Even a kid can do these proofs if he knows a lot of properties
#groups-rings-fields
lethal dune
u still need help
lethal cipher
Okay Chmonkey, I need some help understanding. So even though a poly has distinct roots in a splitting field, I am guessing an extension being separable does not necessarily mean it will be the splitting field of every min poly which contains an element.
untold basin
u still need help
I feel dumb x)
lethal cipher
All that seems to matter is that there is a splitting field s.t. the roots are all simple. The separable extension does not necessarily need to be the splitting field. Am I understanding that right @next obsidian?
woven delta
Lol I never use definitions when doing this sort of thing
I proceed mostly by intuition
untold basin
Lol I never use definitions when doing this sort of thing
yeah
untold basin
I proceed mostly by intuition
I feel like it's more like asking yourself what do you want and make links with what you know
Idk if there is any visual intuition here
woven delta
Intuition isn't necessarily visual
woven delta
I feel like it's more like asking yourself what do you want and make links with ...
I would describe this as using intuition
next obsidian
All that seems to matter is that there is a splitting field s.t. the roots are a...
I mean idk what simple is but
Like yeah
You take a in your extension
lethal cipher
simple just means the root is distinct
next obsidian
Take a splitting field for the min poly
If those roots are all distinct
Then a is separable
So I guess the answer is yes
You have to take a splitting field
lethal cipher
Okay, so let's say f=(x-a1)...(x-an) and we are given that F(a1) is a separable extension. So I think this makes sense because there is a splitting field for f: namely, F(a1,...,an) s.t. f splits completely, but F(a1) is not guaranteed to be said splitting field.
next obsidian
Yeah
lethal cipher
So here F(a1) is not guaranteed to be normal, since a2,...,an may be independent of a1, but it is separable
Okay, and normal would not imply separable since it is possible to have a min. poly with multiple roots.
lethal dune
I feel dumb x)
so InnG ~ G/Z(G)
InnG subgroup of Aut G so cyclic
meaning G/Z(G) cyclic
so abelian
untold basin
Yep
untold basin
so InnG ~ G/Z(G)
I meant that I understood directly after Emma's tip
It wasn't difficult
kind jungle
next obsidian
There’s only 3 c to test
Because they draw from Z_3
So if you know about the fact that the quotient by an ideal is a field iff the ideal is maximal this tells you what to do
And if not, you could still do it
It’ll just be a bit more complicated
mint gulch
Are there UFD of any cardinality? (Infinite) 👀
next obsidian
Yeah, because there’s fields of any cardinality
Look at Q(x_i) where you adjoin kappa variables for any cardinal kapp
mint gulch
Hmm. I think you're right, but I need them to be non fields
next obsidian
Uhhh
I think you can just do
Q[x_i] then
I think that’s still a UFD…
My gut instinct there is that you should be able to factor everything using the variables just like in a normal polynomial ring
Because every element is a finite sum of monomials of finite degree
mint gulch
If you take the index to be aleph_alpha, then Q[x_i] should have that cardinality, right?
tribal moss
Or let K be a sufficiently large field, and consider K[X].
Not to say that Q[x_i] doesn't work, but it might be faster to explain why Q(x_i)[Y] works.
upbeat swift
If J is an ideal in a commutative ring A, prove that J[x] is an ideal in A[x]. How do I start this?
next obsidian
You should first say what your definition of J[x]
Then grab two elements in there, I assume J[x] = {a_0 + a_1x + … + a_nx^n | a_i is in J}
So you grab two of those, write them out as like Sum a_ix^i, and Sum b_ix^i
Add them together, and show that is still an element of J[x]
Then grab an element in J[x], and then an arbitrary element in A[x]
Multiply them, and show that’s still in J[x]
I guess you also need to show 0 in J[x], but that is a lot easier than the other two
upbeat swift
so i could grab two elements like a(x) and b(x)?
tribal moss
They should be arbitrary elements of J[x].
I.e. they are not really elements you autonomously grab, they are generic elements you're preparing for your enemy to select for you to work on.
upbeat swift
alright
upbeat swift
So you grab two of those, write them out as like Sum a_ix^i, and Sum b_ix^i
i'm a bit confused at this step then
Also, idk how J being an ideal comes into place, like whats the relation between J and J[x]?
next obsidian
well you want to combine them
you end up with Sum (a_i + b_i)x^i
and you need to know this is still in J[x]
so you need to know something about a_i + b_i
and that's where the relation to J comes in
upbeat swift
oh okay
obsidian sleet
hello all
im not really understanding the last part of this question
to exhibit an F-algebra isomorphism between these two algebras
i keep running into like
transpose of product is not the product of the transpose and idk why
so ok i guess we need background
umm so the map phi -> phi* is the map which sends the linear map phi into the map which takes in functionals f\in V* and does f\circ phi, which is another functional in V*
and so viewing phi and phi* as matrices they should be transposes of each other
and im just not really sure how to make the vector space iso play well with the multiplication
like i tried something like uhh
tribal moss
Forget they are duals. You just have two vector spaces of the same dimension and want to show their endomorhism rings are isomorphic ...
tribal moss
Yeah.
tribal moss
Even better: You have two isomorphic vector spaces and want to show their endomorphism rings algebras are isomorphic.
obsidian sleet
a linear map is determined entirely by its action on the basis
i can just relabel the basis ?
tribal moss
Yeah, or in other words relabel the entire vector space.
lethal cipher
So I could use a little help. I need to show given a field F of characteristic p, if a min. poly is non-separable, then it takes the form $x^{p^e}-a$ for some $e\geq 0$. I am not too sure how to go about this
cloud walrusBOT
dackid
lethal cipher
Actually, I don't think I need to go that deep here. This is the question I have to work on. I am working on part (b), and I still need to use the fact that F has characteristic p. My guess is to show that K=/= L, otherwise the claim would not be true.
