You want to consider subfields Q(\zeta_{dn}) of Q(\zeta_n)?
#groups-rings-fields
dim widget
dim widget
Oopsies
Yeah those correspond to the congruence subgroups corresponding to the divisor.
So just the stuff in (Z/n)^* congruent to 1 mod m, which you can compute prime by prime.
dim widget
no worries 
Yes as long as N is finitely presented or R is noetherian and N is finitely generated.
Depends on the generality for R, is R still a semisimple ring?
south patrol
I will slow down lol
glossy crag
Depends on the generality for R, is R still a semisimple ring?
No assumptions made on R in the theorem, but the section is on semisimplicity.
Thanks.
dim widget
I will slow down lol
But it's definitely interesting to think about, you may know that every abelian galois extension of Q comes from a subgroup of Z/n^* in this way for some n
Not necessarily coming from a divisor, might be something more complex
south patrol
Yeah it's cool
white oxide
how is it that every factor group being of order p directly implies that each group is abelian? i know that every group of prime order is cyclic and hence abelian, but im wondering how we can deduce directly from p being a prime that such a group is abelian
lethal dune
prime order groups are cyclic and hence abelian
for a nontrivial element, look at the subgroup it generates
white oxide
no yea i know that
but why did they say actually cyclic
ig that's confusing me, their phrasing implies that a group of prime order is abelian automatically without using the fact that it's cyclic (otherwise i thought they might've wrote "where the factor groups are of order p, and hence cyclic and actually abelian")
woven obsidian
Because what is important is that they are abelian
lethal dune
ig that's confusing me, their phrasing implies that a group of prime order is ab...
yes they are cyclic and hence abelian
glossy crag
Is semisimplicity local like ACC/DCC (M/N and N have X property => M has X property)?
dim widget
Is semisimplicity local like ACC/DCC (M/N and N have X property => M has X prope...
No
Z/p^2 is not semisimple but Z/p is
white oxide
any hints for trying to show that a subgroup of order 35 is normal in a group of order 35(47)?
chilly ocean
willing to bet $5 it has to do with the fact that 47 is prime
white oxide
yea definitely it's some sylow shit
trying to show that any group of order (7)(5)(47) is abelian and cyclic lol, already showed that if H is of order 35 and K is of order 47 then H x K is abelian and cyclic, just trying to prove that any group G of order 35(47) is isomorphic to H x K (which requires that I show that K is a normal subgroup of G)
south patrol
Sorry 5 and 7 I meant lol
white oxide
all g
white oxide
Well like an element count should finish it up
wait wdym by element count
i was considering just taking the union of the sylow 7-subgroup and sylow 5-subgroup??
oh wait oops we could just consider HK
south patrol
Yeah so what I originally said but adapted with the right numbers (lol):
Don't you get n_5 = n_7 = 1
And that is p much enough to conclude
white oxide
i'm using a proof involving the commutator subgroup actually now
dim widget
@white oxide one way: the subgroup H of order 329 is normal because it’s index is the smalles prime dividing the order of the group. So the group is a semi direct product of a group of order 5 and one of order 329 which must be cyclic since 7 doesn’t divide 46. The automorphisms of Z/329 do not contain an element of order 5, so we see that in fact the group is cyclic
And every subgroup of a cyclic group is normal
tropic bear
Poorly worded question: What is the intuition I should keep about semidirect products of groups? I never really understood them when I took algebra. The notion of direct product is easy enough. Is there a source you think provides good intuition?
next obsidian
Direct product but one side is sort of twisted
rustic crown
split extensions 
(since you asked the question, i know i shouldn't be saying stuff to confuse you more >.<)
one motivation comes from asking the question, how do you put together the simple components of a group back to form it. for example, after loads of effort, people are able to classify all finite simple groups. how far are we from classifying all finite groups.
in general, if you're given groups N and H. then the question would be how to find all the groups G such that N is a normal subgroup and H is then isomorphic to G/N.
this would be very hard, to solve. but if magically H was also a subgroup of G such that NH = G, then you're in luck!
tropic bear
I see, so the constraint is "Whatever decomposition I find, I must do it by quotienting pieces by normal subgroups"?
rustic crown
i don't fully understand the question, sowwy >.<
tropic bear
Well, if I understand what you're saying, we're trying to decompose an arbitrary group G as some two pieces N and H, such that N is normal in G, and NH=G
So we're not decomposing G as a direct product, but rather as this pair (N, H), and then recurse to decompose N and H respectively in the same way. Is that right? Is that a decomposition one wants to achieve?
rustic crown
yep
by jordan holder theorem, the final gruops you get in the end won't depend on how you broke the group up
but the information how these were put together is totally lost
tropic bear
Is there some sort of guarantee here, of the form "If some_property(G), then G has a unique decomposition of this form"?
rustic crown
yee
delicate orchid
by jordan holder theorem, the final gruops you get in the end won't depend on ho...
first known use of the jordan holder theorem in human history
delicate orchid
when
south patrol
Uhh
delicate orchid
exactly
tropic bear
by jordan holder theorem, the final gruops you get in the end won't depend on ho...
From reading that theorem's statement, possibly naïve question: Why do we only list the normal subgroup chain, and don't bother decomposing the intermediate quotients? That is, if we decomposed G into N and H this way, where N is normal, don't we also need to decompose H?
south patrol
So if A is a left artinian & left noetherian ring then applying JHT to the left A-module A shows that there are finitely many simple left A-modules
south patrol
So in particular there are only finitely many k-linear representations of a finite group without assumption on characteristic &c.
south patrol
Ugh spent a while trying to do a proof in a nice way
But it came down to just like "one field is contained in the other and they have the same degree over Q hence equal" lol
delicate orchid
wait a min
nvm I'm equating irreducible and indecomposable in my head which is not true for modular reps
lucid pilot
Since this server is filled with people that are experienced with math can anyone give me a recipe for meth?
*math
delicate orchid
I was thinking "how could every kC-module be left/right artinian when there are finite groups with wild representation types"
lucid pilot
Because the last ones I've been given didn't work. And only produced shards when I wanted numbers to eat.
elder wave
...
delicate orchid
have some creativity to it
obsidian sleet
banned
delicate orchid
banned for being boring
long geyser
math sounds like meth lol amirite guys
elder wave
i hate superreactions
obsidian sleet
anyways proceed eveyone
elder wave
mine was on accident as well
paper aurora
what is meant by 'prime density reasons' in this stack exchange post? for context, we just finished proving that for p a prime that is 3 mod 4 not dividing the discriminant of the elliptic curve E from earlier, that the order of E(F_p) is p+1, and that for such primes p, we have an injection from Phi to E(F_p). also Phi is a finite group. for more context, the link is https://math.stackexchange.com/questions/3765573/points-of-finite-order-on-y2-x3dx
delicate orchid
is it the thing where primes are equally distrubuted among all possible values mod n is that the thing
hmmm
summer path
wow i left for 5 minutes and missed something worth superreacting to 
delicate orchid
since the statement doesn't hold for only finitely many there must be an infinite number of multiples of 4 (p+1 = 0 mod 4) that |Phi| divides, therefore (because Phi is finite) |Phi| divides 4?
is that the doodad? the thingimajig?
glossy crag
This is probably a stupid question, but I don't get how exactly $U$ is the sum of the isogenous components contained in itself. The proof shows that if $S\subset U$ is simple of type $\tau$, then $M_\tau\subset M$, but I'm having trouble showing that the sum of these $M_\tau$ is actually all of $U$.
cloud walrusBOT
Ocean Man
glossy crag
I tried the following (and failed): if $x\neq0$ in $U$ is arbitrary, then $Rx\subset M_{\tau_1}\oplus\cdots\oplus M_{\tau_n}$ for some $\tau_j$ (since $M=\bigoplus_\tau M_\tau$) and if $S\subset Rx$ is simple (such an S necessarily exists), then $S$ is wlog of type type $\tau_1$, so $S\subset M_{\tau_1}\subset U$, but this doesn't show that $x\in M_{\tau_1}$.
paper aurora
since the statement doesn't hold for only finitely many there must be an infinit...
hmm i dont see it. say |Phi| divides 7+1, 11+1, 19+1, and generally p+1 for p that is 3 mod 4. why does this mean |Phi| have to divide 4?
cloud walrusBOT
Ocean Man
paper aurora
since the statement doesn't hold for only finitely many there must be an infinit...
okay was looking around some more and found this paragraph in koblitz elliptic curve book. here i think m = |Phi|
boreal inlet
Okay uh. If you have an arbitrary family of A-modules (A is a commutative Unital ring), how do you define direct sum and direct product?
