dont really have any idea how to show a max in sigma is also max in the ring
#groups-rings-fields
wooden ember
Im trying to do f) here but I feel I may be misunderstanding the question. Ive managed to do the first part for any non-trivial automorphism of $P$, not just those with order coprime to $p$ by appealing to c). Simply take a minimal generating set of $P$, ${x_1,...,x_r}$. If our automorphism $\sigma$ is nontrivial then it permutes this set so that $\sigma(x_i)=x_{\pi(i)}$ where $\pi\in S_r$ acts nontrivially on ${1,...,r}$. But then the induced automorphism $\overline{\sigma}$ on $\overline{P}$ acts on the basis ${\overline{x_1},...,\overline{x_r}}$ by $\overline{\sigma}({\overline{x_i}})=\overline{\sigma}(x_i\Phi(P))=\sigma(x_i)\Phi(P)=x_{\pi(i)}\Phi(P)=\overline{x_{\pi(i)}}$ so that $\overline{\sigma}$ is nontrivial on $\overline{P}$
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
here $P$ is a $p$-group, $\Phi(P)$ is the Frattini subgroup of $P$, and bar notation refers to passing to the quotient by $\Phi(P)$ so that $\overline{P}$ is an elementary abelian group of order $p^r$
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
my argument must be wrong or im misunderstanding the question, would love a pointer
maybe the text is instead meaning to speak of fixed point free automorphisms of P instead of nontrivial automorphisms of P?
wooden ember
right i think im just stupid there's not reason sigma should permute the minimal generating set
wooden ember
okay I think I have a working proof
im confident about my first step which shows that if sigma has prime order $q$ with $q\neq p$ then the result is true, so I wont show it here. What im less confident in is my second step, which is inductive. I take $|\sigma|=q_1^{\alpha_1}...q_r^{\alpha_r}$ and by induction $\sigma^{q_1}$ is non trivial on $P$ and so $\overline{\sigma}^{q_1}$ is nontrivial on $\overline{P}$. Hence $\overline{\sigma}$ is non trivial on $\overline{P}$ as required
cloud walrusBOT
𝓛ittle ℕarwhal ✓
cloud walrusBOT
𝓛ittle ℕarwhal ✓
upper cape
Does anyone know how i could check if a quartic homogeneous polynomial in two variables is reducible?
wooden ember
solved this but im curious to know if there's a nice way to construct these maximal subgroups: I can only think of two maximal subgroups of P
ah no nvm
im a bit stupid lol
prime cloak
Hello guys I need help on an exercise about linear algebra.
Let E be a vectorial space.
Let f be an endomorphism of E .
Show that there exists a projector p and an automorphism g such that f=gop.
I was told to make a drawing and make sure that the statement is true but I just can't do it
<@&286206848099549185>
south patrol
i think doing a drawing is a good idea - consider E = R^3 and think about what happens for f with different ranks
in particular, what's the link between the image of f and the image of p, given that g is an automorphism?
dusty river
f is surjective
Why is that?
prime cloak
nvm
Thought using gof surj => g surj
Idk how to draw linear algebra tbh and idk where to learn
prime cloak
What do I do with this
dusty river
Rank (gop)=rank(p)=rank (f) ?
so in 3 dimensions, can you visualize what maps of different ranks looks like?
or even in 2
prime cloak
Yeah you have a line a plane and the whole space
dusty river
Yeah
And now you have to write this random f as first a projection then some automorphism
prime cloak
And every object usually used contains 0
dusty river
Yeah
prime cloak
What I did is
I took a supplementary of Ker f let's say S and called p the projection on S
dusty river
Nice!
prime cloak
But from there I can't get g
dusty river
There will not be a single choice of g, even once you fix a choice of p
prime cloak
Ok
dusty river
So don't try to construct a g explicitly because you won't be able to get necessary conditions on it
prime cloak
Ok
dusty river
instead think of extending it from something you already know
dusty river
f restrain to Im(p)= g
In particular
prime cloak
Let's say I have a base of imp and I complete it to get E ?
But then we need to make sure that g of these new vectors and the one before is a base
Dim(imp)= rk f
dusty river
well g on the im(p) is already defined and injective right? You have to just define it outside. Use the extended basis
Does the problem say that E is finite dimensional?
prime cloak
Yes
dusty river
ok nice, it doesn't seem to be true for infinite dimensions
prime cloak
well g on the im(p) is already defined and injective right? You have to just def...
I don't understand why it's injective on this
dusty river
kernel will have to 0
prime cloak
Ok
dusty river
because ker f = ker p, if f restricted to im(p) also had some kernel, then ker(gop) = ker f would be larger than ker p
south patrol
oh nice, my way of thinking was just that we can ||project E onto a subspace of E of the same dimension as Im f|| and then ||all (finite dim, over same field) vector spaces of the same dimension are isomorphic, hence g, which we can extend arbitrarily to an automorphism|| that works right?
dusty river
Yeah except you don't need finite dimensions to say that all vector spaces of the same dimension are the same
south patrol
ah, i just added that to be safe
dusty river
You need it to say that an isomorphism of subspaces extends to an isomorphism of the whole space
dusty river
because codimensions may not be equal in infinite dimensions
south patrol
sure :)
urban acorn
hmm, I think I see what's going on with that
like, in the F-space freely generated by N, the subspace generated by all even numbers probably can't be transformed via automorphisms to the subspace generated by all numbers bigger than 5, even though they have dimension
it also has the same dimension as the entire space, but it's a proper subspace
hidden haven
Yeah consider v_n ↦ v_n+1 where v_i is the ith basis vector out of denumerably many
novel parrot
what does A^m x n actually mean
what are the generators?
the cartesian product?
urban acorn
what does A^m x n actually mean
formally, you can think of the elements of A^(M x N) as functions M x N -> A whose image is 0 on all but finitely many elements of M x N
you can define a A-module structure on this by pointwise operations
This works with any set X replacing M x N, I don't know why they are looking at our set as given by the cartesian product of two other sets here.
urban acorn
you can define a A-module structure on this by pointwise operations
(which has a crucial property that for any other A-module B, an A-module map A^(MxN) -> B is the same as a set map MxN -> B)
urban acorn
Is this taken out of a proof of existence of tensor products?
