id argue quals could be fun
#groups-rings-fields
pastel cliff
dusty verge
Less fun though cause I don't feel like I'm helping people
pastel cliff
without the pressure of actually having to take them ofc
dim widget
Man I want a phd, but it's hard
y?
dusty verge
Hard to get in.
delicate orchid
Yeah I got super lucky with mine
dim widget
no but y do you want a phd?
dusty verge
cause then people will call me dr
dim widget
spoiler: they probably wont
delicate orchid
They will.
dusty verge
But then I get to go on that cliche rant, and that's always fun
dim widget
They will.
Real doctors know ochem, don't believe the hype
delicate orchid
Real doctors know ochem, don't believe the hype
I know some IUPAC naming schemes I can fake it til I make it
dusty verge
Real doctors know ochem, don't believe the hype
md more like smd
dim widget
md more like smd
h
dusty verge
phd is way cooler
dim widget
Eh getting credentials just to get them is pointless. But the 5 years getting paid to do math and nothing else is pleasant.
delicate orchid
getting paid
Oh no no no
dim widget
>getting paid
Oh no no no
I'm sure if you haggle with your PhD program they'll let you work for free.
delicate orchid
I’m paying to do mine
dusty verge
I mean the actual reason I want one is a mixture of the prestige of it plus the doors it opens, plus the opportunity to study at a high level without financial pressure to work on other things
Plus a professor told me that I wasn't enough of a "detail oriented perfectionist" to get one, so fuck him
Hey here's a cool little thing. Since the positive rationals are generated by the primes, you can construct an isomorphism between them and the ring of polynomials with integer coefficients
Except in this case, the addition of this ring ends up being multiplication in the rationals, and the multiplication ends up being some wonky shit
But it does get you a pretty unique structure, where you have a ring + a semigroup, such that the addition of the ring distributes over the semigroup operation
Where the semigroup is addition in the rationals
If you can find a way to relate the multiplication in the ring with the semigroup operation, all kinds of number theory shit pop out for free.
Have people studied the extension of the natural numbers where you allow for an infinite number of prime factors?
long nebula
Have people studied the extension of the natural numbers where you allow for an ...
Do you mean like 2^n1 3^n2 5^n3 ... for an infinite number of primes? If so, that is just the space of sequences of natural numbers! :)
dusty verge
Do you mean like 2^n1 3^n2 5^n3 ... for an infinite number of primes? If so, tha...
Yes, and yes. The reason I was asking was reated to my previous things
Basically you can show that Q+ is iso to integer polynomials, which is a subring of formal power series, which lets you invert some things
So I guess my question would be better phrased as "have people studied what happens to addition if you extend the natural numbers to allow for an infinite number of prime factors?"
long nebula
Oh I see
dusty verge
My initial guess is that there would be, unfortunately, infinitely definitions that would be compatible
long nebula
What is the way that you want to define addition?
dusty verge
I mean it's already defined for finite sequences, so just trying to figure out how many ways to extend
Let's see, there's uncountably many integer sequences, right? I'm not being dumb am I?
long nebula
I guess you could work with hypernaturals in which case you could define hyperfinite prime factorizations 
dusty verge
Cause you can just 1-1 them with decimal expansions between 0 and 1 edit: this doesn't quite work, but you can just diagonalize
long nebula
yes N^N is uncountable
dusty verge
Yeah, just wanted to double check without googling it lol
Yeah, there should be infinitely many permutations that fix the naturals then, right?
And that would imply infinitely many definitions of addition that are compatible with normal addition.
Is there a name for a ring where the additive group is a semigroup instead of a group?
agile burrow
like a semiring?
barren sierra
not sure what this is saying
"fixing a 2-by-2 matrix A over k"
then what? How does this give a module structure?
thorn delta
then what? How does this give a module structure?
I think they just mean a k[x] module over k^2
like, the scalar multiplication ring action is a ring hom k[x] —> End(k^2). Let this be the homomorphism which maps x to A, and k to k(Id)
a polynomial acts on vectors like: p(x).v = p(A)v
solar shore
Let $G$ be an Abelian group with order $p_1p_2 \dots p_k$, where $p_i$ are distinct primes. Prove that $G$ is cyclic.
cloud walrusBOT
blanket
barren sierra
a polynomial acts on vectors like: p(x).v = p(A)v
ah ok
that's clearer
also is showing Hom_R(M, N) satisfies ACC and DCC the right way to go about this? Having troubling doing either of those but I can't think of a different direction.
ah maybe I should use finite generation?
wooden ember
also is showing Hom_R(M, N) satisfies ACC and DCC the right way to go about this...
Yeah you want to show Hom(M,N) is noetherian and artinian, and for that you’ll want to use finite generation to try and extend the result from free modules to your situation if I remember correctly
dim widget
Here's an interesting question I had, what are (nontrivial) examples of modules M over a Dedekind ring R such that End_R(M) = R? I know such things exist if R = Z, but I'm curious if there can be nontrivial examples if R is say some algebra over an uncountable field which is Dedekind.
dusty verge
Let $G$ be an Abelian group with order $p_1p_2 \dots p_k$, where $p_i$ are disti...
Do you have fundamental theorem of finitely generated abelians yet?
dusty verge
Here's an interesting question I had, what are (nontrivial) examples of modules ...
What's a Dedekind ring?
Nevermind found it, that's not bad at all
delicate orchid
Do you have fundamental theorem of finitely generated abelians yet?
do you really need such a strong result?
dim widget
Let $G$ be an Abelian group with order $p_1p_2 \dots p_k$, where $p_i$ are disti...
By Cauchy's theorem there are two elements a, b the first of which is order p_1 the second is order p_2, then a + b has order p_1*p_2 since p_1, p_2 are coprime, so it generates the entire group.
dusty verge
do you really need such a strong result?
No, but it gives it in a sentence
dim widget
It is probably used implicitly in the classification of torsion abelian groups.
formal ermine
Bro everything follows from the definitions 
just don't use , and . and you'll be able to do it in a sentence
dim widget
So maybe a bit disingenuous to prove this using such a strong result.
dusty verge
Just don't use coma and.? I don't understand that sentence
ivory trail
i've forgotten all the easy results and just remember the strong result that implies all of them
dim widget
What's a Dedekind ring?
It's a one-dimensional Noetherian domain such that all prime ideals are maximal. I think that should be enough
dusty verge
Oh I saw that it a Dedekind domain is an integral domain such that ideals factor to prime ideals
That the same?
dusty verge
By Cauchy's theorem there are two elements a, b the first of which is order p_1 ...
This only works for two primes. You can't induct directly on it since Cauchy's theorem is only for primes
dim widget
Actually my definition is terrible, one-dimensional is the same as all prime ideals being maximal.
dusty verge
Honestly I don't know how the dimensions of rings are defined
frigid lark
Something something krull?
dim widget
This only works for two primes. You can't induct directly on it since Cauchy's t...
Yes but similar ideas work to prove the chinese remainder theorem, which is usually an input in classifying fgabs.
formal ermine
Honestly I don't know how the dimensions of rings are defined
maximal prime ideal height iirc
ivory trail
Yes but similar ideas work to prove the chinese remainder theorem, which is usua...
the proof of classification of finite abelian groups looks like it goes through chinese remainder anyway
dim widget
Oh I saw that it a Dedekind domain is an integral domain such that ideals factor...
My definition should be equivalent to yours.
formal ermine
which coincides with the krull dimension of the noetherian topological space of the closed set you're looking at or something
dim widget
the proof of classification of finite abelian groups looks like it goes through ...