Atiyah makes it confusing
coral spindle
The direct product comes first.
The direct product's underlying set is just the ordinary set-theoretic cartesian product.
You may happily think of elements of the direct products to be very large tuples of elements in your A-module
The direct sum lives inside the direct product. It consists of tuples whose entries are 0 for all but finitely many (we sometimes say 'cofinitely' many) entries.
The action of the ring and the addition are done in the (dare I say) obvious way.
boreal inlet
You may happily think of elements of the direct products to be very large tuples...
oh so, finite tuples, but there's no limit in how long those can be
coral spindle
That's right
boreal inlet
wait so
glossy crag
Yes as long as N is finitely presented or R is noetherian and N is finitely gene...
I didn't ask then, but what does finitely presented mean here? For my purposes I need this to hold for N cyclic and R arbitrary (which after sketching a proof, I assumed it did), but the definition on Wikipedia is "there is an epimorphism R^n->N with the kernel finitely-generated", so the condition doesn't apply for R absolutely general.
boreal inlet
(a, b) and (c, d,e) can reside together?
coral spindle
Well we would really say (a, b, 0, 0, ....) and (c, d, e, 0, 0, 0, ....) if I'm interpreting what you're saying right
But yes
coral spindle
Ah no it's not
boreal inlet
hmm
coral spindle
We are looking at infinite tuples, but they're just zero in most places
boreal inlet
so for direct sum it's precisely the ones with finite non zero entries in the tuples
coral spindle
That's right
boreal inlet
and in case of direct product it doesn't have to be finite
coral spindle
Yup
boreal inlet
damn
coral spindle
Of course this means that if your family of A-modules is finite, there's no difference between the direct sum and product
boreal inlet
now I wonder why there's an distinction between these two
coral spindle
Right this is an interesting point
It comes from how we want these objects to behave.
boreal inlet
oh hmm..
coral spindle
The desired property of the direct sum/product can be expressed in what we might call category-theoretic terms. Don't let this put you off, it's quite simple.
Sorry this is going to take a second to type up
cloud walrusBOT
Boytjie (never-to-be-glomed)
coral spindle
This is obviously a bit technical...
boreal inlet
hmmm.... i got the projection thing
coral spindle
You can think of it as saying the following: the product is the 'least restricted' module that has each M_i as an image.
boreal inlet
OH
is it like the uniqueness you establish in case of quotient objects as well
i mean the 2nd statement
coral spindle
Yeah it's similar
For the sum, we have a similar property, but everything is reversed: we have inclusions M_i → S and a unique map eta : S → N.
boreal inlet
damn
coral spindle
And as it turns out, this describes the sum and product up to isomorphism.
Any object that satisfies the property above is isomorphic to the product, I mean.
boreal inlet
ahh
coral spindle
So I mean again this is a bit technical and strange
boreal inlet
i mean it is technical
coral spindle
But to kinda make a long story short, it turns out that we need the cofiniteness condition on the direct sum because in the unique map Sum M_i → N that must exist, we can't add together infinitely many things.
boreal inlet
the first time I saw this stuff for Quotient groups and rings I was shocked too
coral spindle
Haha yes well you do get used to it
I would really recommend trying to prove that the direct sum and product satisfy the properties I claim here. It ought to really explain what I'm getting at with this 'summing together infinitely many things' explanation
Hope that explains things in any case
boreal inlet
thank you very much 🥰 let me actually try prove what you said so maybe I don't forget about it
formal ermine
Hope that explains things in any case
maybe a simple and intuitive explanation would be even better
boreal inlet
coral spindle
har har
coral spindle
Well I give both the technical definition and the intuition behind why it causes things to be that way
so I hope to think that it really is approaching both
Hilbert be damned, I can't explain it to a person on the street
formal ermine
what's the best response to a harmonic analysis joke? "hardy haar haar"
coral spindle
That's p good
formal ermine
what's an algebraic group? can't find any definition online that isn't like "a group object in the category of k-varieties"
formal ermine
where
chilly ocean
nah dude
chilly ocean
learn to google
formal ermine
I overlooked that line
chilly ocean
literally how
formal ermine
I am illegally blind
G x G -> G being regular means G^2 -> G being regular?
or wait
that doesn't make sense
cuz we can't talk about the coordinate ring then, no?
I was thinking like product topology
what does it mean for G x G -> G to be regular then
chilly ocean
the product of two algebraic varieties doesn't get the product topology but something slightly different
formal ermine
I don't know how you define the product of varieties, I'm guessing you could just multiply their vanishing ideals together?
chilly ocean
i don't know what algebraic geometers like to call it
formal ermine
X*Y = V(I(X)*I(Y)) but this is probably wrong lol
rustic crown
you take the tensor product of the coordinate rings
chilly ocean
TENSOR PRODUCTS ARE ESSENTIAL TO THE EXISTENCE OF BLACK HOLES
rustic crown
chilly ocean
do algebraic geometers call the topology on product varieties the zariski topology
chilly ocean
might be a goofy question lol, just asking since i know it's not the product topology
shows what i know!
rustic crown
AG people play with many topologies tho, fppf, fpqc, etale, tho idk much about them 🙈
formal ermine
don't forget fnjc
rustic crown
hearing that for the first time
formal ermine
yeah because I just made it up
rustic crown
formal ermine
you take the tensor product of the coordinate rings
regarding this: are there any nice results on tensors of polynomial rings
rustic crown
AG people play with many topologies tho, fppf, fpqc, etale, tho idk much about t...
(but i think these are like topologies on a grothendieck site and not directly open subsets of X)
rustic crown
regarding this: are there any nice results on tensors of polynomial rings
k[x] ⊗_k k[y] = k[x, y] lol
formal ermine
wait so
we take the tensor product of our coordinate rings
how do we identify it with a variety (up to iso)
rustic crown
umm like it's still a reduced separated finite type blah blah algebra over k
rustic crown
and the categories of varieties over k and these are anti-equivalent
(i hope i'm right)
chilly radish
Yea like your new Algebra is still some quotient of a free Algebra
The corresponding variety is the vanishing locus of the ideal
It should correspond to something involving the 2 varieties but uhhh I don't remember
Ok yes
It's just the coordinate ring of the product
rustic crown
I just wanna understand the notion of a regular G x G -> G map
if G is affine, you just flip things around and try to understand ring maps A --> A⊗A
formal ermine
X*Y = V(I(X)\*I(Y)) but this is probably wrong lol
wait so this was right????
chilly radish
Idk why * means in this context
formal ermine
Idk why * means in this context
product
chilly radish
k[V] (x) k[W] = k[V x W]
rustic crown
if G is affine, you just flip things around and try to understand ring maps A --...
now G being a group is same as A being a hopf algebra
idk how to work with non-affine G tho
formal ermine
I'm afraid that I don't understand the definition of a hopf algebra either
rustic crown
it's the definition of algebra but you flip all the arrows around
so instead of having a multiplication A⊗A --> A
you have a comultiplication A --> A⊗A
and then you phrase all the good stuff like associativity, etc as commutative diagrams
for example, k[x] is a hopf algebra
the map is k[x] --> k[x] ⊗ k[x] sending x to x⊗1 + 1⊗x
rustic crown
this is same as saying the affine line A^1 is a group under +
rustic crown
for example, k[x] is a hopf algebra
the map is k[x] --> k[x] ⊗ k[x] sending x to...
the function x in k[A^1] gives you the x-coordinate of a point and so x⊗1+1⊗x is the function on k[A^1 x_k A^1] which adds up the coordinates of the two points.
chilly radish
and then you phrase all the good stuff like associativity, etc as commutative di...
