Start with $a \in \ker\gamma$ which is in the kernel of the snake map, then $a \in C$, so we can pull it back to a $b \in B$ up to something in ker $g$. Then $\beta(b) \in \ker g'$ so that there exists some $c \in A'$ such that $f'(c) = \beta(b)$, now by assumption $a$ was in the kernel of the snake map, so that $c$ maps to 0 when we push it into coker $\alpha$, so there exists some $d \in A$ such that $\alpha(d) = c$. Note then that $\beta(f(d)) = f'(\alpha(d)) = f'(c) = \beta(b)$. This means that $\beta(f(d)) = \beta(b)$
#groups-rings-fields
next obsidian
cloud walrusBOT
Mathemagician:
vestal snow
Okay
next obsidian
So now notice that $f(d) \in \ker g$ by exactness, so we can replace $b$ with $b - f(d)$, but notice that then that puts $b \in \ker\beta$ since $\beta(b) = \beta(f(d))$
cloud walrusBOT
Mathemagician:
next obsidian
Then you're done, since $b$ is the element of $\ker\beta$ which maps to our original $a$
cloud walrusBOT
Mathemagician:
next obsidian
Yeah, so like
When defining the snake map, showing it's well-defined is annoying precisely because you pull things back along maps up to something in the kernel of that map
but to show exactness, that's exactly the thing you exploit
vestal snow
Thanks
next obsidian
π
vestal snow
One quick question
So if we pulled back to coker \alpha from b and b - f(d), we might get different elements right?
oh no wait nvm
next obsidian
I'm not sure what you mean
Is there something you're still confused about, or did you clear it up yourself haha
vestal snow
I'm not sure
one sec
I was asking if b and b - f(d) can map to different stuff under beta
which seems to be true
woven delta
Does f(d) have to map to 0?
vestal snow
but the preimages of both of them by f' would lie in the same element of the Coker
woven delta
Which diagram are you looking at?
next obsidian
They can't
woven delta
The one with the snake lemma or the one you posted
next obsidian
You chose d such that alpha(d) = c
and c was chosen such that g'(c) = beta(b)
By commutativity beta(f(d)) = g'(alpha(d)) = g'(c) = beta(b)
Oh wait
Yes!
b and b - f(d) do map to different things under beta sorry lmao
beta(b - f(d)) = 0
always
vestal snow
yeah that makes sense
next obsidian
But this doesn't matter since g(b) = g(b - f(d)) = a
vestal snow
yup
next obsidian
π
chilly ocean
what is the difference between order of a group and order of an element? moreover are order and period the same thing?
hot lake
well a group is not an element
hot lake
well yeah there is a definition for that
the order of a group is the number of elements in the group
scenic sage
and it is connected with order of each element
but order of element is n such that x^n = 1
n > 0
pallid ember
*e
scenic sage
1 = e
chilly ocean
not always
depends if the group is additive or multiplicative
let me see if I understood
pallid ember
who tf uses 1 as notation for identity
somber rivet
multiplicative identity
chilly ocean
you can write 1g nothing wrong with it
somber rivet
if the operation of the group is denoted by the multiplication symbol, then 1 is usually used
chilly ocean
yes
scenic sage
i mean nothing prohibits u treat additive identity which is zero as 
how would you draw sully during an exam
@somber rivet
":sully:"
pallid ember
going against standard notation is a felony
solemn hollow
using your blue and black pens
chilly ocean
ok now, if I have a group (Z8,+) , the order of the group will be 8, while the order of [4] for example will be 2, because [4]^2=[16]=[8]=[0] because e=[0] , am I correct?
mild laurel
yes that's right
scenic sage
in cyclic group <x> of order n, order of x^a is n/(n, a)
chilly ocean
π―
hot lake
you can say that the order of an element of a group is the order of the subgroup of that group generated by that element
olive mirage
@hot lake It took me a while to realize this, but it makes it feel a little less silly
next obsidian
Let's say you have some sort of algebraic object, let's just go with a ring for the sake of example. Let's say you have some set S, and two binary operations on it which are addition and multiplication, but let's say you don't know that every axiom to be a ring holds (i.e. say, you don't know multiplication is associative). Let's say we have a ring R, and we can create a ring homomorphism R -> S which is bijective, now ai priori calling it a ring homomorphism isn't correct, but it's multiplicative, sends 1 to 1, and is additive. Is that enough to show that S is actually a ring?
If so, this should imply all you need is a surjection right? Since you could still just quotient out by the kernel and get your isomorphism?
Again, if true, besides having all the operations, are them some axioms you must know beforehand in order to get this result? I feel like multiplication distributing over addition might be necessary, but perhaps you can deduce it from the "isomorphism"
solemn rain
i dont think so no
you can construct something that isnt a ring and does what you say ( i hope )
this argument is like similar to like
the all normal subgroups --> abelian argument
it just doesnt work cuz there is some counterexample (Q_8 here) that fucks it up
idk tho
just my opinion :d
@next obsidian
next obsidian
I don't see how all normal subgroups --> abelian is a parallel to this
solemn rain
abelian --> all normal subgroups
you said there is something nice about something
soj that something must be the osmething that always makes it nice
makes*
xd lmmfao but i meant like
next obsidian
But this actually says it looks like something that is a ring
solemn rain
yes id guess it does look like something that is a ring
but it may not be a ring
a ring is not only something that satisfies homomorphism properties
is it
and i think u can think of some counterexample
consturcting a map that says all these nice things
but goes from a ring to not a ring
just like explicit shit
any 'homo'
between a ring to a group
should work
it works for a group homo
it has the nice things us aid
but it isnt a ring its only a group
right?
next obsidian
A group doesn't even have a second binary operation defined
It doesn't make sense to talk about a multiplicative map to a group
solemn rain
oh right yea
mb
maybe try plaaying with an already ring homo
and play with the set
till u still have the nice shit with homo
but u lose being ar ing
lmfao playing with associativity maybe
idk tbh
woven delta
I think you can definitely say that the image under the homomorphism is associative
But outside of the image you can't really say much
next obsidian
Yeah I was thinking about it more and thought so too
I think you can show pretty much every axiom of a ring
woven delta
Like probably if you had such a set A and a ring map N \to M you can define a "ring map" N \to MxA
And then you would be done
Idk if you need associativity of A to verify all the other ring properties of MxA though
Off the top of my head
next obsidian
I was thinking even more lowlevel like
take a,b,c in A
suppose you had some surjection from a ring R to A which is multiplicative
call this map f
solemn rain
yea i think u need
surjection
if u have surjection u cna just play with the ring elements
woven delta
Wait aren't the octonions an example
solemn rain
when they are mapped
next obsidian
then you have a',b',c' such that f(a') = a, f(b') = b, f(c') = c
next obsidian
Hmmmm
next obsidian
Is it surjective?