novel parrot
yes
i didnt understand much of watch you said
but its isomorphic to M cross N?
well the set that we are taking as basis
@urban acorn
urban acorn
yeah, M x N generates A^(MxN)
urban acorn
i didnt understand much of watch you said
point out a specific part you didn't understand and maybe I can explain it better
novel parrot
yeah, M x N generates A^(MxN)
okay good
i got it
i understand the actual construction
is with functions
urban acorn
but you can think of them like the elements of MxN are "atoms", and the elements of A^(MxN) are A-linear combinations of them
so, you might have x, y in MxN
novel parrot
but you can think of them like the elements of MxN are "atoms", and the elements...
yup
urban acorn
then something like "2x - y" is an element of A^(MxN)
and we can use any coefficient from A
novel parrot
yupp
theres no motivation for tensor products here lol
another book has a really good explanation of why we are interested in it
here it says just quotient by this subgroup 
urban acorn
here it says just quotient by this subgroup <:kekw:752021756650455061>
submodule but yeah
that construction is very weird
novel parrot
yup submodule i mean
urban acorn
you shouldn't think of it really using that construction
novel parrot
yeah
urban acorn
okay, so let's say you're interested in bilinear products V x V -> V
or U x V -> W for that matter
which there are many reasons to be interested in them
the cross product is an example for a bilinear product on a 3d real vector space
novel parrot
ive never actually studied linear algebra
urban acorn
the inner product on a real inner product space is an example of a bilinear map into F
novel parrot
all i know about bilinear maps is just thats a homomorphisn at both points
urban acorn
yeah, that's what it is, when you fix one of the arguments, it's linear with respect to letting the other one vary
okay, so let's say you're considering linear maps V x V -> V
you might wonder if you can construct some space V', such that your maps V x V -> V will turn out to just be maps V' -> V.
and actually you can.
The tensor product V cross W is characterized by the fact that a linear map V cross W -> U is exactly the same thing as a bilinear map from V and W to U
novel parrot
oh
urban acorn
if in the future you'll learn category theory, you'll learn that these sort of characterizations of the structure-preserving maps of an object must characterize it up to isomorphism
novel parrot
do these tensor products have any connection to tensors
tensors as in generalized matrices
urban acorn
I have no idea what tensors are about
novel parrot
oh
urban acorn
but I think they are related to this
fossil shuttle
if V is an n-dimensional vector space, and W is an m-dimensional vector space, there is a natural isomorphism V*\otimes W \cong L(V,W)
so if you choose bases for V and W then you can identify V* \otimes W with the set of nxm matrices representing linear maps from V to W
of course if you choose basis for V then you can identify V with V*
more generally if V1,.... Vk are vector spaces of dimensions n1,...., nk then the iterated tensor product V1.... Vk is of dimension $n_1n_2 \dots n_k$ and so their elements can be represented exactly by these higher-dimensional analogues of matrices in $\mathbb{R}^{n_1\times n_2\dots\times n_k}$
cloud walrusBOT
diligentClerk
fossil shuttle
i think that's why engineers think of these as higher-dimensional matrices
or physicists
south patrol
in physics it always annoyed me slightly how they'd call stuff e.g. the inertial tensor and then define it by its matrix representation
tall jay
Is the identity element in a group unique?
Like if ae=a, e is the only element that makes it true?
tall jay
Bc in thedefinition, it looks like it only requires one element
it didnt say anything about uniqueness
scarlet estuary
proof: ||suppose f is an identity element, so a^-1 * a = f. then ae = a implies a^-1 * ae = a^-1 * a, but this simplifies to fe = f. but f is an identity, so fe = e, and hence e = f.||
tall jay
Oh interesting
scarlet estuary
invertibility is powerful
tall jay
How come the definition doesn't just say it's unique
scarlet estuary
because it's easy to prove
it could say that, but it doesnt have to
and we generally prefer our axioms to be "minimal"
tall jay
I see
scarlet estuary
(less things to check each time)
urban acorn
I had an idea about something recently. \newline \newline
So, as a simple special case, let's look a finite fields. We know they're all extensions of $F_p$, so consider some finite extension $E$ of $F_p$. \newline
For each element $x \in E$, we know $F_p(x)$ is a much easier field to study, because we can just look at the homomorphism $\phi : F_p[y] \to E$ given by sending the $F_p$ in $F_p[y]$ to the $F_p$ in E, and sending $y$ to $x$. \newline
But then, here's the cool part, we have to be able to get E by repeatedly taking extensions like that, because E is finite.
cloud walrusBOT
All groups are abelian
urban acorn
So the cool idea I had, is I think this can be generalized to arbitrary field extensions using transfinite recursion.
urban acorn
I had an idea about something recently. \newline \newline
So, as a simple speci...
I mean, and then $F_p(x)$ should be isomorphic to the field of fractions of $F_p[y]/ker \phi$
cloud walrusBOT
All groups are abelian
tall jay
proof: ||suppose f is an identity element, so a^-1 * a = f. then ae = a implies ...
Wait how did you do this thing where it covers the proof
urban acorn
\|\|your text here\|\|
I did an additional thing to make it render as the text with | characters in this message
hidden haven
How come the definition doesn't just say it's unique
You don't need invertibility either, it is also true in a monoid. Multiply the 2 identities, and that should give you both the identities
dusty river
yes
dusty river
yes
dusty river
Yeah the first step fails, 2x1 = 1x2
novel parrot
ok
dusty river
but that just proves that this proof doesn't work
so it's not a proof of it being non zero
dusty river
You have to show that the relations that you quotient 2Z x Z/2Z by, don't generate (2,1) (which is a hard problem)
Alternatively look at proposition 2.12
novel parrot
2.12 of atiyah?
dusty river
yes
dusty river
That is called the universal property of the tensor product. What that means is that if any module satisfies that property then it will be the tensor product, and it is almost always better to work with that universal property than to get into the annoying details of the construction
novel parrot
the details of the construction i dont completely understand
dusty river
What part?
novel parrot
a couple of them
The two modules
they are both left and right modules?
