Could a spectrum of a ring have infinitely many clopen singleton sets(basically a ring with infinitely many prime ideals and all prime ideals are maximal), if so how exactly is it compact
#groups-rings-fields
golden pasture
hmpf apparently spec of infinite product of ring isnt trivial
some se comments seem to suggest it isnt possible cuz spec is always compact feels strange tho
spice bay
@golden pasture look at Lemma 00EE on Stacks, that might help
This at least lets you reduce to the following statement - such a ring will have the property that each prime ideal is of the form V(1-e) where e is an idempotent
golden pasture
hm ye that lemma does make sense tho i'm more like wondering why shouldn't a spectrum with infinitely many points and is totally disconnected exist(totally disconnected should implie not compact if there are infinitely many points cuz the covering by taking every single point as a open set has no finite subcovering)
upper pivot
is the the A/nil = absolutely flat question? cause i had the same question, and its cause totally disconnect means each singleton {a}, {b} is open in the subspace {a,b}, not the whole space
upper pivot
it is in {a,b}, if every singleton were open in the Spec then it would have the discrete topology
golden pasture
hm tru
golden pasture
apparently not https://math.stackexchange.com/questions/1529988/spectrum-of-infinite-product-of-rings
@ Bananensaft
ah right ok i had a error in my proof that ended with every singleton is clopen oops
golden pasture
kek
spice bay
@golden pasture ah I think I see why that fails - you can definitely take an infinite product of rings, and then take its spectrum, the problem is that you can't break this up into an infinite disjoint union since the category of affine schemes is not cocomplete, and therefore may not have infinite coproducts. If it did, you would get something that is not quasi-compact, but as you point out taking the singletons gives you an unrefinable cover, so that's a contradiction
golden pasture
yea
spice bay
so you can still have Spec(R) have infinitely many clopen points, it just may not be totally disconnected
if won't* be totally disconnected
mild laurel
This isn't the right channel
indigo arch
topaz solar
I didn’t expect some stone čech compact stuff, that’s interesting
golden pasture
yea that came out of nowhere
cloud walrusBOT
mart:
chilly ocean
does this work?
upper pivot
yeah, but you can be more concise with your counter example probably. like just say "Let A=1/2 I" i guess
chilly ocean
ok ty
tribal pasture
Shouldnt this be Z/gZ?
latent anvil
Yes
topaz solar
Is there some nice characterization of what partially ordered sets can be embedded into a partially ordered ring
In particular, what partially ordered sets with a bottom element can be embedded into the positive cone would be helpful to know (with the bottom element mapping to 0)
chilly ocean
Exercise is the following
We know that $EZ$ is chain homotopic to the identity, this implies that $H_{i}(S(X)\otimes S(X))\simeq H_{i}(S(X\times X))$.
We recall now that if our chain complexes its free we have: $H^{i}(Hom(Y,\mathbb{Z}))\simeq Hom(H_{i}(Y),\mathbb{Z})$ in our case, using the fact that all the $H_{i}(X)$ are finitely generated free abelian groups, we have then:
$$\bigoplus_{i+j=n} H^{i}(X)\otimes H^{j}(X)\simeq H^{n}(Hom(S(X)\otimes S(X)),\mathbb{Z}))\simeq Hom(H_{n}(S(X)\otimes S(X)),\mathbb{Z})\simeq_{EZ}$$ $$ Hom(H_{n}(S(X\times X)), \mathbb{Z})\simeq H^{n}(Hom(S(X\times X), \mathbb{Z}))\simeq H^{n}(X\times X)$$
the requested isomorphism.
cloud walrusBOT
Stephen:
chilly ocean
this is my work, does this look ok? moreover i tried the following approach in search for a less theoretical and more "evident" isomorphism
An attempt in finding a more explicit formula for the isomorphism:
\begin{ocg}{name_ext}{name_int}{1}
an element $\varphi \in H^{n}(S.(X)\otimes S.(X))$ correspond to an homomorphism:
$$\varphi: S_{i}(X)\otimes S_{j}(X)\to \mathbb{Z}$$ where we let $i,j$ be all the integers that sum to $n$.
For every element in $S_{n}(X\times X)$ we have a correspondent element in $(S.(X)\otimes S.(X)){n}$ via EZ. The composition $\varphi\circ EZ$ is then an homomorphism from $S{n}(X\times X)\to \mathbb{Z}$ and we define $\alpha_{n}(\varphi)=EZ*\varphi$. This gives us a controvariant group homomorphism:
cloud walrusBOT
Stephen:
Compile Error! Click the 
reaction for details. (You may edit your message)
chilly ocean
$$H^{i}(X,\mathbb{Z})\otimes H^{j}(X,\mathbb{Z})\to H^{i+j}(X\times X,\mathbb{Z})$$
cloud walrusBOT
Stephen:
chilly ocean
but I'm failing to find anything useful, my EZ map is the AW map.
@sour plume maybe? He looks like the only one helping here lol
sour plume
I think the idea is mostly güd; You shouldn't say that EZ is "homotopic to the identity" though, because EZ and the identity map act between different complexes. I think you want that EZ induces a homotopy equivalence of the two spaces
Also, might wanna add a short sentence as to why $\bigoplus_{i+j=n} H^{i}(X)\otimes H^{j}(X)\simeq H^n(\text{Hom}(S(X) \otimes S(X)))$
cloud walrusBOT
Lartomato:
sour plume
Your attempt at giving a more explicit formula would probably work, but I think most of the "explicit" information is hidden in the definition of the EZ-map, so if you keep that as a blackbox, you're not really giving the reader much more intuition
@chilly ocean mebbeh that helps
chilly ocean
@sour plume isn't that isomorphism because of definition? Ez is not homotopic to the identity but chain homotopic and because of that it induces identity homomorphism on homology no?
This was my attempt in expliciting the map but I stopped idk if because I made mistake or I just don't see how to go on
Sorry to bother you but i've been literally stuck for days
sour plume
Hm, I'm not sure what definition you're appealing to? The space $H^n(\text{Hom}(S(X) \otimes S(X)))$ is defined as the cohomology of a certain complex, the space $\bigoplus_{i+j=n} H^{i}(X)\otimes H^{j}(X)$ is a direct sum of other cohomology spaces. But perhaps you've already proven that they're isomorphic in your lecture notes? In any case, it's not immediate from the definition, but it's probably an easy proof
cloud walrusBOT
Lartomato:
sour plume
And two maps can only be chain homotopic to one another if they're defined as mapping between the same complexes; EZ maps between differeent spaces, the identity maps from one space to itself, so you can't say they're chain homotopic
I'd look back into your lecture notes/book to see what exactly has already been proven about the EZ-map. Probably that it is a homotopy equivalence of the two complexes. That's a different thing.
