say F2² we have v=(1,1) then ‹v, v› = 2 =0
#groups-rings-fields
rustic crown
if it's an ordered ring, then we can probably talk about symmetric billinear inner products as > 0 means something
i don't think finite fields can be ordered 
lethal dune
rustic crown
like positive elements should at least contain all the squares, and then also be closed under addition.
lethal dune
idk never worked with anything other than R/C or A.c.f. s
rustic crown
can we talk about hermitian inner products for stuff other than C? (and its subfields)
hidden haven
Ye finite field or in general positive characteristic fields can't be ordered
chilly ocean
https://tenor.com/view/trollge-troll-the-day-of-reckoning-gif-19655212
When you "accidentally" put all the Italians in the kernel
tfw the homomorphism is not injective
sorry this memeing is basically off topic at this point my bad
untold cloud
Hi, guys, i saw it on mathstackexchange, but i don't understand it, why representation theory of a lie algebra can be viewed as this
why all possible, why one enveloping algebra is not enough?
dawn kiln
@hidden haven sorry to @, but would you mind explaining the GCD thing you were talking about yesterday wrd to this example
so would the idea be to take the gcd of 3, 5, 11, -5, 7, 9 (ie. 1) and arrange that to the upper left corner
then just subtract multiples of that to eliminate the other numbers
or wait, is that just what you do for the first row and column
as you end up with a whole diagonal of non-zero entries
hidden haven
Ye so you get the gcd in the top left corner somehow (I can't rember how but it should just be by maybe doing rowwise gcds and then taking gcd of all the rowwise gcds?
But once you have that then make all the other first row and first column entries 0 by subtracting
Then induct
dawn kiln
yh so i understand the second step
but i guess im just trying to work out which elements im GCDing
hidden haven
Ye so first put gcd of first column on top left
That is just some column operations and is easy
With some bezout
dawn kiln
just these ones?
dawn kiln
oh i see
alright, imma work through a few examples just to see the machinery in action
whats the name for the form that this gets it in?
ie this
hidden haven
Smith normal
dawn kiln
The first goal is to find invertible square matrices {\displaystyle S}S and {\displaystyle T}T such that the product {\displaystyle SAT}{\displaystyle SAT} is diagonal. This is the hardest part of the algorithm. Once diagonality is achieved, it becomes relatively easy to put the matrix into Smith normal form
was the bit from the wiki article
hidden haven
My notes just say "get gcd of all entries on in first slot (easy)" 
dawn kiln
lmfaoo
dawn kiln
> The first goal is to find invertible square matrices {\displaystyle S}S and {\...
oh so i guess thats probably just reorganising it so you have the property that each A_i,i divides the A_(i+1),(i+1)
hidden haven
Yeah
dawn kiln
i guess i might as well ask while you're here, so my end goal is to solve a system of linear diophantine equations, and i was told that computing SNF is the first step
hidden haven
😵💫 I don't know much about diophantine equations
dawn kiln
is that generally a good way to go about doing it, or is there something simpler (i only need to verify that a solution exists, rather than compute it)
dawn kiln
😵💫 I don't know much about diophantine equations
only linear ones, so not as spooky
hidden haven
much = at all 
dawn kiln
hah fair
everyone raving about integer programming but no one seems to want to solve lin dioph eqs 😔
anyway, cheers for your help man, i rly appreciate it
hidden haven
Btw do you just want the SNF or also the operations that get you there
dawn kiln
unsure yet
hidden haven
Because there's a fact that you can use to find it directly sort of
terse crystal
Don’t understand why you keep asking… it’s all written in the book…
hidden haven
The first diagonal entry is gcd of all 1x1 minors (entries)
nth entry of SNF is gcd of all det of nxn minors
oh whoops
No
The det of the top left nxn submatrix in SNF is the gcd of dets of all nxn minors
So you can figure them out in succession
You just gotta compute a lot of dets
Found old assignment with example lol
dawn kiln
The det of the top left nxn submatrix in SNF is the gcd of dets of all nxn minor...