That is what I was saying.
ivory trail
that's the first idea i had, and that's what wikipedia seems to suggest too
yeah so basically it's easier to prove the classification theorem and then show that the problem follows
dim widget
No
The classification theorem is hard work, chinese remainder easily implies this problem and is an input for the classification (but not the only input), thus the classification is unnecessarily harder.
ivory trail
yeah i see
dim widget
yeah i see
If you like this question is like step one of classifying torsion abelian groups, which is part of the classification.
dusty verge
Let $G$ be an Abelian group with order $p_1p_2 \dots p_k$, where $p_i$ are disti...
Sylows theorem gives subgroup of order p_i for each i. Since they're prime order they're all cyclic. Define x_i to be the generator of subgroup p_i, then induct on number of x_i, applying that the fact that the orders of \sum_{i=1}^k x_i and x_{k + 1} are comprime
Oh fuck off discord one sec, gotta make that readable
ivory trail
If you like this question is like step one of classifying torsion abelian groups...
yeah the only other part is structure of abelian groups of order p^n
basically
dim widget
yeah the only other part is structure of abelian groups of order p^n
Yeah, essentially step 1 is CRT, step 2 is classifying abelian groups of p-power order. Then you're done.
dim widget
Sylows theorem gives subgroup of order `p_i` for each i. Since they're prime ord...
You don't need Sylow's theorem, Cauchy's theorem is fine for this specific problem.
But it's all the same.
dusty verge
I mean you just need the existence of a subgroup of order p
Oh that is Cauchy's theorem, my b
dim widget
Yeah the case of two primes is just as hard as n primes.
dusty verge
Oh yeah for sure, I just wanted to give a complete answer
I wouldn't be on a math discord if I didn't have a little voice in my head telling me to nitpick every little point
glossy crag
Is it true that for $L/K$ purely inseparable the norm of $x$ is $(-1)^{|L:K|}(-1)^{|L:K(x)|}$? My thinking was $N_{L/K}(x)=N_{K(x)/K}(x)^{|L:K(x)|}$ and $N_{K(x)/K}(x)$ is $(-1)^{|K(x):K|}$ times the constant coefficient of the minimal polynomial of $x$, which is $-1$.
cloud walrusBOT
Ocean Man
dim widget
This cannot be correct because what you've written can only be +/- 1, what is your definition of the norm for a general extension? Usually for a purely inseparable extension it should just be some pth power and for a separable extension it is defined as usual.
glossy crag
This cannot be correct because what you've written can only be +/- 1, what is yo...
Strictly speaking this is a p-th power, because since L/K is purely inseparable both exponents will be divisible by p, but that's beside the point.
If this is wrong, where is the mistake? Because I don't see it.
dim widget
Why did you define that Nm(x) = (-1)^{[L:K]} (-1)^{[L:K(x)]} for any x? this makes no sense
glossy crag
I did not define anything.
dim widget
For instance you just said Nm(0) = +/- 1
glossy crag
Firstly, N_L/K is equal to a power of N_K(x)/K
dim widget
You are assuming perhaps that x is of a specific form?
glossy crag
That works in every extension.
glossy crag
You are assuming perhaps that x is of a specific form?
No, I'm not
dim widget
Then the constant term of the minimal polynomial of x need not be -1 so your reasoning is incorrect
glossy crag
Then, since L/K is purely inseparable, the minimal polynomial of x is of the form x^{p^m}-1 for some m
dim widget
That is false.
glossy crag
Oh shit, I just saw it
it's x^{p^m}-a for some a in K
not -1
nvm, same thing basically, just replace -1 with a
dim widget
Yes but that is a crucial difference.
dim widget
Lol, potato, definition of the norm which is not multiplicative....
glossy crag
I was just doing some cyclotomic stuff before, hence why x^{p^m}-a turned into x^{p^m}-1 for me
dim widget
That makes sense.
Important to point out that x^{p^m} - 1 never generates a purely inseparable extension since F_p is perfect.
glossy crag
True
dim widget
So maybe the best way to say it is that the norm along a purely inseparable extension is just taking the p^m th power, rather than what you said which is pretty inexplicit.
glossy crag
Wdym, N_L/K(x)=x^p^m, where p^m is the degree of x?
dim widget
Yes.
glossy crag
Is it really that simple?
dim widget
That's the same as your definition, using the fact that x satisfies its own minimal polynomial
It's just cleaner.
glossy crag
If x^p^m-a is the minimal polynomial, then N_K(x)/K(x) is (-1)^p^m(-a)
dim widget
Yes if you notice either p is odd then your signs cancel or p is even and -1 = 1.
So you're just saying N(x) = a
glossy crag
Oh, right
dim widget
But x^{p^m} = a by definition
glossy crag
So N_L/K(x)=x^|L:K|
dim widget
When the extension is purely inseparable, yep.
glossy crag
Right, thanks
dim widget
No worries 🙂
glossy crag
I'm thinking how to prove transitivity of norm for a general extension
Separable and purely inseparable is easy
Mixed I'm kind of wondering
Breaking the tower up in stages is a little messy
dim widget
Yes that makes sense that it is confusing. So in general: every L is of the form L/K/F where K/F is separable and L/K is purely inseparable right?
glossy crag
Yeah, that's what I'm trying to do
dim widget
Okay so take an x \in L, what does its characteristic polynomial look like in terms of the characteristic polynomial of x^{[L:K]}?
glossy crag
Okay so take an x \in L, what does its characteristic polynomial look like in te...
L/K being purely inseparable? Char. poly. of x^|L:K| should be (y-x^|L:K|)^|L:K|, since x^|L:K|\in K and the charpoly is a power of minpoly.
dim widget
I mean the characteristic polynomial of that element over F not over K.
glossy crag
I need to step out, so I'll think about this later. Appreciate the help.
dim widget
No worries!
paper flint
Does there exist a finitely-generated group which has elements of all finite orders, that is, for each natural n, there exists an element of order n?
I've been trying to search for examples online but one or the other condition is compromised (neither Prüfer groups nor a solution to Burnside's problem directly suffice here)
wooden ember
Does there exist a finitely-generated group which has elements of all finite ord...
For one it can’t be abelian right?
paper flint
Yes, that sounds right
sly storm
Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct (with straightedge and compass) a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?
lusty marlin
Does there exist a finitely-generated group which has elements of all finite ord...
I get the feeling that such a group can't exist, but I'm not able to prove it at this moment.
paper flint
Yeah, this feels too good to be true in a way
wooden ember
Aight so
You have a surjective map to the direct product of Z/pZ for all p
Composing that with a surjective map from a finite gen free abelian group would give you that this direct product is finite gen
So you just have to show this isn’t the case
Which should be doable
dim widget
Does there exist a finitely-generated group which has elements of all finite ord...
yes, you can take \oplus_{i = 1}^\infty \oplus_j^{\infty} Z/p_i^j
wooden ember
Oh yeah it isn’t by fund theorem of finitely generated abelian groups
wooden ember
or just Q/Z
Not finite gen
paper flint
Finitely-generated is key here
dim widget
Then it's provably impossible
wooden ember
I think my proof works?
dim widget
You can only have torsion of bounded order.
Thanks
dim widget
Oh yeah it isn’t by fund theorem of finitely generated abelian groups
Yes you have Z^r \oplus Z/p_i^j and all torsion elements are killed by \prod_i p_i^j .
Ah sorry I misread the question twice
You were asking about an arbitrary group?
paper flint
Yes
wooden ember
You were asking about an arbitrary group?