or one can be left and other right
dusty river
Everything in AM is with commutative rings so all modules are 2 sided
novel parrot
ok
dusty river
I am not sure how tensor product works with non commutative rings
novel parrot
in dummit and foote the ring is commute
but module doesnt need to be both left right
dusty river
hmm I don't see how that would be the case, could you send a screenshot?
novel parrot
and 2 x 1 = (1 + 1,1) - (1,1)
dusty river
in Z x Z/2Z?
dusty river
sure
novel parrot
i mean generally
novel parrot
ok in an arbitrary modules
dusty river
Yeah ok
novel parrot
dusty river
I think they are implicitly assuming non commutative rings there when they say "the general tensor product"
Because the axiom that breaks over non commutative rings is a(bv) = (ab)v, which becomes v(ab) = (va)b when you just turn this into multiplication from the right
novel parrot
i see
are these elements 0?
because it says that the map from free module of M x N sends those to 0
wooden ember
there's something im misunderstanding here I think... Doesnt the result follow pretty trivially from the fact that G is nilpotent? Then sylow subgroups of the Hall-pi subgroup are Sylow-subgroups of G and so are normal so that any Hall pi-subgroup is just the direct product of the sylow p-subgroups for p in pi, and hence unique???
oh sorry @novel parrot
novel parrot
nw
novel parrot
because it says that the map from free module of M x N sends those to 0
and i dont know whyit would do that
$\overline{f}: A^{M \times N} \rightarrow P$
dusty river
by bilinearity
cloud walrusBOT
ActiveChapter
novel parrot
oh yep
ok
why not just abuse notation and say u x z = (u,z)
i dont like using cross
dusty river
should be fine if its unambiguous
if you are talking about direct product or tensor product
novel parrot
lol
so the important thing about this really was
multilinear functions can be factored by the map into tensor product
but how does that solve my initial question about 2 x 1 being non zero
next obsidian
Are you asking how to show a specific simple tensor is non-zero?
novel parrot
yes
next obsidian
So say you wanna show a x b is nonzero
Find a bilinear map f:M x N -> L which has f(a,b) is non-zero
Then by universal property you get a map g: M (x) N -> L
Such that liek the triangle commutes
The map M x N -> M (x) N sends (a,b) to a x b
Then g(a x b) = f(a,b) is non-zero
So a x b is non-zero because if it was zero then g(a x b) is zero
This is basically the only way to do it because the relations in the tensor product are incredibly fucked
novel parrot
Then g(a x b) = f(a,b) is non-zero
i dont get this
next obsidian
The universal property of the tensor product
For a bilinear map from M x N to L
You get an induced map from M (x) N -> L making the triangle commute
novel parrot
yeah
next obsidian
The map M x N -> M (x) N sends (a,b) to a x b
So evaluate f at (a,b)
And then call the map M x N -> M (x) N like say… h
Then f(a,b) = g(h(a,b)) by commutativity of the diagram
But g(h(a,b)) = g(a x b)
So g(a x b) is non-zero because we assumed f(a,b) is non-zero
next obsidian
novel parrot
So g(a x b) is non-zero because we assumed f(a,b) is non-zero
is f, the bilinear function 0 if and only if h(a,b) was 0
if $f:2\mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z} \rightarrow \mathbb{Z}$ $(2,1) \mapsto 2(1)$
cloud walrusBOT
ActiveChapter
dusty river
f = f' g
if f is non zero at some point then f' and g must be too
at that same point
dusty river
Yeah so g(a,b) != 0 in particular
and g is the map from the product to the tensor product
so a \oplus b != 0
novel parrot
how do we know f' is zero if input is zero
dusty river
Homomorphisms map 0 to 0
dusty river
ye'
novel parrot
okay nvm it does lol
dusty river
lol
dusty river
**ActiveChapter**
Are you defining this as (2a, b) maps to 2ab?
dusty river
(4,1) should map to the same thing as (2,2)=(2,0)
And if it were bilinear you could use the same argument to say that 2 \otimes 1 is non zero in Z \otimes Z/2Z which is false
novel parrot
(4,1) should map to the same thing as (2,2)=(2,0) <:catThin4K:811354320460578896...
wdym?
why 2,0
dusty river
Your map sends 4,1 to 4 but 2,0 to 0
novel parrot
yep
dusty river
but for it to be bilinear both of these should map to the same thing
because (2*2,1) = 2(2,1) = (2,2) = (2,0)
because the second coordinate is mod 2
novel parrot
oh rightt
i forgot mod 2
whops
i just gotta find a bilinear map that is always 0 yeah?
dusty river
No you need to prove that all bilinear maps are 0 everywhere 
dusty river
and this is using the fact that the natural map from the direct to the tensor is surjective
Because thats a bilinear surjective map so must be 0
or you can show that the 0 module satisfies the universal property if you know that all bilinear maps out are 0
novel parrot
okey
next obsidian
You can also show that R/I (x)_R R/J ≈ R/(I + J)
I think you can probably even just show every simple tensor is 0 manually
novel parrot
finding bilinear maps kinda hard
i think z/3z x z/2x -> z/6z by a,b = ab
is bilinear
but is this the only one
hmm
oblique river
you shouldn't be finding bilinear maps
you're trying to prove that every bilinear map out of Z/n x Z/m is zero
i.e. "let A be a Z-module (abelian group) and let f: Z/n x Z/m --> A be a bilinear map"
dusty river
try to continue from that statement and figure out conditions on f
dusty river
novel parrot
idk if im wrong or right
but
0,0 + 0,0 = a + a = 0,0 = a
so a + a = a
a = 0
so im right i think
hidden haven
Yeah homomorphisms map 0 to 0
hidden haven
You might have a slightly easier time thinking if you try to prove that 1 tensor 1 = 0 tensor 0
novel parrot
hm
f(1,0) and f(0,1) are always 0 right?
if f(1,1) = x, f(1,1) + f(0,1) = x + y = f(1,1) = x
this must be wrong
novel parrot
ok
next obsidian
Yeah homomorphisms map 0 to 0 <:catThimc:732011802459635713>
This mf switched from an alt to main and sent messages from each in the span of a minute
wooden ember
for 38 ive managed to show that the rank of ${z\in Z(M)|z^2=1}$ is $\leq 2$ but how does this extend to $Z(M)$?