So then how do I ensure a non-separable poly exists in our field
woven delta
Nope, the point is that K is the seperable closure of F in L
So K is the maximal seperable extension of F inside L
lethal cipher
I do understand that. But if K does equal L, then K subs L is definitely separable. so part (b) would not hold
lethal cipher
There should of been a slash in between. I meant to say K does not equal L
I see it removed it :p
woven delta
Okay that makes more sense
lethal cipher
So it seems that char(F)=p is why we can ensure that there is a poly that is not separable.
lethal cipher
I'm not sure what composition means here
I'm guessing you mean $F\subseq K$ and $K\subseq L$ are both separable, then $F\subseq L$ is separable?
woven delta
Yes
lethal cipher
Ah yes, that is the theorem it is referring to
woven delta
Oh okay lol
lethal cipher
So it seems that char(F)=p is why we can ensure that there is a poly that is not...
Once I can get here, I can take the contrapositive of that theorem and the rest is easy
So what I need to do is show that $\exists a \in F \exists \alpha \in L\setminus F$ s.t. $\alpha^n= a$.
cloud walrusBOT
dackid
lethal cipher
But how do I ensure that there is an alpha not in F that satisfies this
woven delta
Is being a purely inseperable extension the same as your seperable closure being yourself
Is that what we're proving
lethal cipher
no, if K=L, then L would be separable over K
woven delta
No L is purely inseperable over itself
I guess verify this with the definition in your book
lethal cipher
I see why, but then I don't understand why char(F)=p matters here
lethal cipher
for characteristic 0
woven delta
Yes
lethal cipher
Although it isnt given, p is always prime in this book
woven delta
Yes
For char 0 every algebaic extension is seperable
So this theorem is trivially true
lethal cipher
Well, once you prove that of course
lethal cipher
What I am getting at is that char(F)=p does not seem to be relevant at all in the proof itself
lethal cipher
Hmm, alrighty then. Gotta love getting more info than needed. I see why they mentioned it, but they also made me overthink it way too much :p
woven delta
Actually it may be a little relevant in terms of the initial definition they used for purely inseperable
I guess write it up and see what happens
lethal cipher
Actually no, it really doesn't. An alpha being in L\K means it's min poly is not separable. Then we can use the thm it mentions to say K subs L is not separable, and we can quickly conclude that it is purely inseparable since it is true for all alpha in L\K
woven delta
Okay cool
I wasn't sure if they used a different definition of purely inseperable or not (maybe one that exploits char p in some way)
lethal cipher
Nah it's the same, it just didn't mention that the trivial extension is purely inseparable. But that's not needed in the definition. If L=F, then F\F is the emptyset. So every element in F\F (which is none) has nonseparable polynomials
chilly ocean
Does every PID have unity
woven delta
Yes
chilly ocean
Why..
chilly ocean
Definition: An integral domain D is a principal ideal domain (abbreviated PID) if every ideal in
D is a principal ideal.
shell brook
i forgot all of basic gt but every proof I see online about S5 being generated by an arb 5 and 2 cycle assumes (wlog) that the 5 cycle is (12345) and the 2 cyc is (12)
why can you do this
coarse stag
what does it mean by "taking non identity values on a finite set of points?"
obsidian sleet
see: https://math.stackexchange.com/questions/772294/how-do-i-prove-how-s-5-is-g...
my guess is that its because conjugacy classes of elements of Sn are determined by cycle type
but its sleepy hrs for me so i might be wrong
shell brook
that (12345), (12) is fine to assume wlog?
obsidian sleet
yeah
woven delta
I'm confused, is this question asking about being generated by some 5 cycle and some 2 cycle, or is it saying any 5 cycle and any 2 cycle
woven delta
So the second one
obsidian sleet
u could turn (12345) into some other 5 cycle by conjugation by a transposition or something
so like
yeah
woven delta
You can also apply an automorphism which permutes the elements
Which is probably what you're saying
obsidian sleet
do all the automorphisms of Sn appear as inner autos?
somehow that sounds familiar
brain melting
obsidian sleet
right
woven delta
Actually idk
obsidian sleet
yeah cus T(a,b,c,..)T^{-1} = (Ta,Tb,Tc,...) something
i had this as a hw exercise or sth
shell brook
okay i have another more chunky question now i want to show that f(x) = x^2 - 5 is irreducible in E[x], where E[x] splitting of x^3 - 2. If you assume f(x) reducible, then let \beta be a root of f in E, we have like Q -- K = Q(\beta) -- E. And I know the splitting field of E is fully given by E(cbrt(2), wcbrt(2), w^2cbrt(2)), where w = 1/2(-1 + sqrt3).