Arbitrary finite gen but you can map into an abelian group
dim widget
Ah that is more difficult.
dim widget
Arbitrary finite gen but you can map into an abelian group
Not necessarily
wooden ember
Not necessarily
Why not?
Send everything to identity and an element of order p to 1 gives a map to Z/pZ
Then take a direct product
That’s abelian
dim widget
Well for instance there exist nonabelian groups in which no two elements commute unless one is a power of the other, and there exist nonabelian groups in which the abelianization is trivial
wooden ember
But not finite gen
dim widget
So in general you cannot expect there to be lots of maps either to or from abelian groups.
wooden ember
Well for instance there exist nonabelian groups in which no two elements commute...
Sure yeah
dim widget
The only maps you can expect are the trivial Z \to G by n \to g^n for some g \in G.
dim widget
You'll have to explain what you're trying to do?
dim widget
What I’m doing clearly doesn’t work
Okay, that's what I thought. But of course you're not an idiot.
wooden ember
Aight let’s think some more
dim widget
Yes so I think the best way to do this is quite complicated
I would use some topology if that's alright.
wooden ember
Oh neat
dim widget
Hmm actually now I'm starting to think its possible
It will take longer to construct an example.
wooden ember
Hmm actually now I'm starting to think its possible
Can’t you like glue disks to a Hawaiian earring by taking increasing degree maps, and look at the fundamental group of that or smth?
If you glue nicely enough (say in R^3 by making them smaller) you should get something compact and locally simply connected no?
dim widget
I am not sure, it will probably not be locally simply connected unless you get rid of an infinite amount of the fundamental group and then you might as well just have a finite number of circles.
wooden ember
I am not sure, it will probably not be locally simply connected unless you get r...
Ah damn I wasn’t sure if that was an issue or not
dim widget
I have moved away from the topology idea, I originally thought we could say something about the fundamental groups of a bouquet of circles but then I realized that the statement sounds like it's true.
I.e. there probably does exist such a group.
wooden ember
But yeah you’re right gluing the disks isn’t meant to make the small ones simply connected 🤦♂️
carmine fossil
Does quotienting by commutator subgroup preserve order?
dim widget
No
For instance for simple groups the commutator subgroup is everything.
Or nothing.
But if they're nonabelian its everything.
Okay so here is close to an answer.
So there is a map from Sl_2(Z) \to \Prod_{p prime} Sl_2(Z/p) which surjects onto Sl_2(Z/n) for any n a squarefree integer.
(viewed inside of the product by the chinese remainder theorem)
wooden ember
So there is a map from Sl_2(Z) \to \Prod_{p prime} Sl_2(Z/p) which surjects onto...
By just reducing coeffs mod p and taking the product?
dim widget
Yes, it is a little tricky to show the claim about the surjection thing but that's the map.
wooden ember
Aight sure
dim widget
So Sl_2(Z) is well known to be generated by 3 elements (probably 2, but I am not feeling smart enough to work out what the minimal generators are vs the minimal generators of PSl_2)
carmine fossil
Can't you do a similar thing for non square free n too
dim widget
-1, (0 -1 | 1 0) and ( 1 1 | 0 1)
dim widget
Can't you do a similar thing for non square free n too
Yes you can but the exponents always have to be bounded which is frustrating
Anyway this quotient of Sl_2(Z) gives you a finitely generated of group with elements of p-power order for all p, and the exponents can be as large as you like.
Still cannot do all n though....
wooden ember
That’s real close though
dim widget
But I am thinking it's possible.
carmine fossil
Why can't you do all n
dim widget
Because the quotients that you get if you reduce mod larger powers of p are different groups.
So for each sequence n_i I construct a group G_{{n_i}} which has an element of p_i^{n_i} power order for each prime p_i.
But they don't all occur in the same group.
AHA
I think I see it...
fast stratus
$\ker A \oplus \text{im } A^T = V$, where $A: V \to W$ and $\oplus$ is direct sum and $V, W$ are vector spaces
is this correct a identity?
cloud walrusBOT
Helium
dim widget
Why can't you do all n
Ah okay I think I can, I was being silly.
dim widget
Does there exist a finitely-generated group which has elements of all finite ord...
Look at the map from Sl_2(Z) \to \prod_{i, j \in \mathbb{N}} Sl_2(Z/p_i^j) for all i,j, and p_i some enumeration of the primes. The map is given by reducing an element mod p^j for every j in each factor. The image of this map is some subgroup G. One can show that the map Sl_2(Z) \to Sl_2(Z/n) is surjective for all n \in Z, thus G contains an element of order p^j for any p and any j (just look at a unipotent matrix in Sl_2(Z/p^j)). Thus G has elements of arbitrary order.
Because Sl_2(Z) is finitely generated (in fact two-generated, it;s just Z/4*Z/6) so is G.
paper flint
Sick! 
dim widget
$\ker A \oplus \text{im } A^T = V$, where $A: V \to W$ and $\oplus$ is direct s...
Yes if they are finite dimensional vector spaces. Otherwise you have to be a bit more precise about what you mean by A^T.
In general you can say more: from the definition Im(A^T) is the orthogonal complement of ker(A)
I'm assuming that you want to prove this for vector spaces over R or C right?
fast stratus
I'm assuming that you want to prove this for vector spaces over R or C right?
yes R^n, i used it in a proof... just double checking if it's true
slim kayak
I am given a group G which is a semidirect product of the integers and some group H. Let H^{ab} and let denote the abelianized factor group of H, with [h] standing for h[H,H]. Let Z(x+-1) be the ring of Laurent polynomials, I am supposed to show that this (see picture):
Defines scalar multiplication and turns H_{ab} into a module over the laurent polynomials.
It seems like a simple calc and check exercise but I have no idea how this would work out. "Adding" two elements [h] and [k] gives me h[H,H] k[H,H] = hk[H,H] and the subsequent "sum" would look like \sum z_i [g^{i} hk g^{-i}] (since multiplicative notation is already used and H has only one operation since it is a group the sum should be interpreted like a product IMO). I have no idea how this result would "distribute" to \sum z_i [g^{i} h g^{-i} ] (operation symbol of H of choice) \sum z_i [g^{i} k g^{-i} ]
So... is there some wizardry that makes it work or is the question just very ill-posed?
slim kayak
An element from the pre-image of 1 of the canonical map from G to Z.
slim kayak
I think the sum has to be interpreted as the group law on Hab
Ooooh, yeah that makes sense then.
hot lake
So you need to show that f.g.[h] = fg.[h] ?
slim kayak
I wanted to start with f.([h] x [k])=f.[h] x f.[k]
Although I probably got it from there since I can finally parse it. Thanks
wooden ember
Look at the map from Sl_2(Z) \to \prod_{i, j \in \mathbb{N}} Sl_2(Z/p_i^j) for a...
that's sick
i hate that this exists, and what more is given by only two generators
dim widget
I think worse you can embed any countable group into a 2-generated group.
It's some famous result from the 70's
dim widget
i hate that this exists, and what more is given by only two generators
Here's a very simple 1-paragraph proof: https://www.jstor.org/stable/pdf/2324618.pdf?refreqid=excelsior%3A04897f26a477855d6e4bd1bbb4bf7d40&ab_segments=&origin=&initiator=&acceptTC=1
delicate orchid
I think worse you can embed any countable group into a 2-generated group.
every countably generated free group appears as a subgroup of the free group on two generators, and obviously you can embed any countable group into a countable free group - for those who were wondering why this is true
dim widget
You cannot embed any countable group into a countable free group.
dim widget
every countably generated free group appears as a subgroup of the free group on ...