woven delta
Obviously not
next obsidian
Oh yeah, I think you need at least a surjection
solemn rain
yea yea
woven delta
If it was surjective the octonions would have to be associative
next obsidian
Yeah
woven delta
But I'm saying they are an example for your earlier question
woven delta
Oh lol
solemn rain
lmfao really
next obsidian
Then I thought you can weaken it to just surjection
solemn rain
i thought u meant hom
next obsidian
Oh lmaoo
solemn rain
lmfaao sry
woven delta
Oh I just parsed it as a ring map
solemn rain
yea my bad
next obsidian
lmao
next obsidian
I wonder if given a surjection there are any ring axioms that fail to descend via a surjection
next obsidian
Or in more generality just general algebraic stuff
woven delta
I think for the underlying group it's enough
solemn rain
yea why not
woven delta
Yeah
woven delta
No
next obsidian
Oh sure actually I see
solemn rain
stupid question : f(a)f(a^-1) = f(aa^-1) = f(1)
next obsidian
I thought not knowing presence of inverses wouldn't be enough to say that if gh = h for some h in S that g is the identity
solemn rain
if not injectivve then aa^-1 may not equal 1 ?
next obsidian
but by surjectivity you could enumerate over all h
woven delta
What mo3men
solemn rain
nvm
next obsidian
Is this what they call universal algebra?
woven delta
No
next obsidian
Just kidding lol
chilly ocean
all I wanna say, wikipedia is a gift
Socratica's videos too
they really help when you feel stuck with too much rigor and can't grasp a concept
thanks for reading my Ted Talk
scarlet estuary
define "squaring"
it's just multiplication, right?
so suppose you have a = b
then you can write, say, aa = ba
via right-multiplication
but a = b so substituting that into the "ba" side
aa = ba = bb
hence a=b implies a^2 = b^2
shy bluff
Yea
Like Im' trying to prove that this
And so far I have like this?
b
And yea like multiplication
Err sorry supposed to be a 1 on the left on that last line
thorn delta
i am trying to go through serge lang and artin too rn.
opal vale
does anyone want to go through an abstract algebra book with me?
@chilly ocean which part?
knotty mason
@chilly ocean ive been reading Lang - Algebra
some of the exercise are too difficult for me
vestal snow
It says that A_p can be identified with the ring of all rational functions which are defined on almost all points of V. Shouldn't it be the rational functions which are defined on at least one point of V?
Because if f(x)/g(x) is in A_p, then that means that g is not in p and thus g(x) is not 0 for at least one element of V
Oh wait nvm
golden pasture
defined means g(x)!=0
vestal snow
I just realized that even if g is not in p, it can still be 0 at all points in the variety
I still don't get how A_p is being identified with that ring
Is this something that the book just mentions w/o proof to show why this stuff is important or can I prove it from what I know so far?
golden pasture
if g is 0 everywhere then it is in p
vestal snow
Why?
golden pasture
cuz the 0 function is in every ideal
vestal snow
but why is g the 0 function
V can be finite, and g(x)=0, for all x in V does not imply g is the 0 function
it's only the 0 function over V
vestal snow
np it's all good
golden pasture
but like basically V contains all the points that vanishes in all functions of p and p actually also contains all functions that vanishes on V(one of the nullstellensatz variants)
which means that A_p is well defined on V
vestal snow
Oh okay
So we do need something the book hasn't developed yet right?
I'm using atiyah-macdonald and this is chapter 3
vestal snow
I stilll don't get one thing
say that f(x)/g(x) is in A_p. Then g is not in p and by what you just said, g must not be equal to 0 for some x in V
Why does the book use "almost all" instead of "at least one"
golden pasture
note that V is the set of points that vanishes on all the functions
hm actly
lemme slowly wake up brain
yea should be like at least one point
golden pasture
yea g has only finitely many roots
vestal snow
So g can only be not defined at finitely many points of V
golden pasture
but V can also be finiteΒ―_(γ)_/Β― tho usually it is either one point or a whole plane of points
vestal snow
The places in V where g is not defined are finite. And there is at least one place in V where g is defined. So g is defined at almost all places in V
Does this make sense?
chilly ocean
is there a group Zn equal to < [6]18 > ? was thinking Z3, but I'm not sure
also another question
how do you visualize geometrically a homomorphism?
kindred rivet
i don't
chilly ocean
there must be a way to visualize it, it can't only be rigor
kindred rivet
i mean
what exactly is necessary for it to be visual
like
how would visualization improve your understanding
vestal snow
Can someone verify this for me real quick: To see if a module N is exact, it suffices to show that M' ---> M ---> M'' is exact implies M' (x) N ---> M (x) N---> M'' (x) N is exact
chilly canyon
Hi ! Does anyone know about a book or lecture notes on Lie groups, Lie algebras, Dynkin diagrams and Root systems ? :)
next obsidian
Can someone verify this for me real quick: To see if a module N is exact, it suffices to show that M' ---> M ---> M'' is exact implies M' (x) N ---> M (x) N---> M'' (x) N is exact
@vestal snow a module cannot be exact
Perhaps you mean flat, in which case it suffices to check that for an exact sequence 0 -> M -> Mβ that 0 -> M (x) N -> Mβ (x) N is exact
vestal snow
Ah my bad. Perhaps I can phrase my question more generally
Given a functor F, F is exact if N' -> N -> N" is exact implies F(N') -> F(N) -> F(N") is exact
Now that I think about it, this follows from the definition of a sequence being exact iff it's exact at every point
So I guess it was pretty trivial
Anyways, thanks for the help
next obsidian
Tensoring with a module is automatically right exact though, all you need to do is show it preserves injections
Those which do preserve injections, i.e. modules F such that F (x) - is exact are exactly the flat ones
olive mirage
I'm not sure exactly what you mean, but I suspect the answer is no in general, since you can't even determine if the group is trivial.