I also know Gal(E/Q) iso to S_3, and so it has degree 6. So I think |E:K| = |Gal(E/K)| = 3, and so it's isomorphic ot Z_3.
woven delta
My grad algebra professor did a proof of this in class
shell brook
And then idk what to do, maybe something blows up
woven delta
The exceptional one
obsidian sleet
one of my homeworks was to show that for all of the other Sn there are no nontrivial outer autos or something it was a mess
i think
i dont remember exactly but we had to omit S6
super bleak
shell brook
okay i have another more chunky question now i want to show that f(x) = x^2 - 5 ...
i thought maybe by THE FUNDEMENTAL THEOREM OF GALOIS THEORY i could get a contradiction b/c it says that we must have Z_3 normal in S_3 but thats actually just true so no contradiction
for ref:
woven delta
Huh I think you're overthinking this
shell brook
me too
woven delta
Unless this is intentional
woven delta
So for degree 2 if something is reducible that means the roots of it are in your field already
shell brook
here is Q btw
shell brook
shell brook
woven delta
It's about degrees of extension
shell brook
I think maybe I'm misunderstandign the exntesions here
woven delta
So if you add in cuberoot 2 that's a degree 3 extension
Now suppose it contained sqrt(5)
shell brook
is it like
E
/ |
Q(\beta) |
\ |
Q
If you're still discussing this, the way you show S_n is generated by (12) and (12345) is by conjugating the transposition with the 5 cycle
shell brook
okay I'm misunderstanding something b/c this is the picture I drew before
where wrong
shell brook
If you're still discussing this, the way you show S_n is generated by (12) and (...
ya this is the proof i ended up with
shell brook
wcbrt(2), w^2 cbrt(2), where w = 1/2(-1 + sqrt(3))
the only degree i'm not sure about in the picture is E/Q(\beta) but idk why wrong
I mean it follows cuz
Q(\beta)/Q is 2 and E/Q is 6
but maybe thats the contradiction?
lethal cipher
What's the contradiction here? I'm not sure I see it
shell brook
yeah I dont think there is one lol
lethal cipher
but maybe thats the contradiction?
Ah, got confused by this
shell brook
I was just saying that was the only place I could see one popping up so far sry
woven delta
Yeah that was confusing me
lethal cipher
The "apply Galois theory" bit makes me think we wanna look at the Galois group
woven delta
Okay so what's the Galois group
shell brook
which one
woven delta
For x^3-2
shell brook
it's S_3
woven delta
Okay
shell brook
yeah
woven delta
Yeah just want to verify
Before I'm too confident
Oh a order 3 subgroup would have index 2 so should be normal
woven delta
No that's okay
One sec
You just need to find some other degree 2 that has roots in there
Do you have that?
shell brook
wdym
woven delta
Well we know there's only one order 3 subgroup
shell brook
Right
woven delta
So there should be only one degree 2 sub extension
shell brook
of E?
woven delta
Yeah
shell brook
hm
shell brook
well we only have three choices
woven delta
Between the lattice of subgroups
woven delta
Lol
lethal cipher
So roots get sent to roots. So maybe we can try and show there is no automorphism that sends sqrt(5) to -sqrt(5)
woven delta
I think we got it
shell brook
actually wait I do not really understand fullyh why having two things that are both deg 2 breaks it
like Q(\beta)/Q and Q(w)/Q
woven delta
So there's a correspondence between subgroups of the Galois group and sub extensions of E
woven delta
Okay and so we know that in S_3 there are 3 proper subgroups
shell brook
but okay why cant they be isomorphic
like theres an obvious isomorphism between Q(\beta) and Q(w)
woven delta
Isomorphic isn't enough
shell brook
woven delta
Okay look
shell brook
Q(w) wont have a root for the x^2 - 5
woven delta
We have the lattice of subextensions
woven delta
Q(w) wont have a root for the x^2 - 5
Yes they won't even be isomorphic
And we have the lattice of subgroups
And they are the same lattice
So the index of the extension is going to correspond to the order of the group
There's only 1 subgroup of order 3
Therefore there's only one extension of index 3
shell brook
ahh
I see
but we have shown there are TWO cause E/Q(\beta) and E/Q(w) are both deg 3
💥
woven delta
Yes the index is [E:Q(beta)]
shell brook
there being two identical extension chains with diff intermediates is the contradiction?
woven delta
Sure
woven delta
So yeah the theorem is cool
chilly ocean
Hi I have a quick question: If we have K<H<G. Would lagranges theorem still hold for H and K as in |H|=[H:K]|K|?
I’m pretty sure it does but just a sanity check
thorn delta
sure. I mean, H is a group, and K is a subgroup of H, so lagranges theorem applies directly
@chilly ocean
frail perch
Question: does this proof not implicitly assume we are wprking over a domain? Since otherwise we could have that the image of y_s vanishes in S^-1(M) but not in (s)^-1M
next obsidian
I don’t understand what that has to do with anything
All you’re doing is saying that we can express the generators of M (the y_i) in terms of the x_j using finitely many denominators
So if we localize at just the product of all of those, we have access to all of the denominators we need in order to still express the y_i in terms of the x_j
If it helps, if the collection {y_i} generates M over A, then {y_i/1} generates S^-1M over S^-1A for any multiplicative set S. This is just a really simple computation
frail perch
Yea, but like by definition y=m/s in the localization if and only if t(sy-m)=0 for some t in the multiplicative closed set
So dont we also need to account for the t in the case of R having zero divisors?
next obsidian
I see, I guess that is a bit of an issue. You can solve it by noting that there are only finitely many t involved in those expressions, so you can let the s in the lemma actually be the product of the s_ij along with the t_i, or maybe it’s t_ij, but I think you actually only need one t for every y
Leave a comment on the page, Johan will probably thank you for it in x weeks-time and your name will be added to the credits page :)
broken stirrup
How can i prove that Q is not projective Z-module?