WHAT ABOUT THE ANTIPODE
They always forget the antipode
white oxide
i still don't get what why a = 0, because if a^2 is congruent to +-7 (mod 5) then 0 is congruent to 2 mod 5 which is clearly not the case
rustic crown
yea that's weird
they probably meant go mod 7
then either b = 0 mod 7, which implies a = 0 mod 7. or (a/b)^2 = 5 mod 7. which is not possible.
if you go mod 5 then there are no solutions for a^2 = +-2 mod 5
in either case you get that 7 is irred in Z[sqrt5]
south patrol
if you go mod 5 then there are no solutions for a^2 = +-2 mod 5
Wait ye lo this nice
Weird reasoning afterwards ig lol
dim widget
@glossy crag in general if M is finitely presented iff Hom(M, -) preserves colimits as a functor (as Chmonkey mentions, only filtered colimits). However I suppose to just preserve direct sums M only has to be finitely generated it seems
glossy crag
Any D-vector space V is End_D(V)-simple, regardless of dimension, right?
rustic crown
D is a division ring i'm guessing? as long as V is non-zero that sounds good
glossy crag
D is a division ring i'm guessing? as long as V is non-zero that sounds good
Yes, D is a division ring.
glossy crag
D is a division ring i'm guessing? as long as V is non-zero that sounds good
Yeah, just wanted to double-check.
rustic crown
next obsidian
<@233934386863669248> in general if M is finitely presented iff Hom(M, -) preser...
No, it’s finitely presented iff this commutes with all direct limits (or filtered colimits if you want to use that language)
You always commute with limits by definition of what a limit is
wet phoenix
hey y’all, I just can’t figure out what the operation in example 9 is doing exactly. I know it’s simple but im not seeing it.
I see that it’s not commutative b * a =a but a* b=c can someone please explain the operation
wet phoenix
Why is a *a =b
wet phoenix
yes row entry * column entry
next obsidian
Okay, then it’s because b is in the square associated to (a,a)
That’s it
There’s nothing more you can say
It’s just defined that way
wet phoenix
I thought there was something I wasn’t getting, that’s just how it’s defined I see now
past knot
So I am trying to show that every element of $\mathbb{Q}[\omega]$ is representable as $a_0 + a_1\omega + a_2\omega^2 + ... + a_{p-2}w^{p-2}$ where all $a_i$ are in $\mathbb{Q}$. The hint is that I have to show that $\omega$ is a root to $f(t) = t^{p-1} + ... +t + 1$, and show it is irreducible by considering $f(t+1) = ((t+1)^p - 1)/t$, and I think I've got a proof, but I'd like to get it checked
cloud walrusBOT
mirzathecutiepie
past knot
$w$ is a root of the polynomial because it can be rewritten as $(t^p - 1)/(t-1)$ and when we plug in $w$, which I forgot to mention is the pth root of unity where p is an odd prime, we get $(w^p - 1)/(w-1) = 0$
cloud walrusBOT
mirzathecutiepie
next obsidian
How do you know f(t+1) has that form?
Also showing it’s a root, that’s indeed how you’d do it
Oh never mind sure, you’ve written down what f is equal to so f(t+1) follows from that
Yeah this proof works, I assume you’re showing f(t+1) is irr by Eisenstein?
past knot
Now we need to show that it is irreducible. The book suggests using the eisenstein critereon. We know that $Q$ is a field and that that means there is only one maximal ideal namely ${0}$, here $f(t+1) = \frac{(t+1)^p - 1}{t}$. We can consider for now just $t f(t+1) = (t+1)^p - 1$, if this is irreducible over $\mathbb{Q}$ then $f(t+1)$ will also be irreducible. The book specifies that the first and last elements of a polynomial have to be outside the maximal ideal while the elements in between are not, which works here, since all the in-between elements are 0
cloud walrusBOT
mirzathecutiepie
past knot
**mirzathecutiepie**
This one is the iffy part for me
Anyways if $f(t+1)$ were reducible then $t f(t+1)$ would be too, deriving a contradiction, so therefore $f(t+1)$ is irreducible as well. Now we did all of this so that we could take the quotient by the ideal and show the following equality $\mathbb{Q}[\omega] = \mathbb{Q}[X] / (f(t))$
cloud walrusBOT
mirzathecutiepie
next obsidian
No, this step is not valid
tf(t+1) is not irreducible, it’s divisible by t
Eisenstein requires a prime ideal where all coefficients except the leading coefficient are in it, and the constant is not in the square of the ideal
past knot
ahh
next obsidian
You can look at f(t+1)
And then see that in the numerator the constant is 0
And everything is divisible by t, so you can remove that, that’s the /t handled
So the constant of f(t+1) is the degree 1 term of (t+1)^p
past knot
hm
next obsidian
Or really just, use binomial expansion
past knot
which is p
next obsidian
To see why Eisenstein is applicable
past knot
Okay but before that can we see if the last step is correct
next obsidian
What last step
Ah sure
Yes you can write it that way and then apply known field theory results
Or you can just unwind why that works and why Q[t]/(f(t)) is finite dimensional
With basis 1,t,…,t^n-1 for f of degree n
past knot
hm i realize my last step is iffy too
next obsidian
I think it’s nicer to do it the latter way
No, it works
You know F(a) is just F[x]/(m_f(x))
And you demonstrated that f was the min poly of w
But I think it’s worth expanding on how you get the basis
You showed that
w^p-1 = stuff in degree <= p-2 in w
So Q(w) is spanned by 1,w,…,w^p-2
past knot
i wanted to show here that every element would be able to be written as (w-r_0) ... (w-r_{p-2})
next obsidian
Why?
past knot
Then expanding it would give the result right
next obsidian
Which result?
Oh I see, okay it’s worth expanding on why your isomorphism you wrote does it for you
past knot
that everything can be written as$ a_0 + a_1 w + ... + a_{p-2}w^{p-2}$
next obsidian
But we can just do it directly
next obsidian
You showed that
w^p-1 = stuff in degree <= p-2 in w
I had degrees off by one, so please read this
past knot
You showed that
w^p-1 = stuff in degree <= p-2 in w
what does stuff in degree mean here
next obsidian
Anything in Q(w) is a polynomial in w, but this relations means you can always take things which require more than p-2 ws and replace p-1 ws with something with <= p-2 ws
Like
It’s a polynomial in w
Of degree p-2
The fact that f(w) = 0 tells you this
Knowing that f = t^p-1 + … + 1
past knot
hmm
past knot
w^p-1 = -w^p-2 - … - 1
could you explain this a little more
next obsidian
Take f(w) = 0
past knot
okay
past knot
oh
next obsidian
Good?
past knot
right okay
past knot
Yeah, that makes sense
next obsidian
So really for your statement that’s enough
But really you want to say this is a basis
past knot
We removed the dependency so it's a basis
next obsidian
This is where linear independence is needed
And why you needed f to be irreducible
If there’s a dependence then there’s a lower degree poly w satisfies, so it divides f
And then f isn’t irreducible
past knot
If there’s a dependence then there’s a lower degree poly w satisfies, so it divi...
right, makes sense
past knot
but also don't we need f to be irreducible to be able to take the ideal and thus the quotient
next obsidian
It won’t be an isomorphism
past knot
ah right
next obsidian
Or uhhh
past knot
it wont be a field
past knot
could we go over this definition of eisenstein too
next obsidian
Sending t to w won’t be injevtive
Is the problem
Sure
The most general form of Eisenstein I know is this:
Let A be a UFD, and F its field of fractions
past knot
Let $M$ be a maximal ideal in a ring $R$ and let:
$f(x) = a_n x^n + ... + a_0$ where $n \geq 1$
be a polynomial over $R$ such that $a_n \not \in M, a_i \in M$ for all i < n, and $a_0 \not \in M^2$. Then f is irreducible over $R$
next obsidian
If f = a_nx^n + … + a_1x + a_0 is a polyomial in A[x] with a prime ideal p < A such that a_n not in p, a_0,…,a_n-1 in p, and a_0 not in p^2, then f is irreducible in F[x]
cloud walrusBOT
mirzathecutiepie
next obsidian
You need a bit more assumptions I think, but maybe the maximal ideal-ness saves you?
In particular I imagine you need at least domain
next obsidian
If f = a_nx^n + … + a_1x + a_0 is a polyomial in A[x] with a prime ideal p < A s...