But if your presentation has a reasonable form, you could make an algorithm for it
yeah, in general it is the word equivalance problem that is hard in a provable sense.
knotty mason
most questions about group presentations are impossible for algorithms to solve in general
olive mirage
it's one of those weird things where most cant' be answered in general, but of course you never encounter the general case, you always encounter specific cases.
cloud walrusBOT
Have a Banana Bitch:
vestal snow
I'm not sure because $S^{-1}A$ does not necessarily contain $A$
cloud walrusBOT
Have a Banana Bitch:
next obsidian
The line even above that doesn't hold
just because 1/a * b = 0 doesn't imply b = 0
it means there's some u in S such that sb = 0
You need an integral domain to conclude that 1/a * b = 0 implies b = 0
vestal snow
Can't you multiply on the left by $\prod s_i t_i$ to get the second last line?
cloud walrusBOT
Have a Banana Bitch:
next obsidian
No
equality in a localization is defined the following way
a/b = c/d iff there exists some u in S such that u(ad - bc) = 0
If zero-divisors don't exist then u(ad - bc) implies either u = 0 or ad - bc = 0
but if you localize at a set with 0 in it then the localization is always trivial
so if you're not in that case then you get ad = bc
vestal snow
1/a * b = 0 implies that there is some u in S such that sb=0 . Therefore, b/1 = 0/1
because again u(b-0) = 0
next obsidian
Oh wait
vestal snow
Am I missing something?
vestal snow
"1/a * b = 0 implies that there is some u in S such that sb=0 . Therefore, b/1 = 0/1
because again u(b-0) = 0"
next obsidian
I thought that was an equality stated in M (x) N
vestal snow
Which one are you referring to?
next obsidian
the third line
If that equality is taking place in S^{-1}(M (x) N) then yes
But if it's in M (x) N then no
vestal snow
Ah frick
I used phi instead of f
which is why you got confused
Here I was trying to prove that f/phi is injective
next obsidian
Okay so if I'm following correctly
what we're saying is that if
sum m_i/1 (x) n_i/1 gets sent to sum m_i (x) n_i = 0
is sum m_i/1 (x) n_i/1 = 0?
Correct?
vestal snow
and the first tensor is happening in A and the second in S^-1 A
I think we should label the tensors so that there is no ambiguity where it's happening
next obsidian
Holy shit uhhhhh I want to say you can conclude this
but jeez this is really fucking hard to think about what's going on here haha
vestal snow
It is
vestal snow
Give me 5 minutes and I'll write out everything in a clearer way
Next, I want to show that $\phi(\sum \frac{m_i}{s} \otimes_B \frac{n_i}{1})=0$ implies that $\sum \frac{m_i}{s} \otimes_B \frac{n_i}{1}$ is $0$
cloud walrusBOT
Have a Banana Bitch:
next obsidian
So, tbh I reaaaaally don't see trying to show injectivity by elements like this to be a very successful endeavor
Tensor products don't really every lend themselves to that sort of stuff
vestal snow
It seems like I'm almost there though
next obsidian
The only way I can see you doing this is to sort of emulate the proof of corollary 2.13
The issue is the only way you know it's equal is if these pairs of elements lie in some set D
which is being quotiented out
And it just ends up really really messy really quickly IMO
vestal snow
$\phi(\sum \frac{m_i}{s} \otimes_B \frac{n_i}{1})=0$ implies that \sum \frac{m_i \otimes_A n_i}{s} = 0
next obsidian
Yeah but now you're dealing with (sum m_i (x) n_i)/s = 0
and trying to extract meaningful relations out of some sum of tensors being 0 is really really hard
Like even for a simple tensor of the form a (x) b
This being 0 doesn't meann that like
there is some u such that ua = 0 or ub = 0
cloud walrusBOT
Have a Banana Bitch:
Compile Error! Click the None reaction for details. (You may edit your message)
next obsidian
Is there a reason you're trying to verify this like "by hand" rather than using tensor rewrite rules?
If you do it that way at each step you know what's going on and can just figure out the composition of the functions defining each step
vestal snow
Tensor rules don't seem very convincing to me for some reason
next obsidian
I mean they're really the only way to go
Verifying stuff like this becomes an absolute nightmare, if even possible at all
vestal snow
I'll keep that in mind
next obsidian
If you believe each tensor rule like
M (x)_A A iso to M
and like S^{-1}A (x)_A M iso to S^{-1}M
then doing something like this is just a chain of stuff, somewhat like noting some composite function is continuous since you can show it's the composition and sum of a bunch of continuous functions
vestal snow
Can one conclude from $\sum {m_i \otimes_A n_i = 0$ that $\sum {m_i \otimes_{S^{-1}A} n_i = 0$?
cloud walrusBOT
Have a Banana Bitch:
Compile Error! Click the None reaction for details. (You may edit your message)
next obsidian
Like, i can't even really give you an answer for that. I want to just say yes and that it's obvious and be done with it, but if you dive straight into the definition of tensor product it gets all confusing and stuff
vestal snow
Ah I see
next obsidian
Maybe some other people have had better luck trying to peer into the abyss, but I gave up very quickly trying to deal with tensor products like this
vestal snow
Well, it's the only missing piece of the proof if we do it this way. Anyways, can you show me the slicker way of doing it?
next obsidian
$S^{-1}M \otimes_{S^{-1}A} S^{-1}N \cong S^{-1}M \otimes_{S^{-1}A} S^{-1}A \otimes_A N \cong S^{-1}M\otimes_A N\cong S^{-1}A\otimes_A (M\otimes_A N) \cong S^{-1}(M\otimes_A N)$
cloud walrusBOT
Mathemagician:
next obsidian
I think this is right
The point here is you can sort of factor out two S^{-1}A's, but you only want one
But you can kill one since tensoring with S^{-1}A over S^{-1}A does nothing
next obsidian
Maybe if you write it like
$(S^{-1}M\otimes_{S^{-1}A} S^{-1}A)\otimes_A N\cong S^{-1}M\otimes_A N$
cloud walrusBOT
Mathemagician:
next obsidian
It's more clear
vestal snow
Hold on
Tensor products only assosciate when you're tensoring over the same ring
Here the two tensor products are over different ring
next obsidian
I don't think that's true?
vestal snow
I'm pretty sure it is
Like you need some additional condition when tensoring by different products is assosciative
Hold on
next obsidian
Isn't this what I did?
I think?
If not then I'm really really clueless as to how you'd show it
vestal snow
Yeah but we will then have to show the additional conditions right?
next obsidian
I mean the only condition is that S^{-1}A is an (A,S^{-1}A) bimodule
next obsidian
And I think that's true, it at least better be true lol
vestal snow
Wait
I just realized something
Showing that S^-1 A is a bimodule is equivalent to showing the "missing piece" in my proof, isn't it?
next obsidian
Maybe??