I know that there exist a free Z-module and a Z-module K such that F= K + Q (direct sum)
hidden haven
One way is to use the fact that Z-submodules of free Z-modules are also free (this is true for all PIDs)
frail perch
I see, I guess that is a bit of an issue. You can solve it by noting that there ...
Yea I thought so too, thank you!
hidden haven
Another is to give an example of a surjection of abelian groups and a map from Q to its codomain that doesn't lift
broken stirrup
One way is to use the fact that Z-submodules of free Z-modules are also free (th...
yes i realized this one but we haven't covered that fact so I assume that I'll have to go with the second one
broken stirrup
I know that there exist a free Z-module and a Z-module K such that F= K + Q (dir...
what about divisibility + F= K+Q?
hidden haven
Yeah you should be able to use that without having to use the heavy fact
Suppose Q sits inside some free Z module
Take any non zero element x of Q
As an element of the free Z-module (which is isomorphic to a direct sum of copies of Z), it has finitely many non zero entries
Take n = max of its entries
Then x/(2n) exists in Q
but there is no such element in the free module
dusty cipher
Absolutely stuck on this mess of a problem, can anyone point me in the right direction?
broken stirrup
How do i show that when |X|=1, F=0?
broken stirrup
How do i show that when |X|=1, F=0?
so there won't be a unique homomorphism right?
maiden ocean
When |X| = 1 there's only one map from F to X (the constant one), and likewise for maps A to X. so any two homomorphisms f, g : A -> F agree under F -> X and by uniqueness must be equal
broken stirrup
in fact for any function iota and f any homomorphism will satisfy it, functions are constant
broken stirrup
and its zero homomorphism
so basically we can't have non-zero homomorphism and thus F must be zero
thanks
broken stirrup
and this when |X|>=2
cool problem i think
so co-free'ness is actually too strong condition that we don't have much on it
broken stirrup
how is it any different to the definition? Don't we just only replace arbitrary ring with unitary ring an R-module with unitary R-module
lethal cipher
Can someone help me confirm if the logic in the converse is right so far?
I really wanna just make sure that I can draw the conclusion that $\sigma(\alpha)=\alpha$
cloud walrusBOT
dackid
woven delta
This doesn't make sense to me
But let me look a little closer
@lethal cipher yeah I don't understand what you're doing at all
Maybe you're omitting something important here
Because I understand how you get sigma in Gal(E/K) from $\tau\circ \sigma \in Gal(E/K)$ but I don't get how you got $\tau\circ \sigma \in Gal(E/K)$
cloud walrusBOT
Emma
woven delta
It seems like you're assuming the result
Unless there's a theorem you're implicitly invoking that I'm missing
I can see $\sigma \tau \sigma^{-1}$ fixes $\sigma(K)$
cloud walrusBOT
Emma
woven delta
And that the conjugation by sigma gives you an isomorphism between $Gal(E/\sigma(K))$ and $Gal(E/K)$
cloud walrusBOT
Emma
cloud walrusBOT
dackid
lethal cipher
We know [ \sigma (\alpha_1 \dots \alpha_n) \sigma^{-1}=(\sigma(\alpha_1)\dots \sigma(\alpha_n)).] And you get a similar result for any permutation
cloud walrusBOT
dackid
toxic zephyr
is $x\equiv_ny$ a common alternate notation for $x\equiv y\pmod{n}$?
cloud walrusBOT
nix
toxic zephyr
my professor used it a couple times and i was wondering if it was common enough to not be confusing for people outside my class
woven delta
I got that since $\sigma \circ \tau \circ \sigma^{-1}=\tau\circ \sigma$
I don't know if that's true
That seems off
Like let's say tau commutes with sigma
Then that's very much not true
lethal cipher
**dackid**
I understand what you mean. I was initially thinking this is the same as tau o sigma
Okay, so how do I go about fixing this. I know sigma o tau o sigma^-1 fixes K by assumption
woven delta
**Emma**
Maybe think about this
I can give you the answer in like 4 words
But I probably shouldn't
lethal cipher
I am not sure how an isomorphism ends up giving us an equality
woven delta
Oh I mean the isomorphism is an equality
Its from an automorphism
By congugation
$\sigma^{-1} Gal(E/\sigma(K)) \sigma= Gal(E/K)$
Right
Or something like that
Is that it?