(okay the proof for the tensor product thing isn't easy, but it should probably have been done in your lecture notes or so)
chilly ocean
I'm sure about the chain homotopy, probably my teacher used and abuse of notation, I will look into Hatcher, thanks for the help
last ether
Is there an easy way to find a ring with exactly 5 elements that satisfy x^n=x?
chilly ocean
Let G be a group and H a subgroup of G. Let x e G. Let xHx^(-1) be the subset of
G consisting of all elements xyx^(-1) with y e H. Show that xHx^(-1) is a subgroup of G
smoky cypress
What are the three axioms for a subgroup
Verify each of them
For the subset xHx^{-1}
chilly ocean
i've proved for e and that inverse is in xHx(-1) however i am stuck proving that product of elements in xHx(-1) is also in xHx(-1)
solemn rain
suppose a in subset and b in the subset
a=xyx^-1
b=x_1y_1x_1^-1
ab=xyx^-1x_1y_1x_1^-1
@chilly ocean
is x fixed
chilly ocean
x is in G
solemn rain
lmao yea so the subset is {xyx^-1 | y in H}
for some x in G
so x and x_1 are the same mb
so to repeat
a=xyx^-1
b=xy_1x^-1
ab=xyx^-1xy_1x^-1
can u continue
@chilly ocean
chilly ocean
i got to that but then (x^-1x) has to be in H and (y_1x^-1) has to be an inverse of (xy) right?
solemn rain
np
chilly ocean
sorry, i thought that x itself is not fixed, that what confused me
smoky cypress
It doesn’t matter if x is fixed or not
For any element x in G, x^{-1}x is always the identity
By definition
chilly ocean
i get that, my remark was to the exercise in general)
smoky cypress
Ah ok
haughty hamlet
chilly ocean
scarlet estuary
sigh
chilly ocean
shy timber
@chilly oceanSo no Abhyankar's conjecture?
stone fulcrum
So I have a ring, and special values b in that ring that for a choice of x,
bx = x
Almost like x "eats" b.
I want to notate them in a set like b ∈ Dom(x) as if to suggest that b is an element that x "dominates". I'm not sure if this is the best way to say this though haha. Anyone got anything?
upper pivot
this is 1+ann(x)
stone fulcrum
Oh fuq that's just directly related to the annihilator right
I even noticed that too and didn't go with it
Eeh on second thought, even thought the annihilator is probably the more common object, I want to really exploit the particular form I have above haha
It would be nice if I didn't call it "1 + ann(x)" every time
Actually, just the annihilator itself might work for what I'm trying to do, hmm.
upper pivot
well if you want there are some nice properties u could find
like the immediate one is that if a is an ideal, then 1+a is closed under multiplication
golden pasture
it's a stabalizer too
grim umbra
upper pivot
oh hey ari you are doing the localization chapter right? try this neat problem
If p is a minimal prime ideal of A, then p contains only zero-divisors
golden pasture
isnt that a direct corollary of like one of the exercises?
like the exercise was like let S_0 be the set of all non zero divisors, then all maximal multiplicative sets contain S_0
tho all exercises are direct corollaries of each other
upper pivot
true oof
i didnt remember that, it was another problem in ch 4 lol
which used this idea
golden pasture
oh cute
golden pasture
kek
upper pivot
tb fair its hard to remember ch 3 exercises
cause literally all problems
have 5+ parts
lel
golden pasture
yea
latent anvil
wait so if p is a prime ideal of A containing some non zero divisor x, let S be the complement of p and T the set {x^n s : s in S}. Clearly T is multiplicative and doesn't contain 0, since if 0 = x^n s then 0 = s in S (as x isn't a zero divisor).
Now let T be any multiplicative subset not containing 0. If we localize wrt T we get a nonzero ring, which has a prime ideal, and this corresponds to a prime of q of A disjoint from T. Taking T as above we get that q is disjoint from S <= T, so q is contained in p, but doesn't contain x, so it's properly contained in p. Thus p isn't minimal
Does that look right for the problem you mentioned? I remember doing this in AM but it's been a couple months since I looked at that book
upper pivot
right should work, my approach was to directly localize with respect to the minimal prime ideal p, which only has 1 prime ideal hence, which must be the nilradical, and then the rest follows
latent anvil
ahh that's nice
upper pivot
yeah thats why i liked this one
latent anvil
I was thinking about like the saturation of a multiplicative subset
But didn't quite remember how that works
delicate bloom
yep
chilly ocean
ha i found an error in artin 
scarlet estuary
yeah sadly i dont think theres a collection of artin errata out there
most of the book is pretty good on that though
chilly ocean
oh well i'll just have to be careful
latent anvil
it'a kind of cool that any finitely generated abelian group is iso to the underlying additive group of some ring
Is this true for any abelian group?
unital ring ofc, preferably commutative
Suppose Q/Z had a ring structure
Then since every element is torsion, we'd have |1| = n for some n > 0. But then nx = 0 for all x, which is impossible, e.g. take x to be the coset of 1/(n+1)
granite compass
if in a ring i have that for all x there exists y such that , xy=x and yx is x can i concclude that y is identity without knowing that this is ideal domain
knotty mason
what is the definition of the identity element?
and is it unique?
(rhetorical questions asked to guide you)
granite compass
ah yes it is unique so this must hold xD
knotty mason
yeah
granite compass
true true :3
tribal pasture
Say I have a transformation T from W to V. Is it always possible to pick a basis for W and V such that T maps the basis to the basis?
chilly ocean
what if it is the 0 map
tribal pasture
What about for a Non-trivial transformation then? 😅
I think yes, because for any w_i, I can let T(w_i) be the basis element of V.
provided ofcourse, if it's not zero.
delicate bloom
I guess the question you should be asking yourself next, what's the technical word you should be using that specifically means "non-trivial transformation"
tribal pasture
Any transformation whose domain isn't equal to its kernel I guess?
delicate bloom
closer but not entirely
imagine a projection from R^3 to R^2
the domain isn't equal to its kernel but it would still break making a basis in the way you describe
I was more thinking something along the lines of surjective, injective, bijective
tribal pasture
Surjective would be enough then right?
delicate bloom
no not quite
think about the projection I described, it's surjective
T(w_1), T(w_2), T(w_3) would be what you have to map your basis to from W to V
but V is 2 dimensional
tribal pasture
Ah I see your point! So I presume injectivity is required since if we have injectivity then T(w_i)\neq T(w_j) and thus {T(w_i)} can be picked as a basis for target space
delicate bloom
yeah, good but actually what about the reverse inclusion mapping from R^2 to R^3?
tribal pasture
True, bijective then xD
delicate bloom
it's injective but not surjective, so it doesn't span the space and so we don't get the full basis
haha yeah
cool, yeah just thought it was worth talking through since "non trivial" was kind of a mysterious condition
tribal pasture
It was just that I had to prove something about a given transformation T:W to V and I had to arbitrarily define its operation on the elements. So I just stuck with w_i to T(w_i) xD
Was hoping if I could just do w_i to v_i by saying pick appropriate basis but that's not possible now I see
chilly ocean
i was about to give the T(w_i) answer but then i was like wait
fallen bluff
Could someone recommend me a good text summarizing Emmy Noether's contributions to math that is also at least a bit formal and features definitions and theorems?
stone fulcrum
I've ended up doing a lot of work with an algebra (which I understand it as a vector space with an associative vector multiplication - mine also is commutative and has identity)
Is there a reasonable way to put a basis on something like this? I've thought up a "multiplicative basis" where the multiplication of two basis vectors is another basis vector, but I'm wondering if there's other notions.