oh thats handy
i don't know yet if i need the P and Q matrices, but ill bear this in mind
rustic minnow
If I have a commutative ring with multiplicative inverses defined for all non zero elements do I need to have a nonzero unity in order for it to be a field? For context I have a set with two binary operations defined which I have proved is a commutative ring, however I get a zero element as the identity element for the second of these binary operations.
hidden haven
0≠1 is a field axiom
rustic minnow
right, so I must have made a mistake
But what I have is :$x \otimes y = x + y +xy$, so I should get $e \otimes x \otimes y = x \otimes y \otimes e = x \otimes y$ which is satisfied when $e=0$.
cloud walrusBOT
Fredrikpiano
dawn kiln
0≠1 is a field axiom
i always thought it was derivable from the other field axioms
hidden haven
The 0 ring satisfies all fields axioms other than that
hidden haven
**Fredrikpiano**
Ye so I guess you get the 0 ring lol
rustic minnow
0 ring = null ring?
hidden haven
But if it's on Z then there must be some other axiom not satisfied
Ye ring with just 1 element
But if underlying set is larger than there's some other problem as well
Like distributivity
rustic minnow
Its on R and the other operation is $x\diamond y = x+ y+ 1 $
hidden haven
Oh then 0≠1
0 here just means additive Identity
Additive identity is different under diamond
It is -1
rustic minnow
yes, I found it to be -1, but I thought unity meant identity under the second operation i.e $\otimes$
cloud walrusBOT
Fredrikpiano
rustic minnow
So as long as the additive identity does not equal 1, I have a nonzero unity for $(\mathbb{R}, \diamond, \otimes)$ ? Im confused as well
cloud walrusBOT
Fredrikpiano
hidden haven
What do you mean by "non zero unity"
I'd assume that that means additive identity ≠ multiplicative identity
In this case, -1 and 0 respectively
rustic minnow
oh, I first thought there could be no multiplicative identity equal to zero
hidden haven
0 means nothing more than additive identity in an arbitrary ring
rustic minnow
The formulation from my textbook has the following definition: A field F is a commutative ring with nonzero unity such that the set of nonzero elements of F is a group under multiplication.
terse crystal
If 1=0 then this ring is just a singleton,too trivial
rustic minnow
You mean {0} where 1 = 0 ? (0+0 = 0 and 0 * 1 = 0*0 = 0)
terse crystal
yeah
rustic minnow
Good. I have a hard time reading 0 vs O apparently )
long obsidian
This is interesting thank you
It's clear to me that if you have a finite subgroup H<G then gHg^-1 is a subgroup of G as well. But it's not clear to me that the cardinality of this conjugate set gHg^-1 is of the same size as H. I know that cosets of H are the same size but is it sensible to just say "this is the right coset "(gH)g^-1" of the left coset gH and that's why it is of the same size?
I feel like this argument presumes the coset gH is a group
viscid pewter
clearly gHg^-1 can't have more elements than H
if it has less elements, then gh1g^-1 = gh2g^-1 for some h1, h2
but then that's only true if h1 = h2
so you'll never get less elements
so it has to have the same number of elements
chilly ocean
This is cool you bring this up cause this is exactly the argument you need in part to show that joins of normal subgroups are also normal
wooden ember
Somebody pin this: it solves all questions on homomorphisms
On morphisms in general to be fair
long obsidian
if it has less elements, then gh1g^-1 = gh2g^-1 for some h1, h2
So can I say something like $f:H\to gHg^{-1}$ via $h\mapsto ghg^{-1}$ is a bijection since it is clearly surjective. Then by an argument similar to what you just said it's injective?
cloud walrusBOT
fajitas
viscid pewter
yes
long obsidian
This is cool you bring this up cause this is exactly the argument you need in pa...