Oh I was right the first time
lethal cipher
Normally, Gal(L/sK) is on the other side and sGal(L/K)s^-1 is whats being conjugated
cloud walrusBOT
Emma
woven delta
Yeah
lethal cipher
But in truth, it doesnt make much of a difference
woven delta
Okay I was right the first time
Okay so what does that tell us
Lol I have a hard time making myself look into claims I make sometimes
woven delta
Yes
lethal cipher
So whatevr fixes K fixes sigma K
woven delta
Yeah
This gives us Gal(E/K) = Gal(E/sigma(K))
Sorry I've been using E instead of L
lethal cipher
So then sK should be a subset of K
And then since it's an equality. So the other inckusion holds as well
lethal cipher
So then sK should be a subset of K
So why wouldnt it be larger?
woven delta
Because it goes in the other direction as well
As you said
$Gal(E/\sigma(K)) = \sigma Gal(E/K)\sigma^{-1}$ and $\sigma Gal(E/K)\sigma^{-1}= Gal(E/K)$
cloud walrusBOT
Emma
cloud walrusBOT
Emma
woven delta
And we have facts about fixed fields
Right
There's a one to one correspondence
Between intermediate extensions of fields and subgroups
So we are using the one to one correspondence
lethal cipher
Sp Gal(E/K) maps to the subgroup K with our 1-1 correspondence
woven delta
The fixed field K, yes
lethal cipher
And since Gal(E/K)=Gal(E/sK) K and sK must be equal
woven delta
Yeah
Galois extension was part of the problem
And this is where it comes in
Btw I didn't know for sure this approach would actually work lol
This is just what my intuition told me to do and I filled in the details lmao
And you asking questions made me know what details I had to fill in
lethal cipher
That's good. I'll have to relook at the Galois correspondence. We just learned it recently, so it's still kinda fuzzy to me
lethal cipher
It is. I've been raking while we've been talking, so I've been a bit distracted 😅
woven delta
Oh lmao
Raking
I live in the city so that's not something I ever do
But it fits your bob Ross vibes
lethal cipher
Final question: how do I know Gal(L/K) maps to K instead of some other field?
woven delta
That's what we were just talking about
The Galois correspondence
Is what tells us that
Because K maps to Gal(L/K)
So Gal(L/K) maps to K
Do you want to go over the statement of the theorem?
I would like to actually
As a review
@ me if you'd like to
lethal cipher
@woven delta sure, I think it would be helpful
lethal cipher
Didn't know D&F had Galois stuff
woven delta
Okay so yeah if K is a Galois extension over F, then we have this correspondence
Okay so first off for every intermediate extension E we have K Galois over E
However we have E Galois over F iff Gal(K/E) is normal in Gal(K/F)
Let's discuss why that's the case
It's very related to the problem we've been looking at I think
lethal cipher
Ooo, there was a problem about normalizers in the book, but we never really discussed its relation to Galois groups
You have my interest
woven delta
Lol that's good
Okay so what happens if we can congugate Gal(K/E) to another subgroup
I guess before that
Lets define a Galois extension
Can you write a definition for me?
Cause there's a couple of equivalent ones and I want to make sure we're on the same page
lethal cipher
An extension L/F is Galois if one of the following is true:
- L is a splitting field of a separable poly over F.
- L_G= F, where G=Gal(L/F)
- L is a normal and separable extension
If one of them is true, all are true. We just need to check one
lethal cipher
Yea. The fixed field of Gal(L/F)
woven delta
And by Gal(L/F) do you mean Aut(L/F)? Because in D&F they only say Gal(L/F) when you know L is Galois over F
But that's not really relevant
lethal cipher
No, because not all automorphisms are in the Galois group.
Either way, L is Galois over F in our problem
woven delta
Wait so how do you define Gal(L/F) if not as the automorphisms of L that fix F?
lethal cipher
That is how we define it
lethal cipher
I'm gonna have to pause this for a bit. I have a zoom meeting in 4 minutes
woven delta
So suppose we have E with Gal(K/E) not normal
Okay sure
@ me when you're free again
I am using this as a review also btw
lethal cipher
Will do. Thx for the help Emma
lethal cipher
Go nuts :p
woven delta
Okay so suppose that Gal(K/E) is not normal, so there is a congugate of it. You can show that if you have two distinct subgroups of Aut(K) they will have distinct fixed subfields of K
It all comes down to this theorem
This theorem says that if G is a subgroup of Aut(K) then K is Galois over K_G
And that G = Gal(K/G_G)
Which is kinda surprising a priori
woven delta
Okay here's a nicer way of looking at it actually
Consider the map from Gal(K/F) to Gal(E/F) by restriction.
barren sierra
Ok so I'm dumb but I cannot figure out how to get any invarient factor form stuff from this.
Z[i] is a PID but Z[i]/(5) is not so idk what to do from there. All I can really say for sure is that for all such M we have that (5) is a subset of Ann(M) but otherwise I am stuck.
I genuinely do not know how to apply the structure theorem of modules over PIDs to this
next obsidian
They have no free part because if so 5M ≠ 0
barren sierra
ye
next obsidian
So they look like a direct sum of R/(p) for various prime elements (note that in Z[i] this isn’t just prime numbers)
barren sierra
so it's just a direct product of a bunch of Z[i]/(a_i)'s
next obsidian
And you want 5R < (p)
And then you’re done
So you need to find all the p for which this is true
If it helps, it suffices to just show that 5 is in (p)
So you want to find all p for which that’s true
barren sierra
hm ok
I'll go look back and remember the conditions that create primes in Z[i] cause I remember there being some sort of classification of those
terse crystal
You know structure of finitely generated modules over pid?