This is super far away from the tensor product setting
so it's really easy to show
once you start going into the tensor product setting you're quotienting out by infinitely many relations and I start to lose track of what's going on
next obsidian
Sure thing, sorry I can't help you show your initial thing
vestal snow
It's no problem. I'll let you know if what i said earlier about the two statements being equivalent is true
next obsidian
Sounds good
chilly ocean
how do I prove that a function that goes from (Z9, +) to (S7,Β°) with [n] that goes to s^n, is a well defined function? (S is the symmetric group of permutations)
hot lake
you have to show that if [n] = [m] then s^n = s^m
chilly ocean
but how do I show that 2 permutations are equal
solemn rain
wdym equal
they are equal
if they are equal
send each element ot the same
they permute the same
permutations are functions after all
chilly ocean
you have to show that if [n] = [m] then s^n = s^m
i meant how do I show this?
please I don't need vague answers
this subject is already frustrating, every thing I look online answers vaguely
stone fulcrum
@chilly ocean
I assume s is just any permutation? Well, let's talk examples. What's f([0])? What's f([9])?
Note that [0] and [9] are the same thing in (Z9, +), so if f is well defined, f(0) and f(9) should be the same too.
chilly ocean
but I have to prove f is well defined
elder valley
i'm reading Silverman's arithmetic of elliptic curves. here he's talking about DVRs and i think the wording is implying that he's taking Y as a uniformizing parameter, but i'm confused about the justification. he says Y generates M_P / M_P^2, but i thought the uniformizing parameter just needs to generate M_P. why is he looking at the quotient?
chilly ocean
ok, I am puzzling over a few things
- SU(2) representations, which can have many dimensions, seemingly corresponding to spins in physics --> why is this called a "representation"? SU(2) is 3 dimen
s
i
on
a
l
ok I don't know why that happened
SU(2) is 3dimensional, what is the point in a 500 dimensional "representation" of it?
The only answer I can come up with myself thus far is that an arbitrary function on SU(2) -> R would have a taylor expansion involving an infinite sum over all these representation spaces (as each polynomial rank would correspond to a spinor? and spinors of multidimensions map 1-1 into the representations of SU(2))
elder valley
<@&286206848099549185>
sturdy marsh
@elder valley The quotient is the cotangent space at that point. If the cotangent space needs more generators (as a vector space) then the curve is not smooth at that point.
in the second case I think you need both X and Y to generate M/M^2
at (0,0)
elder valley
i don't know about tangent spaces, he hasn't talked about that yet. it just seems we're trying to find a uniformizer in order to calculate those orders, so i'm trying to understand how that quotient relates to uniformizers
this is the only other time he's talked about that quotient
elder valley
in the above case C is a curve so is 1 dimensional, so M/M^2 is 1 dimensional K vector space. he's saying that M_P/M_P^2 = { kY + M_P^2 : k in K}, and that implies Y is a uniformizer in the localization at P?
neat ginkgo
Im trying to understand normal subgroups, but this book just randomly referenced the term "conjugacy class"
the only thing defined is a conjugacy relation between 2 elements
also, im reading this in my native language so there's kind of a language block when im trying to research online :/
can someone explain to me what a normal subgroup and/or conjugacy class is please
elder valley
conjugacy class of an element g is the set of all elements that are conjugates of g
so under conjugacy relation, it's all elements related to g
stone fulcrum
"Taking the conjugate" is like a binary operation one can do with two group elements. x conjugated with g is g'xg.
Where I'm using ' to represent "inverse".
Two elements are "conjugate", or "of the same conjugacy class" if you can get from one to another using conjugation. That is, a is conjugate to b if g'ag = b for some g.
Normal subgroups are subgroups that are also closed under conjugation. If h is in a normal subgroup, then g'hg for any g will also be in that normal subgroup.
@neat ginkgo
neat ginkgo
oh my god
you said "closed under conjunction" and suddenly it all made sense
thank you
both
stone fulcrum
Np feel free to ask if you need anything else haha
stone fulcrum
That's a spicy isomorphism
Most straightforward way, provide the isomorphism itself. What function maps D12 β A4ΓZ2, and captures the group structure?
That Z2 seems to encode "which side is up" if you ask me
Hmm, maybe not though
elder valley
what is your definition for D12?
stone fulcrum
Do you know for sure they're isomorphic groups?
elder valley
How do you define it as a set though
stone fulcrum
Yes they're not iso lol
Maybe your friend is messing with you? Or perhaps wanted a proof they're not iso?
elder valley
Or the friend is confused π
neat ginkgo
apparently im supposed to break down D12 into two groups such that they are isomorphic to A4 and Z2 respectively
(???)
Yes they're not iso lol
@stone fulcrum u sure?
thorn delta
yea, i was thinking that you'd write D12 as a semi direct product
stone fulcrum
Well, this website could be wrong
thorn delta
but idkidk
stone fulcrum
Oh wait is that not a direct product? Haha
neat ginkgo
i think it is
elder valley
you can definitely show it's the semi direct product of Z2 with something else
Z12 i believe
neat ginkgo
ok nevermind they're not isomorphic
sigh
thank you for your efforts
lol
and sorry ...
stone fulcrum
Nah fam it was an interesting question and I spent a fair amount of time thinking about how those two could possibly be iso
Not iso is also an interesting result. That's two groups of order 24
summer path
why does $g \left(\bigcap N_i\right) g^{-1} = \bigcap \left(g N_i g^{-1}\right)$ show that $\cap N_i$ is normal?
cloud walrusBOT
Tubular Cat:
steady axle
$gN_ig^{-1}=N_i$ so $g \left(\bigcap N_i\right) g^{-1} = \bigcap \left(g N_i g^{-1}\right) = \bigcap N_i$
cloud walrusBOT
bertwit:
steady axle
remember N_i is normal
summer path
oh right... thanks
also for the first half regarding $\langle N_i \rangle$, how does $gHg^{-1} \supseteq N_i = gN_i g^{-1}$ go to that $\langle N_i \rangle$ is normal?
cloud walrusBOT
Tubular Cat:
summer path
since H could contain elements outside of <N_i>?
unless H means $\langle N_i \rangle = \bigcap_{K \supseteq N_i, K \subseteq G} K$?