Actually I don’t think you need to introduce F. You can always stay in A and then Gauss’s lemma handles it, so maybe you don’t even need UFD
I think all you need is domain
So this is how the proof goes
past knot
So if I'm looking at $\mathbb{Q}$ then the only maximal ideal this guy has is ${0}$ right, so then I can only consider polynomials with the biggest and smallest degree terms
next obsidian
No
cloud walrusBOT
mirzathecutiepie
next obsidian
You’re showing it’s irreducible over Z using the prime /maximal ideal (p)
And then applying Gauss’s lemma to see that irreducible over Z => irreducible over Q
Eisenstein’s criterion is useless over a field, because a_0 in p but not in p^2 is impossible
Since p is forced to be 0
past knot
ahh i see
next obsidian
Well it’s useful if you can apply Gauss’s lemma
past knot
okay
next obsidian
But you have to go through a ring
past knot
this is pretty cool
next obsidian
Swag
past knot
Alright makes sense, that bit about the Q[w] as a vector space over Q really made sense, thanks
past knot
hm how
next obsidian
The proof I gave shows this field has basis {1,t,…,t^n-1} for f being of degree t (I mean the class of t in the quotient ofc)
And the iso to Q(w) sends t to w
Now f is of degree p-1
So you win
past knot
w8 dont we want degree p - 2
next obsidian
No the min poly is degree p-1
Because that let’s us write
degree p-1 = stuff in degree <= p-2
boreal inlet
Let's assume A is a finite direct product of rings A_i
boreal inlet
Gotta show that this is an ideal of A
The additive subgroup part is obvious (it's even isomorphic to A_i)
But there's another condition to check
For any $x \in A$ and $y \in \alpha_i$, we have $xy \in \alpha_i$
cloud walrusBOT
Riku
boreal inlet
how to show this? I don't know how the multiplication is defined in the direct product
this is probably just a guess but does componentwise multiplication work?
okay nvm
This is what Lang says
if that does work, $xy = (0,0,.....,x_ia_i,....0)$ so it's in $\alpha_i$
cloud walrusBOT
Riku
ebon pine
rather was rather easy no?
ebon pine
I see, whenever in doubt do the natural thing
boreal inlet
dim widget
No, it’s finitely presented iff this commutes with all direct limits (or filtere...
wow that's really interesting tell me more
next obsidian
You originally wrote down limit which is the only reason I said anything
Although it is also important to have the word filtered otherwise the result can still fail spectacularly
dim widget
You originally wrote down limit which is the only reason I said anything
yes i meant to say "direct limit"
glossy crag
Is this proof of R,S left-semisimple rings (semisimple as left modules over themselves) => RxS left-semisimple valid:
R,S left-semisimple => R,S direct sums of some minimal left ideals. As left ideals of RxS these remain minimal (since left ideals of RxS are products of left ideals of R and S; I assume this holds the same way it does for 2-sided ideals) and RxS is a direct sum of these => RxS left-semisimple. Same way with right instead of left.
dim widget
@glossy crag yes
glossy crag
<@233934386863669248> yes
So this way I can show via the Wedderburn structure theorem R left-semisimple => R-right semisimple, since R is isomorphic to a finite product of matrix rings over division rings and these are right-semisimple.
Now if I want to show the converse (right ss=>left ss), is there some simple way of reducing this to (left=>right) via R^op somehow?
Or do I just go "the theory developed for left ss rings holds exactly the same way for right ss rings, symmetry, blah blah".
dim widget
@glossy crag you can show that if R is a matrix algebra over a division ring then R has a canonical involution. Using Artin wedderburn this makes the right iff left very trivial
But yeah you could also argue that however you proved Artin wedderburn using left semi simplicity the same argument would go through for the opposite rings
glossy crag
<@233934386863669248> you can show that if R is a matrix algebra over a division...
So what's the canonical involution for M_n(K), K a field? Surely not transposition, that's an antiisomorphism?
dim widget
You exactly want an anti-isomorphism
glossy crag
But shouldn't an involution be a ring homomomorphism such that f^2=1?
dim widget
It depends on your definition
I want to transfer structural results about left ideals in A to those about right ideals in A^op so I want an anti-isomorphism
Maybe you’d prefer anti-involution as terminology
glossy crag
I want to transfer structural results about left ideals in A to those about righ...
I suppose, yeah.
Thank you.
Another dumb semisimplicity question: I know that if R is semisimple there are only finitely many isomorphism classes of simple R-modules. Now I want to show that in the direct sum decomposition of R every class is represented, i.e. if \tau_1,\dots,\tau_k are all the classes, then R\cong M_1^n_1\oplus\cdots\oplus M_k^n_k. where M_i is of type \tau_i. How to do this? I know that every simple R-module is isomorphic to a submodule of R, but how do I know they are all represented in the direct sum. Equivalently, how do I show that the isogenous component of type \tau_i is a finite sum.
Actually nvm, I know that if M=\sum_i N_i with N_i simple, then every simple N\subset M is isomorphic to one of the N_i, therefore if R is the direct sum of some simple submodules, every type \tau_i has to be represented in that decomposition.
next obsidian
yes i meant to say "direct limit"
Okay, that makes sense
dim widget
@glossy crag I don’t think your argument is enough if I understand it correctly. You have to show that a simple ring has a unique nontrivial irreducible left module at some point.
glossy crag
<@233934386863669248> I don’t think your argument is enough if I understand it c...
Wdym?
dim widget
I think your statement is slightly wonky
glossy crag
Gimme a moment, I'm clarifying.
dim widget
For instance if R is M_n(K) you have to show that K^n is the unique simple module for M_n(K), that’s what your statement amounts to for simple rings
The statement for semisimple rings reduces to doing this for K replaced with an arbitrary finite dimensional division ring
glossy crag
Gimme a moment, I'm clarifying.
That a (left) semisimple R has only finitely many isomorphism classes of simple modules is a classical result, not my invention. If R is semisimple, it is the sum of some minimal left ideals => sum of finitely many, since 1 belongs to finitely many, so I can write R=I_1\oplus\cdots\oplus I_m with I_j minimal left ideals. Now my claim was that every isomorphism type is represented here: it is known that every simple R-module is isomorphic to a simple submodule of R and if you have a decomposition of M as a sum of simple N_i, then every simple submodule of M is isomorphic to one of the N_i (again, none of this comes from me), so if S simple of type \tau, it is isomorphic to one of the I_j, so every type \tau is represented in this decomposition.
dim widget
Okay, I though you wanted to prove the statement that every simple R module is isomorphic to a sub module of R
Since that seemed to me to be equivalent to what you were asking
glossy crag
Okay, I though you wanted to prove the statement that every simple R module is i...
That holds for R semisimple and I already digested that, I think. IDK if it holds for R general.
dim widget
But yes upgrading from the proof for R simple to R semisimple is trivial
delicate orchid
woaw!
glossy crag
But yeah you could also argue that however you proved Artin wedderburn using lef...
Could I say here that $R$ as a right module over itself is the same thing as $R^{op}$ as a left module over itself, so right semisimple $\implies R_R$ semisimple $\implies\prescript{}{R^{op}}{R^{op}}$ semisimple $\implies R^{op}_{R^{op}}$ semisimple $\implies\prescript{}{R}{R}$ semisimple to complete my proof of left semisimple iff right semisimple?
cloud walrusBOT
leave_no_norm
coral spindle
How do you justify R^op being left semisimple implying it being right semisimple? This seems circular to me.
hollow mica
after applying the induction assumption how does the expression simplify to that?
this is for proving the generalized associative law for groups
glossy crag
How do you justify R^op being left semisimple implying it being right semisimple...
That part I already covered with the Wedderburn structure theorem (a left semisimple ring is isomorphic to a product of matrix rings over division rings and such products are right semisimple), I was looking for a simple way of doing the converse.
rustic crown
Could I say here that $R$ as a right module over itself is the same thing as $R^...
idk how to show left semi-simple => right semi-simple since i only worked with f.d. alg/k alg closed. but i'm guessing the wedderburn-artin works in higher generality as well?
but yea your argument shows that to show left iff right it suffies to show left => right
rustic crown
That part I already covered with the Wedderburn structure theorem (a left semisi...