I've seen joins in hungerford but I haven't seen an application of joins. A join is essentially creating a group out of the union of two subgroups right? I mostly see intersections of subgroups pop up.
chilly ocean
yeah "join" is just the lattice-theoretic term for the subgroup generated by some two, or a number of other, subgroups
flat treeBOT
long obsidian
I am encountering a specific way in to proof that $A_n$ for $n>5$ is simple by using the fact that $A_5$ is simple. The hint says to show that if $N<A_n$ is a nontrivial normal subgroup then there is a normal subgroup $H\cong A_5<A_n$ that has nontrivial intersection with N. There is a further hint that says to construct a nontrivial permutation $\tau$ that fixes at least n-5 points and is contained in N. I believe that such a permutation exists.
In fact I believe that such a permutation $\tau$ "feels" like an element in $A_5$ embedded in $A_n$ since it only acts on at most 5 elements. But it's not clear to me how this leads to such an H. In particular I believe $<\tau>$ feels like a subgroup of $A_5$ (or a copy of). But I'm not sure how this helps me find such an H which is isomorphic to $A_5$ it only feels like I found a subgroup.
cloud walrusBOT
fajitas
next obsidian
You can find a copy of A_5 by looking at the permutations which only refer to 5 of the numbers
So all even permutations which only permute say, 1,3,6,7,9 inside of A_10 or something
You can alternatively frame this as saying all permutations which fix n-5 elements, the ones which the permutations don’t refer to
If you can find this tau inside of N, if it fixes all numbers except n_1,n_2,n_3,n_4, and n_5 (technically tau could fix even more than n - 5 things, in which case some of the n_i can be chosen to just be whatever you want), then you can look at the subgroup of all the permutations which only permute those 5 numbers, and that’s the H you’re looking for
To show that the subgroup of elements only permuting 5 elements is actually isomorphic to A_5, I think the easiest way is to let it act on those 5 elements. This is a faithful action so it gives an embedding H -> S_5 whose image is the even permutations
long obsidian
To show that the subgroup of elements only permuting 5 elements is actually isom...
Thank you that clears up a lot!
lavish nexus
I got AB+BA = 0
How do I prove they are both zero
Do I need to show they commute?
Also I haven't used anything about char K !=2
next obsidian
Well if you want to prove it using the fact that they commute to conclude that since 2AB = 0, then AB = 0, that requires char K != 2
but what's a projection linear map?
Is that just some silly terminology that means a linear transformation?
no, there's no way, since the sum of two linear transformations is also a linear transformation
gritty sparrow
it means PP=P
also the proof should be ABA=-AAB=-AB (by swtching the second A and the B) but also ABA=-BAA=-BA (by switching the first A and B). hence AB=BA, but we know AB+BA=0, so we get 2AB=0 hence AB=0
lavish nexus
also the proof should be ABA=-AAB=-AB (by swtching the second A and the B) but a...
thank you
tribal moss
Where do you get the assumption (?) that A and B commute, though?
lavish nexus
this doesn't use A and B commute
AB+BA=0 so AAB+ ABA = 0 and AB + ABA = 0
also ABA + BAA = 0 so ABA + BA = 0
tribal moss
Ah, I see -- you're multiplying out the LHS of (A+B)²=A+B. I thought the above reasoning was instead of that.
lavish nexus
yes I got AB + BA = 0 from that then saketh pointed out the rest
tribal moss
Right, don't mind me. :-)
lavish nexus
if I have a matrix in GL group over a field F with minimum polynomial p(x)
is it true that the centralizer is isomorphic to F[x]/(p(x))
gritty sparrow
That doesn’t really make sense to me, the centralizer is supposed to be a group. What group do you mean when you say F[x]/(p(x))?