Because a subgroup of a free group is free (good exercise)
So the only countable groups you get are Z and the trivial group inside of any free group.
delicate orchid
are we using different definitions of countable here
I took countable to mean "has a countable generating set"
dim widget
Even then it doesn't work, because you cant have any torsion inside of a free group.
delicate orchid
embed was probably the wrong word
dim widget
But you can get any "countable" free group (since there is just one) inside of a 2-generated free group.
delicate orchid
you take a countably generated group and then throw away the relators you get a countable free group which is in F_2, then add the relators back in
that's what I meant
u get a little square of morphisms how wholesome
dim widget
Yeah but that doesn't show the result, since those relations may or may not extend to a normal subgroup of F_2
carmine fossil
How evil/annoying is it to check if a polynomial is irreducible over GF(2^192)?
dim widget
What is GF? is that the finite field of that size?
carmine fossil
Yes
dim widget
American country voice: sure do be sounding quite evil/annoying
I mean are you looking at a specific polynomial? What is its degree?
carmine fossil
Well so I am looking for degree 192 polynomial with minimal number of non zero terms which is irreducible over GF(2^192)
dim widget
Hmm, well you'll be relieved to know that the minimal number of nonzero terms is 2
carmine fossil
Is that like unrealistically hard
carmine fossil
Hmm, well you'll be relieved to know that the minimal number of nonzero terms is...
But I suppose that would be reducible?
dim widget
Nope!
carmine fossil
Is that like unrealistically hard
Because this is supposed to be part of a cryptography standard
dim widget
If I'm not messing something up (I am pretty tired): 192 has only 2, 3 as prime factors
dim widget
So we know that 3 does not divide 2^{192} because we're smart
white oxide
can somebody give me a hint to show that a + bsqrt(2) has a multiplicative inverse if a, b nonzero? i wrote 1/a + bsqrt(2), multiplied by the conjugate and I got a - bsqrt(2)/a^2 - b^2(2). but then 1/a is not an integer
dim widget
We also know that 3 does divide 2^{192} + 1 because 3 divides 2^n + 1 for all n (all n ODD!)
So 3 is coprime to 2^{192} - 1 which is the order of the multiplicative group of F_{2^{192}}
Obviously 2 is also coprime to this number
barren sierra
can somebody give me a hint to show that a + bsqrt(2) has a multiplicative inver...
why do you need multiplicative inverses?
carmine fossil
We also know that 3 does divide 2^{192} + 1 because 3 divides 2^n + 1 for all n ...
That is (-1)^192+1=2 mod 3 tho
white oxide
why do you need multiplicative inverses?
i thought it's because the real numbers are a field, so every nonzero element has a multiplicative inverse
barren sierra
subring
white oxide
ah yeah i suppose that's true
dim widget
That is (-1)^192+1=2 mod 3 tho
Ah I was being tired, it's only true if n is odd as you say
barren sierra
ring
barren sierra
Z is a subring of R which is a field but plz tell me a multiplicative inverse of 2 in Z
lethal dune
1x0=0 because 1 is the identity or 0 is the zero element 
barren sierra
yes
dim widget
That is (-1)^192+1=2 mod 3 tho
Yeah I dunno then, doing it by hand is clearly impossible.
dim widget
But I suppose that would be reducible?
Ah wait no, this is still fine.
I think there is another approach you can take to making a solution
Unsure how in the world you'd ever prove something is minimal though...
carmine fossil
Well I am looking at a crypto implementation and they are like "implement it yourself"
white oxide
can somebody give me a hint for this question? my strategy was to break it up into two cases: R has unity, or it doesn't. if R doesn't have unity then I showed that phi must be the trivial homomorphism (phi(a) = phi(a x 1) = phi(a)phi(1) = phi(a)0' = 0' for any a in F). i'm struggling a bit with the injective case tho, and would appreciate a hint
dim widget
can somebody give me a hint for this question? my strategy was to break it up in...
What would the kernel look like?
dim widget
The kernel has to be an ideal of F right? How many ideals are there.
(including the trivial ones)
white oxide
i have no clue how many ideals in F there are tbh
celest furnace
i have no clue how many ideals in F there are tbh
Try proving it: either the ideal contains something nonzero, or it doesn't
dim widget
Well I am looking at a crypto implementation and they are like "implement it you...
Why are you interested in it having a minimal number of nonzero terms?
It seems like you just want any irreducible polynomial no?
dim widget
i have no clue how many ideals in F there are tbh
For a general ring R, which ideals contain units?
carmine fossil
It's a crypto standard thing
Well ig I am gonna BS this assignment since it's like quite literally impossible to compute such a polynomial
white oxide
ok i think i got it?
If $R$ is a ring w/o unity, then for any $a \in F$, $\phi(a) = \phi(a) \cdot \phi(1) = 0'$ since $\phi(1) = 0'$ since $R$ does not have unity. Otherwise, suppose $R$ does have unity $1'$ and $\phi$ is not the trivial homomorphism. Then, for any $a \neq 0 \in F$, $\phi(a) = \phi(a) \cdot \phi(1) = \phi(a) \cdot 1' \in R \implies $\phi(a) = 0'$. But then $\phi$ would be the trivial homomorphism, contrary to assumption. Hence $a = 0$, and so $\phi$ is injective.
idk if that works, it should??
cloud walrusBOT
okeyokay
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
dim widget
Well ig I am gonna BS this assignment since it's like quite literally impossible...
There could be some clever trick, or some well known pre-compiled database of such things. All I'm saying is that it's completely absurd to brute force.
carmine fossil
Yeah I think that's like the point
Like it's supposed to be pre compiled
But no one does it
hot lake
what are you trying to do ?
dim widget
But no one does it
But it can't be that everyone is using the same polynomials? Which protocol is this?
carmine fossil
But it can't be that everyone is using the same polynomials? Which protocol is t...
Yeah it's standardized
dim widget
Then surely the point isn't that "you can't guess the polynomial which is hard-coded into everyone's implementation" that wouldn't be secure.
carmine fossil
Well my prof isn't very bright
She probably thinks you can
Like once she expected us to run 2^47 iterations on a computer
And print everything on paper
dim widget
And print everything on paper
Maybe she hates trees.
carmine fossil
Probably
dim widget
Probably
Anyway my point is that I don't think the cryptographic security of AES is predicated upon the fact that your specific irreducible polynomial is hard to generate or hard to guess, that doesn't make much sense (so I think the algorithm works quite differently from what you think, you might want to read more about it). My impression is that everyone uses the same fixed irreducible polynomial of degree 8 (x^8 + x^4 + x^3 + x + 1) over F_2 for key-scheduling, and the only difference between the various instantiations is how many round keys are produced from the initial key.
carmine fossil
Ok this is not aes per se
This is CMAC
idk why it is called cmac aes everywhere
But yea the polynomial is fixed here too
dim widget
Yes and I really doubt it's of degree 192 over F_{2^192}...