cloud walrusBOT
Tubular Cat:
elder valley
i think i answered my own question from yesterday while lying in bed last night
let $R$ be a DVR and $M$ its maximal ideal, with $x \in M$. Then $(x) = M$ if and only if $x + M^2$ is nonzero in $M / M^2$
cloud walrusBOT
Auvera:
elder valley
the forward is easy since $M^2 = (x^2)$ and $M \not= M^2$ since $R$ is DVR
cloud walrusBOT
Auvera:
elder valley
since R is a DVR we have M/M^2 being a 1 dimensional vector space over the field of fractions of R
i thought i had the reverse direction but now i don't see it
oh wait. R is DVR so M is principal, say generated by y. then since x is in M we have x = ry for some r in R, and M^2 = (y^2). so if r were in M we would be able to write r=sy for some s in R, and so x=ry=sy^2 in M^2, which is a contradiction. so r can't be in M. since R is local, M is the set of nonunits of R, and so r is actually a unit. so we have y = r^-1 * x and x in fact generates M
now i understand Silverman's logic xD
cloud walrusBOT
Manas:
If M and N finitely generated over a local ring A , then $\mu (M\oplus N) =\mu (M)+\mu (N) $ where $\mu(M) $ is number of minimal generators in M
is that in reference to my problem above?
opal vale
is that in reference to my problem above?
@elder valley no
half nebula
I'm a bit stuck here. Like just in general how do I found out which are functions? I would think only C since it has 4 elements but I'm not sure if that's how this works.
elder valley
what's the set theoretic definition of a function?
half nebula
That there's some kind of rule where we can map each element in A to B right?
Or something like that.
elder valley
it's a subset of AxB satisfying some properties
half nebula
Okay. What would be the property in this case?
elder valley
call the subset R. then R is a function if and only if $(x,y) \in R$ and $(x,z) \in R$ implies $y=z$
cloud walrusBOT
Auvera:
half nebula
Okay. And what are x, y, and z from?
elder valley
R is a subset of AxB, so x is in A and y and z are in B
half nebula
Okay. So in that case it would just be a, b, and c would be functions. Because they all fit that.
Like all the x are in A and all the y are in B
elder valley
yes. but not all are functions
the second condition is that for every x in A, there is a y in B such that (x,y) is in R
half nebula
Okay so I was right in what I said earlier. It's only c. Because not every x in A is used for a and b.
half nebula
Oh it can use them more than once
elder valley
for that second condition yes, but look at the first
half nebula
Oh wait I feel dumb now. I forgot what N was XD
I got N and R confused.
Okay yeah so it's not b then
And I think d works too but I'm not sure. Because wouldn't that be {(-1,1),(0,3),(1,5),(2,7)}?
elder valley
yeah that works
pallid ember
this isnt really AA
elder valley
true
half nebula
What would it be under? This is for an analysis class but It's for the "review chapter" on AA
chilly ocean
#proofs-and-logic most likely
half nebula
Okay
pallid ember
shy bluff
isn't this basically one of the isomorphism theorems?
elder valley
basically, but X doesn't have to be an ideal
elder valley
i think of them as "things you can quotient by"
elder valley
because you won't get a ring unless you quotient by an ideal
elder valley
R is a ring, and say J is an additive subgroup that's closed under multiplication. then R/J is a group, but it's only a ring when J is an ideal
shy bluff
And J is ideal when it contains all products of rj, where r is an element of R and j is an element of J?
shy bluff
Ah the question is basically this but for rings
mmmm don't think that we learnt that yet sadly
solemn rain
whats the problem
shy bluff
Ah I just don't really understand ideals
And the latter half of this assignment is all about ideals
solemn rain
an ideal is a subring
with a special property that it is closed under elements from the ring
call L a subring of R
and let a be in L
elder valley
assuming his subrings don't have to contain 1
solemn rain
if xa is in L for any x in R then L is a left ideal
the same definition with right ideals
ax is in L for any x in R --> right ideals
an ideal would satisify both left and right ig
thats it
you can quotient with ideals
just the same as you can quotient with normals
same reasons same shit
@shy bluff
shy bluff
Hrm
solemn rain
why would something like this be satisifed
shy bluff
Idk
shy bluff
Oh wait quotient with ideals like normals not that they are normal
shy bluff
Ok so a left and right ideal is basically a left and right coset?
solemn rain
a left ideal is a subring that is closed by left multiplication from ring elements
a right ideal is a subring that is closed by right multiplication from ring elements
shy bluff
It's left multiplied just like a left coset right?
solemn rain
an 'ideal' would be both
idk what you mean
do you have any problems with what i said above
shy bluff
Mmm when we have left/right cosets we have like a group G, a subset H of it, and then we have gH being a coset that consists of H and all the products of all elements of g with all elements of H?
solemn rain
okay maybe this is getting out of hand
can we not draw parallels?
do you get the definition?
if x is in I --> rx is in I for any r in R --> I is a left ideal
thats it
shy bluff
Kind of? It's a subring that requires all the products of an element of itself and any element of R are still in it
shy bluff
But what does it look like? Like with cosets I can see that it partitions the group
solemn rain
an example would be pZ of Z
where p is prime
and Z is set of integers
a more explicit one :
{0,2,4} in Z/6Z
pick any element in Z/6Z --> multiply it by any element in the set --> ur still in the set
thats an idieal
ideal*
got it ?
bold coyote
One of the easiest examples is just 2Z : any sum of even integers is even and any integer times an even integer is even, so the properties of the definiton are satisfied.
solemn rain
no
we say aRb or a=b mod H if a=bh for some h in H
the equivalence classes of this equivalence relation are called cosets --> partition the set
what equivalence relation would you suggest
to define on these rings
notion of ideals are more number theory'ish
ig
if a divides b --> b=at for some integer t --> b is in the ideal generated by a
the ideal generated by a and b is the ideal generated by the gcd of a and b
cool shit u will learn about ideals later
the ideal generated by a is the set of multiples of a basically
(a) ={ra | r in R}
so those are the things 'ideals' do and obv more but idk about
got it?
solemn rain
cool
whole basalt
Hey y'all. I would appreciate if someone could give a quick look over my solution I've written for a problem (which I also wrote), and make sure I haven't made any dumb mistake.