Any good literature on algebras, btw?
cloud walrusBOT
mart:
chilly ocean
@chilly ocean
cant follow ^^
i.e set of 2x2 invertible matrices over the real numbers
ah the 2 is just saying that this is a 2x2 matrix?
yup
ty
smoky cypress
$GL_n(\bR)=\brc{A\in M_n(\bR):\det(A)\neq0}$
cloud walrusBOT
Whoever:
tribal pasture
Can somebody explain the therefore part?
stone fulcrum
I don't know what x, ρ(x), or λ1 are
Or what ^(-1) means in this context
But I'm thinking this is a proof that the quintic can't be solved?
tribal pasture
G is a group in which x is an element. p is a map from G to GL_n(C). So p(x) is a complex-valued matrix. Lambda_1, lambda_2 are its eigenvalues
How does p(x) and p(x)-1 being conjugates imply lambda_1 lambda_2 = lambda_1 ^{-1}lambda_2 ^{-1}
red imp
is there a nice way of generating Cayley tables for rings
right now I'm just writing down random equations until I can equate stuff
it's really painful
tribal pasture
@stone fulcrum Any idea whether this is a property of unitary matrices?
Ah nvm, taking the determinant was all that was required
brisk granite
Is it true that you can have at most n subgroups of order 2^n-1 in a group of order 2^n?
latent anvil
I don't think so, take C2×C2
I'm not sure off the top of my head how many subgroups (C2)^n has but it's a lot. Should be solvable by thinking of it as an n dimensional vector space over F2 and doing linear algebra
brisk granite
I see, thanks
mild laurel
you need subgroups of order 2^(n-1) sham
latent anvil
yeah, 2^(n-1) = 2 in the klein four case
and n = 2
but we have 3 subgroups
of order 2
brisk granite
Yea
latent anvil
Oh were you talking about the other thing? The linear algebra comment?
latent anvil
for the general case of how many index 2 subgroups there are in (C2)^n, that's the same as the number of order 2 subgroups by linear algebra (i think?) and so you're just asking about how many elements of order 2 there are in (C2)^n. All elements have order 2 except 0 so you get 2^n - 1
edit: better proof. Given a subgroup G of (C2)^n of index 2, define χ : (C2)^n -> C2 by χ(g) = 0 if f in G and 1 if g in C2. Clearly Χ determines G and vice versa, so the number of index 2 subgroups is the number of nontrivial maps (C2)^n -> C2. Then Hom((C2)^n, C2) ≈ Hom(C2, C2)^n ≈ 2^n, so there are 2^n - 1 subgroups of index 2
Does that sound right?
(and n < 2^n - 1 in general, lol)
@red imp what are you doing with Cayley tables of rings?
I've pretty much never seen those used
red imp
@latent anvil For an assignment I have to show that every unital ring of order 4 is commutative
I figured it probably wouldnt be too hard to just construct a cayley table for the possible scenarios
and just go "hey look they're all symmetric down the diagonal"
brisk granite
I've been stuck on this problem for a while. Could I have a hint for it pls?
sick inlet
haven’t worked it out but my guess is that you want to show that each element of Q adjoin that list of roots of primes takes the form of “integer plus a sum of roots of products of those primes all over another integer”, and then show that if sqrt(p_i) takes that form you get a contradiction (?)
golden pasture
sketch
||suppose a linear relation exists||
||write it explicitly as like sum c_isqrt(p_i)=d||
||multiply by sqrt(p_i) and take traces||
||conclude that all c_i are 0||
latent anvil
I think that works Ariana although you won't necessarily get a linear relation, right?
rain escarp
can anyone help me out with understanding what an ideal is. I'm having trouble understanding the concept, I think I understand what affine varieties and that there are some relations between the two.
solemn rain
i dont know about affine varieties
but an ideal in a ring R ,I,is a subset of R such that rI=Ir
rI= {ri|i in I}
upper pivot
Anuj do you know normal groups from group theory
upper pivot
hmm what about homomorphism of rings
rain escarp
ive heard a little about isomorphisms but nothin about homomorphisms
latent anvil
where are you coming across ideals and varieties?
upper pivot
yeah hmm the best route would be to go ahead study some intro algebra
asking what an ideal is right now wont be helpful to you
scarlet estuary
the idea behind an ideal is that it's a subset of a ring such that multiplication "stays in" the subset
upper pivot
before you get some foundations in algebra
scarlet estuary
for example, given the ring Z
its ideals are sets nZ
so for example, 2Z is the set of even integers
rain escarp
a prof suggested the book Ideals, Varieties, and Algorithms, An introduction to computation algebraic geometry and commutative algebra by cox little and o'shea to me
scarlet estuary
and it is an ideal of Z
upper pivot
no lol
latent anvil
Yeah so like, learn ring theory
upper pivot
that sounds advanced
latent anvil
Before commutative algebra
latent anvil
It is a pretty basic AG book
And I've heard good things
But like
You are not ready if you don't know what a normal subgroup is
rain escarp
ok
latent anvil
I am surprised a prof recommended this to you
scarlet estuary
the book doesnt claim to require abstract algebra as a prerequisite
of course we know authors tend to underestimate but
still
latent anvil
Oh really? Wild
upper pivot
interesting
scarlet estuary
latent anvil
Actually I take back the surprise at the professor
scarlet estuary
surprises me too
latent anvil
a prof told me to read hartshorne before I knew any commutative algebra
rain escarp
some of the exercises say that they need abstract algebra so i just skip those
latent anvil
which turned out extremely poorly
latent anvil
Anuj I would strongly strongly recommend learning algebra first
upper pivot
like, why learn Algebraic geometry before algebra right, specailly since algebra is in the name
rain escarp
uh im doin a research project in continued fractions and he said reading the book would be useful
upper pivot
hmm interesting
anyhow, my recc for intro algebra books would be either artin or jacobson
latent anvil
You're not stupid
Stupid would be ignoring this advice
And trying to read a book because you think it makes you look smart
rain escarp
thanks for the advice everyone, ill look into those books
scarlet estuary
hmph
latent anvil
did you look more at the book?