lavish nexus
OK so I'm looking at a counting argument. I have this 3*3 matrix over F_5 with minimal polynomial x^3-1=0
so
0 0 1
1 0 0
0 1 0
and try to count the centralizer. I did it in a cumbersome way and I got the comment
The centralizer is the ring F[x]/(x^3-1) which is the product of a field of 25 elements with a field of 5 eleements so has 24 x 4 =96 invertible elements
gritty sparrow
Oh so i guess you mean the group of units of some particular ring then, not the ring itself
lavish nexus
I think that is what he meant here
gritty sparrow
In that case the comment is false, for example the identity matrix has centralizer =Gl(n,F) but the group of units of F[x]/(x-1) = F is just F\{0} so if n>1 these are two different groups
lavish nexus
👀 hmm but the result is the same as I got
I listed all impossible centralizers using the fact the centralizer in its general form has a determinant that is symmetric
and subtracted them off and got exactly 96
gritty sparrow
The result may not be true in general, but maybe for this matrix it just so happens that there is some nice argument
gritty sparrow
Np
keen sparrow
There’s a visual abstract algebra book that is trending on Twitter. I think the focus on visualizing things is the wrong approach.
Students should learn to internalize abstract concepts in their own way. We can teach them how to formalize those intuition.
delicate bloom
that's their punishment for going on twitter
chilly ocean
What is the name of this book?
tribal moss
No, that's knights-and-knaves puzzles, not abstract algebra.
iron vessel
What’s the book called?
terse crystal
I rarely talk about math on Twitter cause I haven’t seen any serious twitter post about math yet…mostly boring memes,except one post introducing a book called diagrammatic algebra seems interesting…I wonder what kind of social apps you use to talk about math except discord.
At least Twitter isn’t one of them…
I also followed the topic physics but only to find some bots or weirdo keep posting some aphorisms people like Einstein had said…
fallow plume
I rarely talk about math on Twitter cause I haven’t seen any serious twitter pos...
4chan obviously: https://haruhi.fandom.com/wiki/The_Haruhi_Problem
i still wanna know who the fuck had the nerve to anonymously post a proof to an open problem on 4chan for a fucking anime problem 
terse crystal
I don’t understand the problem… what does string mean?
fallow plume
let's move to #combinatorial-structures
shrewd ravine
Hi, can someone help me understand this
It can be rotated through pi/2 about the centre so the rotation permutation is (2 3 4 5) but does it have a line of symmetry
because that would cut through the centre 1
does it matter as 1 is fixed anyway?
rustic crown
if you flip it vertically then 2 and 3 are swapped and so are 4 and 5
so if you like simplify that figure it's kinda same as symmetry of a square which is D4
shrewd ravine
so if you like simplify that figure it's kinda same as symmetry of a square whic...
for the rotations, are (2 3 4 5)(1) and (2)(3)(4)(5)(1) the same permutation? it should have three rotations right?
or three rotations + the identity
rustic crown
yep, and these are just the four powers of rotation by pi/2
shrewd ravine
if you flip it vertically then 2 and 3 are swapped and so are 4 and 5
yeah I got that bit, so the two reflections are (2 3)(4 5) and (2 5)(3 4) but I'm having trouble presenting them
rustic crown
there are two more reflections
shrewd ravine
the ones that change 2,4 and 3,5 but keep the other two fixed?
rustic crown
yep, flipping diagonally
shrewd ravine
I see
and to find the presentation I can just take any and see how many times I can permute it by itself until I get the identity?
or do I know that already?
rustic crown
well, if you know about dihedral groups, then most of the work is already done there.
This group of symmetries is isomorphic to D4, and there is a standard presentation for dihedral groups.
rustic crown
yep!
shrewd ravine
but there are four reflections
shrewd ravine
the coset is {1, b}?
shrewd ravine
oh
rustic crown
because bab = a^-1
by raising to nth power you also get
b a^n b = a^-n
this shows (a^n b)^2 = 1
rustic crown
yep!
shrewd ravine
thanks
shrewd ravine
yep!
hi, if possible could I also get help understanding part b and c of this?
ii and iii I mean
so if (a,b) and (c,d) are in the same oribt, they are conjugate subgroups of GxG, and so (a,b) = g(c,d)g^-1
is that right?
and by that (a,b) = (c,d) so ba = dc?