So I read the documentation for AES-CMAC, the way messages are encoded is just done with the usual AES algorithm. I don't see any reason why you would need such a huge polynomial for doing any of this (or indeed any polynomial at all except for the standard one used in the MDS matrix part of AES), I would revisit your textbook and see if you have understood the algorithm.
gilded osprey
Are the elements of $\mathbb{Z}_3[x]/(x^2)$ ${[0],[1],[2],[x],[x+1],[x+2],[2x],[2x+1],[2x+2]}$ ? Just double checking I am thinking of ring quotients properly
cloud walrusBOT
JacobHofer
rustic crown
gilded osprey
Why F_3? That was just the format I was given
coral spindle
Z_3 is often used to mean the 3-adic integers
a more standard ring-theoretic ways of denoting that quotient of Z is just Z/3Z, or Z/(3), or my personal preferred shorthand, Z/3.
If your course requires you to write Z_n though, this is what you should write for that course.
coral spindle
heyo
charred bison
Let $f(x) \in \bZ[x]$ be a primitive polynomial. How do you show that $f(x)$ is irreducible over $\bZ$ iff it is irreducible over $\bQ$
cloud walrusBOT
CoolShot
coral spindle
Gauss' lemma
formal ermine
beat me to it
charred bison
damn i also have to prove gauss's lemma
coral spindle
Gauss' lemma is precisely the statement you're looking for
charred bison
i mean the other statemnt
dim widget
Yeah but just saying Gauss's lemma doesn't help you prove gauss's lemma
Gauss allegedly tried shouting "Gauss's lemma" at the problem for years before he figured out the proof
coral spindle
Yeah but just saying Gauss's lemma doesn't help you prove gauss's lemma
Which is why I linked the proof :)
dim widget
but that story might be apocryphal
gilded osprey
Also, that ring from before is not a field correct? Since [x] has no multiplicative inverse?
coral spindle
Have you proven that x has no multiplicative inverse?
Because if so, yes that would mean it's not a field, by definition
gilded osprey
I mean, the question is asking us to construct tables, so I've essentially exhaustively shown that there is no element a such that [x]a = [1]
coral spindle
If you've done so then I think the answer is clear
There is an easier way to prove that x has no multiplicative inverse, which doesn't require drawing a table. Hint: consider x^2.
gilded osprey
Would it be that x is a zero divisor since xx=x^2=0, and so it cannot be a unit?
coral spindle
Yes
barren sierra
Can I get some help on this one?
I know I want to somehow show that Hom(M, N) has ACC and DCC
so I can get ACC probably via finite generation somehow
not sure how to get DCC
nvm got ACC, not sure how to do DCC
coral spindle
Can we maybe distinguish submodules of Hom_R(M, N) by looking at the span of the images of the submodules of M?
Just guessing here, I don't recall the proof :)
barren sierra
I wanted to try something like that but wasn't sure how to go about that
dim widget
So can you describe what modules you think should occur in the composition series for Hom(M, N)?
barren sierra
well they should be simple meaning they have no proper submodules but I'm not sure what this really means in terms of module homomorphisms
I know that since M is finitely generated, morphisms from M are defined by where they map the generators but I don't think that's super helpful
kinda spitballing here
coral spindle
Topos mentioned the composition series, not the composition factors ;)
dim widget
Usually with these things about finite length modules it's a good idea to induct on the length, have you tried that?
barren sierra
Topos mentioned the composition series, not the composition factors ;)
oh hm
dim widget
Like I would say 99% of things you ask about artinian modules reduce to checking them for simple modules by inducting on the length.
coral spindle
I think I can see some nice submodules of Hom_R(M,N)
barren sierra
what would I even try to show by induction?
dim widget
I guess you could try to do double induction on the length of M, N to show that Hom(M, N) is finite length
Using these submodules of Hom(M, N) that we are hinting at.
lethal dune
did someone say Artinian 
dim widget
This thread has escalated from "all rings are Noetherian" to "all rings are zero dimensional"
lethal dune
wym all rings are already 0 dimensional
lethal dune
This thread has escalated from "all rings are Noetherian" to "all rings are zero...
is this some non commutative joke I'm too commutative algebraist to understand
barren sierra
So can you describe what modules you think should occur in the composition serie...
I don't really know what I should expect the actual series to look like
dim widget
what would I even try to show by induction?
I mean you don't really need to for the problem, but it might've pointed you in the right direction.
Like if it were me I would think about what happens with finite dimensional vector spaces or maybe reduceable but not decomposable representations of a group or smth
But maybe just see if you can use the composition series for N and M to relate Hom(M, N) to some smaller modules?
And thus do some kind of induction
dim widget
is this some non commutative joke I'm too commutative algebraist to understand
I'm now confused about the bit lol. The krull dimension of artinian rings is zero but there are rings like Z which are one-dimensional... or are you making another layer of joke that I don't follow?
lethal dune
lol it's the latter
cloud walrusBOT
lethal dune
sonic coral
is 18b possible?
lethal dune
trivial
sonic coral
if the constant was odd i could apply mod 2 test
lethal dune
Eisenstein
lethal dune
what's the only p that satisfies all the axuoms,
sonic coral
but 13 doesn’t divide the constant
lethal dune
doesn't?
barren sierra
nope
sonic coral
i tried that
barren sierra
that was my first thought as well lol
1, 2, 4, 5059, 10118, and 20236.
these are the factors
sonic coral
it looked easy until i realized 13 didn’t divide it
barren sierra
lethal dune
Lol
lethal dune
ok let me think now
summer path
those are some real terrible numbers for an exam problem 
sonic coral
since 2 and 3 didn’t work
and i couldn’t use 5
maybe it should’ve been an odd constant?
which was the point of making it huge
barren sierra
should have been a constant divisble by 13
sonic coral
that too
barren sierra
lol
sonic coral
but not by 13^2
sonic coral
and then for c, is there a way to do it without knowing that -4 is a zero
or is that the intended way
to just factor it since it’s degree 3 or less
sonic coral
well yeah, intended to be odd
lethal dune
which one?
sonic coral
oh b
lethal dune
I mean what's the supposed correct problem statement then
sonic coral
wouldn’t it just have to be the same thing with any odd constant
then mod 2 test works
lethal dune
prolly, ask the instructor
dim widget
Could be that you were meant to reduce mod 3
sonic coral
it didn’t help me. i tried it
unless i made a mistake
i’ll ask him
thanks for looking at it though
lethal dune
ye doesn't sound right
patent girder
Can any padic group be viewed as its analogue lie group manifold or is there some nicer way to find the topological structure?
prisma ibex
Can any padic group be viewed as its analogue lie group manifold or is there som...
you can regard p-adic groups as p-adic manifolds, it's just that there are some ways in which p-adic manifolds are maybe not so nicely behaved
chilly ocean
Im new in abstract algebra and I have a question. If we have a monoid (S,•) and a in S and a^2=a then a=e?
coral spindle
Im new in abstract algebra and I have a question. If we have a monoid (S,•) and ...
No; try to come up with an example where this is not true.
You will be very familiar with one
chilly ocean
Yes (N,+), e=0 and 1^2=1
coral spindle
No, in (N, +) 1^2 = 2.
barren sierra
^^
barren sierra
(close though)
coral spindle
That's right
chilly ocean
Im dumb )) thanks
coral spindle
Dumb? You got a correct answer, I don't see how that's dumb
chilly ocean
This is my first day on abstract algebra it looks so different and I know it is alot of theory. I hope I can understand it.
barren sierra
good start for a first day lol
chilly ocean
The question is related to a problem that a finite monoid where a^2≠a for every a≠e is a group
And from matrices problems I said that a^i=a^(i+k) for some i,k>=1
And I got a^i=a^(i+tk) for every t>=1
coral spindle
very good start
coral spindle
The question is related to a problem that a finite monoid where a^2≠a for every ...