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
also note it means to say P/M in the hint not G/M
wooden ember
Reeeeeee this is the last exercise I need for this section it's pissing me off
okay im stupid as fuck
i literally showed the kernel of the pth power map has the same rank as the starting set a few sections ago
AAAAAAAAA
barren sierra
I struggled with this concept back in Abstract Lin Alg
what does it mean for a function to be dual to another function
thorn delta
what does it mean for a function to be dual to another function
maybe youre thinking of a dual map in linear algebra?
any linear map f : V -> W induces a linear map f* : W* -> V* given by f*(w) = wf
otherwise "dual" might mean something more informal, like "corresponding" or "closely related"
barren sierra
so someone in my AA class said that the logical operation or is dual to and
and I can't find something about that at all @thorn delta
thorn delta
You can’t find anything about dual maps? https://en.m.wikipedia.org/wiki/Transpose_of_a_linear_map
thorn delta
so someone in my AA class said that the logical operation `or` is dual to `and`
Also, that’s wack I’ve never heard that before
chilly ocean
hidden haven
so someone in my AA class said that the logical operation `or` is dual to `and`
Whenever you have a partially ordered set, and some notion defined on it, you can define a dual notion by reversing the partial order and defining the original notion on this reversed order. That would be called the dual
Here logical statements would be the elements in the partial order and the order would implication (a<b iff a → b)
Then or is the least upper bound of 2 elements and and is the greatest lower bound
hidden haven
More generally there's a notion of duality in category theory in which you just formally reverse all arrows in a category and dualize all your statements and they remain true
And posets are just glorified thin skeletal categories 
wooden ember
@hidden haven I see you laughing at me 
urban acorn
huh, so you know how (a, b) = {xa + yb | x, y in R}
you can prove this by checking that it's an ideal and that it contains a and b
and then noticing that by the properties of an ideal, every ideal containing a and b contains all of these
but I thought of a different way to think about this
which - although it's more complicated from a certain perspective - I like it more, and it feels more natural in some sense
so, the idea is, consider R as an R-module. Ideals are precisely submodules. Then, given a, b in R, we get a map from the free module on "a" and "b" to R. Then the submodule generated by a, b is the image of that map.
and the free module consists of expressions "xa + yb"
and sending them across the map consists of evaluating them in R
potent briar
hmm what's a good idea for trying to find bases of vector spaces?
this exercise is asking me to find two bases in C4 that have two vectors in common: (0, 0, 1, 1) and (1, 1, 0, 0)
brazen mesa
C4 = $\mathbb{C}^4$?
cloud walrusBOT
Anomalocaris
potent briar
yeah sorry
brazen mesa
are the scalars complex or real numbers?
brazen mesa
just making sure
you know two vectors that are in the basis already and the dimension is 4, so you need to find two other vectors for the basis
try finding something that's linearly independent from (0,0,1,1) and (1,1,0,0)
potent briar
ok i think a good one is (1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 0, 1), and (0, 1, 1, 0)
chilly ocean
(0, 1, 1, 0) = (1, 1, 0, 0) + (0, 0, 1, 1) - (1, 0, 0, 1)
potent briar
right
chilly ocean
so this isn't a basis
potent briar
yeap
im looking for one
ok here's one
(1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 0, 1), (0, 0, 1, 0)
i actually checked this time
anyway my question was more along the lines of
are you expected to brute force it
or is there a trade secret heuristic
urban acorn
given any linearly independent non-spanning set, add a vector outside its span, and you've got a larger linearly independent set
keep doing this until it spans the space and that's how you get a base
potent briar
oh yeea
urban acorn
and by making different choices of vectors outside the span, you can make as many different bases that you like
potent briar
thats the proof of a theorem in this book
silly
ok in a related question
how do i check if a given l.i. set generates the space?
brazen mesa
if you know the dimension of the space it's easy (assuming finite dimension)
potent briar
sure, if the set is l.i. and its cardinality = the dimension of the space, it's a basis
but how could i write it out with like a system of equations
brazen mesa
which means it generates the space, i.e. its span is the entire space
scarlet estuary
how do i check if a given l.i. set generates the space?
take an arbitrary vector in space and show somehow (usually through explicit computations) that you can generate that vector
for example, if you want to prove a set spans $\mathrm{C}^4$, you'd choose a vector $\begin{pmatrix}a\b\c\d\end{pmatrix}$ with $a, b, c, d \in \mathbb{C}$ and choose coefficients $\lambda_i$ (based on $a, b, c, d$) for your basis vectors $v_i$ such that
[
\sum_{i=1}^{4}\lambda_{i}v_i = \begin{pmatrix}a\b\c\d\end{pmatrix}
] and youd demonstrate explicitly that the sum on the left is indeed equal to the right hand side
cloud walrusBOT
Namington
scarlet estuary
let me do a really simple example
lets say i wanted to show $\begin{pmatrix}1\1\end{pmatrix}, \begin{pmatrix}1\0\end{pmatrix}$ spanned $\bR^2$
cloud walrusBOT
Namington
scarlet estuary
so i take an arbitrary $\begin{pmatrix}a\b\end{pmatrix} \in \bR^2$ and then observe that, if i choose coefficients $\lambda_1 = b$ and $\lambda_2 = a - b$, we get:[
b\begin{pmatrix}1\1\end{pmatrix} + (a-b)\begin{pmatrix}1\0\end{pmatrix} =
\begin{pmatrix}b\b\end{pmatrix} + \begin{pmatrix}a-b\0\end{pmatrix} = \begin{pmatrix}b + a - b\b + 0\end{pmatrix} = \begin{pmatrix}a\b\end{pmatrix}
]
cloud walrusBOT
Namington
scarlet estuary
now, as for finding those coefficients (scalar values of lambda_i)
you would indeed typically set up a system of equations
can you think of how you'd do that?
