fair quartz
yeah the way we were taught this looked something like this:
We were taught eigenvalues on the most basic level in linear algebra
then every course that followed it up stated "yeah yall already know everything about eigenvalues from LA, so here are these random definitions without further explanation"
coral spindle
With all due respect, your lacking in understanding of prior content is not necessarily the fault of the instructor. That being said, we now seem to have filled in the gaps, or at least I hope.
delicate orchid
not explicitly mentoning the spectrum is the set of eigenvalues is dog water pedagogy and I refuse to blame ludicrous for that
relying on students to recognise something that important is silly
coral spindle
Not necessarily – if that's the definition of eigenvalues they're working with then it would be standard. I agree with you that it's a bad definition, but it may be consistent with the course.
fair quartz
a source of confusion is that we were also introduced to the "Eigenvalue spectrum" as a subset of the spectrum
but its also known as the "point spectrum"
coral spindle
And now we see that some context has been missing
delicate orchid
oh.... lol?
fair quartz
delicate orchid
mfs leaving out context for the sole purpose of making me look silly once again
coral spindle
So I imagine for the infinite-dimensional case these things can differ.
fair quartz
either way, being made aware of this property definitely helped a lot, so thank you guys! 😄
south patrol
Yes this is common stuff for functional analysis
coral spindle
So I imagine for the infinite-dimensional case these things can differ.
Yeah indeed, if we have a vector space with basis v_n for integers n>=0, then the map M sending v_n to v_{n+1} is not invertible (doesn't have an inverse on one side) yet has no nontrivial solutions. (unless I'm mistaken)
summer path
This was all just an extended setup to make wew look silly 
delicate orchid
This was all just an extended setup to make wew look silly <:KEK:586240877358350...
the demiurge has done it again
coral spindle
Mean >:(
summer path
D:
delicate orchid
moderators ban this user for gang stalking
coral spindle
quick everyone sully him
coral spindle
Yeah indeed, if we have a vector space with basis v_n for integers n>=0, then th...
Anyway @fair quartz the point I'm making here is that when your vector space is not finite-dimensional, this gives an example of when the eigenvalue spectrum is not equal to the spectrum as defined previously, where we generalise it correctly to infinite-dimensional spaces.
fair quartz
ohh i see
coral spindle
extra points if you can work out what the elements of the two are, but no pressure
hollow mica
thank you for the insightful answers Merosity and jagr2808!
coral spindle
@delicate orchid do you know any nice quick references to get the lowdown on Brauer characters?
You're the first person who comes to mind on this so forgive me for pinging you
delicate orchid
bobby fusions covers them and that's where I learnt them but uhh it's very dense
kinda like reading fulton-harris for an intro rep theory course lol
I'll check da archives
delicate orchid
really though, it's not analogous to regular character theory as far as I've seen
coral spindle
yar
delicate orchid
as in, it seems to be much less useful
coral spindle
My purposes are very narrow. I just need a couple of results
delicate orchid
although I haven't worked much with them I just do the funny blocks and the defect groups
coral spindle
Mfw when the
when the defect group is Abelian
delicate orchid
when the princple defect group is trivial (I am working over C)
hollow mica
y’all know how an equivalent definition for a vector space V over a field F is an abelian group V plus a ring homomorphism pi: V -> End(V), where then scalar multiplication c * v can be defined by pi(c)(v)… if we just replace F with an arbitrary commutative ring then what we get are modules right?
coral spindle
You mean a homomorphism pi : F → End(V), but yes :)
hollow mica
yea oops
delicate orchid
commutative
no need for that
hollow mica
but then it’s left and right modules which is weird
coral spindle
C.f. G-sets being maps pi : G → Sym(X)
delicate orchid
it is weird but you can still get two sided ones and everything is wholesome
delicate orchid
C.f. G-sets being maps pi : G → Sym(X)
sets are modules over F_1
delicate orchid
nobody does
hollow mica
ok nice, weird how wiki only mentions this definition for vector spaces but not modules
coral spindle
I guess it is a bit of an omission, but it's not really the greatest way to view these things
hollow mica
but if I want to remember the definition
delicate orchid
modules, or as I like to call them - ideals
hollow mica
this is ideal
delicate orchid
the definition is just a bunch of stuff that says "yeah this works like how you think it should"
hollow mica
Bro is sniping me
coral spindle
Well yeah honestly if that helps you personally, that's fine
But at least personally, once you learn more module theory you want to think of R-modules as algebraic structures in their own right
Anyway I'm distracting myself from the wholesome chungus modular rep theory
hollow mica
Ooook fine
hidden haven
it is weird but you can still get two sided ones and everything is wholesome
Don't you get left modules
Right modules would be R^op → End V
hidden haven
Also I prefer this definition of a module over the "function R ⊗ M → M satisfying a couple more things"
untold turret
we're in some integral domain. the ideal (3a+c, b+c) that is a subset of the maximal ideal (a,b,c) can be written as some sort of space of vectors with basis (3,0,1), (0,1,1), and the bigger ideal can be written as some sort of space of vectors (1,0,0), (0,1,0), (0,0,1)
is there any definite statement that formalizes the above?
some sort of isomorphism between the ideals (3a+c, b+c) and (a,b,c) and some sort of product space
untold turret
a space where things look like vectors, i'm trying to give a definite statement
coral spindle
We can't analyse a typical integral domain as a vector space unless you mean that your integral domain is a field
untold turret
no i know that
coral spindle
Oh ok wait you're saying that (3a+c, b+c) = (3a+c) + (b+c)
coral spindle
It's a sum, not a product
I'm trying to think of some nice way to do this
Yeah indeed, the elements x,y are R-linearly independent iff (x,y) = (x) + (y) is a direct sum (i.e. the intersection is empty)
I can't really see what you're trying to say beyond that so I'll just say that
untold turret
that's a start, thanks
rocky cloak
Is U a unitary operator on a finite dimensional space or an arbitrary hilbert space? The statement is true either way, but it the proof will be a little different.
fair quartz
arbitrary hilbert space
rocky cloak
I see, do you know how the spectrum interacts with taking adjoints / inverses?
Also I guess this is functional analysis, so I dont know if it should be here or in #advanced-analysis . Guess it doesnt matter
slim kayak
nobody does
Are we sure F_1 isn't just a big prank
delicate orchid
Are we sure F_1 isn't just a big prank
this is an open problem
hollow mica
wikipedia says: an algebra A over a field F is a vector space A over F that comes with an additional binary multiplication operation that is distributive and compatible with scalars… but then the associative algebra page claims that the algebra turns into a ring because of associativity of multiplication; don’t we need there to exist a multiplicative identity for that to happen?
oh they’re assumed to be unital
coral spindle
Yes.
lethal dune
yes
coral spindle
The word 'algebra' is a bit overloaded. Not all algebras in the first sense are those in the second
A Lie algebra is not a ring, for example.
Don't confuse these two.
hollow mica
A Lie algebra is not a ring, for example.
because it’s not unital?
lethal dune
algebra can also mean a textbook
hollow mica
or like completely different
coral spindle
It's not even associative
hidden haven
The word 'algebra' is a bit overloaded. Not all algebras in the first sense are ...
I've been studying physics and these notes say "after a little bit of algebra we get this" and it's just equation manipulation
hollow mica
well “algebra over a field”, at least according to wikipedia, does not require multiplication to associative
hollow mica
ok so the main discrepancy between naming is unital
south patrol
I've been studying physics and these notes say "after a little bit of algebra we...
ye that's what algebra meant to me in physics lol
coral spindle
ok so the main discrepancy between naming is unital
No, not even
rocky cloak
The definition of ring also changed from field to field
coral spindle
I've been studying physics and these notes say "after a little bit of algebra we...
This is the greatest sin of applied math 😔
jk but also it is annoying
rocky cloak
You can tell whether someone is doing analysis or algebra by where a ring has 1
hollow mica
what do analysts deal with that doesn’t have 1
formal ermine
I've been studying physics and these notes say "after a little bit of algebra we...
what's the corresponding isomorphism?
rocky cloak
what do analysts deal with that doesn’t have 1
Continuous functions with compact support
coral spindle
There are two main senses of the word 'algebra'
(1) a vector space with a bilinear binary operation.