Other than changing it to say "all prime powers that divide 60", since otherwise it makes it sound like you could have an element of order 8. I've already fixed that.
whole basalt
k awesome.
solemn rain
im a beginner so i think im good judgement haha
but i think i would maybe be abit weirded out
by 'does not necessarily have '
that's for me ig
whole basalt
Well, this is for a GRE subject test practice
solemn rain
oh so its a well known word/phrase there?
whole basalt
So it really is meant to be done with process of elimination
And basic theorems about subgroups
solemn rain
yea yea
whole basalt
It is guaranteed to have 2, 3, 4, and 5
elder valley
@whole basalt maybe i learned the sylow theorems differently but i dont think the first theorem alone gives subgroups of every prime powers. i think that's using another theorem about p groups
olive mirage
I usually state the Sylow theorems that way for undergrads: "if a prime power divides the order of a group, then the group has a subgroup of that order"
whole basalt
I've seen it stated different ways in different books
Some books I see that it only has the highest powered prime
Others I've seen it stated that it has all possible prime powers up to that highest powered prime
olive mirage
yeah, I would say the standard version is only the highest power becasue it is a statement about a group action
whole basalt
Yeah
Though recently I realized that's not true about element orders, only subgroup orders, derp!
opal vale
Yeah
@whole basalt are you started Math Gre study group?
whole basalt
Yup
On a Discord server, we just did our first session which was free
I'm doing 7 more
summer path
How do you read the / in a quotient group?
summer path
ok thanks
oblique river
phi^(-1)(r_0 mod p) is going to be more than just one element in general
for example, in Z/(4), the preimage of 3 is the set {3, 7, 11, ..., -1, -5, ...}
I think you're making it too hard on yourself though. To say it's surjective just means you need to find some element of R whose image under phi is r_0 mod p
and you have one: it's r_0
I'm not sure what to say other than: yes
you've given the definition: phi(r) = r mod p
stepping back for a second, I know that the definition of "ideal" is pretty weird, but in my opinion a good way to think about ideals are "things that you can take quotients by"
saying that I is an ideal of R is exactly the same as saying that R/I is a ring
and the map R --> R/I is always surjective
this is just something good to know/remember
so you don't have to reprove it everyt ime
cloud walrusBOT
AdamCM:
elder valley
"well defined" just means that the thing you defined is actually a function. typically you need to address this only when elements in the domain can be represented in more than one way, and the mapping you're defining uses a particular representation of the element to determine its image
fallen bluff
Universal algebra. Could anyone help? I don't understand this proof of THM on locally finite varieties (can send definitions)
chilly ocean
hi
in the group Ξ4, r is the anticlockwise rotation of Ο/2.
H=<r> is the subgroup generated by r.
How do I prove that H is a normal subgroup of Ξ4??
vestal snow
What is an idempotent ideal?
Is it an ideal where every element is idempotent or an ideal such that I = I^2
opal vale
Is it an ideal where every element is idempotent or an ideal such that I = I^2
@vestal snow i think last one i.e. I^2=I
opal vale
Ah okay
@vestal snow can you give the full question?
vestal snow
A is absolutely flat iff A_m is a field for all maximal ideals m
Though i used a theorem about idempotent ideals here
shy bluff
Is the entire ring an ideal of itself?
solemn rain
yes
shy bluff
Ok thats' what I thought
smoky cypress
There are always two trivial ideals: the whole ring and the additive identity itself {0}
elder valley
Does the theorem that a morphism of curves is either constant or surjective follow from the image being a variety, so is either dimension 0 or 1?
olive mirage
well, the statement is false as given
A_1 includes into P_1
and that is not a surjection
so you need a whole stew of adjectives there to get a true theorem
royal harness
(Am sorry to disturb the discussion, I just want to ask you: where does this image come from @elder valley ?)
elder valley
Silverman's Arithmetic of Elliptic Curves
royal harness
Thank you! π
shy bluff
Is the quotient map always a homomorphism?
shy bluff
Yea ok that's what I thought
stone fulcrum
That's the first homomorphism theorem
Every normal subgroup generates a surjective homomorphism, every surj homomorphism can be represented by a normal subgroup
shy bluff
Hrm
Yea bc I'm trying to prove uh this
So far all I've got is that q satisfies the fact that $\phi(X) = 0$ and that q is a surjective homomorphism
cloud walrusBOT
Liria ^(;,;)^:
elder valley
you need to state what psi is
shy bluff
Well idk what psi is? Nor do I know how to construct psi?
elder valley
there aren't many options on how it would be defined
shy bluff
Like do I just say that if $\psi$ is a homomorphism and $\psi(x) = 0$ for all X then we see that clearly X is in its kernel... and that's where idk what nebt to do
cloud walrusBOT
Liria ^(;,;)^:
elder valley
psi maps from R/<X> to S, not R to S
shy bluff
Oh sorry, I met phi
If $\phi$ is a homomorphism from R to S and $\phi(x) = 0$ for all x in X
elder valley
but regardless the problem is only asking to show that a psi exists which satisfies phi = psi q
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
Yes but how do you show that psi exists
elder valley
you have to define it. elements in its domain look like r+<X>. so psi will need to take something of that form and give an element in S
elder valley
the elements of R/<X> are cosets that have the form r+<X> for some r in R
shy bluff
can we necessarily say that there's a psi like that though?
elder valley
the purpose of the problem is to show that there is
shy bluff
ok so we can see that for any r in R, $\phi(r) = \phi(r + <X>)$?
cloud walrusBOT
Liria ^(;,;)^:
elder valley
no, the domain of phi is R, not R/<X>
shy bluff
yes, domnai of phi is R, but X is in kernel of phi, so phi(<X>) just goes to 0 right?
elder valley
well, technically yes i suppose. applying phi to anything in the coset r+<X> will map to phi(r)
shy bluff
Yea, so we can define a $\phi$ as being the composition of two functions f and g, such that f(g(X)) = 0
elder valley
it's not really proper because r is an element and r+<X> is a set. more properly it would be phi(r+<X>) = {phi(r)}
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
and then we see that naturally a function that maps X to 0 would be the quotient homomorphism, and so we see that g is just the quotient homomorphism?
elder valley
you mean psi? what is f?
shy bluff
f and g are arbitrary functions
like if I just say psi and q I will get confused
So I say f and g so that they're not already defined as in the question
i.e, let f and g be functions that we don't know anything about, except that we know that phi = f o g
elder valley
you have to say what they are, otherwise you can do trivial things like f=identity and g=phi
i think you are making this much harder than it should be
shy bluff
I don't understand how to construct this function at all
elder valley
you just need to define $\psi : R / <X> \rightarrow S$ by giving some formula of the form $\psi(r+<X>) = $ [put something here]
cloud walrusBOT
Auvera:
elder valley
but you need to determine the right thing to put there so that it's in S and you get phi=psi q
shy bluff
Can't it just be the inclusion map?