(this isnt required in the proof itself, but its required to actually, you know, come up with the proof)
(assuming you cant just figure it out by inspection)
urban acorn
he specifically said the set is linearly dependent, and he's working with finite dimensions
scarlet estuary
(to use a metaphor from analysis, its like picking delta in an epsilon-delta proof; you're usually gonna have to do some work first to determine an appropriate delta)
urban acorn
so there's no need to explicitly construct a linear combination for arbitrary vectors
scarlet estuary
i didnt?
im assuming they didnt know the dimension
just that its finite
hence we need to show the set is spanning
an alternate approach is to show the vector space has a certain dimension
through some OTHER basis
(say a standard basis)
urban acorn
okay, sure
scarlet estuary
and then linear independence + counting elements works
urban acorn
if you don't know the dimension, then yeah
scarlet estuary
"the standard basis for ℝ² has 2 elements, (quick proof that it is in fact a basis), so the dimension of ℝ² is 2. this set has 2 linearly independent vectors. so its a basis"
this will TYPICALLY be faster for sets where there's an "obvious" basis
like, F^n for a field F for example
but i gave a slightly more general approach when you cant necessarily easily determine the dimension
both work.
urban acorn
yep
scarlet estuary
technically you dont have to find a basis of 2 elements, just a spanning set of size 2
ie you dont need to prove its a basis, just that it spans
since that gives an inequality on your dimension, but you have the other direction of your inequality from the fact that a set of 2 elements (or however many) is linearly independent
so you know the dimension is ≤2 and ≥2
hence = 2
(replace 2 with whatever number and this still works)
(as long as its finite dimensional)
alternatively, if youve seen it in class, you could just apply this theorem:
for a field $F$, $\mathrm{dim}(F^n) = n$
cloud walrusBOT
Namington
scarlet estuary
the complex numbers form a field so $\mathrm{dim}(\mathbb{C}^4) = 4$ and hence any l.i. set of size $4$ spans $\mathbb{C}^{4}$
cloud walrusBOT
Namington
scarlet estuary
(and is therefore a basis)
you can probably tell that theres a bunch of approaches here
but they're all similar "in spirit"
just sometimes you have faster techniques to your destination
terse crystal
Any way unless X and Y are path-connected you still need to specify their base points respectively…
hidden haven
Any way unless X and Y are path-connected you still need to specify their base p...
Wrong channel? 
terse crystal
Oh yes lol
maiden ocean
Suppose i fix a field k and have two extensions K | k and L | k both finitely generated as fields over k of transcendence degree 1
then an inclusion K -> L implies L | K is algebraic
is L | K necessarily finite?
hidden haven
k(x) and k(all rational powers of x) should be a counterexample
oof finitely generated
Yeah L | K should be finitely generated by taking the finitely many generators of L | k
And finitely generated + algebraic = finite
terse crystal
Any element r=f(x)/g(x) from k(x), The minimal polynomial of x in k(r) is f(X)-rg(X), so yes algebraic
hidden haven
You're proving that k(x) | k(r) is algebraic?
hidden haven
But they need not be purely transcendental
k(all rational roots of x) | k also has transcendence degree 1
terse crystal
Maybe I can find a commutative diagram. L is algebraic over L’=k(x) and K’ is the intersection of K and L’
L is algebraic over K’ therefore algebraic over K. I am checking
Oh nvm…
L over L’ is not necessarily finite…
Though L’ over K’ is
maiden ocean
semi related follow up question:
suppose we have extensions L | K | k as before (so L and K are finitely generated fields of trdeg 1 over k and L | K is finite) and take some f in K transcendental over k
consider A, the integral closure of k[f] in K, and B, the integral closure of A in L
can we express B as the integral closure of k[g] in L for some g in L?
I suspect the answer is no which will make this construction not as nice as i had hoped, F
hidden haven
L is algebraic over K’ therefore algebraic over K. I am checking
If L | K | k is such that L and K have the same finite trdeg over k, then L | K is algebraic because if you take some tr basis of K | k then everything in L is algebraically dependent on this
I think that's what moth used in the beginning
maiden ocean
Yea
i mean
trdeg(L | k) = trdeg(L | K) + trdeg(K | k) lol
man
i hope that this works out because then you get something really elegant where pulling back the affine cover on a proper normal curve gets you a very similar affine cover...
Its workable regardless but less nice
maiden ocean
suppose we have extensions L | K | k as before (so L and K are finitely generate...
@next obsidian do u know the answer to this
This sounds like something you would know
novel parrot
if A - S is union of primes
any xy in A- S
(xy) is contained in some prime
so S is saturated
does that work
next obsidian
<@!146419489083228160> do u know the answer to this
Dude what
Just take f = g?
Why doesn’t that work?
maiden ocean
Uhhh why would that work
maiden ocean
does x integral over A imply it is integral over k[f]
next obsidian
Yes
maiden ocean
Yes
next obsidian
This is how you find the integral closure of stuff in extensions often
maiden ocean
I have
maiden ocean
I thought i would do this at first and then i tried writing it out to be sure but it wasnt immediate
next obsidian
maiden ocean
So i gave up
next obsidian
Yeah it isn’t immediate
next obsidian
Because it’s like polynomial over polynomial
Anyway I think you can get it from trasnivtivity of finiteness
maiden ocean
i think u can just use that given sum a_i x^i with a0 = 1 the ai are integral so A[a0, ..., an] is a finitely generated A-module
next obsidian
Yes that’s exactly what I was saying
next obsidian
Or had in mind
maiden ocean
Man
next obsidian
I read it
maiden ocean
Anyway thanks everything works out then
next obsidian
And almost let you gaslight me by asking this question into thinking this wouldn’t work
next obsidian
It is your fault
maiden ocean
Its gaslighting me
maiden ocean
Thats what gaslighting is!!!
next obsidian
THEY CALL ME MISTER COMMUTATIVE ALGEBRA
maiden ocean
Thank u qween
maiden ocean
ok in honor of u i will let u decide convention
next obsidian
mathbb
maiden ocean
mathbb or mathbf for affine and projective lines
next obsidian
100%
maiden ocean
Poggers
next obsidian
Actually
maiden ocean
Thank you for affirming what i already did 
next obsidian
200%
novel parrot
do you know if im correct?