(oh oops 🙈)
rustic crown
this is for proving the generalized associative law for groups
pull out the a1 and rest would follow inductively
that product is (a1 * (stuff)) * (other stuff) and by associativity, this is same as a1 * ((stuff) * (other stuff))
hollow mica
nvm I figured it out
rustic crown
That part I already covered with the Wedderburn structure theorem (a left semisi...
oh so do you just say each M(n, D) is left-semisimple as it's a direct sum of its columns and by appealing to transpose, it's then right semi-simple as it's a direct sum of rows.
wraith cargo
is $\mathrm{Tor}_1^{k[x,y,z,w]}(k[x,y],k[z,w])=0$?
cloud walrusBOT
Irony Incarnate
wraith cargo
I'm trying to figure out if $(x,y)\cap (w,z) = (x,y)(w,z)$ (I strongly suspect this but I want to be sure)
cloud walrusBOT
Irony Incarnate
rustic crown
is $\mathrm{Tor}_1^{k[x,y,z,w]}(k[x,y],k[z,w])=0$?
yep
i can see it using the koszul complex with the k[x,y]-sequence (x, y)
wraith cargo
thanks!
Eisenbud exercises can be rough
I still have no idea how the hilbert function works 
rustic crown
i still have no idea how commutative algebra works
wraith cargo
I thought you were the algebraic geometry type hmmmm
rustic crown
i directly started with learning scheme theory as you need to know only like tensor prods and localizations to get started
but i'll have to come back to commutative alg now >.<
will try reading matsumura
wraith cargo
I wanna be a commutative algebra person before going deeper into algebraic geometry
wraith cargo
will try reading matsumura
I really don't like matsumura lol
Idk the way it's formatted is not my taste
Literally feels like a list of results
Eisenbud I really like
It's long af but his writing style is phenomenal tho sometimes it can be somewhat overwhelmingly dense
rustic crown
my motivation for doing matsumura
formal ermine
comm alg feels like rusty metal to me
wraith cargo
breaks your brain I get that
rustic crown
and this 😂
#algebraic-geometry message
wraith cargo
my motivation for doing matsumura
If you read Eisenbud you'll get brain swelling and you'll have to go to the hospital
In conclusion Eisenbud good
(warning: I am a person that likes books other people are known to dislike)
you could say I go against big math :smug:
rustic crown
i like cute books >.<
i hope matsumura gets cuter >.<
i like just started and it already hurts my brain
rustic crown
in second section he decomposes modules into direct sums of countably generated submodules or something
in order to show projective/local = free even in non f.g. case
wraith cargo
Damn eisenbud started talking abt like how localization is connected to sheaves in ch. 2 lmao
rustic crown
sheaves are nice
wraith cargo
I think he talks abt projective modules later in some like homological algebra setting
wraith cargo
but wanna be a commie alg person before starting that
so much field theory
tbh like the nt I've read up to this point isn't very comm alg heavy
I can imagine ti becomes later tho lmao
rustic crown
idk, when i did some alg-nt like 1.5 years ago a lot of arguments felt like they would be very natural geometrically but are phrased purely in terms of commutative alg and so look shit
even proving like equivalent conditions for being a dedekind domain was weird
wraith cargo
idk, when i did some alg-nt like 1.5 years ago a lot of arguments felt like they...
what did you read
rustic crown
we had a first course in alg-nt
wraith cargo
ah
Idk I like Neukirch a lot
rustic crown
so like stuff about dedekind domains, and a little geometry of numbers
rustic crown
oh i see
wraith cargo
He even does a bit of algebraic geo to present how it's used in nt
rustic crown
i knew no commie alg back then, so neukirch was hard
wraith cargo
Oh yeah lol that is true haha
rustic crown
and knew some basics, so marcus felt like reinventing the wheel
wraith cargo
Idk where the boundary of comm alg and like other alg is tho
my algebra class just kinda mixed a lot of stuff
and did some more explicit comm alg stuff towards the end of the class
rustic crown
will read neukirch after matsumura maybe
wraith cargo
Yeee
Neukirch also has fun problems
chilly ocean
Hey guys quick question
Primitive element theorem for finite extensions of finite fields
The proof is just that we say the extension field is finite, hence has cyclic unit group, and we adjoin a generator of this to the base field, right?
glossy crag
oh so do you just say each M(n, D) is left-semisimple as it's a direct sum of it...
Yes, pretty much.
south patrol
The proof is just that we say the extension field is finite, hence has cyclic un...
Yes
chilly ocean
Ok nice
south patrol
Well, that works, but there is another way you can do it i think
Well tbh using a generator is probably best lol
formal ermine
I'm not sure if this is the right channel for this
but given a cone
how can we determine the linear forms that for its positive locus
I've noticed that if we're working in Q^2
then it seems to be just swapping around the vector components and negating some of them?
not sure if this is just a coincidence and that I've been looking at too well behaved examples
waxen mural
What is a cyclic permutation? The book talked about permutations in an earlier chapter but it's not said anything about a cyclic permutation.
formal ermine
yeah
waxen mural
yeah
ion know bro im a bit confused
how can ~ be an equivalence relation
how is (x1, x2, x3) ~ (x1, x2, x3)
true
cause that means f(x1) = x1, f(x2) = x2, f(x3) = x3, and thats (x1)(x2)(x3)
coral spindle
The identity is typically thought of as a 'trivial' cyclic permutation.
Writing (x1)(x2)(x3) is silly, though. We could also write (x1 x2)(x3) and then loudly proclaim, "look, it's not just one cycle!"
It is still one cycle.
In any case I think it is far simpler to think of this as just an action of the cyclic group. Then the equivalence relation is a particular case of a general thing.
formal ermine
not sure if this is just a coincidence and that I've been looking at too well be...
another question on a similar topic: if I know that a conic is 0-sharp then how do I find the corresponding functional?
tribal moss
In this context I think a "cyclic permutation" means something different from the general meaning @coral spindle and @formal ermine describe.
Note that it speaks about a "cyclic permuation" of a particular p-tuple. In that case it is clear that it speaks about the relation between
(x_1,x_2,x_3,...,x_p) and (x_{k},x_{k+1},...,x_p,x_1,x_2,...,x_{k-1})
specifically, for some k with 1<=k<=p.
formal ermine
if the conics generating vectors lie in the positive spectrum then it's pretty clear that we can just choose any absolutely positive functional, no?
coral spindle
You're of course right Tropo, my mistake
waxen mural
In this context I think a "cyclic permutation" means something different from th...
Ah thank you, that makes the problem much easier now.
sullen sand
. Let S be the set of all integers 0, ±1, ±2, ..., ±n, · · · . For a, b in S
de ne ∗ by a ∗ b = a − b. Verify the following:
(a) a ∗ b = 1 = b ∗ a unless a = b.
(b) (a ∗ b) ∗ c 6= a ∗ (b ∗ c) in general. Under what conditions on a, b, c is
(a ∗ b) ∗ c = a ∗ (b ∗ c)?
(c) The integer 0 has the property that a ∗ 0 = a for every a in S.
(d) For a in S, a ∗ a = 0
coral spindle
Do you have a question or are you just sharing a problem?
sullen sand
I need help with that
formal ermine
!status
flat treeBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
OK, post your answer then and we'll be happy to check it.
hollow fjord
Hi guys!. I have a question on linear representation theory. We have that for finite abelian groups every irreducible complex representation has degree 1. This is a consequence of Shur's lemma. I want to know if this holds for compact groups (we have an analogous Shur's lemma). I'm reading the proof for the finite case and i don't see where we use that G is finite, so... somebody knows if there is a problem with just taking that proof and replace finite abelian group for compact abelian group?
coral spindle
Well yes we do have a Schur lemma, but it really has nothing to do with the groups.
Schur's lemma for rings states that if M is a simple module, then End(M) is a division ring
For k-algebras where k is an algebraically closed field, and M is fin. dim., we have End(M) is isomorphic to k
So since an irreducible, finite-dimensional representation of any group G – whether finite or infinite – is in particular just a f.d. simple kG-module, we obtain the lemma.
delicate orchid
Does a simultaneous diagonalisation argument work for the infinite case
coral spindle
It really ought to, just by induction on the dimension of the spaces, right?