elder valley
inclusion of what? R isn't contained in S necessarily
or R/<X> for that matter
the only inclusion you have is X in R
i don't want to just give you the answer here
shy bluff
I am not understanding how to do the question at all
Like we're trying to prove psi has to exist
But we can't just assume that we can have phi = psi o q for some psi
and for q being the quotient map
elder valley
no, that's exactly what you're trying to prove: that there exists a psi which has that property
shy bluff
Yes
We want to prove it
So all that we know is that phi(X) = 0
And that phi is a homomorphism from R to S
elder valley
yup. and that will help you to define psi
shy bluff
Wait can I just use the first isomorphism theorm
Wait no we dont'know the entire kernel of phi
Nor can we say anything about X
elder valley
yeah i don't see that it will help you
shy bluff
yea so idk how this is supposed to work
Like if X were an ideal then we could say something but it's not
elder valley
here's a different approach
you want psi to satisfy the equation phi = psi q right? so apply an element to both sides of the equation and look at what happens
shy bluff
$\phi(r) = \psi(q(r)) = \psi(r + <X>)$?
cloud walrusBOT
Liria ^(;,;)^:
elder valley
exactly. take that as the definition of psi
same thing i was hinting at above
so $\psi : R/<X> \rightarrow S$ as $\psi(r+<X>) = \phi(r)$
cloud walrusBOT
Auvera:
shy bluff
But because it's a homomorphism we have to have that $\psi(r + <X>) = \psi(r) + \psi(<X>)$?
cloud walrusBOT
Liria ^(;,;)^:
elder valley
no. first, you haven't shown it's a homomorphism yet. second, psi is define on R/<X> not on R, so you can't apply it to individual elements. it only makes sense to apply it to a coset r+<X>
shy bluff
ok but what can you do to r + <X>
elder valley
your proof should start out by defining psi with that equation. then you need to show: psi is well defined, psi is a homomorphism, and phi=psi q
i'm not feeling well so i'm going to sleep. someone else should be able to help you if you need it
shy bluff
hrm ok
oh wait so that's what I have to show?
I will go try that... I just wasn't understanding what I had to show
elder valley
you don't do anything to it. the elements of R/<X> are of the form r+<X>. it doesn't simplify any further than that
shy bluff
So define psi with phi = psi o q?
And then show that psi is well defined, a homomorphism, and finally that phi = psi o q?
elder valley
no, as psi(r+<X>) = phi(r)
shy bluff
Oh
Ok
So it should go something like
Let psi: R/<X> to S be defined by psi(r + <X>) = psi(r)
Followed by showing that it's well definde, that it's a homomorphism, and finally that phi = psi o q
elder valley
yes exactly
my impression is that you are struggling with quotient rings and that is what is making this difficult for you
shy bluff
That's probably true
I struggled with quotient groups too
But I am also just struggling with what/how to show this... if it had said to show that psi was well defined, a homomorphism, and that phi = psi o q then I would've probably been ok
cloud walrusBOT
AdamCM:
shy bluff
The wording isn't clear to me what it's asking for :x
elder valley
Can you link the question again
@sharp peak any single element set is linearly independent, but it doesn't generate the whole module so it can't be a basis
Actually that statement was false, so it's not a basis for both reasons
shy bluff
elder valley
2/3 things I said up there the questions asks you to show. The well defined is introduced because you are defining a function on a quotient ring
elder valley
Homomorphism and the equation
shy bluff
Oh so it's already well defined because it's a function ona quotient ring?
elder valley
No you have to show that
shy bluff
Ah ok so teh things I have to show are still well definedness, homomorphism, and equation
elder valley
You were saying you weren't sure what you needed to do and I was just pointing out that 2 of the things you need to show are stated in the question
Yes
shy bluff
Like that makes more sense to me now that I know that that's what I'm proving
Like now that I know that we can assume that psi exists, and that we have to prove that it's well defined, a homomomprhism, and that it satisfies the equation, I think I know what to do
elder valley
Well you aren't assuming it exists, you gave a definition for it so you know it exists
shy bluff
Err yea, you define it and then you go and show that it satisfies those . properties
those 3 properties*
shy bluff
Wait, for showing that psi is a homomorphism, our definition is this
Then in order for psi to be a homomorphism, wouldn't we need that R/<X> is a ring?
shy bluff
And as X isn't an ideal we don't have that R/<X> is a ring?
bleak abyss
X isn't an ideal but <X> is
woven delta
<X> literally means the ideal generated by X
shy bluff
Wait, X is a subset of R, but the generator of it is an ideal?
bleak abyss
Not the generator of it, the ideal generated by it
delicate bloom
X is an element of R, not a subset
(err I didn't read what was written earlier tbh so I could be wrong on second though lol)
kindred rivet
I think lattices are a good way to think of ideals but Iβm biased cause I just read chapter 11 of artin and that is about lattices and ideals
sacred pier
Hello guys, what is the deal of f: R->R on line 1?
Can I get some keywords so I can google it? Thank you?
chilly ocean
this isnt abstract algebra
you can just say "f is a function defined on the nonnegative reals"
if you want to be pedantic, a "real-valued function on the nonnegative reals"
chilly ocean
of addressing what?
sacred pier
R>0 -> R
chilly ocean
idk
sacred pier
lol
cloud walrusBOT
AdamCM:
timid hull
Wait, X is a subset of R, but the generator of it is an ideal?
$\left <X\right>$ is defined as ${r\cdot x | r\in R , x\in X }$ which is "clearly" an ideal in other words $\left<X\right>$ is the "smallest" ideal containing $X$ . Btw the problem you are trying to prove is actually a "slight generalisation" of 1st isomorphism theorem for rings ...
cloud walrusBOT
WhyamIsohot?:
vestal snow
Can someone help me out with this one?
I'm proving 1) implies 2)
Here's what I've done so far
Let $p$ be a prime ideal. Suppose that $\frac{a}{t} \cdot \frac{x}{m} = 0$. Then $\frac{ax}{ts}=0$ which implies $uax=0$ for some $u$ in $A-p$.
royal harness
try \cdot ?
cloud walrusBOT
Have a Banana Bitch:
vestal snow
I want to conclude that either a=0 or x=0
I also need similar help with 3) implies 1).
Suppose $ax=0$. If $a$ is a unit, we're done. If not, $a \subseteq m$ for some maximal ideal $m$. Then we have that $\frac{a}{1} \cdot \frac{x}{1} = 0$ which implies either $a=0$ or that there exists a $s \in A-m$ such that $sx=0$
cloud walrusBOT
Have a Banana Bitch:
latent anvil
@vestal snow how can you use the assumption that M is torsion free with uax = 0?