novel parrot
.
next obsidian
(xy) is contained in some prime
Wat
maiden ocean
thonk
maiden ocean
you want to assume xy is in S and then find a contradiction if x or y is not in S
assuming xy is in A - S will not really be helpful
novel parrot
Wat
the ideal
we take xy in A - S
A - S = union of prime
so ideal generatedby (xy) is in union of primes
that works? no?
maiden ocean
I dont see why you are taking xy in A - S
novel parrot
so i can go by contradiction
novel parrot
A - S = union of prime
so we know then its contained in some prime ideal
maiden ocean
Yes chmonkey
maiden ocean
1 is not contained in a union of prime ideals 
novel parrot
?
next obsidian
Okay but that tells you one of x or y is in a prime right
novel parrot
yea
novel parrot
im confused
novel parrot
but then we take the negative of statement
novel parrot
xy is not in the prime, x and y are not in prime
next obsidian
What is your starting assumption?
maiden ocean
Yeah its not clear what ur trying to do exactly
novel parrot
A - S is union of primes, i want to show S is saturated
maiden ocean
The negation of "S is saturated" means "there exists an xy in S with x, y notin S"
novel parrot
negation of xy in A/S
next obsidian
What does that have to do with anytbing
novel parrot
cuz if its not in A/S, its in S
next obsidian
But why are you negating that statement?
How does that prove S is saturated
You’ve started by assuming A\S is a Union of primes
Next you said xy is in A\S
novel parrot
buy showing that if xy was in S, then x and y gotta be in S
novel parrot
(xy) is contained in atleast 1 single prime
next obsidian
Okay
novel parrot
so xy is contained in a prime
next obsidian
Oky
novel parrot
so x or y is in a prime
next obsidian
And then
novel parrot
x or y is in the union
next obsidian
Okay
Like I have no idea what you’re hoping to accomplish by negating xy is in A\S, and from what you have so far you can only ensure x or y is in the Union
You need both to be in the union
novel parrot
union = A/S
maiden ocean
next obsidian
What? Now you’re assuming S is saturated?
novel parrot
no
next obsidian
Am I just being pepega brained rn
novel parrot
so weve shown xy is in A/S then x or y is in A/S yeah?
next obsidian
Yeah
maiden ocean
are you trying to take the contrapositive of "xy is in A/S, then x or y is in A/S"
next obsidian
What
maiden ocean
Hurb
novel parrot
if xy was not in A/S, then not (x or y is in A/S)
next obsidian
Yes
novel parrot
ok so i just have one part left
next obsidian
But that literally only translates to x,y in S then xy in S
It isn’t saturated from this
weary terrace
Let $G$ be a finite group and $\rho$ a one-dimensional representation of $G$. Denote $N:=\ker \rho$ and we know that $G/N=\mathbb Z/n \mathbb Z$. Take another irrep of G, $\sigma$ and observe $\text{Res}_N \sigma$. What can we say about $\text{Ind}_N^G (\text{Res}_N \sigma)$?
cloud walrusBOT
RiesZ
maiden ocean
I thikn you should just do it directly
(@ active not riesz)
like assume xy is in S
what happens if x is not in S
novel parrot
ugh
novel parrot
I thikn you should just do it directly
wdym exactly?
iwas thinking x or y isin A/S -> xy is in A/S
and contrapositive it
nvm
urban acorn
so you know sometimes it's nicer to work over N rather than Z?
like gcd's and prime factorisation are unique
I wonder what a generalization of this to other rings would look like
maybe something like a set $S \subset R - {0}$ that contains exactly one element out of each orbit of the action of $R^\cross$ and is closed under addition and multiplication
cloud walrusBOT
All groups are abelian
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
urban acorn
**All groups are abelian**
Compile Error! Click the <:errors:504266556054962196>...
I mean, that describes $\mathbb{N} \subset \mathbb{Z}$ pretty well
cloud walrusBOT
All groups are abelian
urban acorn
Can we find something like that in $\mathbb{Z}[i]$, for example?
cloud walrusBOT
All groups are abelian
urban acorn
We necessarily have $1 \in S$ and $-1, i, -i \notin S$
cloud walrusBOT
All groups are abelian
urban acorn
I think that if you just require closure under multiplication then this equivalent to choosing a unit for each Gaussian prime
I just realized we need to have all naturals because we have 1 and closure under addition
so for all real Gaussian primes, the unit is 1
mild laurel
I mean, there's an equivalence relation on elements of a ring that two elements are equivalent if and only if they differ by a unit. This amounts to picking an element from each equivalence class, but I think in general rings there's no straightforward way to do this
urban acorn
yeah, but what structure do we give this set? we can multiply on it but we can't add
I think that a set with the properties I specified doesn't exist in Z[i]
next obsidian
The nilradical of A/a corresponds fo the image of r(a)
You assumed r(a) is maximal so the nilradical of A/a is maximal
novel parrot
ok ty
wooden ember
im not sure i see where the last line comes from? Why does every Sylow 3 subgroups normalize some Sylow 7 subgroup? Ive included more content because i think it's related to the fact the normalizer of a sylow 7-subgroup has order 21: this means every sylow 7 subgroup is normalized by some sylow 3 subgroup. But i dont see how this is true the other way around. Here G is any simple group of order 168
next obsidian
im not sure i see where the last line comes from? Why does every Sylow 3 subgrou...
I think this idea works?
Let P be a Sylow 3 subgroup
there’s 8 Sylow 7s, so we can act on the set of them via conjugation
The orbit can only be size 3 or 1 by orbit stabilizer
So there must be a size 1 orbit
next obsidian
It took 5ever for me to figure out how to prove it tho lmao
wooden ember
weird that the step is skipped though there must be some more obvious way 
wooden ember
or im stupid and forgot some theorem we used a few times lmao
wooden ember
is there a proof of schur zassenhaus that doesnt use cohomology 
the theorem seems so innocent it feels weird that it needs such advanced machinery
next obsidian
I think my old TA explored this quite deeply
I think he couldn’t figure out a way or find one that didn’t involve cohomology?