(2) a ring with a natural vector space structure.
The two definitions on two different wikipedia pages are these.
(2) implies (1), but (1) does not imply (2).
hollow mica
because no 1 in (1)
coral spindle
That's true, but it also may not be associative for example
delicate orchid
see lie algebras
hollow mica
eeee forgot about that
delicate orchid
or R^3 with the cross product lol
or R^7 with a cross product if you want to smoke crack
hollow mica
wait where’s the wiki entry for (2)
rocky cloak
or R^3 with the cross product lol
That's a lie algebra
hidden haven
what's the corresponding isomorphism?
coral spindle
wait where’s the wiki entry for (2)
delicate orchid
Z
hollow mica
ahhh and they assume them to be unital there
wait what if you assume multiplicative inverses too
coral spindle
That is called a division algebra
hollow mica
that’s like a field over a field
coral spindle
E.g. C, H
delicate orchid
O, S, T
coral spindle
Nah those aren't division algebras in the associative sense
Those are other things
delicate orchid
trying to think what comes after T
rocky cloak
that’s like a field over a field
Don't need to be commutative
So more like division ring over a field
hollow mica
mhm
rocky cloak
Field over a field is usually called field extension
coral spindle
R, C, and H are the only three fin. dim. real division algebras btw
thought I'd mention that
hollow mica
H is an element bruh
coral spindle
H is the quarternions.
cloud walrusBOT
jagr2808
south patrol
what i find funny is like
normed division algebras aren't necessarily assumed to be associative, so like O is a normed division algebra but not a division algebra (in some terminologies)
rocky cloak
surely people who dont require normed divison algebras to be associative, dont require that for division algebras either, right?
vague granite
I just realised
Is it the case that the induced group operation on the quotient group is well defined iff the quotient map is a homomorphism?
I suspect these mean the same thing
I've just always desliked the term 'well defined' since ironically enough I think that term isn't well defined enough
rocky cloak
what does it mean for a map to be a homomorphism if its not between groups?
south patrol
Well defined would mean it doesn't exist
delicate orchid
I've just always desliked the term 'well defined' since ironically enough I thin...
it's an incredibly clear definition. x = y => f(x) = f(y)
vague granite
I guess you'd restate is as f(g*h) = f(g)*f(h) where the * on the RHS is the induced binary operation
vague granite
The notation f(x) already assumes f is a function right
yes
which is the main issue
open sluice
you could rewrite it in terms of sets
delicate orchid
what
open sluice
function as a set of ordered pairs
and as long as each x corresponds to exactly one y you have a well defined operation
delicate orchid
u people are some of the most pedantic mfs on the planet
coral spindle
A priori we only have a relation when we write something like f([x]) = g(x). Well-definedness states that the relation in question is a function.
vague granite
to speak of f(x) means you already have a set of ordered pairs where there is a unique y st (x,y) in f
south patrol
I mean otherwise it doesn't really make sense
open sluice
I mean you have to check
delicate orchid
could map to the set of all possible values
vague granite
Is a function by definition not such a relation?
delicate orchid
I'm actually going to put my fist through my screen
rocky cloak
I guess you'd restate is as f(g\*h) = f(g)*f(h) where the \* on the RHS is the i...
But that assumes there is an induced binary operation
delicate orchid
why do I fucking bother sometimes
south patrol
Anyway so like the well-defined bit usually comes up in this situation: we have a set X with an equivalence relation ~ and we define a function X/~ -> Y (Y another set, say) by sending [x] -> something dependent on x i.e. through a function X -> Y
then saying it's well-defined just says that if [x] = [x'] then x,x' have teh same image under that map X -> Y
delicate orchid
oh but if you set that equivalence relation to be "=" then the entire fucking chat dog piles you
south patrol
Well you said it was incredibly clear and then said something unclear lol
vague granite
But that assumes there is an induced binary operation
Define it as $aH*bH:=\cup{(xy)H:x\in aH, y\in bH}$
cloud walrusBOT
philka.
south patrol
Here I'm saying the function X -> Y is already well-defined
rocky cloak
Define it as $aH*bH:=\cup\{(xy)H:x\in aH, y\in bH\}$
Okay, but then it's always a homomorphism
vague granite
Okay, but then it's always a homomorphism
Is it?
If H is not normal then it will have strictly larger cardinality than H no?
vague granite
Here's one for the fans. $\text{id}_S = =|_S$.
cloud walrusBOT
philka.
mighty kiln
Regarding the proof: How is B' guaranteed to be maximally linearly independent? Doesn't one run into problems of the following sort if v is not a successor:
{v1, v2, v3, ...}
{v1+v2, v2, v3, ...}
{v1+v2, v2+v3, v3, ...}
...
{v1+v2, v2+v3, v3+v4, ...} ← does not span V
cloud walrusBOT
hausdorff
mighty kiln
Regarding the proof: How is B' guaranteed to be maximally linearly independent? ...
Also are there any other references to the stronger result that is proved here? Namely, that if S is linearly independent and B is a basis then one can replace |S| elements of B with elements of S to get another basis
This is from Aluffi's Algebra: Chapter 0
median pawn
**hausdorff**
my bad, it's (X+1)(X^3 + X + 1)(X^3 + X^2 + 1)
coral spindle
Factorising polynomials over a finite field has a simple algorithm: go through every polynomial in order of ascending degree and check if it's a factor, if so divide, and continue.
This works because we can enumerate polynomials over a finite field very easily.
Now if you're asking for some kind of trick to see it easily, I don't think there is such a thing I'm afraid.
timber cloud
There's also the algorithm with square free factorization into distinct degree factorization into equal degree factorization though
lusty marlin
my bad, it's (X+1)(X^3 + X + 1)(X^3 + X^2 + 1)
x⁷+1=(x+1)(x⁶+x⁵+...+1)
Any factor of (x⁶+x⁵+...+1) must have a nonzero constant term, and one can rule out x+1 as a factor
Then you are left with two degree 2 polynomials to check, x²+1 and x²+x+1, but x²+1=(x+1)² so we can rule it out instantly. Then you only need to check for factors of degree 3
Can't think of a 'quick' method though
hollow mica
isn't the last condition redundant?
f(a) = f(1_R a) = f(1_R) f(a) and f(a) = f(a 1_R) = f(a) f(1_R)
open sluice
no
rocky cloak
no, f(x)=0 for example
hollow mica
and clearly there is only one multiplicative identity in a ring
open sluice
beware of division by zero
(it would be redundant in a field)
wait
no f = 0 would still kill it nvm
karmic moat
`f(a) = f(1_R a) = f(1_R) f(a)` and `f(a) = f(a 1_R) = f(a) f(1_R)`
this assumes cancellation, which only holds in specific rings
hollow mica
then it's the same right
hollow mica
this assumes cancellation, which only holds in specific rings
where'd I use cancellation
median pawn
Now if you're asking for some kind of trick to see it easily, I don't think ther...
yep indeed
formal ermine
can we find a polynomial in n = 2 variables such that the lexicographic, right lexicographic, homogeneous lexicographic, and homogeneous right lexicographic valuations wrt minimal valuation are all pairwise different? for n > 3 you can find one easily, but I doubt such a polynomial exists for n = 2
karmic moat
f(a) = f(a) f(1_R) implies 1_S = f(1_R)
median pawn
x⁷+1=(x+1)(x⁶+x⁵+...+1)
Any factor of (x⁶+x⁵+...+1) must have a nonzero constant...
i see, thanks!
hollow mica
my point was that since your first equation holds for all elements a in the ring, the element f(1_R) is an identity in S, and since identities in rings are unique, f(1_R) is the multiplicative identity
hollow mica
ok how about if we we say `f(1_R) ≠ 0_S`
but I guess I assumed that f(1_R) ≠ 0_S, because we cant have 0_S = 1_S (fixed the typo above)
so that's why I think that if you replace the condition of f(1_R) = 1_S with f(1_R) ≠ 0_S it is equivalent
open sluice
wouldn't you still need an integral domain for that
hollow mica
why is that
open sluice
ok f(a) = f(a) f(1) for all a in R right
so f(a)(f(1) - 1) = 0
but how do you know that f(1) - 1 isn't a zero divisor
hollow mica
what's wrong with f(1) - 1 being a zero divisor
open sluice
let me look over your claim again
you want to use uniqueness of the identity in R and S here
karmic moat
ok i think the mapping identities condition is redundant
some definitions omit it, even when considering rings w/ identity
open sluice
karmic moat
and i take back what i said, you did not assume cancellation above
and i believe your proof holds
hollow mica
I'm looking through Dummit Foote and it's so infuriating; they define rings as not necessarily having multiplicative identity
open sluice
f = 0 tho
hollow mica
then f(1) = 0 so not a homo
open sluice
ig if it sends everything to 0 we can call it the trivial homomorphism but i don't think people like that
open sluice
I'm looking through Dummit Foote and it's so infuriating; they define rings as n...
it makes me sad
hollow mica
so in DF they're fine with homomorphisms where f(1) ≠ 1 ?
karmic moat
ugh
hollow mica
that's actually messed up wtf
karmic moat
always the functional analysists gotta ruin everything
rocky cloak
but I guess I assumed that `f(1_R) ≠ 0_S`, because we cant have `0_S = 1_S` (fix...