scarlet estuary
i skimmed the text briefly, it looks decently approachable with no algebra background but absolutely not approachable with no proofs background
you need to have the mathematical maturity to immediately pick up and reason on the definitions
rain escarp
i had a calc and lin alg full year course with proofs and stuff
we had to use definitions to prove stuff
upper pivot
hmm i still dont see the point of doing AG w/o alg
scarlet estuary
the point is CS
upper pivot
i see
rain escarp
CS?
latent anvil
I'm not sure how AG applies to continued fractions (not saying it doesn't, I'm just ignorant here)
Computer science
But it didn't sound like that's what your context was anuj
scarlet estuary
¯_(ツ)_/¯
latent anvil
huh there's a section on automatic theorem proving
rain escarp
i really wanna practice creating proofs by myself to ready myself for 2nd year
latent anvil
you should!
upper pivot
try reading a book and doing the proofs of some theorems before reading the books proof
upper pivot
try it with like, an intro alg text since that seems to be your next step
latent anvil
yeah I'm looking more at this book and my recommendation is still that it's not a good idea to read for you right now
But like, introductory abstract algebra is a lot of fun
And would prepare you for something like this later
rain escarp
ok ill just set it aside for a while and try one of those texts that u peeps suggested
upper pivot
btw artin is a more LA heavy intro alg textbook, While jacobson is more directly AA
rain escarp
also, nice aerodactyl
upper pivot
i always personally recommend artin cause i personally enjoyed it a lot
placid solstice
hey all, is there any significance to elements of a field (i'm working in a finite field) that satisfy the property that their additive AND multiplicative inverses are the same? in other words:
a^-1 = b mod N
-a = b mod N
Well, the element must be a 4th root of unity
placid solstice
i'm thinkin about this because I just got a problem on my final wrong because I dropped a negative sign on -5 mod 26
(which, 5 and 21 mod 26 satisfy this property)
mmm, a 4th root of unity?
wdym
ahh
so it's an equivalent condition to say
a^4 = 1 mod N
?
mild laurel
Uh sure
N has to be prime here if you want it to be a field
But there are finite fields that aren't just mod
placid solstice
ik N has to be prime
i just came out of a cryptography class so all of our questions were under mod
why's that hold true? i'm messing with it on some scratch paper and i can't figure it out
mild laurel
It's not equivalent, but the conditions you stated imply that it must be a fourth root of unity
since a^{-1} = -a
placid solstice
oh!
how's that implication work then?
ahh wait
the 4th roots of unity themselves form an abelian group (?)
mild laurel
just move stuff around, multiplying by a on both sides gives you that 1 = -a^2
placid solstice
talking to a friend who knows more abstract algebra rn
we're trying to prove a^-1 = -a <=> a is a 4th root of unity
but we found out by working in GF(4) ~ field of 4th roots of unity that a^-1 = -a ONLY works for element 2
(as in like, it only works for 2 on {0,1,2,3})
which means!
a^-1 = -a <=> a is a 4th root of unity AND a is a 2nd root of unity
😄
oblique river
careful
GF(4) is not the field of 4th roots of unity
in fact, the only 4th root of unity in GF(4) is 1 itself
Also, GF(4) is not {0,1,2,3}
under any standard interpretation of {0,1,2,3}
I think the proof is easier than you're making it out to be. Suppose that a^(-1) = -a. Multiplying both sides by a, we get 1 = -a * a, or equivalently, a^2 = -1. Now square both sides to get a^4 = 1
The converse isn't true
and a = 1 is a counterexample
1 is a forth root of unity, but 1^(-1) is not equal to -1
(unless you're in characteristic 2)
scenic tide
Hey guys I think you may be in the wrong chat for this
chilly ocean
Why
upper pivot
what the hell is going on here
upper pivot
ily2
scarlet estuary
fuck off
candid comet
What are the instructions
upper pivot
<@&268886789983436800> it seems we have trolls
vital quail
yeah, you need help man
upper pivot
all of yall need help for sure
vital quail
go get some help
candid comet
Yeah you;re wasting our time
vital quail
sniped
candid comet
@moderaters
tulip barn
👀
candid comet
@Cohomologay @simple valley @deft spade @tulip barn we got some trolls on our hands
tulip barn
Don't troll @peak frost
candid comet
@tulip barn he's doing it again
chilly ocean
what even was that
tulip barn
You don't need to ping me. I'm right here
chilly ocean
was it a joke or something
candid comet
Idk someone @tulip barn
solemn hollow
did they leave out "identify this shape"
tulip barn
Doesn't matter. it was dumb.
raw moth
clearly they were asking for D_8
candid comet
I don't know if he was trolling though maybe he was in the wrong channel
raw moth
troll or idiot
candid comet
@tulip barn
tulip barn
Stop pinging me
scarlet estuary
im pretty sure this is a group of users
who organized to raid #groups-rings-fields for
some reason???
chilly ocean
scarlet estuary
considering bruhment was
asking high school geometry questions
why are they checking #groups-rings-fields
regardless, water under the bridge
woven delta
seems like a strange channel to raid tbh
scarlet estuary
idk its just weird to me that
4 randoms would check a channel
completely past their level
all at once
candid comet
oh no @scarlet estuary
tulip barn
👀
woven delta
oh btw there's a translated copy of galois' notes in Edwards galois theory book
it's pretty interesting
@scarlet estuary
candid comet
Oh yeah I was looking at that the other day! Really cool stuff.
woven delta
don't know if you've seen it
maiden ocean
wtf is happening
woven delta
well I was pinging you because I was addressing you
but yeah the language is interesting
maiden ocean
namington is ur name "oh no" in reference to the excellent "oh no" comics
woven delta
really makes me appreciate how much work has been done to clean things up and make everything slick
scarlet estuary
maiden ocean
woven delta
lol
red imp
latent anvil
dumb question time:
take an ideal I of k[x1,...,xn]
assume I = (f1,...,fr) for homgeneous degree d polynomials f1,...,fr
intuitively I feel like we should have deg g >= d for each g in I
but I don't see a good way to prove it
gritty fractal
Well, can u just say that multiplying a polynomial of degree d by another polynomial never decreases the degree, since degrees always add when multiplying, and then for addition, it takes the max, so it can never go below d
brisk granite
Could I have a hint for how to solve part (a). I am able to do the others given part (a) but Idk how to start (a).
mild laurel
Uh, have you learned the fact that all symmetric polynomials can be expressed in terms of the s_i?
sour plume
not quite, right? the question asks for expressing all symmetric rational functions with elementary symmetric polynomials. first proving that all symmetric polynomials can be expressed with elementary symmetric polynomials may be a good start
but i'm not sure yet; just saying there's a difference
brisk granite
oh, I see
ok
So, how might I start proving this?