I think there’s some weird transfer shit that you can use which makes it simpler? I don’t really remember the details at this point
wooden ember
weird
still on simple groups of order 168: the book has just shown that normalizers of sylow 3-subgroups are isomorphic to S_3 and then conclude that there are no elements of order 6 in G?
im not sure how the conclusion follows
the whole thing so far for reference
wooden ember
nvm im being dumb as usual
novel parrot
how r they using 1.11
like the cap of those primes is contained in the union so its contained in a single one yes?
urban acorn
nvm im being dumb as usual
that's what doing math be like always
maiden ocean
hmm suppose i have a finite extension L | K of fields with discrete valuation rings S and R of L and K, and a local homomorphism R -> S
Is this local homomorphism necessarily injective
gritty sparrow
hmm suppose i have a finite extension L | K of fields with discrete valuation ri...
Does the map from R to S commute with the localization to K and L? If yes I think so, and that seems like a natural condition
maiden ocean
Contextually S and R are stalks so i think they should
the full situation is that i have a surjection of zariski riemann curves Y -> X and S is a point of Y with image R
Wait what do you mean the localization to K and L, at the 0 ideal?
gritty sparrow
I mean, the localization of R at 0 is K in this situation right?
gritty sparrow
Yeah
And the field map is also injective so then we must get that the R to S map is injective
maiden ocean
Yeah
Im not 100% sure this is true but it sounds like it should be because the diagrams of restriction maps are
gritty sparrow
I mean if a in R maps to 0 in S, it maps to 0 in L one way so a maps to something in K mapping to 0 in L which has to be 0, and the only element of R napping to 0 in K is 0
maiden ocean
No yeah im just not sure that the diagram actually commutes
maybe you can apply universal property of colimits somehow
gritty sparrow
Well, in our situation, isn’t the map from K to L defined as the localization of the map from R to S?
Or is there some other map at play
gritty sparrow
But the stalks are R and S
maiden ocean
Theyre all stalks
the points of a zariski riemann space are discrete valuation rings including K and L itself
with the local ring at R just being R
yea this is kind of a fucky construction which is why i tried to avoid using the geometry 
gritty sparrow
I see
maiden ocean
another way you could say this is that like
K is dense in X
so its the limit of all open sets of X
same for L
and we have an induced map K -> L
R and S are limits of a subcollection of open sets
does this mean that R -> S commutes with K -> L
gritty sparrow
Yes I think so
maiden ocean
yea i think if u do everything from this perspective its easier
the inclusion map R -> K is given by like
K is the colimit of a collection C of open sets, R is the colimit of a collection C' of open sets, and C' subset C
I wonder if there is a source that writes this all out explicitly
maiden ocean
Well theres this book called szamuely on etale shit
And im trying to verify a proof
gritty sparrow
I see
novel parrot
what does this mean
this definition feels circular to me
if p belongs to a
so p is already in sigma
?
next obsidian
weird
So I asked my TA and he actually formalized a proof in lean without using cohomology, in the abelian case it uses some shit called transversals but idk what that really is, then you generalize to the non-abelian case
maiden ocean
Sigma is not necessarily all the prime ideals belonging to a
it is a collection of them
Hello chmonkey
novel parrot
Sigma is not necessarily all the prime ideals belonging to a
o
my mistake
thank you
maiden ocean
Oh this is good can i ask you a probably elementary geometry question
@next obsidian Suppose we have a morphism Y -> X of locally ringed spaces and a collection C of open sets of X containing a subcollection C' so C' subset C and all that
where the colimit of the C is a stalk A_X of a point and the colimit of the C' is a stalk B_X of a point
similarly we have A_Y and B_Y taking the preimages of the C and C'
next obsidian
What in the fuck are you saying
gritty sparrow
Yes the stuff we want above will work
next obsidian
Why A_X
gritty sparrow
Because commutativity holds on the open sets
next obsidian
Just call it like O_X,x and O_X,y
next obsidian
Okay so
gritty sparrow
And then, given an element of R, express ut as the image of a germ from some open set
maiden ocean
do we get inclusions of the stalk from C' into the stalk at C
gritty sparrow
Check commutativity there, and we will be done
maiden ocean
and do these commute with the maps of stalk coming from Y -> X
gritty sparrow
Check commutativity there, and we will be done
There is probably a way to do this with just categorical nonsense
next obsidian
Yeah but the fact you get a stalk
gritty sparrow
Is what I’m saying wrong?
next obsidian
Suggests to me there’s enough open sets
maiden ocean
uhh well in the case im working with the stalk from C' is a DVR R and the stalk at C is a field K
maiden ocean
sorry saketh i am trying to read multiple things at a time 😵💫
Yes
it is
im just not sure if the map u get from the above construction is necessarily the inclusion
next obsidian
Wait
I think it is
Because direct limits are exact in R-mod
Or someshit
Like the map comes from the direct limit of a direct system of rings via identity
maiden ocean
So true
next obsidian
And so if you can prove the map is the same as like say, the map you get considering them all as abelian groups
And taking the colimit as an abelian group or some shit
Or maybe just directly show that the direct limit of a system of injective ring maps is injective
Use like an explicit construction
maiden ocean
suppose we have an injection A -> B of dedekind rings and p is a prime ideal of A factoring into q_i^e_i in B
can we get conditions for k(q_i) | k(p) to be separable?
i guess for context im considering a finite surjection Y -> X of affine normal curves corresponding to A -> B and putting riemann surface structures on Y and X
where the e_i are the ramification indices of the q_i and thus the e_i = 1 if and only if Y -> X is a cover in a complex nbhd of p
but etaleness should require that the k(q_i) | k(p) are separable in addition to the e_i = 1 so im trying to see why in this case being etale at p is equivalent to being a topological cover at p
or maybe separable is implicitly used to show that the e_i are 1
idk
gritty sparrow
Ok there may be some extra context here but since you are talking abt riemann surfaces, the curves are over C right? In this case bc everything is char 0, separability should be no issue
@maiden ocean
maiden ocean
Oh my god lmfao
next obsidian
Pwned
wooden ember
So I asked my TA and he actually formalized a proof in lean without using cohomo...