Consider the ring of 2x2 matrices and $f\colon R \to M$ by $f(r) = \begin{pmatrix} r&0\0&0\end{pmatrix}$
cloud walrusBOT
jagr2808
lusty marlin
I'm looking through Dummit Foote and it's so infuriating; they define rings as n...
What is infuriating about this definition?
hollow mica
What is infuriating about this definition?
I don't think it's inherently worse, I just don't like having to juggle around different definitinos for the same thing in my head
hollow mica
**jagr2808**
here f(1) = 1 right?
lusty marlin
I don't think it's inherently worse, I just don't like having to juggle around d...
Ah
open sluice
hmm
f(x) = f(1)f(x) for all x in R
you want to say that since f(1) satisfies this for all f(x) in S, it must be the identity
maybe if f is surjective
but what if f is not
rocky cloak
here f(1) = 1 right?
no
hollow mica
oh I thought your ring was just matrices of the form [r 0 ; 0 0]
rocky cloak
no all matrices
hollow mica
Ahh
open sluice
ok it looks like this is where you need surjectivity
hollow mica
so surjectivity is the killer then
yea
(DF)
I would think that not requiring ring homomorphisms to send 1 to 1 would significantly change a presentation of ring theory
rocky cloak
It does yes
hollow mica
but looking at the table of contents, DF proves the same stuff?
south patrol
I'm looking through Dummit Foote and it's so infuriating; they define rings as n...
Ye this made me avoid lol
rocky cloak
However for any nonunital ring R, there is a unital ring ZxR such that the representation theory of ZxR coincides with that of R.
More specifically the forgetful functor from unital rings to nonunital has an adjoint
So the general theory is still similar
hollow mica
Ok I'll take your word for it then
rocky cloak
So it's pretty intuitive for a ring R you make the new ring consist of elements of the form n+r where n is an integer and r is in R. With addition and multiplication defined in the obvious way.
Then a unital ring homomorphism, must map n to n*1 and can map r as any nonunital ring homomorphism
rocky cloak
Yeah, so if R is nonunital and S is unital. Then there is a bijection between nonunital ring maps R to S and unital ring maps ZxR to S. This is what it means for two functors to be adjoint.
barren sierra
quick sanity check, if two subgroups are conjugate their normalizers are equivalent right?
or at least isomorphic?
ivory trail
quick sanity check, if two subgroups are conjugate their normalizers are equival...
yeah you just think of the conjugation as an inner automorphism
rocky cloak
So the normalizers would also be conjugate
barren sierra
ok cool that's what I thought
untold turret
suppose an ideal A is not a subset of an ideal B, and neither is B a subset of A. Do we have A/B = (1)?
elder wave
I'm not sure what you're asking
wraith cargo
suppose an ideal A is not a subset of an ideal B, and neither is B a subset of A...
ur quotient is ill defined
untold turret
or, is the image of A under the natural quotient map to R/B equal to the entire quotient ring?
untold turret
hm.. right, thank you
wraith cargo
I'm curious what the condition needed for that to hold is
Since for R = Z, A=(5) and B = (4) it holds
ig an element of A has to be a unit mod B
untold turret
i want to say (4 mod 6) is not an ideal of Z_6 because 2 mod 6 * 4 mod 6 = 2 mod 6, and we'd want to say 2 mod 6 is not in (4 mod 6). but in what sense is 2 mod 6 not in (4 mod 6)? since if (4 mod 6) is the additive group generated by 4 mod 6, clearly it contains 2 mod 6 = 4 mod 6 + 4 mod 6
coral spindle
I'm going to write just 2 instead of 2 mod 6 for the element of Z/6, because it is a schlep to do so otherwise.
(4) is most certainly an ideal of Z/6. I don't know why you think otherwise
It is - by definition! – the ideal of Z/6 generated by 4
Now note that 4 + 4 = 2 in Z/6
formal ermine
bowte beat me to it
coral spindle
So indeed 2 is in (4)
I think there is a severe lack of understanding going on here.
tribal moss
That sounds very dramatic ...
untold turret
failing sucks, but at least you can fail dramatically
vague granite
Confusingly enough in our linear algebra course they didn't insist on ring homomorphisms fixing 1, but in our course in rings and modules they did, and every course subsequently after that
I think it has something to do with the fact that not all authors demand rings to have a multiplicative unit
hollow mica
Yeah homomorphism should be a structure preserving map, so if your ring has not 1 then certainly you don’t need to preserve it
formal ermine
non unitary rings aren't real they can't hurt you
vague granite
whats the deal with rngs anyway
hollow mica
idk ask dummit and foote they seem to really like them
vague granite
I guess they have a nice definition
open sluice
@dummitandfoote what are your thoughts
vague granite
(R,+,*,0) such that (R,+) is an abelian group with identity 0 and every g_r:s\mapsto rs, h_r:s\mapsto sr is a homomorphism of (R,+)
coral spindle
Yeah we know the definition lmao
vague granite
Yeah we know the definition lmao
coral spindle
Idk really why people talk about rngs. Having an identity is very nice
hollow mica
Consider the ring of 2x2 matrices and $f\colon R \to M$ by $f(r) = \begin{pmatri...
I just realized I made the exact same error a while back when I thought that functors necessarily map identity morphisms to identity morphisms
tribal moss
Non-unital algebras arise in all kinds of places, and they naturally become rngs if we forget the scalar multiplication.
formal ermine
Non-unital _algebras_ arise in all kinds of places, and they naturally become rn...
spotted the phyisisicts
vague granite
I guess 1 just acts as the identity homomorhpism then
but you might want to exclude such a map
chilly radish
I just realized I made the exact same error a while back when I thought that fun...
but they do
i've never seen someone define a functor without saying that it preserves the identity
vague granite
From what I understand you can always give a ring a unit so it doesn't matter what you specify
hollow mica
but they do
no like I thought that preserving compositions alone implied identity morphisms were preserved too
chilly radish
oh
hasty sinew
Mapping identities to identities is part of the definition of functor
chilly radish
ah yea, that is not correct
rocky cloak
I'm curious what the condition needed for that to hold is
Since for R = Z, A=(5)...
The condition is that A+B = R. For R=Z we have (m) + (n) = (gcd(m, n)) so this happens when m and n are relatively prime.
vague granite
what's your favourite ring (you can't say F_1)
hollow mica
ah yea, that is not correct
thought it does hold if the functor is surjective on hom-sets
untold turret
The condition is that A+B = R. For R=Z we have (m) + (n) = (gcd(m, n)) so this h...
this seems like some sort of bezout for ideals. do we have A+B = R for proper ideals A,B iff A,B maximal?
rocky cloak
what's your favourite ring (you can't say F_1)
Upper triangular matrices
tribal moss
For example, you can have a not-a-functor from Grp->Grp that maps every group to itself and every homomorphism to the zero map. That preserves composition, but doesn't preserve identities.
hasty sinew
what's your favourite ring (you can't say F_1)
The ring of formal power series on the reals
hollow mica
thought it does hold if the functor is surjective on hom-sets
similar to how a surjective function between rings that preserves addition and multiplication necessarily fixes 1 (and is thus a homomorphism of unital rings)
coral spindle
this seems like some sort of bezout for ideals. do we have A+B = R for proper id...