It’s finding the factors of 5 in Z[i]
Any finitely generated module M over a pid D is or the form :direct sum of D/(x_i) for 1<=i<=n. Where x_1|x_2|…|x_r, x_r+1=…=x_n=0
Now your r=n this case, and all x_j are factors of 5, like 2+/-i,1+/-2i……
next obsidian
barren sierra
It’s finding the factors of 5 in Z[i]
Ya I just didn't make the connection of factors of 5
Cause I'm slow
chilly ocean
It says that the mapping of S3 to +-1 is a homomorphism. My question is how do we choose which cycles go to - 1 and which to +1? Why is (1 2)-->-1 for example and (1 2 3)-->+1?
thorn delta
It says that the mapping of S3 to +-1 is a homomorphism. My question is how do w...
a permutation gets sent to 1 if it is composed of an even number of transpositions (aka 2-cycles), and -1 if it is composed of an odd number of transpositions.
If you represent S3 as the group of 3x3 permutation matrices, then this homomorphism is actually just the determinant
shell brook
here is Q btw
continuation of this question from yesterday: What do I need to do to show E(sqrt5)/Q is Galois? I know it's like enough to say that it's the s.f. of f(x) = x^2 - 5 which is irreducible, and since Q is char 0, f(x) is separable over Q and then by some theorem E(sqrt5) is Galois. But I dont want to use that fact. Can I say gcd(f, f') = 1 and so no multiple roots in any extension field, and so separable, and so E(sqrt5)/Q is Galois?
this is just a sanity check I guess cause I think im right lol
chilly ocean
if E is the splitting field of x^3-2 then E(sqrt5) is not the splitting field of x^2-5
shell brook
why
chilly ocean
because E contains cbrt(2) or a primitive third root of unity for example
shell brook
i do not understand what makes this not a splitting field though
Q(sqrt5) is an sf for x^2 - 5 surely?
chilly ocean
the splitting field is "smallest" field in which your polynomial splits completely
shell brook
Ah
right
then how do you argue that E(sqrt5) is Galois
do I need to get another polynomial
chilly ocean
how did you define normality?
chilly ocean
well then how did you define what a Galois extension is?
shell brook
Field where |Aut(E/F)| = |E:F|
So the theorem I'm using is that if E/F is a sf of a separable polynomial then |Aut(E/F)| = |E:F|
chilly ocean
how did you define splitting fields? as a field extension generated by the roots of some polynomial?
shell brook
chilly ocean
ok, so you have to find a polynomial st E(sqrt5) is generated by its roots
what are some generators of E(sqrt5) over Q?
shell brook
(x^3-2)(x^2-5)?
chilly ocean
yes, that works
shell brook
I see okay so then I need to argue that this is separable
can I just show the gcd(f, f') is 1 still? Is that adequate for separability
chilly ocean
how did you define separability for a field extension?
shell brook
f in F[x] is separable if it has no multiple root in any sf E/F
I just dont want to have to prove that char0 & irreducible => separable
lol
chilly ocean
that wouldn't work either since your polynomial is not irreducible
chilly ocean
I'm confused, are you trying to prove that f is a separable polynomial or that E(sqrt(5))\Q is a separable field extension?
shell brook
I am trying to show that f is separable
so I can show it's splitting field is Galois
the goal is to show E(sqrt5)/Q is Galois
chilly ocean
ok, so computing the gcd works
sorry but I am not familiar with the definitions you were given
chilly ocean
2x+2 is an irreducible in Z[x] and not a primitive
shell brook
what is D
chilly ocean
D is UFD
shell brook
2x+2 is reducible
2(x+1) and neither are units in Z
i don't know why this is true exactly, I think every irreducible element in a ufd is prime so maybe tahts why
chilly ocean
Why a unit rather an unit
shell brook
are you asking why the english phrase is "a unit"?
chilly ocean
Yes .
shell brook
I think the rule is that you use "a" when the word that comes after starts with a consonant, and "an" when it starts with a vowel
"an owl" sounds better than "a owl"
idk if that even works for Unit lol
woven delta
What is a primitive polynomial?
shell brook
its a polynomial st gcd of coeffs is 1
woven delta
Oh okay
chilly ocean
U is a vowel
shell brook
yeah con 1
@chilly ocean im actually not sure why, it just sounds better to say "a unit" to my native english speaker brain
ask @scarlet estuary
lol
woven delta
Anyway yeah this sort of thing is probably up to a unit
That's what they mean
Oh oops
Lol
Anyway 2x+2 is reducible
Because 2 is not a unit
That's your answer
Lol
Oh you said that
Jesse
Oops
Okay yeah anyway this is obvious then
Because if you have content not a unit then you are reducible
Because the content can be factored out
Does primitive mean content 1 up to a unit?
shell brook
surely
woven delta
Okay so yeah
The contrapositive is obvious
If you have non 1 content then you are reducible
scarlet estuary
Why a unit rather an unit
unit starts with a consonant sound /j/
a/an doesnt care about spelling
it cares about pronunciation
it's "an honest kid", not "a", because the h is silent
shell brook
I have a silly question: Let \sigma be an automorphism of E(sqrt2) where E = Q(a,b,c) or something. Suppose sigma(E) = E also. Is sigma identity? Or is s(sqrt2) = -sqrt2 valid?
i hope this is true bc it makes my life easy :V
tough raven
sigma(E) = E doesn't mean sigma fixes each element of E, right?