I see, yeah the cohomology stuff was only used in the abelian case I think the extension is quite elementary. Thanks for checking it out
next obsidian
Yeah, I would send you his proof but it’s… well… in lean
hot lake
yes
upper pivot
hint is m>m^2>m^3>....
upper pivot
huh
upper pivot
yeah thats it
novel parrot
btw
when squaring an ideal
its the same as squaring the generators?
or cartesian product of generators?
novel parrot
(2,t)^2 = (4,t^2,2t)?
next obsidian
Yes
upper pivot
Do you know what the square of an ideal is
novel parrot
vaguely
upper pivot
right so the product of two ideals I,J is the ideal generated by the products ij with i in I and j in J
novel parrot
yeah
upper pivot
do you see how that implies what C.H. monkey said
novel parrot
yes
i was kinda thinking that caresian product of the generators
its like cartesian product
times everything
by everything
for showing a is that decomposition
i can just show that the generators lie in each other yes?
novel parrot
im sure
but
just sometimes need a sanity check
i dont wanna do these problems and then it being that i did them wrong
novel parrot
how do i find the isolated primes?
isolated primes are just minimal primes?
of the decomposition
minimal meaning doesnt contain any other prime in the decomposition?
wooden ember
solved this one using Hall subgroups (which ended up giving that the subgroup of order m was in fact characteristic) but it feels a little overkill and was wondering if there was a simpler more fundamental property of solvable groups that would solve this one more easily (and perhaps only straightforwardly show normality)
gilded trellis
So I found this proof for the simplicity of An for n>=5 : https://math.stackexchange.com/questions/3481899/simple-proof-of-a-n-being-simple
But I'm having difficulty understanding some things (confused about a few things).
isn't $$ \alpha = (12)(34)(56) $$ not an element of $$ A_n $$ ?
cloud walrusBOT
Der Gegenstand ist einfach.
chilly ocean
yo araragikun :3
chilly ocean
ye is not in An'
gilded trellis
I see you hangout mainly in #groups-rings-fields ^^
chilly ocean
ye
wooden ember
where in the proof is this written though?
wooden ember
yeah just a mistake
gilded trellis
I tried to prove the statement and the first thing that came to mind was showing that if G is normal in An and G != {id} then it has a 3 cycle
wooden ember
the correct argument is simply that if alpha is an even product of disjoint 2-cycles it permutes more elements than a 3 cycle
gilded trellis
yes, but it would only invalidate the minimality of alpha if the normal subgroup already had a 3 cycle
wooden ember
oh alpha is in the normal subgroup right
gilded trellis
so we need to discuss multiple cases depending on the "structure" of alpha
wooden ember
they mistyped that i see
gilded trellis
just noticed it too :""
so what are the possible structures of alpha ?
like, how many possible forms are there for it to take ?
wooden ember
lemme think
gilded trellis
okay ^^
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
$\alpha = (a_1, a_2)...$ where we dont actually have to care about all the 2 cycles after the first one
cloud walrusBOT
𝓛ittle ℕarwhal ✓
gilded trellis
yes
shouldn't we take the first 4 ? if alpha = (ab)(cd) tau then we proceed like the proof
wooden ember
consider $\beta = (a_1,a_2,a_3)$ where $a_3$ doesnt occur in any of the 2-cycles composing $\alpha$. Then $\beta\alpha\beta^{-1}=(a_2,a_3)...$ where everything after the first 2 cycle is unchanged. So then $\beta\alpha\beta^{-1}\alpha=(a_1,a_2)....(a_2,a_3)...=(a_2, a_3, a_1)$
gilded trellis
yes
cloud walrusBOT
𝓛ittle ℕarwhal ✓
gilded trellis
so we construct a permutation that permutes less elements than alpha
wooden ember
yeah, but which is in the normal subgroup too
wooden ember
my only issue here is how we can assure there is an a_3 not in the cycles of alpha
because alpha could permute every element of {1,2,....,n}
gilded trellis
aah
wooden ember
probably easy way to construct a smaller permutation in this case too
since it permutes everything
gimme a sec to think about it
yeah lmao just take beta to be a 2 cycle
a product of 2 2-cycles sorry
gilded trellis
short circuit it ?
gilded trellis
it's okay :"
(currently googling derangements)
(to see if I can find any interesting property... cuz I'm kinda weak in knowledge when it comes to symmetric groups..)
wooden ember
right okay got it
in the case where $\alpha$ permutes every element of ${1,...,n}$ we can write $\alpha=(a_1,a_2)(a_3,a_4)...$ and $\beta=(a_1,a_2,a_3)$ this will still result in $\beta\alpha\beta^{-1}\alpha$ being a smaller permutation when $n>4$ since it will be a product of 2 2-cycles
cloud walrusBOT
𝓛ittle ℕarwhal ✓
wooden ember
honestly proofs of simplicity of A_n are all kinda nasty in my experience, it's a decent amount of case handling
i like the one i saw in dummit and foote
gilded trellis
lol
wooden ember
it doesnt go by showing A_n is generated by 3 cycles
wooden ember
for A_5 it's not too hard
gilded trellis
and then does some weird stuff to go back to the case A_5
wooden ember
nvm A_5 is nasty too
well i mean it's easy to understand but not particularly pleasant looking
it's exhibiting conjugacy classes of A_5
gilded trellis
:" D
yeh
so right now you've shown that alpha must be written as a product of transpositions that cannot all be disjoint right ?
wooden ember
yeah
gilded trellis
I honestly don't have an idea of what can come next :"
wooden ember
yeah i dont quite follow the rest of the argument
gilded trellis
why are the only two cases left (1 2 ... n) and the other one
wooden ember
oh nvm i guess i do see
gilded trellis
:"
wooden ember
you can have alpha as an odd length cycle
or you can have it as a product of some disjoint n-cycles for n>2
in the first case you can always get an appropriate choice of beta to do as before and get a shorter cycle, and so can you in the second case i guess
again it's doable just nasty
since you construct those betas
im an anti constructivist lmao
should i give you the proof i know of
i find it a little less nasty than the others
gilded trellis
I'd be happy to learn about the second proof :"
wooden ember
starting at bottom left