No, I mean in the first place we could have A = B maximal. In general the sum of maximal ideals need not be the whole ring: take for example the maximal ideals (2) x (3) and (3) x (3) of Z x Z.
vague granite
Upper triangular matrices
Could you embed every ring into that?
What kind of a ring is $M(\omega, R)$.
cloud walrusBOT
philka.
coral spindle
No, I mean in the first place we could have A = B maximal. In general the sum of...
Wow this was way wrong! My bad.
Those aren't maximal, but they are prime.
rocky cloak
this seems like some sort of bezout for ideals. do we have A+B = R for proper id...
If A and B are maximal and distinct, then yes. The condition A+B=R is sometimes called comaximal. But also coprime or relatively prime
untold turret
thanks!
tribal moss
What kind of a ring is $M(\omega, R)$.
Probably that would mean the endomorphism ring of an omega-dimensional vector space?
untold turret
No, I mean in the first place we could have A = B maximal. In general the sum of...
thanks for the example
hollow mica
can anyone give a tldr of why the “field with one element” has a gigantic wikipedia page
coral spindle
It was wrong!
lean imp
Hi
hollow mica
like what is there to study lmao
coral spindle
can anyone give a tldr of why the “field with one element” has a gigantic wikipe...
Because it's not a field, nor does it have one element
It's not a ring at all
Keep reading the article
rocky cloak
Could you embed every ring into that?
For any ring R, consider the upper triangular matrix ring with coefficients in R. Boom boom
vague granite
For any ring R, consider the upper triangular matrix ring with coefficients in R...
🙀
tribal moss
F_un has very good publicity for something that doesn't exist.
hollow mica
I fell for it
lean imp
Who among you guys are mathematicians?
vague granite
For any ring R, consider the upper triangular matrix ring with coefficients in R...
Sorry meant to say integer matricies
vague granite
Who among you guys are mathematicians?
Not me pshsh 👀 💦
rocky cloak
Sorry meant to say integer matricies
Then I fairly certain you cannot
coral spindle
I mean literally any uncountable ring obviously cannot be
hasty sinew
If you allow infinite-dimensional matrices, perhaps. If not, certainly not
vague granite
Then I fairly certain you cannot
Yeah I see how that messes with functional rings, but I'm sure there's a eutopia where it works
formal ermine
what's your favourite ring (you can't say F_1)
Q(zeta_p)
hasty sinew
Cyclotomic fields are indeed nice
vague granite
Consider instead $M_\omega(\mathbb{Z}^{\cdots^\mathbb{N}})$
formal ermine
Cyclotomic fields are indeed nice
I find regularity in primes fascinating
cloud walrusBOT
philka.
vague granite
could you then do it?
coral spindle
No; in the first place there are no torsion elements in this group
hasty sinew
I mean, even if you can, I doubt it would be a particularly interesting way to look at rings
coral spindle
So clearly rings such as Z/2 cannot be realised
vague granite
No; in the first place there are no torsion elements in this group
Fine pick a subring of $R\leq M_\omega(\mathbb{Z}^{\cdots^\mathbb{N}})$.
cloud walrusBOT
philka.
vague granite
$M_\omega(M)$ where M is a Z module
chilly radish
if you require your embeddings to preserve 1 this is never going to work
since there's rings of different characteristics
cloud walrusBOT
philka.
vague granite
mwahaha
coral spindle
That doesn't make sense. What's the multiplication?
Anyway, conversely, every ring is a quotient of a polynomial ring Z[x, ...] for some (possibly very large) number of indeterminates
Well, commutative.
If we want non-commutative ones, we need Z<x, ...>
where the indeterminates do not commute
rocky cloak
Every ring is a subring of End(M) where M is some Z-module. More specifically if M = the ring itself
coral spindle
Since rings can be arbitrarily large, we cannot find one single such Z<x, ...> for which every ring is a quotient
vague granite
This might be a stupid question but isn't every R module with R is a PID a ring??? By the structure theorem
Or is that wrong
coral spindle
Who's fighting?
coral spindle
This might be a stupid question but isn't every R module with R is a PID a ring?...
Yes, but not in a canonical way. You can define different ring structures on the same Z-module for example.
open sluice
no he's asking who so he knows who to bet on
coral spindle
You must specify the multiplication
rocky cloak
This might be a stupid question but isn't every R module with R is a PID a ring?...
Structure theorem only applies to finitely generated modules
vague granite
Yes, but not in a canonical way. You can define different ring structures on the...
So you can have different multiplication functions on the same PID module
coral spindle
Yes
vague granite
Right yes
Oh I wonder if people have studied the group of multiplication maps
Sounds interesting
next obsidian
Every ring is a module over Z, so you can see this in the extreme by looking at non-isomorphic rings with the same underlying abelian group
rocky cloak
Oh I wonder if people have studied the group of multiplication maps
The unit group?
coral spindle
Yeah idk what that means either
vague granite
The unit group?
As in ${\pi: M\times M\rightarrow M \text{and }(M,+,\pi)\text{ is a ring}}$
rocky cloak
As in $\{\pi: M\times M\rightarrow M \text{and }(M,+,\pi)\text{ is a ring}\}$
And how is that a group?
cloud walrusBOT
philka.
coral spindle
How is it even a ring?
Oh right you're saying thtat (M, +, pi) is a ring
not the whole thing
vague granite
yes
coral spindle
Now I want to know too how it is a group
vague granite
I'm sure there's a natural way to turn it into a group
coral spindle
I don't think so
vague granite
what about $\pi\cdot\tau:=\pi+\tau$
cloud walrusBOT
philka.
vague granite
there's certainly one way to make it into a group
vague granite
does associativity not follow?
coral spindle
That's what I'm asking
vague granite
I don't see why not
coral spindle
That's not typically good enough for a proof
vague granite
At the very least you can study ${\pi}$ as an object with a group action acted on by $\text{Sym}(M)$
cloud walrusBOT
philka.
coral spindle
It's certainly not going to typically be a unital ring, at least. You can check it by just seeing what the definition gives you for the product in Z with itself
coral spindle
Furthermore, you will have to include the map pi(m,n) = 0
vague granite
but still again it may have underlying geometry
coral spindle
and that certainly doesn't work as a multiplicaiton
hasty sinew
The trivial ring is terminal, getting rid of it is probably a bad idea
coral spindle
Just a heads-up: when people say F_1, they don't mean the trivial ring.
They don't mean a ring at all.
I know it is sometimes called 'the field with one element'. Don't be fooled!
hollow mica
I got excited
coral spindle
It doesn't exist
vague granite
Well I say it does
rocky cloak
I know it is sometimes called 'the field with one element'. Don't be fooled!
The polynomial ring is a PID, Z is a PID, hence Z is the polynomial ring over some field. Which field is it? I dunno, let's call it F1
coral spindle
That's not what people mean when they talk about the so-called "field with one element"
It's a very weird idea in algebraic geometry. There is no such field F_1
vague granite
$F=({0}, {(0,0)},{(0,0)},0,0)$
cloud walrusBOT
philka.
coral spindle
That is the trivial ring, correct
hasty sinew
Doesn't satisfy that 0 =/= 1, and doesn't have nice field-like properties besides
coral spindle
when people say F_1 they are talking about something completely different
We were just talking about this earlier
hollow mica
Ah so fields are required to be nonzero
rocky cloak
Ah so fields are required to be nonzero
The definition usually includes 0 \neq 1, yes
hollow mica
ideals and modules aren’t the same thing right (like how abelian groups and Z-modules can be thought of). Sure, every ideal I of a ring R has a natural R-module structure, but an R-module in general has no corresponding ideal?
coral spindle
Every ideal is a module, but not every module is an ideal, yes.
In the first place there are only so many ideals of a ring, but there are a proper class of modules.
And not just in general – there is a proper class of isomorphism classes of modules.
next obsidian
Every ideal is a module, but not every module is an ideal, yes.