Well actually in any case we get that the action of each element is a homothety anyway, so it doesn't matter
long nebula
Can someone explain this to me
I know what the words mean... individually
But wtf?
karmic moat
oh we gotta start teaching this one in elem lin alg
long nebula
(I don't have any context for this, I just saw it in a meme)
alpine island
For any (nonzero) group, is it possible to define another operation such that it becomes a field?
long nebula
For any (nonzero) group, is it possible to define another operation such that it...
No
What are the possible orders of finite fields?
alpine island
Right, and you can have a finite group of any order
alpine island
What about the infinite case?
delicate bloom
sort of related, it's not even necessarily unique, the fields F_3(x) and Q have the same multiplicative groups
well, I mean from the perspective of starting with just an infinite group for multiplication, you could attach an addition to it to make either field
dim widget
Also any infinite field has to have infinitely many generators of its multiplicative group so that’s a big restriction
alpine island
Yeah, it's clearly true for a lot of examples
But like, if you took an arbitrary infinite set, is there a binary operation (let's say addition for now) that you can define on it such that there is no other binary operation you could define (multiplication) where it becomes a field?
lusty marlin
But like, if you took an arbitrary infinite set, is there a binary operation (le...
If you consider your arbitrary infinite set with a binary operation to be any non-abelian group, then there is obviously no other binary operation one could define on it such that the set considered along with the two operations is a field.
near star
can anybody explain how this is even possible
What if H intersect K = 1? Is it impossible to have two disjoint subgroups of an infinite group that each have a finite index?
Since G is a possibly infinite group, wouldnt |G : H intersect K| be infinite, which is not <= mn
ebon pine
@glossy crag can I talk to you for a min, I had few questions about knapp
long nebula
What if H intersect K = 1? Is it impossible to have two disjoint subgroups of an...
Well the thing you're being asked to prove seems to answer that question haha, maybe it would help if you thought of some examples for what G, H, and K could be?
meager fractal
How can i find [Q(ζp):Q] where ζp is the pth primitive root of unity?
rustic crown
show that the p-th cyclotomic poly is irred over Q
meager fractal
cyclotomic?
rustic crown
Phi_p(x) = 1+x+...+x^(p-1)
rustic crown
there are many descriptions of this, when p is a prime stuff follows easily from eisenstein's criterion
rustic crown
and how is ζ a root of this?
for this, notice that Phi_p(x) = (x^p - 1)/(x - 1)
so any non-trivial pth root of unity is a root of Phi_p(x)
(and in general, you define Phi_n(x) as the polynomial whose roots are all primitive nth roots of unity)
rustic crown
they're really fun
meager fractal
should be simple enough to show irreducibility
rustic crown
yep, it's a very standard eisenstein trick
meager fractal
they're really fun
i think beyond the scope of this course probably gonan find them in advanced alg structures
rustic crown
but can be tricky if you've never seen it before
meager fractal
do I use the fact that f irreducible iff f(X-b) is also irreducible?
rustic crown
i think beyond the scope of this course probably gonan find them in advanced alg...
you can use them to prove a few fun things, like the arithmetic progression 1+nZ has infinitely many positive primes
another one is that any finite (possibly non-commutative) domain, is actually a fintie field.
oh another cute one is that you can show if G is a finite subgroup of F* where F is a field, then G is cyclic.
south patrol
Oh are you saying can prove that using cyclotomics? How come
rustic crown
because if |G| = n, then all elements of G are exactly the distinct roots of x^n-1.
(this also tells you that if char F = p, then p doesn't divide n)
and since x^n-1 doesn't have repeated roots, any root of phi_n(x) can't also satisfy x^d-1 for some d | n, d < n
south patrol
Hm isn't that just a rephrasing of the normal proof ig
rustic crown
ig yea, but it literally hands you down the generator
pick any root of phi_n(x), that's a generator!
wraith cargo
me when primitive root
rustic crown
the proof i know all seem to have one hard/tricky part involved. here it's argument using cyclotomics, else there is combinatorics of phi(n), one is to use some sort of structure argument to find an element of maximal order
south patrol
I think the combinatorics of phi(n) are just embedded in saying x^n -1 is product of dth cyclotomic for each d|n
But yeahh
Inch resting
meager fractal
uhh so
can i say X^p - 2 in Q irreducible
because 2 is prime
2|2
2 doesnt divide 1
and 2^2 doesnt divide 2
so eisenstein?
i dont see smt wrong it just feels weird
rustic crown
it feels weird because you're also using Gauss' lemma
a primitive polynomial f in Z[x] is irred over Z iff if it's irred over Q
and eisenstein applies for stuff in Z[x]
as there are no (non-zero ) prime numbers in a field
south patrol
Okay I'll try to work out the meme alternative
meager fractal
can I have a hint for this
I showed that assuming there is a root in K all roots are in K
so now I have to show that assuming f has no roots in K ==> f irreducible
but idk how to move forward with that
south patrol
Consider the field formed by adjoining a single root
You need to show it has degree p over K
Which should be immediate from the work you've already done
delicate bloom
I think you can write down its factorization in an extension field and find that multiplying any of the linear terms is gonna give bad coeffs
meager fractal
assume no roots
let a be a root of f
consider K(a)
show that if a in K(a), a+1 is also in K(a)?
as K(a) has characteristic p
chilly radish
This shows that if one root is in all roots are in, but a factorisation needn't have a root
Like you can have f be a product of 2 polynomials, neither is which have a root
chilly radish
It doesn't
Not immediately at least
What mero is saying is that in K(a) you can write f as a product of its roots
meager fractal
:O
chilly radish
And then use that to show that if you can write f as a product gh then one of those has to have coefficients not in K
If both are not constant
meager fractal
i have smt ignore what i sent for now
i have that: all roots of f are in K(α) while no roots of f are in K
so f factorizes as (x-a)(x-a-1)...(x-a-p-1)
from there i have no idea how irred would follow
might be missing some important thm or idea 😢
that I cant remember
south patrol
you can also think slightly Galois theoretically about it lol
delicate bloom
so f factorizes as (x-a)(x-a-1)...(x-a-p-1)
multiply together m of those and then look at the coefficient on x^{m-1}
meager fractal
nooooooooooooooooooooooooooo
frigid lark
Look at (x-a)(x-a-1) = x^2 -(2a+1)x + a(a+1)
south patrol
Like suppose g|f and g is irreducible. You can show g has p distinct roots in K(a) after you adjoin a root a in K(a)
meager fractal
but then how does irred follow 😔
because we consider the polynomial to be able to be factorized this way in K(a) not K
south patrol
By considering the action of the Galois group on a
frigid lark
Any factor of f will have to be a linear combo of the (X - a - i)'s which don't live in K[X]
delicate bloom
the coefficient on the x^{m-1} term will have m*a in it, which is in K only when m=p
frigid lark
because we consider the polynomial to be able to be factorized this way in K(a) ...
A factorisation in k[X] translates to a factorisation in K(a)[X]
meager fractal
ok
i see
ty
so any combination of these factors other than all of them gives us coefficients outside of K thus it must be irreducible as we cannot factor f into g,h in K[x] where h irreducible and degree h =/= f except if h is 1
sleek fable
Hi, I'm trying to show that if A is a factorial ring then for any multiplicative subset S of A, S^{-1}A is factorial
where S^{-1}A is the localized of A at S
to do this, I first tried to found the irreducible elements of S^{-1}A
I guess you disagree with Lang on this one
he mostly uses factorial ring
sleek fable
(in Algebra)
sleek fable
Well in french we use factoriEl
I want to prove that x/1 is irreducible in S^{-1}A iif x is irreducible in A and (x) \cap S = \emptyset
I did prove the right to left implication but I struggle with left implies right
Here is what I started with :
Let $\frac{x}{1} \in S^{-1}A$ be an irreducible element. If $(x) \cap S \neq \emptyset$, $\frac{x}{1}$ is invertible in $S^{-1}A$ thus it is not irreducible. We conclude that $(x) \cap S = \emptyset$.
cloud walrusBOT
valentin9912
sleek fable
Now, lets prove that x is irreducible in A
Let $a,b \in A$ be elements such that $x = ab$. We want to show that either $a$ is invertible or $b$ is invertible in $A$.