In i => ii
vestal snow
Oh
Yeah got it
either ua = 0 or x = 0. If its the former, then a = 0 as we're in an ID
Can't believe I missed that
latent anvil
Np, math hard
vestal snow
Atiyah-MacDonald very hard
I'm worried about Harthshorne because I've heard it's even harder
latent anvil
it is
vestal snow
Did you figure out 3) implies 1)?
vestal snow
yup
latent anvil
because if so, I think I have it
T(M) = 0 iff T(M)_p = 0 for all p iff T(M)_m = 0 for all m
right?
latent anvil
No I'm saying this is true
A module is zero iff it's zero at all primes iff it's zero at all maximal ideals
latent anvil
yee
latent anvil
I worked through the first couple chapters over christmas break on my own
My actual class used Reid's undergraduate commutative algebra
which I used to supplement AM for integral extension stuff
vestal snow
Ah okay
I'm finding A-M to be really difficult
especially chapters 2 and 3
Is this normal with this book or should I switch?
latent anvil
Yeah that's fair. It's a tough book
vestal snow
Is there a list of recommended exercises from Atiyah-MacDonald?
golden pasture
all of them
like a lot of the exercises are super useful idk how to like say what specifically are those must do other than having a massive list
timid hull
Or you can try a book which is released recently "Commutative algebra" by Andrea Ferretti . This book is quite nice but it is a bit long
kindred mist
Silly question: Let F be a field, take L to be a field extension of F (let L = F if F is alg. closed). Let $y \in L$ Is F[y] \cong F[x]$ (where F[x] is just the usual ring of polynomials over F)?
kindred mist
the latex didnt work, yeah one sec
Let F be a field, take L to be a field extension of F (let L = F if F is alg. closed). Let y be in L. Is F[y] isomorphic to F[x] (where F[x] is just the usual ring of polynomials over F)?
That make more sense ^^^?
wind parrot
this is only true if y is transcendental over F
otherwise F[y] is iso to a quotient of F[x]
shy bluff
How does this work?
I'm guessing that there's something to do with the first isomorphism theorem ?
Like I basically want to show that (x^2 - c) is the kernel of some function and that its image is C?
kindred rivet
ya basically
solemn rain
whats ev_c
kindred mist
you send f(x) to f(\alpha), where \alpha is a root of x - c
then the kernel is all multiples of the polynomial [x - c] (I had a typo in what I said there but I fixed it)
shy bluff
Evaluation at c
upbeat juniper
then the kernel is all multiples of the polynomial [x - c]
^ f(x) has a root c if and only if x-c divides f(x)
so f(x) is of the form (x-c)g(x) β (x-c)R[x]
shy bluff
@kindred mist yea I was thinking something with the roots, but can you just say that lol, like let f: R[x]/(x^2 - c) -> C be the function that sends f(x) to a root of x^2 - c?
shy bluff
Wouldn't it be similar for any power of x?
That if we have R[x] mod some polynomial f, then we see that it goes to 0 for any x that's a root of f?
Because then we'd be dividing by 0? hrm that dosen't make sense...
kindred mist
yeah, if we have an evaluation map F[x] -> L, (where L is a field containing F), and we evaluate h(x) in F[x] at say "b", then if p(x) is an irreducible polynomial over F with b as a root, then (p(x) ) is the kernel of that evaluation map
so for example, take the reals R, then take the evaluation map, h(x) -> h(i), R[x] -> C, the kernel of this must be (x^2 + 1)
shy bluff
What is h(i)
kindred mist
h is some polynomial over R (where R is the reals)
so h(i) is h evaluated at i
(where i is the imaginary unit)
if we have a polynomial in R[x], h, and h(i) = 0 then x^2 + 1 must divide h (since x^2 + 1 is an irreducible polynomial over R with i as a root, namely its the only one with a coefficient of 1 on x^2 and least degree), but if i doesn't divide h, then x^2 + 1 can't divide h
hence (x^2 + 1) being the kernel
with me now?
next snow
Cool
upbeat juniper
I believe that argument requires a bit of field theory
alternatively you could demonstrate an explicit isomorphism between residue classes (which can be written in the form ax+b) and C
eg for c = -1 I believe the map ax+b -> b+ai works
timid hull
How does this work?
@shy bluff
Consider the elements $1+(x^2-c)$ and $x+(x^2-c)$ of $S$ note that to give a ring homomorphism $h$ from $S$ to $\mathbb{C}$ it suffices to define $h(1+(x^2-c))$ and $h(x+(x^2-c))$ as every elements in $S$ is just generated by these two . Now define $h(1+(x^2-c))=1$ and $h(x+(x^2-c))=\sqrt{-c}\cdot i$ , This map does the job along with the fact that $x^2-c$ is irreducible over $\mathbb{R}[X]$ when $c<0$ .
cloud walrusBOT
WhyamIsohot?:
elder valley
ehh the power of the first isomorphism theorem is that you don't have to define maps on quotient rings. you just define it through R[x] --> C and the theorem gives you the map on the quotient. if you do it as R[x]/(x^2-c) you're working harder than you need to
woven delta
That's not the first iso thm, that's the universal property of quotient spaces
elder valley
i mean it's probably the universal property also but it's definitely the first isomorphism theorem in this case
i guess here it's first iso because we're getting an isomorphism but you're saying that if you're just trying to get a homomorphism with a given kernel then it would use the universal property?
woven delta
The statement you just said was about defining maps on quotient rings, not isomorphisms
kindred mist
Ok so iirc Z_n is a field iff n is prime (I believe I proved this before) and then I see some pdfs talking about a field "F_p^n" are these fields something other than Z_p^n, since n >= 2 means that Z_p^n isn't a field right?
or is " Z_n is a field iff n is prime" not correct?
kindred mist
hmm I'm having trouble actually finding a definition for F_p^n, (been googling for a few minutes now)
If someone could link me to something that would be really appreciated
I found a pdf on finite fields but I dont see it in there (at least not yet)
scarlet estuary
im not sure exactly what you mean?
finite fields of the same order are isomorphic
Z_2 is the same as F_2
but Z_4 is not the same as F_4 (since Z_4 is not a field while F_4 is)
kindred mist
so is the way of "defining " F_p^n just: "the (or "a") finite field with p^n elements"?
scarlet estuary
yes
kindred mist
I suppose also that there is a unique field with p^n elements right?
scarlet estuary
up to isomorphism
kindred mist
Thank you!