There’s a construction which basically makes any module an ideal
next obsidian
Form a ring A•M with underlying abelian group A (+) M, and multiplication
(a,m)•(b,n) = (ab,an + bm)
coral spindle
Clever
next obsidian
This makes 0 (+) M into a square zero ideal, and you can verify that almost all module theoretic properties of M as an A-module are the same as 0 (+) M as an ideal
I can’t for the life of me write down a formal way in which they’re “the same”
coral spindle
I mean once we restrict scalars back down to A they're the same, right?
next obsidian
But this construction is used in eg Nagata local rings
Uh, basically but then M isn’t an ideal anhmore
The point I think is that
warm wyvern
coral spindle
Well yeah but this is where it's a module
next obsidian
(a,m)•(0,n) = (0,an)
hasty sinew
Nilsquare extensions are very cool
next obsidian
Anyway this is also used for deformation theory as the trivial extension
Anyway, I agree with what you said, but I think this construction is cool so I mentioned it
In nagata’s book he calls it the idealization
Or maybe idealisation I forgor
coral spindle
A and (A,M) ought to have equivalent module categories right?
next obsidian
That’s my thought, maybe?
Or maybe you look at a subcategory of A•M-mod
Probably the latter
Or wait no
I don’t think so
It’s weird
Because it’s like, we’re associating the module M to A•M
So it’s like…
A•- as a functor or something
Idk
I didn’t work out anything formal
coral spindle
I'm gonna look at the hom-sets
hollow mica
Form a ring A•M with underlying abelian group A (+) M, and multiplication
(a,m)...
who is A
next obsidian
The ring
I do commutative algebra where the French have a large influence so A is often used for rings coming from the French “Anneau”
Which means ring in French
hollow mica
Oh I thought you were giving your ring the name A•M
next obsidian
Yes that is the name
It comes from an A-module named M
And you mash the two together
hollow mica
Ohh
tribal moss
But with the M part implicitly scaled by a nilpotent factor so you don't need to reveal you don't know how to multiply m's together? Clever.
white oxide
how does mapping alpha onto a different zero of irr(alpha, F) imply what they wrote?
coral spindle
A and (A,M) ought to have equivalent module categories right?
Yeah idk. The functors I thought would work don't. Oh well.
Maybe I'm just being naive about how I'm going about this
gritty sparrow
how does mapping alpha onto a different zero of irr(alpha, F) imply what they wr...
if a is in the LHS but not the RHS, the above argument gives a contradiction
gritty sparrow
Thank you, your kind words are always appreciated
next obsidian
It is not
next obsidian
gritty sparrow
not appreciated???
not an alt
coral spindle
I know I was being silly
gritty sparrow
I know I was being silly
well, you were close. This is a @hidden haven alt
coral spindle
so true
white oxide
if a is in the LHS but not the RHS, the above argument gives a contradiction
oh it's because the conjugate of irr(a, F) must be in E right
gritty sparrow
oh it's because the conjugate of irr(a, F) must be in E right
well, in my mid the argument went as follows: By the above argument, there exists an automorphism of K fixing E but not fixing a. But this contradicts the fact that a is in the fixed field of all such automorphisms.
white oxide
well, in my mid the argument went as follows: By the above argument, there exist...
ah that makes a lot more sense thanks
formal ermine
So the entries of column i are the entries in vector v_i
ah ok that makes more sense
thanks
indigo ridge
dont know if this is the right channel, but was wondering what do they mean by 'face' here is it reffering to the face of the entire Rubiks cube or a cubie?
celest furnace
dont know if this is the right channel, but was wondering what do they mean by '...
Face as in like there are 6 faces to a cube
Think like faces of a square
indigo ridge
really? isn't that completely useless though just flipping the face of the entire rubiks cube
celest furnace
Well no cuz you can twist one side
If you know anything about rubics cube lingo this is called the F or F' move
rocky cloak
dont know if this is the right channel, but was wondering what do they mean by '...
Have you ever held a Rubik's cube?
indigo ridge
Have you ever held a Rubik's cube?
yeah
rocky cloak
And have you tried to twist any of the faces?
rocky cloak
Yes...
indigo ridge
maybe Im confused on what you're considering a face here
rocky cloak
Wait, so what is your question?
indigo ridge
a face is the entire square on the cube
a side is only a third of the face
and you twist the sides to get new confiugrations of the rubiks cube
rocky cloak
So when you turn the face of a Rubik's cube you move nine of the cubies (or you just move 8 I suppose)
indigo ridge
you don't turn the entire face
rocky cloak
This is the way you scramble the cube
chilly ocean
Hello. In 10 months I have the 12th grade math olympiad and I started learning earlier. The subjects are abstract algebra and real analysis 2. Real analysis is very fun and I like it the way I liked it until now. But Abstract Algebra looks a bit different so far. I am at the beginning (I started one month ago) and now I am at the applications of Lagrange/Cauchy theorems (+center of a group, centralizer, normalizer) in contest problems. It's getting harder and harder and I don't know how to learn more efficiently. Can someone give me some advice? Where can I find problems, explanations, techniques, ideas?
hollow mica
I'm so lost, Hungerford defines rings with unity but then ring homomorphisms don't need to preserve it?
Like bro that's a rng homomorphism
worn wolf
i think you actually cant deduce homomorphism sends unit to unit if you dont a priori assume it right
hollow mica
Yeah we had a long convo about this earlier today
untold turret
what do i want in order to have: prime ideal that is not maximal is principal
i.e. what's a sufficient condition on a ring so that it has this property
warm ember
i dont understand why the corollary is true
tender wharf
it's not omitted tho?
tender wharf
i dont understand why the corollary is true
if p divides either a or b they're 0
because p is 0 in this ring
oh
sorry corollary
just notice that everything except 0 is a unit
immediately follows
warm ember
just notice that everything except 0 is a unit
huh
tender wharf
hmm okay
let's try a more general method
finite integral domains are fields
let a be a nonzero element in a finite integral domain
warm ember
ye
tender wharf
I think this is what quasi-semi-group meant actually
anyway
so consider the sequence formed by a^n
so a, a^2, a^3 ... etc
since it's finite there must be some integers i, j such that i > j but a^i = a^j
tender wharf
(of course there's the trivial case where a = 1)
warm ember
since it's finite there must be some integers i, j such that i > j but a^i = a^j
wait what does this have to do with an integral domain tho
tender wharf
it has to do with being finite
and integral domains have cancellation
so we notice that
a^(i-j) = 1
untold turret
is any nonprime ideal contained in finitely many principal prime ideals, even in non UFDs? (we need finitely generated ideals here, no?)
tender wharf
I can't figure out why the frobenius map is an automorphism on a field
next obsidian
is any nonprime ideal contained in finitely many principal prime ideals, even in...
No, take k[x1,…] and the maximal ideal (x1,…) should not satisfy this
next obsidian
I can't figure out why the frobenius map is an automorphism on a field
It isn’t always
Such a field is called perfect
untold turret
No, take k[x1,…] and the maximal ideal (x1,…) should not satisfy this
do we have it for noetherian rings?
thanks for the answer btw
next obsidian
For finite fields it follows because it’s injective and then pigeonhole principle says injective<=>surjective<=>bijective
Do you want it contained in the Union?
Or in the ideal generated by those primes
untold turret
in their intersection
next obsidian
That doesn’t really make sense
Why say finitely many if you’re taking intersection?
