latent anvil
Well that suggests it should be in your set S, right?
so are you saying (1,1) + H = H and (0,1) + H are all the costs, and that they're distinct from eachother?
chilly ocean
wait
(1,1) + H != H
hmm
our goal is to construct ZxZ/2Z with the coset reps right?
chilly ocean
yeah but I still dont see how (0,0)+<(1,1)> and (0,1)+<(1,1)> dont give us ZxZ/2Z, what would be missing?
latent anvil
I think you're right, they do give you Z×Z/2Z
latent anvil
so are you saying (1,1) + H = H and (0,1) + H are all the costs, and that they'r...
This wasn't supposed to imply it's wrong, I just wanted to make sure it's what you were saying
Can you prove it though?
latent anvil
You do
Problems in higher math by default include "and prove your answer correct"
So what can you say about an element of Z×Z/2Z?
What does it look like?
latent anvil
sure, so (n, x) for n in Z and x in {0,1}
Can you come up with a nice way to say when (n, x) is in (1,1) + H? What about when it's in (1,0) + H?
Something that makes it clear why these are the only two possibilities?
latent anvil
It might help to compare examples of elements in H and in (1,0) + H
latent anvil
isnt it just the remainder?
It's close, and if you can figure out how to say it exactly it should become clear
chilly ocean
thanks for the help btw, its starting to make sense
coral shale
This looks kinda cool. Is there a corresponding cool example(s) :D
delicate orchid
is that the mother frunkin 2nd iso theorem
delicate orchid
no u wot
I see intersections, a product, and an iso symbol.... there's only one suspect that satisfies all of them...
coral shale
it prolly looks like isomorphism cus of what i said here, no?
Let L/M/K galois tower of extensions (L/M, L/K, M/K all galois)
Then
Gal(L/K)/Gal(L/M) = Gal(M/K)
I feel moldi is hiding something from us
hidden haven
delicate orchid
it prolly looks like isomorphism cus of what i said here, no?
that looks like third iso
@hidden haven what's your thoughts on this one
delicate orchid
yeah I dunno what shuri meant by that lol
delicate orchid
the diagram looks like the one I keep in my head for 2nd iso as well
hidden haven
Maybe they do and I just don't know 
coral shale
the diagram looks like the one I keep in my head for 2nd iso as well
im interested
can u show
hidden haven
T!cat
coral shale
i dont have diagram in my head for 2nd
delicate orchid
I will draw it in mspaint
coral shale
ty ty
sharp sonnet
Gal is some isomorphism of categories, so the lattice of subgroups and lattice of intermediate fields are related
maybe you can make some abstract statement about lattices even
delicate orchid
i dont have diagram in my head for 2nd
can't remember which way the arrows went so I didn't draw them
coral shale
is ok, looks beautiful. Will slowly think
sharp sonnet
this is just part of the subgroup lattice, right?
delicate orchid
I think so
sharp sonnet
and so you have to find this again in the intermediate fields just mirrored
as Gal is contravariant
delicate orchid
category theory.... useful?!?!
sharp sonnet
i wouldnt go that far
hidden haven
delicate orchid
I think you can formulate that lattice as some wack set of exact sequences
I can't remember how though
I think it's like this? and another one with H in the middle
and then you do magic (def of exact sequence) and the 2nd iso theorem plops out
delicate orchid
it has to be, moldi
hidden haven
delicate orchid
the ol N intersect H is trivial, N normal, etc.
hidden haven
Are you assuming G = N semidir H
delicate orchid
I don't think so?
hidden haven
NH = G
N ∩ H = 0
N normal
These 3 imply semidirect product
hidden haven
F
delicate orchid
W actually, for I have learnt something today
coral shale
I too have learned something today. The 2nd isomorphism theorem
coral shale
Consider D6 acting on {1, 2, 3} in the 2 possible ways:
- label the locations in R^2 space 1,2,3 (where equilateral triangle vertices live)
- label vertices of equilateral triangle 1,2,3
Which is 'active', which is 'passive'?
This is what these 2 terms distinguish, right (i confuse which is which)?
I believe the first is passive with respect to the triangle
The 2nd is active wrt the triangle
But I am not 100% sure
rain crescent
What is the biggest order of a permutation in S7?
I reasoned that any permutation in S7 can be factored into either 7 1-cycles or 1 7-cycle which means the number I'm looking for is 7, but I'm not really sure If that's true because there might be some permutation which is not a product of disjoint cycles with a bigger order, can this happen?
delicate orchid
take every partition of 7 then take the one with the highest lcm
(123)(4567) has order 12 = lcm(3, 4) - for instance, I don't know why you think they can be factored into 1 7-cycle because that definitely isn't true
rain crescent
hmm I see, I think I was confusing things, thx!
delicate orchid
the lcm thing only hold for disjoint cycles, btw
but if they're not disjoint you can just multiply them together and simplify it down
fallow plume
are all (finite (?)) field extensions generated by a quotient of the polynomial ring of the subfield?
i couldn't seem to Google it well enough to give anything clear
next obsidian
Yes
Suppose that K is a finite field extension, then there exists elements a_1,…,a_n which generate K as an extension of k
So K = k(a_1,…,a_n)
This means there’s a surjection k[x_1,…,x_n] -> K sending x_i to a_i
Now apply the first isomorphism theorem
@fallow plume
Actually, any field extension regardless of the size is a quotient of a polynomial ring
You do this same process, the only difference is that the number of generators might be infinite, so you might need to take a huge number of the a_i, which in turn gives a huge number of the variables x_i
Oh I guess also you need to note that you have to do generation as an algebra, so you take k[a_i] not k(a_i). In the case the extension is algebraic these are the same (so in particular for finite field extensions)
pastel cliff
is this just a matter of pointing out that {1} is obv normal and then first iso?
pastel cliff
can we elaborate on this btw moldi