Hint: this can be rephrased as a^2 = a implies a = e
coral spindle
And I got a^(tk)=(a^(tk))^2
You've skipped some steps here
in general this isn't true either
chilly ocean
Yes I took tk>i and a^tk=a^i • a^(tk-i)=a^(i+tk)•a^(tk-i)=a^(2tk)
coral spindle
Good good
void cosmos
@lethal dune yo
i have a bad request
Z is noetherian but not artinian
some students like me have bad memory
and can remember falsly ur nickname
and lose marks 😠
change ur name to something true plz 😦
chilly ocean
Yes and from the hint a^2=a implies a=e we get a^tk=e and then a is invertible so S is a group
coral spindle
Why is a invertible exactly? What's its inverse?
chilly ocean
Because a•a^(tk-1)=a^(tk-1)•a=e and this is the definition
coral spindle
Yeah, crucially tk >= 1
chilly ocean
Is abstract algebra harder than linear algebra? I finished my 11 grade olympiad two days ago and I got silver medal because I messed something with jordan blocks and now I want gold medal. I have a year to prepare but I heard it is hard and it is alot of theory and this scares me. Some tips?
coral spindle
Linear algebra is, mostly, a part of abstract algebra. I wouldn't say one is harder than the other, you can really get as difficult as you like with either.
My only tip is: study.
barren sierra
linear algebra is probably easier but having a good lin alg background will help with abstract algebra
I wouldn't say that just because you messed something up with like Jordan blocks you have anything to worry about
that's such a small thing
and alot of things in abstract algebra will remind you of things in lin alg
chilly ocean
Yeah. I understand. Thanks. Probably in the next months I will be very active here 🙂
barren sierra
like if you remember linear transformations, that idea will come up again but for different structures like monoids or groups or whatever
lol I basically live in here
hw is hard
chilly ocean
=))
barren sierra
I have my own question now
Ok so
I couldn't prove this true
so I think it's false
my idea is this
find a module M which is not a semisimple module
but it has a semisimple module N, and also M/N is semisimple
coral spindle
Hint: you can find a counterexample in R = Z.
coral spindle
Yes, but this isn't saying much: that's just translating what the SES states
barren sierra
yea but making it more explicit ig
that is literally what an SES is you're right whoops
south patrol
Interesting
barren sierra
0 -> Z/pZ -> (Z/pZ x Z/pZ) -> Z/pZ -> 0 I think works?
south patrol
Trying to think of a non-semisimple module lol
coral spindle
Z/pZ x Z/pZ
This is a semisimple module
coral spindle
it is the direct sum of Z/pZ and Z/pZ
barren sierra
oh wait yea
coral spindle
both of which are simple
barren sierra
yes
coral spindle
Your idea is good, try again
barren sierra
Trying to think of a non-semisimple module lol
this is where I'm stuck also lmao
I know the semisimple Z modules are direct sums of Z/pZ, p prime
south patrol
Your example works if you change the ring I think lol
coral spindle
Bigger hint: ||think about the classification of f.g. abelian groups||
coral spindle
Your example works if you change the ring I think lol
To what ring? I don't see which one you're thinking of
barren sierra
classification of what?
south patrol
Okay I may be wrong let me think
coral spindle
sorry, typo
south patrol
So like
barren sierra
oh
coral spindle
With what action?
south patrol
Okay I am used to thinking of this in terms of representations so it is hurting my head slightly aha
coral spindle
Now if you'd said F_p[S_2] I would've agreed
though I haven't checked that
In fact no, that can't be the case
south patrol
I basically mean you let Z/pZ act on Z/pZ x Z/pZ by like $1\cdot \begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} 1 & 1 \ & 1 \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix}$
coral spindle
Right, I think that might be right
cloud walrusBOT
potato
south patrol
Okay so basically like
coral spindle
Well, there is still a much simpler example in Z-modules if you want to find it
south patrol
A 2-dimensional rep which has a 1 dim subrep which doesn't admit a direct complement
coral spindle
That's right
barren sierra
I should start writing more non-examples of things in my notes...
coral spindle
Also not to criticise your example, but I think it's funny that instead of just one example you've produced an infinite family of examples :)
coral spindle
But just to mention
barren sierra
I forgot direct sum of semi-simple is semi-simple for second there
coral spindle
if it were true that N and M/N being semisimple implies M being semisimple, then every finite-length module would be semisimple.
In that way, just finding a single finite-length non-semisimple module would suffice to disprove this
barren sierra
interesting
barren sierra
if it were true that N and M/N being semisimple implies M being semisimple, then...
I'll try to prove this later
coral spindle
It's not tricky, just pick a maximal chain and induce
barren sierra
and try to work on that question I asked about earlier cause I got nowhere lol
oh hm I think I can see that
coral spindle
Yeah that one is tricky, I've also been trying to do it
barren sierra
ACC isn't bad it's just DCC is hard.
Trying to do stuff with induction on composition series with like Hom(M_m-1, N_n-1) and such and trying to extend it but no luck
I don't even know what a composition series should look like for Hom(M, N)
Like what even is maximal submodule
coral spindle
I think it will have a lot to do with the kernels and images of the maps in the submodule. I have more details but I haven't finished the proof, so I won't waste your time
dim widget
Trying to do stuff with induction on composition series with like Hom(M_m-1, N_n...
You can look at N_{n-1} \to N \to N^n, where N^n is the nth composition series of N
What happens when you apply hom to this?
You get an exact sequence 0 \to Hom(M, N_{n-1}) \to Hom(M, N) \to Hom(M, N^n)
So if we knew that Hom(M, N^n) and Hom(M, N_{n-1}) were Artinian, we'd be done.
Thus we reduce to showing (by induction on length) that Hom(M, N) is artinian when N is simple.
patent girder
you can regard p-adic groups as p-adic manifolds, it's just that there are some ...
is there any other insight that can be taken about the topoligical structure or no? (im guessing the answer is no)
prisma ibex
I mean features of the p-adic topology certainly play a role in the theory, but I'm not sure this necessarily gives any geometric insight
prisma ibex
like in practice when you study (often infinite dimensional) representations of p-adic groups you are studying these through some sort of Hecke algebra, so like a convolution algebra of smooth compactly supported functions on your p-adic groups which are bi-invariant under some subgroup K
the p-adic topology forces such smooth compactly supported functions to be locally constant, so this is quite different from the usual convolution algebras that one studies over the reals for example
so there are things like this where the p-adic topology intervenes
patent girder
ah i see. the compact support helps us integrate over something compact and nice instead of G right?
prisma ibex
sure, usually you're thinking of these as test functions you're integrating against, usually you want your test functions to be smooth with compact support
patent girder
So you're in C_c^infty(G)?
And cause it's a haar measure it becomes workable
Oh
I see
white oxide
why exactly are these two field of quotients isomorphic? my first thought was that any two field of quotients contained with a field are isomorphic, but they're not contained in the same field right?
ah ok got it thx!
patent girder
why exactly are these two field of quotients isomorphic? my first thought was th...