cloud walrusBOT
valentin9912
sleek fable
We have $\frac{x}{1} = \frac{a}{1} \frac{b}{1}$ but $\frac{x}{1}$ is irreducible in $S^{-1}A$ so $\frac{a}{1}$ is invertible or $\frac{b}{1}$ is invertible in $S^{-1}A$
cloud walrusBOT
valentin9912
sleek fable
suppose that $\frac{a}{1}$ is invertible
cloud walrusBOT
valentin9912
sleek fable
Hence there exists $r \in A$ such that $ar \in S$
cloud walrusBOT
valentin9912
sleek fable
this is were I'm struggling
I guess I have to prove that ar = 1 but I can't figure out how to prove it
if someone could help me....
cloud walrusBOT
Croqueta
cloud walrusBOT
Boytjie (never-to-be-glomed)
coral spindle
They could write $\chi_1(g)/\chi_2(g)$
rotund aurora
as 1/chi_2(g) ?
cloud walrusBOT
Boytjie (never-to-be-glomed)
coral spindle
as 1/chi_2(g) ?
yeah exactly
rotund aurora
formal ermine
how is this proven
rotund aurora
very weird notation imo
rotund aurora
how is this proven
its elementary I think
formal ermine
very weird notation imo
sin^2(x)
rotund aurora
let me think
coral spindle
very weird notation imo
I agree. It's familiar if you've seen the inner product of characters in rep theory
formal ermine
I use f^2(x) for f(f(x))
rotund aurora
how is this proven
notice that $\chi_1/\chi_2$ is another character. Denote it by chi. Then what we want to show is that when chi is not trivial then $\sum_{g\in G} \chi(g)=0$ (chi not trivial means chi(g) is not just one for all g in G)
coral spindle
how is this proven
If chi_1, chi_2 are distinct then chi_1(g)/chi_2(g) =/= 1 for some g. This sum is invariant under multiplication by this quantity, hence 0.
cloud walrusBOT
Croqueta
coral spindle
Lol yeah same proof
rotund aurora
I mean yeah, not that you can do a lot of different things
but I didnt provide a full proof, I wanted Illuminator to finish it OwO
formal ermine
ah
coral spindle
I was too eager
rotund aurora
there is a fun problem that uses a similar idea, but requires a bit of other stuff
crystal vine
I'm guessing this is probably not the case, but if K is a field and a is separable, then is K(a) separable?
formal ermine
I'm guessing this is probably not the case, but if K is a field and a is separab...
what is a separable over? K?
crystal vine
separable over K, yeah
rotund aurora
It is true
south patrol
Yeah it's a handy statement
rotund aurora
I had to check my field theory notes
when I first learnt about separability I was like 
no one cares
cuz everything separable 
crystal vine
if m(x) is the minimal polynomial of a, then for any b = c1 + c2*a, the minimal polynomial of b is m((x-c1)/c2) and then go from there?
crystal vine
ah rip
formal ermine
if m(x) is the minimal polynomial of a, then for any b = c1 + c2*a, the minimal ...
what if the minimal polynomial of a has degree > 2?
crystal vine
yeah
crystal vine
It is true
do you have a proof hint?
crystal vine
ok, I only have a very basic understanding of galois theory though, we saw it at the end of one of my algebra courses, so if it requires heavy machinery I might not get very far
doesn't seem like an overly complicated statement though
south patrol
So if you have some Galois theory one way you can argue this is that there's L containing K with L/K Galois
Okay actually this depends on how you develop Galois theory in the first place lmao
crystal vine
where L would be the splitting field of m?
south patrol
Yes
south patrol
I'm guessing this is probably not the case, but if K is a field and a is separab...
So the problem is sometimes this lemma is used to develop galois theory i guess
It wasn't in my course though
crystal vine
mine either
south patrol
Okay good lol
So you can take a splitting field of m and then L/K Galois
Now L/K is separable
So in particular, K(a)/K is
(do you see why those two follow?)
Tbf how sensible this depends on how you're defining Galois lol
I'm basically saying splitting field of separable polynomial => normal and separable
crystal vine
we haven't done infinite degree Galois theory but I assume the splitting field of a polynomial has to be a finite degree extension
south patrol
Yes it does
After all, it is generated by the roots of the polynomial
There are only finitely many such elements and all are algebraic over the ground field
crystal vine
I'm basically saying splitting field of separable polynomial => normal and separ...
yeah, we haven't seen that explicitly (because we only considered finite degree) but we saw the equivalent thing saying that a finite degree extension is galois iff separable and a splitting field
south patrol
That is the same because these are finite degree
crystal vine
yes
ok cool, makes sense
thanks
I get the impression that very few things are not separable
south patrol
There are quite heavy constraints on inseparable polynomials sure
Like an irreducible polynomial f in F[x] is inseparable iff F has char p and f is a polynomial in x^p
crystal vine
yeah, its derivative must be 0
south patrol
Indeed
rustic crown
splitting fields are always normal, so just need to verify separability
crystal vine
.... which would require showing the separability of the subfield K(a) ?
rustic crown
(remind me the question again >.<)
crystal vine
if K is a field and a is separable, then is K(a) separable?
formal ermine
.... which would require showing the separability of the subfield K(a) ?
the splitting field of a separable polynomial is separable
crystal vine
how come?
wraith cargo
isn't that by definition lol
rustic crown
lemme be a lil pedantic here, you don't say separability of a field usually. you can talk about separability of polynlmials but otherwise you always talk about separability over a field
crystal vine
the definition I was given is that an extension is separable if all of its elements are
crystal vine
lemme be a lil pedantic here, you don't say separability of a field usually. you...
ah, my bad
that may be that I was given an unusual definition
rustic crown
nah, that's the standard definition
rustic crown
the easiest way to think about this is by defining the notion of separability degree and thinking about a few of its properties. ofc you can do the whole argument without actually defining it as well
south patrol
how do we know that the splitting field of m is Galois actually?
For me this is by definition lol it depends
On how you define Galois
rustic crown
if you have an algebraic extension F/k, you define [F:k]_s to be the number of extensions ways to extend the map k --> kbar to all of F. or in notation [F:k]_s = |Hom_k(F, k bar)|
and the cool property is that if you have a finite extension F/k then it is separable if and only if [F:k] = [F:k]_s
crystal vine
what is k bar?
rustic crown
an algebraic closure of k
crystal vine
ah
rustic crown
or really any algebraically closed field containing k. it's not super hard to show that this number doesn't depend on the choice of that.
rustic crown
since F/k is algebraic you can also put it inside. but not in a canonical way. the number of ways to put it is exactly the separable degree
crystal vine
ah ok, I think you switched k bar and F in your definition of [F:k]_s
crystal vine
and that corresponds to extending k --> k bar to F? not extending k --> F to kbar?
rustic crown
yea extending the inclustion k --> k bar, to a map F --> k bar.
extend usually means to make the domain bigger.
rustic crown
okie so try to show the following:
- [F:k]_s doesn't depend on the choice of kbar (more precisely, the choice of the map k --> k bar that you're extending)
- use that to show [F:k]_s is multiplicative in towers
- then try to see what [k(a):k]_s is, and show it equals [k(a):k] iff a is separable over k.
- use multiplicativity and the case of simples to show [F:k]_s <= [F:k]
- finally show that show that if F = k(a1, ..., an) is a finite extension of k, the following are equivalent
(1) F/k is separable
(2) each a_i is separable over k
(3) [F:k]_s = [F:k]
i'd suggest to go (1) => (2) => (3) => (1)
crystal vine
damn
rustic crown
this should give you good understanding of separability uwu
crystal vine
and very basic galois theory suffices for all this?
rustic crown
none actually. just need basic field theory, maybe a little experience on how to work with alg closed fields and a little about degree and stuff.
no real galois theory.
crystal vine
okie so try to show the following:
- [F:k]_s doesn't depend on the choice of kba...