Adding more ideals makes it harder to be contained inside of it
untold turret
well, why? this isn't obvious if there are non unique irreducible factorizations
next obsidian
And the answer is no, any height > 1 ideal cannot be contained in a principal prime because then it wouldn’t be height > 1
Since principal primes are either height 0 or 1 by the Hauptidealsatz in a Noetherian ring
tender wharf
It isn’t always
oh sorry I meant in rings of char p
next obsidian
well, why? this isn't obvious if there are non unique irreducible factorizations
Okay but what I’m saying is it’s equivalent to ask if it’s in any principal prime
next obsidian
oh sorry I meant in rings of char p
Frobenius is only a homomorphism in char p, my answer is the same, unchanged
tender wharf
is it not an automorphism if the field has char p?
next obsidian
It isn’t necessarily
untold turret
And the answer is no, any height > 1 ideal cannot be contained in a principal pr...
hm right
thanks
next obsidian
take F_p(x)
Asking it to be an auto morphism is asking for there to be p-th roots
And x has no p-th root
untold turret
Okay but what I’m saying is it’s equivalent to ask if it’s in *any* principal pr...
thanks for this, i realized that i'm instead looking for conditions for an ideal to be able to, instead of actually being, in an intersection of infinite principal primes
next obsidian
But isn’t that still just the same as asking it to be in a single prime? Unless you mean that there exists an infinite list of principal primes, not the same, with x in their intersection
If so, I think yorue basically asking if you can have an infinite “irreducible” decomposition
untold turret
yeah, existence of a choice of infinite primes
next obsidian
And this is possible yes
Not in a UFD as you pointed out
And I can’t produce an example immediately, but you can do it
untold turret
i'll think about it thanks
next obsidian
You can do it in a non-Noetherian ring easily I think
By taking an infinite polynomial ring and modding out by relations which make you divisible by all xi I think
In a non-Noetherian ring you can also be divisible by infinitely many powers of x
As in,
\cap_1^infty (x^n) is non-zero
It can’t happen in a domain though (IIRC)
untold turret
You can do it in a non-Noetherian ring easily I think
of interest: noetherian implies finitely many minimal primes
Krull's principal ideal theorem: if I is principal, in noetherian ring, minimal primes over I have height 1
oh that's what you referred to by Hauptidealsatz
stuck fiber
So I haven't touched fields outside of linear algebra and analysis but I was reading about commutative rings and I was wondering is it true that, R is a field iff every nonzero element of R is a unit (where R is a commutative ring)
lethal dune
yeah that's the definition
stuck fiber
Oh right duh this is just multiplicative inverse exist
tender wharf
good to note its an integral domain as well
mighty kiln
Is this result true for arbitrary vector spaces? Given a linearly independent set L and a basis β disjoint from L, there exists S ⊂ β with |S| = |L| s.t. β ∪ L \ S is a basis.
stuck fiber
It's also cool that all the units form a group
ripe gulch
When can I learn this
No, what would I have to learn/understand to start abstract algebra
coral spindle
You would read a book on abstract algebra
rocky cloak
No, what would I have to learn/understand to start abstract algebra
There arent really any strictly necessary prerequisits to learning abstract algebra, but it is useful to have some experience with proofs.
taking a proof based linear algebra class for example is often a reccomended prerequisite
summer path
taking a proof based linear algebra class for example is often a reccomended pre...
This while not necessary, can be very useful for intuition as well as giving some easy examples
coral spindle
Well when you say 'is' you mean 'is isomorphic to'
So let's be clear here, they're not literally the same
void cosmos
yea
coral spindle
But the approach here is to find a map. How can you convert an element f : M (x) N → P to a function g : M → (N → P), to abuse notation a bit
If you can do that in the most natural-looking way, you've probably got the isomorphism.
From that point on it's not hard to show it is indeed an isomorphism
void cosmos
my thinking was the map f is probably factored out by the universal property of the tensor product and maybe if i can map the map f to one of the factors or somethhing
but i thought this wouldnt work
ig the other way around is easier first
like sending the map g to something in the tensor
or idk im confused
coral spindle
Sure, either way works. Just write down what f is defined to be
Given g, write down the definition of f
frigid lark
Say y is the root of X^(p^2) - a, an irreducible polynomial in K[x], where char K is not p, and p is odd (idk if that helps). Suppose K has p'th roots of unity but not p^2 roots of unity, is there any way to show that if the extension K(y, w) / K has exponent p, then [K(y, w) : K] > p^2, where w is a primitive p^2 root of unity?
void cosmos
okay i think i got it
or wait
f(m tensor n) --> g where g(m) is f( m tensor n ) for n fixed?
like g sends an element in M , m to f(m tensor n )
and this f now since it has its m is a function of n
coral spindle
You can write this much more simply as g(m)(n) = f(n (x) m).
Now you have to prove that g(m) is in Hom(N, P) and that g is in Hom(M, Hom(N,P))
Remember it's not enough for these to merely be functions – they have to be module homomorphisms too.
void cosmos
this n is fixed right?
void cosmos
You can write this much more simply as g(m)(n) = f(n (x) m).
here
coral spindle
No, I'm defining the function g : M → Hom(N, P)
so for some m, we know that g(m) is a function N → P
So it makes sense to write g(m)(n) for some n in N
Is that clear?
void cosmos
yea
so f in Hom(M tensor N , P) gets send to the map g in Hom(M,Hom(N,P))
where f(m tensor n) gets sent to g(m)_n
where g(m)_n is f(m tensor n) ( which is in P )
is that correct
coral spindle
what does this notation g(m)_n mean, I don't understand it
Furthermore why are we sending f(m tensor n) to anything – that's a value in P and we're not considering any functions with P as a domain
We are sending the function f to the function g, where we defined g as above
void cosmos
yea i just wrote it as f(m tensor n) to like show whats m and n
but yea i see what ur saying
void cosmos
where g(m)_n is f(m tensor n) ( which is in P )
this is g(m)_n
coral spindle
That hasn't really helped me understand what you mean by that
Do you mean g(m)(n) really?
void cosmos
yea but i just dont like this notation
coral spindle
The notation g(m)_n is most certainly worse
If you really don't want to write it, you could say that g(m) = h, where h : N → P is defined as h(n) = f(m (x) n)
void cosmos
yea thats better for me
now i want an inverse map h: Hom(M,Hom(N,P)) --> Hom(M tensor N , P )
and then prove all of those are homomorphisms and inverse to each other
maybe sending the element in LHS , call it f
or like
like the image would be
it sends m tensor n to f(m)(n)
so like the map it gent sent to , call it k or whatever
is defined as it takes m tensor n to P by f(m)(n)
or better by f(m) = g , where g:N-->P and then it inputs that n into the g
yea thats so much better
is that correct
coral spindle
You need to check well-definedness when specifying maps out of the tensor product
void cosmos
isnt f well defined
coral spindle
Not a priori, no
For example the map f : Z (x) Z → Z ‘defined’ by f(a (x) b) = a + b is not well-defined
Look back and remind yourself of what it means for a map from the tensor product to be well-defined.
void cosmos
ig the tensor product is just a quotient?
so i would have two pick two equivalent elements a tensor b and a' tensor b'
and show that they map to the same thing right?
void cosmos
yeah
coral spindle
An arbitrary element of the tensor product is a sum of those
coral spindle
Remind yourself of the universal property of the tensor product.
This should be in whatever source you are reading.
void cosmos
would it be to find a bilinear map from M x N to P so that it induces the map f?
void cosmos
is defined as it takes m tensor n to P by f(m)(n)
or this i mean*
coral spindle
would it be to find a bilinear map from M x N to P so that it induces the map f?
Yes
Tattoo this on your forehead: a map from the tensor product is well-defined iff it is bilinear
void cosmos
Tattoo this on your forehead: a map from the tensor product is well-defined iff ...
nah man it wont look good
coral spindle
Face tats are all the rage rn you'll be so cool
formal ermine
Tattoo this on your forehead: a map from the tensor product is well-defined iff ...
as in M tensor N -> any module?
void cosmos
yea but not on tensor products
void cosmos
Tattoo this on your forehead: a map from the tensor product is well-defined iff ...
yea so u mean that i get the well definitness for free cuz the map comes out from the bilinear map from M x N to P?
by the uiniversal property?
coral spindle
That is the universal property
rocky cloak
Say y is the root of X^(p^2) - a, an irreducible polynomial in K[x], where char ...
What do you mean by the extension having exponent p?
frigid lark
The Galois group has exponent p
void cosmos
That is the universal property
yea
and f is already an R-module homomorphism correct
coral spindle
That is the assumption, yes
void cosmos
so its bilinear 😄
but from the product
not from the tensor
and then by commutativity of the diagram it composes with them pa g(a,b) = a tensor b
so we get the map we are after
is that correct
coral spindle
so its bilinear 😄
Prove it.
It is far from obvious that the map (m, n) |-> f(m)(n) is bilinear.
coral spindle
Well what you've written doesn't make sense as-is
But it's a start
Write down the properties you have to prove for bilinearity and prove them.
coral spindle
Well what you've written doesn't make sense as-is
To be clear, it should be (f(m1)+f(m2))(n)
void cosmos
yea forgot that
and i also need linear in the second argument as well
and also af(m)(n) = f(am)(n)=f(m)(an)
i dont see why this is far from obvious
i am 100% missing something
coral spindle
and i also need linear in the second argument as well
You haven't proved linearity in the first argument yet
Write down the proofs!