Cant you just define a ring homomorphism f: Q[x] \to \mathbb{R} by f(p) = p(pi)
the image is Q[pi]
So just consider kernels
karmic moat
If $\varepsilon^k$ is an $n$-th root of unity, how can I conclude that $\langle \varepsilon^k \rangle = \langle \varepsilon \rangle$ iff $k$ and $n$ are coprime?
cloud walrusBOT
anamono for anamono
tawny magnet
consider the order of e^k
tawny magnet
if coprime you can use bezout's theorem to prove theirr orrders are both n
void cosmos
every divisible abelian group is injective as a Z-module proof:
it is enough to show that any module homomorphism from an ideal (submodule) to the module can be extended to a map from the whole ring (Z) to the module
notice that any ideal in Z is of the form (n) where n is an integer
now the map f:(n) --> D is a homomorpihsm
how do i extend it
pastel cliff
what gives us that second iso
agile burrow
how do i extend it
to define an extension g: Z -> D, it suffices to specify where 1 maps to
agile burrow
No, you want the extension to agree with the original map, so g(n) = f(n)
Try using the fact that D is divisible
frigid lark
I've gotten that E = k(a,b) where k(a) = $E_0 \subset E$, the largest separable extension of k, $\subset E$, but I do not know how to proceed from here
cloud walrusBOT
Parrot Tea
frigid lark
Oh yeah I also have that $b^{p^r} \in E_0$
cloud walrusBOT
Parrot Tea
void cosmos
okay so it would be
1 --> x where the x is "divisor" of the images of f
yea it is
f: (n) --> D , there exists x such that nx=f(n) for any n
so define 1 --> x
got it
tysm
yo super bad question
Hom_Z(R,J)
what ring is this a module over
R is a ring with identity
frigid lark
Z
frigid lark
It's just an abelian group
frigid lark
nf = f + f + f + ... + f n times
agile burrow
no
void cosmos
okay this is isnt right exactness right?
void cosmos
no
oh thank god ur here
they arent the same category ig
cuz that would be saying right exact but u told me its not
and i believe u
haha
agile burrow
I don't understand your question
void cosmos
prop 4.3
now this isnt saying that Hom is a right exact
functor
right?
cuz it isnt
also how is this a functor again?
which category does it take and which does it send
modules to abelian?
agile burrow
Yeah, but if R is commutative then Hom_R(A, B) is still an R-module
agile burrow
now this isnt saying that Hom is a right exact
Correct, Hom(-, D) is not right-exact
It is left-exact, but this is giving a sort of converse to the left-exactness of Hom
void cosmos
would this be right exaxctness if this sequence were to be 0-->A-->B-->C?
agile burrow
the direction of exactness has to do with whether the sequence after applying the functor is left or right exact
that's how I remember at least, but I find myself having to google because contravariance is weird
agile burrow
ok
agile burrow
what ring is this a module over
This is a construction known as coinduction (or coextension of scalars) to get an R-module from an abelian group J. The action is given by (r.f)(s) = f(sr)
void cosmos
wait so is it an R-module or is it a Z-module
agile burrow
But more importantly, it has the structure of an R-module
void cosmos
yea
agile burrow
One reason that coextension of scalars is important is because it is right adjoint to restriction of scalars
void cosmos
i was proving that Hom_Z(R,J) is an injective R-module given that J is divisible
agile burrow
Yes, that's very important
agile burrow
Well I mean, the corollaries of that are important
agile burrow
What do you mean
void cosmos
like
call defitionally equivalent =
injective = all short sequences ( from the right? lmfao ) are exact = Hom(-,J) is exact given J is divsible cuz divsible = injective Z-module?
frigid lark
I've gotten that E = k(a,b) where k(a) = $E_0 \subset E$, the largest separable ...
nvm, but damn this solution was cool
agile burrow
I'm not sure about your second equivalence, but yeah the other three are fine for injective abelian groups
void cosmos
an R-module is injective given that for all R-modules A,B the seq 0--> P-->A-->B--?0is split exact
or did i switch up the arrows
agile burrow
That's right, you're just missing a zero on the left
void cosmos
yea
damn okay
i still didnt prove this
i proved it for the dual
like projectivity
and its much easier
agile burrow
It shouldn't be any different really
void cosmos
the text im using is doing this one at the end
agile burrow
Maybe a little different depending on what defn you use lol
void cosmos
its the dual def of the projecivity one i use
the diagram one
like its the same diagram but the arrows reversed
agile burrow
I understand
agile burrow
To an extent
void cosmos
yo there is something fundamental
that i am missing
okay so whats the difference between the direct sum and product?
like capital sigma and capital pi
like an element in the direct sum would be x1+x2+x3...
while the product would be a tuple?
agile burrow
So they are the same thing if you have a finite index set
void cosmos
So they are the same thing if you have a finite index set
how thoo
thats not like linear algebra
at all
or any set theory XD
agile burrow
But in the infinite case, an element of the direct sum has only finitely many nonzero summands while an element of the direct product can have infinitely many nonzero summands
agile burrow
I think you are interpreting the summation too literally
agile burrow
Do you not see how you can identify between tuples in the product and summations in the direct sum
void cosmos
i can
but
its not mentioned
that one does this
or maybe it is
but they are just magically dual to each other for me
agile burrow
Maybe because you skipped parts of the book where they define these notions
But yes, products and coproducts are dual in a sense
agile burrow
Coproduct = direct sum
void cosmos
XD okay
the section i covered was that
the product was defined for me using a universal property
equipped witth the projections
and co product was just the arrows reversed
wait ohh lmfao the arrows reversed would mean i have inclusion
agile burrow
Indeed
void cosmos
so like x1 --> x1+x2 <-- x2
and the universal property goes along but equipped with inclusion
so its rreversed lmfao
agile burrow
I'm glad you've come to this revelation
agile burrow
Happy to help
void cosmos
yea im going to be asking alot these days
especially once i get to tensor products which are the bane of my existence
one last thing , its an exercise but i think it follows trivially:
in a ring R: every R-module is projective <---> every R-module is injective
proof: consider the SES 0-->A-->B-->P where A ,B,P are R-modules
now if P is projective that this would mean that this is split exact
which would mean i can find a function g such that g o f ( f is B-->P) = 1 from P to B ( splitting lemma )
and i can do the same for A-->B ( cuz they are projective by hypothesis )?
so i can just literally reverse it?
so i get that this sequence is "right" exact split?
agile burrow
If it splits, it splits
I recall referring you to the splitting lemma sometime in the past
agile burrow
Oh sorry I didn't actually read your proof lol
But yeah, right split iff left split
agile burrow
Not as you've written it
void cosmos
wdyum
agile burrow
I don't understand what you mean when you say "i can do the same for A -> B"
void cosmos
A and B are R-modules so they are projective so the sequence 0->A->B would be split exact so i can do the same
uh oh somehting wrong
agile burrow
Indeed
void cosmos
1 momento
wait i have 0-->A-->B-->P all of those projective and by the splitting lemma i get P-->B-->A -->0
eaxct
exact
so im done?
P is injective
thats it?
cuz it splits both left and right
agile burrow
I genuinely have no idea what your argument is
agile burrow
I think you're muddying the proof a bit
barren sierra
You can look at N_{n-1} \to N \to N^n, where N^n is the nth composition series o...