I would prefer a hint as opposed to a complete answer
sour plume
Gotta say, I expect this is gonna be a bit technical and unpleasant; I think starting by showing symmetric polynomials are generated by elementary symmetric ones is necessary, and for that, you might wanna try a kind of double-induction thing over the degree of the polynomial and the amount of variables
(that alone, afaik, is gonna be a bit technical, but may your mathematical heart guide you through that pain)
Then, getting all symmetric rational functions from that might not be so difficult anymore, but even about that I'm not 100% sure...
But yeah, perhaps I'm missing something clever. Maybe someone else has good input
Something like "if f(x_1, ... , x_n) is a symmetric polynomial in n variables, then f(x_1, ... , x_n-1)$ is a symmetric polynomial in n-1 variables" could be helpful
mild laurel
I'm surprised this is an exercise. I'm not sure I've seen an easy proof of the polynomial thing I stated above
But once you have that of course, rational functions follow easily but
All the proofs I've seen of the fact are a little complicated or technical
gritty fractal
let G be a finite group, and let R = {x in G | x can't be written as g^n for 0 <= n < |g| and n != {0,1, |g| - 1} for any element in G}, if a, b != a in R, then is ab in R?
the idea im trying to capture with R is that it contains the head and tail of all "maximal" (not a subgroup of any other cyclic subgroup) cyclic subgroups
gritty fractal
shoot edited
upper pivot
hmm take a, a^{-1} that satisfy, this gives 1 and 1 is not in R usually
ofc this assumes R is non-empty
which im not sure is always the case
upper pivot
1^2=1
gritty fractal
the idea im trying to capture with R is that it contains the head and tail of all "maximal" (not a subgroup of any other cyclic subgroup) cyclic subgroups
ykw i should prolly mention that inside the question
edited
upper pivot
this should be a counter example i think
(im sure simpler ones exist but idk this is the first thing that came to mind that works)
(additive group)
gritty fractal
sqrt(2) and 2sqrt(2) cant both be in R
since 2sqrt(2) = sqrt(2) + sqrt(2)
u said additive group, so exponentiation becomes multiplication
upper pivot
yeah sorry ignore me im being dumb
gritty fractal
lol dw ur trying
i just realized my definition doesnt do what i want it to do anyways
so...
brisk granite
is this reasonable?
latent anvil
The degree of the sum isn't the max of the degrees, take x^2 - x and 1 - x^2
But I think I see how to show what I wanted now
By breaking up into homogeneous parts
And cancelation can only occurs within those parts
woeful flint
Guys quick question. If $G$ is a group and $a,b\in G$ are such that there exist $i,j\in\mathbb{N}$ with $a^i = b^j$ can we say anything about $i$ and $j$, some relation, if any between them?
cloud walrusBOT
Kraft Macaroni:
woeful flint
finite
solemn rain
if the order of the group is ifinite
a^i = a^j <--> i=j
if the order of the group is finite say n
a^i=b^j ---> n|(i-j)
woeful flint
isnt i-j gonna sometimes be less than n?
solemn rain
b = a lmao
thats if b = a what i just said up
a^i = a^j that is
and |a| = n
mb
and if i-j<n this would ocntradict
that order a is n
as a^i=a^j ---> a^(i-j) = 1
but n is the smallest integer such that a^n = 1
contradiction
woeful flint
if G is finite and a,b distinct?
solemn rain
so to answer ur question u get relations for i and j if a=b here
woeful flint
to clarify I have no idea if there is any relation, something im trying to prove would be slightly easier if there is any relation tho
solemn rain
so to answer ur question u get relations for i and j if a=b here
wind steeple
wtf, it's (i-j)|n not n|(i-j)
wind steeple
if the order of a is n, not the order of the group
woven delta
@solemn rain "a^i = a^j <--> i=j"
wtf
this is very wrong
you need the subgroup generated by a is infinite not G is infinite
latent anvil
If A <= B <= C are rings and A <= C is finite type and B <= C is finite, is A <= B finite type?
I don't see how to prove it so I think it's false, but I can't think of a counterexample
smoky cypress
In Artin's Algebra some problems are starred, what does that mean?
sharp sonnet
The starred exercises are some of the more difficult ones.)
it's listed on the notation page
steep hull
Make sure you didn’t fakesolve
smoky cypress
Well here's the problem
The set $U$ of $3\times3$ matrices in the form of $$\begin{bmatrix}1&a&b\0&1&c\0&0&1\end{bmatrix}$$ is a subgroup of $\text{SL}_n(\bR)$, then find the center of $U$
cloud walrusBOT
Whoever:
smoky cypress
Fuck I might have fakesolved
Oh nvm I didn't
Well I said the center are all the matrices in the form of $$\begin{bmatrix}1&0&c\0&1&0\0&0&1\end{bmatrix}$$ for any $c\in\bR$
cloud walrusBOT
Whoever:
chilly ocean
how is it the case that y^i=1 iff i|2? I would've thought y^1531=1, but 1531 isn't divisible by 2 right?
idk this is probably a dumb question
chilly ocean
isn't that the definition of cyclic group
oh wait
oh fuck
sorry I thought every n>|G| had x^n=1 in a cyclic group
lmao but then I reread the definition
serious bruh moment
latent anvil
Yeah so x^n = 1 will only happen if n I'd a multiple of |x|
chilly ocean
ya understood lol ty
golden pasture
Does localization commute with sums of modules, meaning $\sum\left(M_i\right){\mathfrak p}\overset?=\left(\sum M_i\right){\mathfrak p}$. If $M_i$ is a directed system or $\sum M_i=\oplus M_i$ then this would be true as tensor commutes with both of those but am not too sure how it would work in general
cloud walrusBOT
ariana:
golden pasture
this is when a software to play with infinite stuff and localization would be helpful heh
golden pasture
heh but there are noncanonical ways it can be isomorphic too
hm actually
If we let $M_n=\mathbb Z/(n)\mathbb Z$ and $M=\prod M_n$, then we have $M_{(0)}\neq0$ with $\left(M_n\right){(0)}=0$. If we let $I$ be a indexing set over all finite subsets of $\mathbb N$ and $M_i=\sum{j\in i}M_j$ for $i\in I$ then $M=\sum_{i\in I}M_i$ but $M_{(0)}\neq\sum\left(M_i\right){(0)}=0$, tho it seems like $M=\lim M_i$ as a direct limit and tensors commute with limit, so shouldn't $M{(0)}=\lim\left(M_i\right)_{(0)}=0$ feels weird
cloud walrusBOT
ariana:
golden pasture
(should be N-{0,1} everywhere)
latent anvil
I think localization commutes with sums. If M = ΣMi, then you have a surjective map π : \oplus Mi -> M. If we localize wrt S this map stays surjective and S^(-1)(\oplus Mi) = \oplus S^(-1)Mi compatibly with the inclusions of the Mi into M, right?