If you say it's obvious and you can't prove it, then I sincerely doubt it's obvious.
void cosmos
yea it is cuz like f(m1+m2)(n) = (f(m1)+f(m2))n ?
then u get f(m1)(n) + f(m2)(n) ?
coral spindle
Yes, by the definition of addition of functions
coral spindle
That's additivity in the first argument, yes
void cosmos
😄 why is this far from obvious
it took me 5 mins to just make sure
im not missing something
haha
cuz i was like how
what did he mean
coral spindle
Does it normally take you five minutes to see something 'obvious'?
void cosmos
no it took me 5 mins after u said it isnt
coral spindle
actually, it's been 13, so idk.
frigid lark
Your doing better than me
open sluice
tensor moment
frigid lark
It took me 3 days to do a counting argument
void cosmos
cool
tysm flat
Now i still have to do that theh sre homomorphims
And inverse to each other
rocky cloak
Hmm, so k(y, w):k = p^2 would mean k(y) = k(y, w). We have G(k(y)/k(w)) is a normal cyclic subgroup, so the only possibility for the Galois group is CpxCp or Cp^2. You want to show that it's not CpxCp.
Consider the automorphism defined by mapping y to yw. This must map w to w^s where s=kp+1 for some k.
Iterating this mapping p times gives y maps to yw^t where t = 1 + s + s^2 + ... s^p-1 = (s^p - 1) / (s - 1)
Calculating s^p -1 we get 1 + pkp + O(p^3) - 1
So t is congruent to p modulo p^2 hence the map y -> yw does not have order p.
Also I jumped over the case k=0 here but you can check that seperately.
@frigid lark
frigid lark
thx
untold turret
consider the ideal (18, 22) in Z. are (2) and (6, 22) the only proper ideals that contain it?
south patrol
Well every ideal of Z is of the form (a) for some a
You can check that (18,22) = (2)
and also (6,22) = (2)
untold turret
oh, that makes a lot of sense, thank you
south patrol
Well every ideal of Z is of the form (a) for some a
i'd also recommend you look at this
proof isn't bad though
rocky cloak
thx
Also I think its just never the case that the group has exponent p. The remaining option would be that k(y,w):k(w) = p^2, and then G(k(y,w)/k(w)) would be cyclic, which contradicts the group having exponent p.
rocky cloak
I wouldnt say that
frigid lark
I knew that was the case, I just needed the k(y,w):k(w) = p case done for a question
And I killed a day trying to get it
But thanks for the help
untold turret
k is a field. in k[x,y], the ideal (x^2, y^2) is contained in (x,y^2), (x^2,y), and (x,y). are there other ideals containing (x^2,y^2)?
rocky cloak
k is a field. in k[x,y], the ideal (x^2, y^2) is contained in (x,y^2), (x^2,y), ...
(x + ay, y^2 ) for a nonzero also contains it
lusty marlin
k is a field. in k[x,y], the ideal (x^2, y^2) is contained in (x,y^2), (x^2,y), ...
(x², y², xy)
untold turret
thanks for the answers!
median pawn
a hint would be nice. smells of eisenstein
pf(X) = p^n X^n + p^2 X + p but eisenstein's criterion doesn't seem directly applicable
rocky cloak
A hint could be that f(x) is irreducible if f(ax + b) is irreducible.
median pawn
ah, right, putting pX = t seems to do it then
coral spindle
Hint: there is a useful theorem you know that assumes that gcd(k, n) = 1
solemn dew
Do you mean that gcd(k,n) = 1 => kx+ny = 1 where x,y is an integer?
coral spindle
That's Bézout's lemma, correct. It is what you want to use here.
untold turret
if normal subgroups are analogous to ring ideals, is there also analogy between prime/maximal ideals and a certain class of normal subgroups?
formal ermine
if normal subgroups are analogous to ring ideals, is there also analogy between ...
the only thing I can think of would be maximal normal subgroups
untold turret
the only thing I can think of would be maximal normal subgroups
do we have that an ideal is maximal if and only if it is also a maximal normal subgroup of the ring's abelian group?
brittle hare
I don't believe the converse is true
untold turret
hm that would make sense
brittle hare
but I can't think of a counterexample immediately
coral spindle
do we have that an ideal is maximal if and only if it is also a maximal normal s...
No. For example the ideal $0 \oplus \bR$ of the ring $\bR \times \bR$ is maximal as an ideal, but it is far from maximal as a subgroup.
Note that all subgroups of an Abelian group are normal, so it is redundant to specify normality here
cloud walrusBOT
boytjie
untold turret
No. For example the ideal $0 \oplus \bR$ of the ring $\bR \times \bR$ is maxima...
thanks for the example
static yew
Okay I am continuing reading my book
And it days: for polynomials in F[x], if an irreducible poly p divides fg, it divides f or g (or both)
Doesn't that imply that all nonzero elements of F_p^k (mod some irreducible poly) multiply to nonzero elements?
Having trouble distinguishing from x^2 + x + 1 in F_2[x] and x^2 + x + 1 in F_8
rocky cloak
the only thing I can think of would be maximal normal subgroups
Worth noting that not every group has a maximal normal subgroup. For example Q under addition.
coral spindle
Okay I am continuing reading my book
And it days: for polynomials in F[x], if a...
Yeah that's right, modding out by (the ideal generated by) an irreducible polynomial gives you a domain.
static yew
Isn't the normal subgroup the kids in my class that weren't weird (and were closed under +)
static yew
Yeah that's right, modding out by (the ideal generated by) an irreducible polyno...
Haven't encountered "domain" yet outside the set of inputs for which a function is defined
coral spindle
A domain is a ring which has exactly the property you described: the product of nonzero elements remains nonzero.
static yew
Wait that sounds a lot like a field
Except I guess in a domain you can have ab=ac for a,b,c nonzero and b!=c
coral spindle
Z is a domain.
And no, the property you list there is equivalent to being a domain.
static yew
Ahhh so they multiply to unique values and in theory could have multiplicative inverses but those inverses aren't part of the set?
static yew
Which means all non field domains would have to be infinite
coral spindle
If you have a commutative domain R, there is a way to construct a field Frac(R) which contains R
coral spindle
Which means all non field domains would have to be infinite
That's exactly right. Another way of putting this is that finite domains are fields
or well
finite commutative domains are fields. It is true that finite domains are fields but the proof is much harder than just showing the commutative case.
We needn't discuss non-commutative rings for your purposes though
static yew
I forget do we need associativity to be a ring
Yeah I dont even deal in infinite fields, I'm a software engineer
coral spindle
so you've said
open sluice
any kind of algebra without associativity is a waking nightmare
static yew
Life is a pain without commutativity too
coral spindle
Not really, we care a lot about non-commutative rings.
There is only so much to say about finite Abelian groups too.
summer path
any kind of algebra without associativity is a waking nightmare
Everyone loves non associative non commutative word problems in this channel 
coral spindle
Mfw face when (when) I can't put my socks over my shoes 
static yew
So wait
Doesn't irreducibity of a poly depend on the field(domain?) its coefficients belong to?