Wait wdym N_{n - 1} and N^n? I feel you mean n is the length of the composition series but you say it's the "n-th composition series" which doesn't make sense to me since composition series unique up to ordering?
agile burrow
The point is that a module P is projective iff every short exact sequence 0 -> A -> B -> P -> 0 splits and a module Q is injective iff every short exact sequence 0 -> Q -> A -> B -> 0 splits
void cosmos
lmfao nvm nvm
yea but the latter seq already splits
By the first seq
Cuz B is projective
agile burrow
Right
agile burrow
I mean it's kind of inherent in saying that right splitting is equivalent to left splitting
void cosmos
yea thats what i was trying to prove i think
agile burrow
Sure, that's the content of the splitting lemma
agile burrow
Happy to help
pastel cliff
what gives us that second iso
bump
agile burrow
I'm just thinking that we have linear maps R[x]/(x^2 + 1) -> C[x]/(x^2 +1) and C -> C[x]/(x^2 + 1), so you get a unique map through the tensor product which you should be able to write down explicitly. If you do so, you can probably define an inverse in a straightforward way
I can write it out in more detail if it isn't that clear
pastel cliff
i understand that it's kinda like extending bilinear maps to homomorphisms
and that a map M x N -> L kinda factors through the tensor product M \otimes N
but i cant put these ideas to use with concrete examples
agile burrow
Sure, have you computed, say, Z/mZ \otimes Z/nZ
pastel cliff
i dont recall but we can do it
i think part of my problem is that im approaching this through the universal map defn
which seems to be more like a property than a definition
agile burrow
I mean, I use the universal property as a definition and then derive the other properties from that
Like the results that make some computations easier are things like R/I \otimes_R N is isomorphic to N/IN and stuff like that
But I guess that has more to do with right-exactness of tensoring
Sorry I can't be more enlightening at the moment, it's a long week for me
pastel cliff
Like the results that make some computations easier are things like R/I \otimes_...
i have this as a hw question lol
it's okay walter ik we all have lives to live too 
do yk any good resource by any chance then?
agile burrow
These are a bit long but I'm sure you can find what's relevant to you
pastel cliff
yeah i was abt to say
they arent usually 59 pages 
how relevant is the construction of the tensor product to it's actual usage
agile burrow
Not very relevant at all
barren sierra
https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf
Conrad my beloved
agile burrow
I think it's a good exercise in understanding quotients, but you'll never actually work with the explicit construction
pastel cliff
phew
agile burrow
That's why I always just remember the universal property
lethal dune
change ur name to something true plz 😦
wym it's true
lethal dune
(*for 0 dim cases)
dim widget
Wait wdym N_{n - 1} and N^n? I feel you mean n is the length of the composition ...
I misspoke if I said that, I meant for N_{i-1} \to N_i \to N^i to be exact so that the N^i are the composition factors of N.
dim widget
wat
Can you explain your reasoning?
Also I assume that Z is the integers and Z_4 is Z/4 is that right?
solar shore
yep
uh
admittedly i was too lazy to check the multiplication
but we were given this table
at the top, it says F is any of Q, R, C, or Z_p
and it says its a group
and then the ad - bc not 0...
is that enough reasoning?
dim widget
So 4 is not prime
solar shore
oh
dim widget
I think when they say Z_p they mean for p a prime.
solar shore
that says Z-p
p is prime woops
i see okay
i wrote that too and i missed my own thing..
dam
dim widget
So I will spoil it and say that none of these are groups
But it
is up to you to see why!
solar shore
oh quickly before you go @dim widget
for the first group, is it matrix multiplication modulo 4?
like is that implied?
dim widget
Yeah, it is matrix multiplication modulo 4, since no other way of multiplying and adding a, b, c, d is well-defined
a priori
solar shore
$\frac pq$ where $p, q > 0$
cloud walrusBOT
blanket
dim widget
Ah so not including 0
solar shore
i believe so
dim widget
Yeah that is more plausible, but I still don't believe it
So the inverse of a rational matrix with nonzero determinant is given by
1/det * (d -b | -c a) and you can easily see that these cannot all be positive
if all of a, b, c, d are positive
solar shore
yep thats what i had
solar shore
oh good point
dim widget
If you start allowing some of them to be zero (there is a difference between the english and european definitions of things like positive, negative, etc. which is sometimes confusing) then the matrix (1 1 | 0 1) obviously has no inverse with all positive entries.
The other two also have this problem: they have all the right properties but not inverses.
solar shore
right okay i kept getting inverses was the thing screwing me over from having them be groups
thank you!
dim widget
No worries!
solar shore
doing a bit of review right now, taking a break from partial diff eqs lolol
feel like im out of touch with the earlier group theory stuff
dim widget
It's always good to practice algebra, especially if you've been doing PDE's 🙂
tough raven
bump (here K = F(α) is a finite simple field extension)
dim widget
bump (here K = F(α) is a finite simple field extension)
What is your question?
tough raven
The message I replied to.
In full:
Let K = F(α) be a finite simple field extension.
For any polynomial p in K[X], let F(p) denote the field generated by F and the coefficients of p.
Let p, q be polynomials in K[X] with α as a root.
Is it true that if p divides q then F(q) is a subfield of F(p)?
If not, is there a subset of all polynomials with α as a root on which this does hold?
dim widget
What is F(q)?
tough raven
Oh oops
dim widget
q has coefficients in K so it doesn't seem to make sense as a standard notation.
Are p, q supposed to have F-coefficients?
is \alpha \notin F?
tough raven
Edited.
dim widget
I see.
tough raven
I introduced that notation the first time I asked the question and forgot just now that it was made up.
dim widget
So I mean polynomials p, q in K[X] with \alpha as a root are just p'(X)*(X - \alpha)^i so maybe this is a bit of a silly thing to write.
tough raven
They are?
What is p' here?
dim widget
p' is some polynomial without a root at \alpha
tough raven
Oh another polynomial
dim widget
Anyway just in general
Take G(X) to be the minimal polynomial of \alpha.
Let's make things concrete
tough raven
Take G(X) to be the minimal polynomial of \alpha.
Over F?
dim widget
Yes, over K it's just X - \alpha
But we can make things even more concrete we can take F = Q and K = Q(\omega) where \omega^3 = 1
Then p = X^2 + X + 1 divides q = (X^3 - 1)(X - \omega) and both have roots at \omega, but Q(p) is just Q and Q(q) = Q(\omega)
I don't know but I think your question mostly just fails all the time.
tough raven
Ah, I see. So you can have q = (X - α)G divisible by G without having coefficients in F in this case.
dim widget
Without a lot more conditions.
tough raven
All polynomials with α as a root was too ambitious, but what about polynomials dividing the minimal polynomial of α over F?
dim widget
Maybe it's better to ask what you were trying to prove in the first place that you got to this point? I think this question probably doesn't have a very interesting answer.
tough raven
I wasn't trying to prove anything in particular, just wondering more about the proof that finite simple extensions have finitely many intermediate fields.
That proceeds by mapping an intermediate field E to the minimal polynomial p of α over E and showing that is injective (because E is the field generated by F and the coefficients of p by a degree argument).
So I wondered if there's a Galois connection between intermediate subfields and some collection of polynomials, with the maps being “take the minimal polynomial” and “take the subfield generated by the coefficients”. If we take “all polynomials which are minimal polynomials of an intermediate subfield” there clearly is, and it's a bijection, but I was wondering if extends to a larger collection of polynomials.
dim widget
What do you get from N being normal?
dim widget
So let U \cong \oplus_i V_i as N-representations with the V_i = W_i^n_i and the W_i irreducible
Then if there's more than V_i there must exist some g \in G such that gV_i = V_j for j \neq i, can you see why?
low void
Is this a typo? Doesn't K[X] have the basis (1,X,X²,....)?
third sentence, in the middle
It's from this book, it is available here: http://math0.bnu.edu.cn/~liuym/Book-algebras-and-representations.pdf
south patrol
Yes seems a typo
low void
I wonder if the argument still works then, I'll see
oblique river
Yes; the element 1 has to act by the identity regardless
dim widget
Yes.
dim widget
I am doing what people usually do and abusing notation: U refers to a representation but also the underlying vector space of the representation.