So you could probably get something like a = sqrt(3) and sigma fixes sqrt(2), b, c but sends sqrt(3) to - sqrt(3)
shell brook
guh identity maybe wrong way to state it sry
the actual thing I care about is just: do we have sigma(sqrt2) = sqrt2
lol
Sigma would just be permuting a,b,c in this case yeah
I think automorphisms must just act like swappy on roots even if they arent explicit elements of the extension field
so s(sqrt2) = -sqrt2 is totally fine
woven delta
I was talking to Dackid about this earlier, he was proving a theorem that for a Galois extension K over F and an intermediate extension E, a automorphism sigma has sigma(E)=E iff sigma congugates Gal(K/E) to itself
shell brook
whats congugate mean here
shell brook
ah
woven delta
That means you are congugated to yourself
shell brook
loosk wierd
woven delta
It's kinda neat
shell brook
im proving yet another step in this ungodly question I've been working on
and I just need this MAP to be an ISOMORPHISM 
woven delta
What's the question
shell brook
shell brook
so (iii)
woven delta
Oh based
shell brook
and we have a thm that says those restriction maps are both surjective homs
woven delta
I know
Because I was talking to dackid
And that got me thinking about Galois theory
And somehow I ended up there
Oh no it's not exactly the same
As what I was thinking of
Lmao
shell brook
hm
woven delta
Okay so if you are surjective
Then you need to show you are injective
Can't you just do finite stuff
Since all of these groups are finite
woven delta
Yes
woven delta
Yes
woven delta
chrivial
tough raven
the actual thing I care about is just: do we have sigma(sqrt2) = sqrt2
Not guaranteed. If sqrt(2) is not in E, there is an automorphism which fixes E pointwise and sends sqrt(2) to - sqrt(2). By composing with that, for any automorphism, there's another one that behaves the same way on E but behaves the opposite way to the original on sqrt(2).
shell brook
@woven delta can you clarify why the composition map is a surjection slash injection
i do not see it
shell brook
yeah both restrictions are surjective
i was trying to show that the composition map is surjective
but i do not need to do that
woven delta
Okay so that makes that map to the product surjective
I mean it's just true for group homomorphisms
woven delta
Yeah
So that's a product of maps
It's easy to show that if you have G\to H, G\to H' each surjective, then the product map G\to H\oplus H' is surjective
And vice versa
So a product map is surjective iff both of the restrictions are
This is just because of how direct sums/products work
Each coordinate is independent
shell brook
It's easy to show that if you have G\to H, G\to H' each surjective, then the pro...
can you point me to a proof of this
i actually cannot work it out
lol
woven delta
I mean, it's just say x in H and y in H' has f(a)=x and g(b)=y
Where f and g are the two maps
Then (fxg)(a,b)=(x,y)
That's the proof
shell brook
right yeah but thats not the map we have
woven delta
Oh is it not
shell brook
shell brook
but the final map is s -> (f(s), g(s))
woven delta
Yes
Thats what the product map I was talking about is
Oh sorry one sec
Yes I see
Oops
shell brook
lmao its okay
woven delta
Okay no this is fine, we can patch this
Maybe I should stop going into problems thinking I should know the solution immediately and acting on that assumption
Because it just makes me overcommit and feel bad
shell brook
kinda based way of problem solving imo
woven delta
But okay, let's look at mapping to (0,x)
Yeah it makes you very invested
That's why I do it
I think you have to check things about the kernels
shell brook
woven delta
Now that I've actually looked at the maps
shell brook
thats so fucked up
woven delta
Lol
shell brook
i should just say they have equal cardinality and move on
shell brook
which is
woven delta
Maybe
Let me actually look at this
And I'll give you a strategy I think will work
Sometimes I try to solve problems without really looking at them based on what I think they should be asking
Sorry about that
shell brook
please tell Kudla to ask better problems 
woven delta
Okay the kernel is actually not hard to describe
For restriction maps
Oh lmao
Yeah just show these are normal subgroups
And you're done
woven delta
Okay good
shell brook
woven delta
No I mean
shell brook
the tower in our current problem is diff names I think
cz this theorem still applies
woven delta
No look
Suppose you have E/F is Galois, then for every subextension K, we have E/K is Galois and Gal(E/K) is normal in Gal(E/F)
Let's say what E, K, F are in our case
shell brook
E = E(sqrt5), K = E, F = Q
shell brook
cloud walrusBOT
jesse
shell brook
ignore the frac.
woven delta
Oh okay my second attempt was wrong also, good
shell brook
yeah this is cringe behaviour on behalf of my prof
woven delta
Okay the kernels actually are easy
Now that I've looked at it
So if we have a surjective map E(sqrt (5)) \to Gal(E/Q), then the kernel is going to be Z_2
The Z_2 factor
And I can tell you why pretty easily
It's because the other copies of Z_2 aren't normal subgroups
But the Z_2 factor is
Because Z_2 isn't normal in S_3
And then for the other one
Hmm am I a bit off actually
Yeah this argument doesn't quite work
Okay fine let's actually explicitly describe the kernel
shell brook
woven delta
It should be the unique automorphism that fixes E
That's it
So that automorphism is sqrt(5)\to -sqrt(5)
So the kernel of that map is fine
Do you get that?