I'm not being very formal so I might have missed a detail
Haven't read your example
Are you saying M_(0) is nonzero since the element (1,1,...) is infinite order?
golden pasture
yea
latent anvil
I don't see why M is the sum of the Mi
In your example
How do you get the element (1,1,...)?
golden pasture
M_i is basically some finitely generated submodule of M
latent anvil
Right but I don't think they're all the finitely generated submodules
golden pasture
yea
latent anvil
So why is their sum M?
mild laurel
macaulay2 has stuff you can do with infinite stuff and localization I think
golden pasture
https://discordapp.com/channels/268882317391429632/496784958430380033/709189166306689044
ok yea the proof should work cuz S^(-1) does commute with inclusion
latent anvil
yeah, it all comes down to the fact that it's exact and preserves infinite direct sums (meaning the natural map oplus F(M_i) -> F(oplus M_i) is an iso)
this proof should work for any such functor
latent anvil
np
rain crescent
Why is Z not an ideal of itself? Dont every products ab (with a in Z and b in Z) belong in Z?
scarlet estuary
where are you hearing this?
in every ring R, the set R forms an ideal
the unit ideal [if commutative]
stone fulcrum
It's not a proper ideal
scarlet estuary
^
rain crescent
Hm i guess I was mistaken then.
scarlet estuary
indeed, proper ideals cant contain 1
stone fulcrum
Where we define proper as "it can't be the ring itself lol" or "it can't be empty"
Or, not empty but the identity alone
rain crescent
Since I'm here, can someone explain me what the direct product of groups is? The plus in a circle. Are the elements like a vector? Where you choose an element from the first and an element from the second?
toxic scaffold
@rain crescent yes one element from each group determines an element of the direct product of the two groups. The multiplication rule is (g, h) ∙ (g', h') = (gg', h h')
rain crescent
So when you have an isomorphism that would be like a 2 variable function right?
rain crescent
Between for example Z2 product Z2 and R4
The isomorphism is a bijection between those groups. But more practicaly if you wanted an example of an isomorphism that would be similar to a 2 variable function
stone fulcrum
I wouldn't think of Z2×Z2 as a "two element group". Simply that every element is a list of two "things"
For example (0,1) is an element of Z2×Z2
What's R4?
rain crescent
vector space
stone fulcrum
So isomorphisms are things that exist between groups
woven delta
lol
brisk granite
if alpha and beta are the two real roots of x^6 + 10x -5, how might I show beta isn't in Q(alpha)?
latent anvil
I think this would imply stuff about the size of the galois group
Like it implies it's bounded by 6*4*3*2 easily
But by symmetry any time you adjoin a root you get another one right?
And I think they're going to couple up somehow
brisk granite
But by symmetry any time you adjoin a root you get another one right?
wait, why?
this works for the non-real roots
latent anvil
well all the field Q(β) for roots β of your polynomial are isomorphic
latent anvil
So that polynomial will have at least two roots in all of them if it has two roots in one of them
So what I expect is that they'll sort of couple
latent anvil
I'm not sure what you're saying. I'm assuming that's true
And trying to get a contradiction
Oh wait this actually does imply things
Okay so let α,β be the real roots
Assume β in Q(α)
Then x^6 + 10x - 5 has at least two roots in Q(α), right?
But the other roots are non-real
So they can't be in Q(α)
So Q(α) has exactly two roots of that polynomial
If γ is any other root, Q(γ) and Q(α) are isomorphic, so x^6 + 10x - 5 has exactly two roots in Q(γ) as well. Thus we get exactly one other root γ' in Q(γ)
Does that make sense?
brisk granite
yep
latent anvil
Okay so we can pair up the roots (α, β), (γ, δ), (ε, ζ) so that if we adjoin one element of a pair, we get the other (and no other roots)
Let K be the splitting field of x^6 + 10x - 5. I claim that the above implies 6 * 4 <= [K : Q] <= 6 * 4 * 2
i.e. 24 <= [K : Q] <= 48
Do you see why?
Actually I claim more, that [K : Q] = 24 or [K : Q] = 48
brisk granite
yep, I see
latent anvil
Okay, we also know that Gal(K/Q) has an index 6 non normal subgroup
Corresponding to Q(α)/Q
brisk granite
right
latent anvil
And if [K : Q] is 48 it also has an index 24 non normal subgroup
For the same reason
I'm looking at a table of transitive subgroups of S6
And trying to rule things out
brisk granite
uh, how can a subgroup with 1/2 the order of the full group not be normal?
latent anvil
Not 1/2 the order, 1/24th the order
brisk granite
oh, I see
latent anvil
That's the same thing as an order 2 element not in the center
So not that hard
oh I think we can probably say that any automorphism preserves our pairs?
Feels right™
Like if r and s are in the same pair, so are σ(r) and σ(s)
brisk granite
ok... not sure why
latent anvil
Yeah me neither
brisk granite
lol
latent anvil
so if r in Q(s), I'm saying σ(r) in Q(σ(s)). This is actually obvious, just write r as a polynomial in s (with q coefficients)
Right?
The expression r = f(s) gives σ(r) = f(σ(s))
brisk granite
I see,
latent anvil
So we have a transitive subgroup G of S6 with this really weird property
Basically we have three blocks
And we can swap the blocks and also swap the things in the blocks
Sounds like S3 × S2 × S2 × S2
Do you see what I'm saying?
brisk granite
Sounds like S3 × S2 × S2 × S2
how do you know this group has this property
and it's the only subgroup of S_6 (with this property)
latent anvil
Sorry I mean it looks like a subgroup of that
Not that it is that
And my proof is that we can partition the roots into three groups
And any automorphism has to maintain those groups
past pendant
how do i do this
brisk granite
yeah, I'm ok with all this
latent anvil
Post it in #geometry-and-trigonometry
brisk granite
lmao
latent anvil
or use etale cohomology
It's one of the two
So we have a transitive subgroup of S6
Which embeds into S3×S2×S2×S2
In a nice way
(and it has order 24 or 48)
Oh well
The order doesn't give us a lot of choices
It's gotta be S3×S2×S2 or A3×S2×S2×S2
Up to isomorphism
Right?
brisk granite
Which embeds into S3×S2×S2×S2
wait, but how do you know it's a subgroup of this?
latent anvil
So label the roots 1,2,3,4,5,6
brisk granite
ye
latent anvil
I showed that if e.g. 1 is sent to 3, 2 must be sent to 4
Or if 1 is sent to 6 then 2 goes to 5
So really we're reordering the blocks {1,2}, {3,4}, {5,6}, and then within each block we choose to swap or not swap
Does that make sense?
latent anvil
The subgroup of S6 of such permutations is S3×S2×S2×S2
Hmm, this subgroup isn't in the table I'll looking at
But I'm pretty sure it's transitive
maybe I've made a mistake
brisk granite
where could I find a table of the subgroups of S_6?
latent anvil
I mean I googled "table of transitive subgroups of S6"
You definitely don't want all of them
Sorry I gotta get back to grading
This is interesting though
I think the stuff I did is useful? Maybe there's a simpler solution I missed
brisk granite
thank you for the help
oblique river
this sounds like wreath products to me
the wreath product of S_3 and S_2 (in some order)
latent anvil
so the subgroup I'm thinking of is generated by like
(13)(24)
(15)(26)
(12)
(34)
(56)
right?