I might give this a try after reading up a little bit on those
rustic crown
okie :3
crystal vine
thanks for the help
rustic crown
i always found separability really hard to understand without introducing separable degree. like think about the equivalent situation of algebraic extensions and the usual degree. showing that k(a, b)/k is algebraic just by knowing a, b are algebraic/k is kinda hard to show directly. but it's immediate once you introduce degree and prove basic facts about it like the multiplicativity in towers.
white oxide
why exactly does hom(f, g) consist of all pairs (\alpha, \beta)? like why do all maps from f to g come in the form of pairs of functions (maybe i'm having trouble visualizing a morphism between two morphisms lol)
rustic crown
you're defining the category D
you get to decide what to call a morphism
remember category is the data of objects, morphisms and how to compose morphisms and the identity morphisms at each object.
you just need a sensible notion of what a morphism is, so that you can naturally compose two such things, if that happens you're happy
white oxide
oh ok that makes sense, so basically if the axioms are satisfied then any definition is valid
rustic crown
for example, here you didn't have to ask for the commutativity, but to make the category D more interesting you did
white oxide
idk maybe i'm not used to this level of abstraction lol and this is weird
white oxide
for example, here you didn't have to ask for the commutativity, but to make the ...
ah ok
that makes sense thx!
wraith cargo
why exactly does hom(f, g) consist of all pairs (\alpha, \beta)? like why do all...
Imagine it as a morphism between diagrams
You want commutativity plus all the objects to map essentially
rustic crown
cats are nice. think about it the other way... objects in a category can get super weird... but cat theory lets you abstract all of that and think of it as a point sized individual object!!
this makes it a lot easier to keep all the data in your head without getting overwhelmed
rustic crown
if you do some homological algebra, your objects would be infinite chains of maps
... --> A_{n-1} --> A_n --> A_{n+1} --> ...
do you really wanna keep all the info in your head?
when you can just call that thing A and think of it as a point without any internal structure :p
imagine dealing with a diagram of these complexes
white oxide
ahh ok that makes sense thanks!
rustic crown
white oxide
maybe it's because i'm used to like working with the structures of objects and not just a general overview of all of them
so it's weird
idk
rustic crown
true true, but you'll get used to it pretty quickly!!
like have you done some field theory?
white oxide
yee
rustic crown
a similar situation happens there
fix a field k and consider the category of algebraic extensions of k
so your objects are maps k --> F
and a morphism between (k --> F) and (k --> E) would be a field map F --> E making that triangle commute
this is nice to do, in this category you don't always have to keep drawing that k --> again and again
it would simplify up your diagrams a lot
for instance, if k --> L is an algebraic closure, then it has this funny property that if F/k is an algebraic extension then there is at least one map F --> L (over k, i.e. in the category we constructed above)
so it's almost like a terminal object, except the uniqueness
if F/k is any extension then [F:k]_s = |Hom_k(F, k bar)| is a nice invariant of F/k. because given any two algebraic closures of k, they're unique. so
Hom_k(F, L1) =~ Hom_k(F, L2) where L1, L2 are algebraic closures
in short, it fun
(and Fun Things are Fun )
white oxide
oh wow okay i'll have to process that
thanks for typing that all out det! ur very helpful :)
rustic crown
and a morphism between (k --> F) and (k --> E) would be a field map F --> E maki...
notice that this was a special case of your example with A = C = k and alpha = id_k
coral spindle
Are you asking what the size of the subgroup <i> of C* is?
Have you tried writing out the elements? it's unclear to me what your confusion is.
vagrant zinc
Are you asking what the size of the subgroup <i> of C* is?
ye
coral spindle
You're using strange terminology
vagrant zinc
but one is already
coral spindle
"the generator element" are you asking how many generators <i> has?
or are you asking how many elements <i> has?
Please be clear.
vagrant zinc
look 19
coral spindle
OK so you are asked to find the number of elements of <i>.
Can you write down all the elements of <i>?
vagrant zinc
<i>=<1,-1,-i,1>
coral spindle
OK you're still using generator notation
What you wrote is technically correct, but this is not what I asked for.
Are you claiming that <i> = {1, -1, -i, 1}?
That makes no sense: i is not included there.
Try again.
vagrant zinc
<0,i,-1,-i>
vagrant zinc
I think I'm wrong, I've realized
coral spindle
When we're writing a set, we use {}.
Also, 0 is not an element of C*, so it cannot be included in any of its subgroups.
vagrant zinc
It looks prettier xD
coral spindle
Well it's also wrong :)
Are you aware of what writing <i> means exactly? Can you write down a definition?
coral spindle
OK I didn't want to know what the elements are, but yes <i> = {1, i, -1, -i}.
vagrant zinc
coral spindle
Do you understand what the notation <a> means in general?
vagrant zinc
yeh
coral spindle
For example, if I said I had a group G and an element g of G, could you explain what <g> is?
vagrant zinc
If means that i is a generator, apart from being a generator, it generates a subgroup
coral spindle
OK so what is the subgroup <g>, can you explain precisely what it is.
If the answer is no, just tell me
vagrant zinc
nopis
Maybe I know, but for security I would like to understand what you are saying
coral spindle
If you have nopis you should visit a doctor, but let's continue
<g> can be understood in a couple of ways
- <g> is the smallest subgroup of G that contains g
- <g> = {..., g^-2, g^-1, 1, g, g^2, g^3, ....}
that second part is very helpful, I'm sure you agree
Is it clearer using part 2. why <i> = {1, i, -1, -i}?
vagrant zinc
Yes I understand, only that I was wrong when operating and when I saw that it was under the multiplication different from 0
"nopis" means no in English je
coral spindle
I understood what nopis means, I was making a joke.
I'm glad you now have a better understanding of what <g> is.
So do you know the answer to the question now?
covert cliff
suppose $m = p_1p_2$ then $(\mathbb{Z}/m\mathbb{Z})^{\times} \cong (\mathbb{Z}/p_1\mathbb{Z})^{\times} \times (\mathbb{Z}/p_2\mathbb{Z})^{\times}$ does it follow that $(\mathbb{Z}/p_1\mathbb{Z})^{\times} \times (\mathbb{Z}/p_2\mathbb{Z})^{\times} \cong \mathbb{Z}/(p_1-1)\mathbb{Z}) \times \mathbb{Z}/(p_2-1)\mathbb{Z}$?
formal ermine
why should it? the orders aren't even the same
formal ermine
yeah ok "i"lluminator3
haha loser
coral spindle
You're adding ^\times where unnecessary
delicate orchid
moreover, why would that statement follow from the first statement
covert cliff
I was being hopeful lol
coral spindle
You don't mean to imply that (Z/pZ)^\times is isomophic to (Z/(p-1)Z)^\times
this is clearly not true
You mean (Z/pZ)^\times is isomorphic to Z/(p-1)Z.
covert cliff
oh p1 and p2 are prime
delicate orchid
yeah, we know
covert cliff
You mean (Z/pZ)^\times is isomorphic to Z/(p-1)Z.
I didn't even notice that 💀
delicate orchid
You mean (Z/pZ)^\times is isomorphic to Z/(p-1)Z.
did you mean this?
covert cliff
my bad, that is a typo
coral spindle
If you fix that blatant error, there's no issue.
covert cliff
did you mean this?
yes
delicate orchid
ah ok
cloud walrusBOT
ForJoke
coral spindle
It is in fact true that if A iso B and C iso D then A x C iso B x D 😱
covert cliff
I was blindly copy pasting cuz of laziness
formal ermine
now prove that phi(p) = p - 1
covert cliff
It is in fact true that if A iso B and C iso D then A x C iso B x D 😱
smh I was making sure, I've never done group theory, this is for my number theory class
coral spindle
@delicate orchid wtf when did it become :mocha:
I refuse glome
and now it is mocha
wtf
covert cliff
now prove that phi(p) = p - 1
isn't that easy?
coral spindle
illum is probably making a reference to the fact that this observation allows you to prove that phi is multiplicative
delicate orchid
if A iso B and C iso D with phi, theta being isomorphisms then (phi, theta) is an isomorphism between AxC and BxD
hopefully this is clear now
covert cliff
illum is probably making a reference to the fact that this observation allows yo...
oh I see i see
coral spindle
and therefore prove the usual formula
covert cliff
if A iso B and C iso D with phi, theta being isomorphisms then (phi, theta) is a...
that does make sense, i think i was just scared of some goofy group theory nonsense happening
does this even follow if one of the group pairs(A and C/B and D) is non abelian?
coral spindle
Yes ofc
covert cliff
now that I say that, I'm not sure why it wouldn't lol
delicate orchid
it works in an arbitary category 