Irreducibility = non-factor-ability
x^2 - 1 cannot be factored in Q but it can be factored in the quadratic closure of Q
x^2 + 1 needs Q[i] at the minimum
coral spindle
Yes
static yew
Woo I actually am starting to grasp this madness
coral spindle
And not coincidentally, Q[x]/(x^2 + 1) is isomorphic to Q[i]
chilly radish
x^2-1 can be factored in Q
coral spindle
I imagine you wrote x^2 - 1 as a typo. Clearly x^2 - 1 = (x-1)(x+1)
chilly radish
Bad example
static yew
Ok not ready for that on
static yew
I imagine you wrote x^2 - 1 as a typo. Clearly x^2 - 1 = (x-1)(x+1)
A thinko but yeah
Was thinking of solutions to x^2-1=0
Err
X^2-2
The square root of 2 😛
coral spindle
Yeah. And again not coincidentally Q[x]/(x^2 - 2) is isomorphic to Q[sqrt(2)]
static yew
Typing math on a phone sucks
Discord mobile has a bug. Using a bluetooth kbd you can only type one character and then the input box loses focus
static yew
Yeah. And again not coincidentally Q[x]/(x^2 - 2) is isomorphic to Q[sqrt(2)]
Trying to picture this and failing
coral spindle
Well look at it this way
static yew
The first one makes the square of sqrt2 equal to zero
No wait it makes (sqrt2)^2 - 2 = 0
Which seems too tautological to be true
coral spindle
Let's consider the element x in Q[x]/(x^2 - 2). I'm writing x instead of x + (x^2 - 2) for ease of writing
Now x^2 - 2 is in the ideal, so x^2 - 2 = 0 in this ring
so x^2 is 2
so x is a square root of 2
static yew
Isn't x just the most basic nonzero polynomial here?
coral spindle
Idk what that means exactly. When I write Q[x] I'm referring to the ring of polynomials in x (which is just a symbol at this point) with coefficients in Q.
static yew
I don't understand ideals yet
This book has yet to cover them and the wikipedia page was too difficult to follow
static yew
Idk what that means exactly. When I write Q[x] I'm referring to the ring of poly...
Wait now I'm mixed up. At least part of my understanding is that F[x] is the field F plus a "transcendental" x that is not the root of any polynomial with coefficients in F
Which is normally gonna be a ring unless you do F[x,y] with xy=1
Like C is basically R[i]
Wait is C a field? i^2 is not unity
And Q[sqrt2]
coral spindle
F[x,y] with xy=1
This is a contradiction with what you just said above!
static yew
I know C has some weirdness going on, like it doesnt have a proper order
coral spindle
When we write F[x], we mean polynomials in the ‘unknown’ symbol x.
Similarly for F[x,y]
Now conversely, Q[sqrt 2] is suggestive notation, but it means something different.
If we have a ring $R$ which is a subring of another ring $S \supseteq R$, given some element $s \in S$ we write $R[s]$ to mean the smallest subring of $S$ containing both $s$ and the entirety of $R$.
cloud walrusBOT
boytjie
static yew
> F[x,y] with xy=1
This is a contradiction with what you just said above!
That's why I said I'm mixed up. My "knowledge" seems to be inconsistent, and since almost all[1] mathematics is consistent, obviously one of my understandings is wrong
coral spindle
I was in the middle of elaborating
static yew
Lemme process that one for a moment
coral spindle
So now my point here is that Q[sqrt 2] is the smallest subring of C, if you like, that contains both all rational numbers* and sqrt 2
(* as it happens every subfield of C contains all rational numbers, so this is a bit redundant, but oh well).
static yew
So my main misunderstanding is that F[x] could refer to either:
Univariate polynomials of x with coefficients in F
Or
F adjoin(?) Some x that basically defines a+x a-x ax and the ring that results
And with x,y people usually mean the former
And with t,i,j,k and explicit numerical expressions they mean the 2nd
static yew
(* as it happens every subfield of C contains all rational numbers, so this is a...
While this is true not all subrings include Q, such as Z
open sluice
F[x] vs F(a)
coral spindle
While this is true not all sub**rings** include Q, such as Z
Yes, this is why I corrected what I had written
coral spindle
And with t,i,j,k and explicit numerical expressions they mean the 2nd
Yes
static yew
Wait the book had that syntax earlier but I forgot what the distinction was, lemme look it up again
coral spindle
People will also typically say "the polynomial ring F[x, ...]" to denote the first.
F(...) is the smallest subfield such that blah blah blah, F[...] is the ring.
And it may also refer to F(x) the ring of so-called 'rational functions' which are fractions of polynomials
As it happens, Frac( F[x] ) is F(x) :)
static yew
F(...) is the smallest sub*field* such that blah blah blah, F[...] is the ring.
Wait does that mean C is R(i)
static yew
Gotcha. [] adds the elements
() adds the elements and their multiplicative inverses
Wait hrm
static yew
Oof
coral spindle
For example (and don't get spooked by this notation) there is a difference between the ring F[x, x^-1] and the field F(x)
The first has only ‘polynomials’ like x^-1 + x^2 + 2
But the second has things like (x+1)/(x+2)
static yew
Wait x here is a variable, right? We're creating a poly ring?
coral spindle
Well like I said don't get spooked by this notation
It's very suggestive but it doesn't fit into what we've seen before
static yew
Is it cuz F[x,x^-1] has negative exponents
coral spindle
This means something different.
Anyway I did describe what its elements look like above.
static yew
Oh you said it already
Damn this tiny screen
coral spindle
In Q[x, x^-1] for example there is no inverse of the element 1+x, but in Q(x) there is. Maybe you'd like to check that
static yew
That doesn't spook me too much, actually
Yeah that makes sense
It's like
F[x, x^-1] is like (N->Z) while F(x) is like (Z->Q)
coral spindle
Sure, if that makes it easier for you.
static yew
It does
open sluice
Wait x here is a variable, right? We're creating a poly ring?
x is called an indeterminate
it’s not a variable in the sense that we don’t assign it a numeric value at any point
you can formalize this by thinking of a polynomial x^2 + 2x + 1 as an ordered tuple (1, 2, 1)
summer path
I feel like it's easier to think about F(x) as the field of rational functions of F[x]
static yew
I've never dealt with rational functions and might never need to, so I'm trying not to get too bogged down in it
static yew
x is called an indeterminate
it’s not a variable in the sense that we don’t assi...
Except aren't polynomial functions a thing
open sluice
polynomials are expressions, polynomial functions are functions
it’s just a technicality but I think I read somewhere it’s important...
coral spindle
Except aren't polynomial functions a thing
Yes. A polynomial is very different from a polynomial function.
For example: the ring of polynomials on F_2 is infinite.
The ring of polynomial functions F_2 → F_2 is finite.
open sluice
ye because the same polynomial can mean different things in different fields
coral spindle
No it's more like polynomials can have the same output despite being described differently
open sluice
;-;
coral spindle
The ring of definition is important too, but that wasn't the point of this example
static yew
Are they the same polynomial? Or are they different polynomials written with the same sequence of characters?
open sluice
o yeah if it’s like x^3 in F_2
that takes on the same values as x^5 but it’s a different polynomial
coral spindle
No it is still very important!
round hull
maybe for a mathematician
coral spindle
OK.
static yew
o yeah if it’s like x^3 in F_2
that takes on the same values as x^5 but it’s a d...
Yeah like in F2
x+1 = x^2+1 right?
coral spindle
For an example of why we care about it, see classical algebraic geometry.
coral spindle
Yeah like in F2
x+1 = x^2+1 right?
Indeed, since x^2 and x are equal for all x in F_2 :)
static yew
maybe for a mathematician
Mathematicians? In the math server? Is nothing sacred
open sluice
blue
coral spindle
maybe for a mathematician
If we did not have the distinction, then we would not be able to construct finite fields, for example.
round hull
wdym
coral spindle
The finite fields of characteristic two are certain quotients of F_2[x]. If we identified these with polynomial functions, this ring would be finite and would only have a couple of quotients.
Understanding this difference is vital to this construction.
open sluice
ye if you take F[x]/(p(x))
the degree of p(x) impacts the number of elements in the field you end up with
static yew
Btw I tried to write a formal mathematical description of IEEE 754 floating point values in abstract algebra terms and it's a nightmare
The only thing you get is commutativity
No associativity
No closed subsets (except maybe {0})
FOUR sets of absorptive(?) elements
Two of which are subsets of the third, one subset more absorptive than the other
Many elements have no additive inverse
Almost no elements have a multiplicative inverse
round hull
can't you embed F_2 in \bar{F_2}
open sluice
Btw I tried to write a formal mathematical description of IEEE 754 floating poin...
floats aren’t real
coral spindle
And how do you construct the algebraic closure of F_2?
How would we even calculate in it?
Come on man. We need to understand the difference. It's very clear.
static yew
floats aren’t real
They aren't even Q, or even Z or N
The nonassociativity alone is a mess
upper pivot
Lmao is this like trolling, who would argue this distinction doesn’t matter?
I also like my,ring,of,polynomials over Fp to be finite
static yew
And how do you construct the algebraic closure of F_2?
summer path
I can't tell if this is intentional at this point
coral spindle
You're talking about yourself
dreamy fiber