I think
oh no because (15)(26) (13)(24) ((15)(26))^{-1} = (15)(26)(13)(24)(26)(15) = (15)(13)(15) (26)(24)(26) = (35)(46)
hmm
I am confused
latent anvil
Oh I made a typo lol. Fixed
quasi mirage
can you help me @latent anvil
latent anvil
did someone tell you to ping me?
sharp sonnet
not here
quasi mirage
where ? @sharp sonnet
quasi mirage
okay , thanks bro ♥ @sharp sonnet
latent anvil
lol
kindred mist
force but your'e in the wrong chat of course
somber bramble
(accidentally posted this in #advanced-analysis first)
but yea so according to calculation 1 we have ρ(στ) = ρ(τ)ρ(σ), according to the second it’s ρ(τ)ρ(σ)=ρ(τσ)
can’t both be true
so one of them has to be false
but which one and where?
delicate bloom
working calculation 2 backwards to follow it like calculation 1 forwards
this step is where things are done inconsistently
in calculation 1 you factor out the tau on the left but in calculation 2 (worked backwards) you put the sigma in the same place "inside" on the v to get u
the little curly arrow is specifically the step from line to line that is inconsistent, hopefully that's clear
sour plume
Which one of the two is false, tho?
delicate bloom
I'm not really paying attention to the problem but probably neither is false, just different choice of convention
do you want rho to reverse the order or not
final gulch
the first one is false
sour plume
Yeah I'm starting to think the first one isn't quite right
final gulch
the step where you apply rho(tau)
sour plume
Yeah, because your tensor must really be enumerated as "v_1 v_2 v_3"; if you numerate it as "v_2 v_1 v_3" instead, applying a permutation by the given formula may give you a different result
Is that the right idea? Because your permutation might keep the number "2" fixed and exchange 1 and 3, for example, and then both numerations give you different results if you use this formula
final gulch
the permutation should act on the vectors in the order they are written; the way you have applied rho(tau) in calculation 1, the permutation acts on some new arbitrary indices that do not reflect the order of the factors
sour plume
yes this pleases me
@somber bramble the mystery has been unraveled
thanks for pointing out the flawed step merosity, that did help
somber bramble
lemme read through this three times
sour plume
that was an absolutely beautiful contradiction though
delicate bloom
👍 nice
rain crescent
How can I prove/disprove that a finite integral domain is a field? I think it is, I just don't know how to prove it
somber bramble
(the English word for “Körper” is field)
also, yea, it is, and what you can do is show that multiplication by a (for a≠0) is a bijection
it then follows that every number has an inverse
rain crescent
So i managed to prove the multiplication is injective pretty easily but I cant prove it's surjective. Any tips?
dawn kiln
i like to think of it as a finite integral domain not having enough "space" for one of the elements to not be mapped to zero under the function he described
i believe you can use the property that an injection from two sets of the same size is a bijection
smoky cypress
That only works on finite set
Which you can in this case
Because it’s a finite integral domain
rain crescent
This is what I did, but I don't know if it's a valid proof.
Let y=as (where s is the constant we are multiplying by and a is the element of the set). We want to show that there's always x such that ax=as. Since it's an integral domain I can use the product cut rule (I dont know what it's called in english) and we get x=s so there is always a solution. Since there's always a solution the multiplication is surjective and bijective so every element has an inverse
somber bramble
So i managed to prove the multiplication is injective pretty easily but I cant prove it's surjective. Any tips?
an injective function X→X is automatically a bijection if X is finite
that’s the trick here
you can’t prove surjectivity without using finiteness because it’s not true in the infinite case
sullen island
i have a question about the accepted answer to this question: why is $\mathbb{C}G$ (the group algebra) dual to the space of functions from $G$ to $\mathbb{C}$? (i.e. dual to $Hom_\mathbb{C} (G, \mathbb{C} ) $ ?) shouldnt its dual be $Hom_\mathbb{C} (\mathbb{C} G, \mathbb{C} ) ?$
cloud walrusBOT
xy:
wind steeple
I guess it's because of the universal property of CG
for every morphism from G to A an algebra, you have a unique morphism from CG to A
sullen island
isnt $Hom_\mathbb{C} (G, \mathbb{C}) \subset Hom_\mathbb{C} (\mathbb{C}G, \mathbb{C})$ ?
cloud walrusBOT
xy:
wind steeple
it's a bijection
sullen island
?! how
one has linear functionals acting on elements in G, one has linear functionals acting on linear combinations of elements in G
😮
wind steeple
$\operatorname{Hom}(G,\mathbb{C}) \simeq \operatorname{Hom}(\mathbb{C}G,\mathbb{C})$
cloud walrusBOT
Zak:
wind steeple
bc for every morphism from G to C* you can construct a morphism from CG to C
sullen island
but elements in G are necessarily elements in CG, not vice versa, right?
wind steeple
btw I think that's C* isn't it ?
sullen island
C* ?
wind steeple
C\{0}
You look at multiplicative morphisms from G to C*
CG is the algebra of formal sums sum ai gi where ai in C and gi in G
sullen island
yup, thats CG
wind steeple
so if you have a morphism from G to C* say f, we have a morphism from CG to C which maps sum aigi to sum ai f(gi)
sullen island
so the dual of CG is the algebra of linear functionals that act on these formal sums
so the dual of CG should also be a set of formal sums , right?
wind steeple
I don't understand, the dual of CG is by definition Hom(CG,C), which is naturally isomorphic (a bijection) to Hom(G,C*)
so one can say that the dual of CG is Hom(G,C*)
sullen island
the dual of CG is by definition Hom(CG, C), consisting of linear functionals that act on formal sums of the form ai gi, gi \in G, right? so
these formal sums dont exist in the group itself
and so Hom(G, C) is only a subset of Hom(CG, C)
wind steeple
yeah I've just read you stackoverflow url and I think he meant function from G to C where f(gh) = f(g)f(h) I guess
and f(e) = 1
sullen island
because elements in hom(G, C) necessarily act on elements in G
but not on formal sums of elements in G
so what im not understanding is how Hom(G, C) is isomorphic to Hom(CG, C)
because thats like saying CG is isomorphic to G (??)
wind steeple
ok idk, maybe someone else could help you. I missread what you've say x) Maybe by "Dual Hopf Algebra" he don't mean the dual vector space
sullen island
ahh okay
was that relation you gave true though?