shell brook
not really
wait oh
uh
there are two ams that fix E
theres that one and theres sqrt2 -> sqrt2
woven delta
Yeah the second one is the identity
Okay and let's look at the kernel for the other one
So the other one has 6 elements
But it should literally be just Gal(E/Q)
Yeah
It is
woven delta
Lol
shell brook
i thought symbol bash was sufficient to show surjecitvity
woven delta
But now we're doing actual work
Wtf
Yeah ik
I was engaging on that level before
But it wasn't working
Okay so we know what the elements of Gal(E(sqrt(5))/Q) look like now
Well we did before
But we weren't paying attention to it
The important thing here is the kernels are disjoint
And if we apply the restriction map for E to the kernel of the restriction map of F
Then we get surjections
In fact we get isomorphisms
Okay I'm going a little fast
next obsidian
Chmonkey
woven delta
Hey chmonkey
next obsidian
Hello Emma
barren sierra
ok so my notes say the Annihilator of a module M is a right ideal but why is it not also a left ideal?
shell brook
Not now saxophone man we are trying to show osmething is surjective
next obsidian
Owned, but also I think it’s a two-sided ideal regardless of if you have left or right modules. Let’s just take left modules, then if b in Ann(M), you have (ab)x = a(bx) = a(0) = 0.
For the other direction, (ba)x = b(ax) = 0
shell brook
referent
woven delta
So Jesse the kernel of the restriction map to F is S_3 right
What if we apply the other restriction map to that kernel
What happens
barren sierra
Not now saxophone man we are trying to show osmething is surjective <:bleak:8770...
k
shell brook
Uh
woven delta
So the elements of the kernel is literally Gal(E/Q), right
shell brook
fuck
woven delta
Huh
shell brook
im just upset my gues was wrong
shell brook
this is a rollercoaster.
woven delta
I know
Okay the kernel of the restriction map of E(sqrt(5)) is Gal(E(sqrt(5)/E)
I'm doing this in my head so I got confused
So the only nontrivial element of it is the automorphism which maps sqrt(5) to -sqrt(5)
Now if we restrict that to Q(sqrt(5))
That's also the only automorphism of Q(sqrt (5)) also
So we get a surjection
Do you see now
shell brook
Now if we restrict that to Q(sqrt(5))
i do not understand this
woven delta
Okay so right now we have an automorphism of E(sqrt (5)) over E
Restricting means restricting the domain
To only be Q(sqrt(5))
So if we restrict the automorphism sqrt(5) \mapsto -sqrt(5) to Q(sqrt(5))
We get the automorphism sqrt(5) \mapsto -sqrt(5)
But the domain is now Q(sqrt(5))
woven delta
So this is surjective when restricted to the kernel
Here's why that's useful
Say x is in the kernel of the restriction to E
Then $(res_E(x),res_F(x))=(0,res_F(x))$
cloud walrusBOT
Emma
woven delta
Now if we can do the same for the other kernel
Then we have for x in the kernel of $res_F(x)$, $(res_E(x),res_F(x))=(res_E(x),0)$
cloud walrusBOT
Emma
woven delta
(by 0 I mean identity)
And then once we have that we know we are surjective
Oh wait
Here's an easier argument
The kernel of the product map is the intersection of the two kernels
And we found the kernels
And they had no intersection
So therefore the map is injective
And therefore surjective
shell brook
hm
let me try to find the thm abt kernel of product
or maybe its easyt o rpove
wait but we dont have product map
woven delta
Yeah, it's easy
Look at this map
If you map to (e,e)
Them that means x maps to e under both restrictions
Which means x is in the kernel of both
That's it
woven delta
So yeah
Just verify the two kernels don't intersect
And use finiteness
The other way than you first thought
And that does it
shell brook
@woven delta but whats the kernel of r_{Q(sqrt5)}? The theorem says its Gal(E(sqrt5)/Q(sqrt5))
isnt this nasty
maybe I wrote it down wrong
woven delta
No, it's basically just Gal(E(sqrt(5)/E))
The important thing is it doesn't intersect the other one nontrivially
And that's easy to see
shell brook
woven delta
Because the only nontrivial automorphism is mapping sqrt(5) to -sqrt(5)
shell brook
The kernel of the r_{E(sqrt5)} map is {id, sqrt5 -> -sqrt5} right?
shell brook
Oh wait why
woven delta
<@!348994533540888586> but whats the kernel of r_{Q(sqrt5)}? The theorem says it...
This
shell brook
hm
I do nto see it
i mean I assume it is cause we are quotienting by something with sqrt5
lol
woven delta
The automorphisms of E(sqrt(5)) that fix sqrt(5)
So they all must fix sqrt(5)
But the only nontrivial element in the other kernel doesn't fix sqrt (5)
So they can't intersect
shell brook
I still do not see it actually
woven delta
Okay what does the kernels intersecting nontrivially mean?
woven delta
Well that's part of the fundamental theorem of Galois theory
shell brook
I get the statement
woven delta
Okay cool