Hom(CG, C) isomorphic to Hom(G, C) ?
wind steeple
Hom(CG,C) the set of C-linear morphisms is in bejction with Hom(G,C*) the morphisms of group
sullen island
hmm okay
red imp
How can I show that every unital ring of order 4 is commutative
I dont have to compute cayley tables for all the different cases do I, because that's all I can think of right now
latent anvil
Okay, so your ring looks like R = {0,1,a,b}
red imp
yep
latent anvil
And you know 0 and 1 finite with everything
So there's only one way things could fail to compute
*commute
red imp
if ab\neq ba
hmm so I could contrapose and maybe just find a contradiction from that
do you have any clues as to where I would search for a contradiction? I think I tried to find one using the distributive property last week but couldnt see one
latent anvil
Yeah so I would do cases on what the possible values of ab and ba are
And maybe a^2 and b^2
red imp
ohkay so it basically does come down to computing cayley tables except we've optimised the problem a bit since we have assumed ab =/= ba
red imp
uhhhhhhhhhhhhh
oh right
then two of them would be the same so it's not order 4 anymore
wait no
what does happen if ab=1
upper pivot
hmm
latent anvil
So if your ring isn't commutative, then 1 = -1
Since otherwise -1 = a or -1 = b and that's impossible since -1 commutes with everything
red imp
:O
latent anvil
Does that make sense?
red imp
yeah that's clever
latent anvil
haha ty
I think this forces a + b = 1
If it's 0, then b = -a = a (contradiction)
If it's a or b, you can cancel the other and get a = 0 or b = 0 (contradiction)
Does that make sense?
red imp
not right now but once I work through it on my whiteboard I'm sure it will
latent anvil
sure. Then b = 1 + a so ab = a(1 + a) = a + a^2 = (1+a)a = ba
Contradicting the assumption that R isn't commutative
red imp
you've been a great help, I'm sure I'll be back once I get stuck again
upper pivot
very nice sol btw
smoky cypress
Every group of whose order is a power of a prime p contains an element of order p
So my thought was that, pick any element a other than the identity, then the subgroup generated by this element is cyclic has order p^k for some k. If k=1 then we’re done. If k>1 then a^{p^{k-1}} has order p
Idk is this the way I’m supposed to do this?
latent anvil
Looks good to me
hot tusk
can someone help me figure this out
why do they only check the order of the odd numbers in Z16?
mild laurel
Because those are the only elements of the set Z_16^x
mild laurel
No, because lots of numbers will not have inverses
like 2 doesn't have an inverse in that case
lime skiff
I'm not sure I understand one assignment I'm given
Let $K$ be a field of characteristic not 2 and let $P\in K[X]$ be an irreducible polynomial of degree 2.
- Show that $P$ has both of its roots in $K[X]/(P)$.
- Show there exists $d\in K^*$ such that $K[X]/(P)\simeq K(\sqrt{d})$.
- If $K=\mathbb{Q}$, show we can take $d$ a positive squarefree integer.
cloud walrusBOT
kuri:
lime skiff
First of all, why isn't 1 trivial?
If you have a root, you can just factor by (X-k) right
And you'll get the other root
As for the second/third part I don't really understand what to do
Should I construct an actual isomorphism?
Like send X to \sqrt(d)?
For instance it's clear that R[x]/(x^2+1)=R[i]
smoky cypress
Do you know lagrange's theorem
Well, it states that the order of any subgroup of a finite group divides the order of the group
So in particular, the subgroup generated by a single element a
The order of the subgroup generated by a single element a is just the order of the element
Let |a|=k, |G|=n, then we have by Lagrange theorem that k|n
and try to complete it from here
Do you get it now?
you can post your proof here if you think you get it
cloud walrusBOT
Mustafa Erol:
smoky cypress
Yes
stone fulcrum
I feel this is equivalent to Lagrange?
No wait, this just makes a statement on all the cyclic groups. Mb it does leave out info
cloud walrusBOT
wstayman:
raven juniper
hello, how do i show that x^2 - 2 is a minimal polynomial in Q? Is it sufficient to show that it is irreducible?
steep hull
Yes
raven juniper
alright, thanks
hot tusk
can someone explain how to do this without sylow
this is a true/false q
does z/z20 x z/z5 work?
latent anvil
@chilly ocean what have you tried?
cloud walrusBOT
wstayman:
latent anvil
Sure. So suppose you have an automorphism σ. What can you say about σ(5^1/4))?
cloud walrusBOT
wstayman:
latent anvil
Why does it have to be mapped to any of those?
chilly ocean
Because it's an automorphism, I thought that meant the set of homomorphisms of a group or field to itself
latent anvil
Well it means an invertible homomorphism from the field to itself, but those four elements you listed aren't all of Q(5^(1/4))
Why can't it map 5^(1/4) to 5^(1/4) - 3?
chilly ocean
True I suppose it could map to any element of the field, but I thought the goal was to find automorphisms between adjoined elements, if that makes sense.
latent anvil
Yeah so this is an important point
You're looking at all field isomorphisms from Q(5^(1/4)) to itself
However it turns out that any of these must send 5^(1/4) to ±5^(1/4) or ±i 5^(1/4)
Let's think about why
(this is probably in whatever textbook you're using, too)
cloud walrusBOT
wstayman:
latent anvil
Yeah, that's the right line of thought
chilly ocean
If its the set of invertible homomorphisms, then doesn't any automorphism have to send "generators" to "generators"
latent anvil
That's not a very precise statement
And we actually get something stronger
Also 5^(1/4) - 3 is a generator
In that Q(5^(1/4) - 3) = Q(5^(1/4))
chilly ocean
Right, that's the problem I was thinking too
latent anvil
So if you have an extension E/F, and an element α in E with minimal polynomial m, any automorphism of R over F sends α to a root of m
Try to prove it
Also I gtg, I can help later if you're still working on this sorry
chilly ocean
All good I appreciate the help you've given!
latent anvil
Okay, so we have this lemma
Which tells us that any automorphism sends 5^(1/4) to one of the other four roots
And you think it can't be sent to any of the imaginary roots
But aren't sure how to justify that
chilly ocean
correct, that is my position currently.
latent anvil
So let's think about why
So we know it's gotta be sent to something in Q(5^(1/4)), right?
cloud walrusBOT
wstayman:
latent anvil
yeah exactly
So let's think about why
Suppose otherwise
We know what elements of Q(5^(1/4)) look like, right?
cloud walrusBOT
wstayman:
cloud walrusBOT
shamrock:
chilly ocean
oh my, sorry wasn't thinking. your second statement is obvious, but why isn't 5^(1/3) in the extension?
latent anvil
What's the degree of the extension Q(5^(1/3))/Q?
cloud walrusBOT
wstayman:
chilly ocean
So the degree extension of Q(5^(1/3))/Q would be 2?