rustic crown
wait i didn't follow that
not sure why you're bringing up subgroups of G x G, it's enough to think of it as just a set
what we actually have is g((a, b)) = (c, d)... this is what it means that (c, d) is in the same orbit as (a, b)
so you have (c, d) = (ga, bg^(-1))
is that okie?
shrewd ravine
what we actually have is g((a, b)) = (c, d)... this is what it means that (c, d)...
oh I see
I don't understand how to show that ba = dc from that
shrewd ravine
so if (a,b) and (c,d) are in the same oribt, they are conjugate subgroups of GxG...
this is a result from my notes somewhere
rustic crown
issokie, you're good >.<
shrewd ravine
for part iii), is it enough to say the stabiliser is {g in G: g((a,b)) = (a,b)} => (ga, bg^(-1)) = (a,b) which is only possible when g = e
shrewd ravine
issokie, you're good >.<
I'm not sure about the last step I did there
rustic crown
for part iii), is it enough to say the stabiliser is {g in G: g((a,b)) = (a,b)} ...
yep, this looks good!
shrewd ravine
damn group theory is starting to look fun once I understand it haha
rustic crown
shrewd ravine
mind if I ask more questions if I don't get something?
rustic crown
yee sure
shrewd ravine
these are all example sheets btw they just dont have answers so I wanna check if what I'm doing is right
rustic crown
ryu hi !
lethal dune
Hi det 
shrewd ravine
yee sure
having some trouble with ii
2 sylows have order 8, w2 is in {1,5,7,35}
5-sylows have order 5, w5 is 1 or 56,
7-sylows have order 7, w7 is 1 or 8
from the question w5 is 56 and w7 is 8
for the question we need to show that w2 = 1 and so its normal in G so G is not simple
but I'm confused as to how to show that
lethal dune
count the number of elements
rustic crown
what do we get when we know that there are an awfully lot of subgroups of order 5, 7
like ryu says, that will put some very serious restriction on the number of elements of order 2
maybe there is only enough place to fit at most 4 sylow 2 subgroups
shrewd ravine
what do we get when we know that there are an awfully lot of subgroups of order ...
what I was thinking is that here are 280 elements but 56 x 8 = 448 > 280 already
shrewd ravine
56 sylow 5-subgroup means 56 * (5 - 1) elements of order 5
I haven't seen this result before
rustic crown
it's not any serious result, we're just doing simple counting.
any two subgroups of order 5 either intersect trivially or overlap completely, there isn't a third choice because 5 is prime.
so what this tells us that if you ignore the identity element then all the 56 sylow 5-subgroups are "disjoint"
which gives you a way to count number of elements of order 5
each sylow subgroup will give you 4 elements of order 5
shrewd ravine
each sylow subgroup will give you 4 elements of order 5
I see
could I also argue that as the orders are coprime their intersections are trivial, supposing that any two 5-sylows and any two 7-sylows intersect trivially. Then the total number of elements supposing w2 = 1 is 8x5x7=280
which means there is no room for any more 2-sylows
rustic crown
right, but you don't really need to say that as we're counting elements
if an element has order 5, it possibly cannot have order 7 as well 😛
shrewd ravine
ah fair enough
it's enough to let w2 = 1 and do the calculation showing there is no more room?
so it can't be >1
rustic crown
so we have 56 * 4 = 224 of order 5
we have 7 * 7 = 49 elements of order 7
how many elements are left?
shrewd ravine
nono it's all good
rustic crown
.<
shrewd ravine
>.<
for this part I'm a bit confused again
a unique 7 sylow is normal and therefore cyclic
G/N has order 40
I know the result that a group of order pq is cyclic
rustic crown
use the map G --> G/N
next obsidian
a unique 7 sylow is normal and therefore cyclic
this doesn't follow from being normal, normality is a property of a subgroup and its relation to the supergroup, being cyclic is a property inherent to the group. The sylow-7 is cyclic because it has order 7 which is prime
shrewd ravine
this doesn't follow from being normal, normality is a property of a subgroup and...
my bad, I wrote it wrong
that is what I intended
next obsidian
ah okay, I just wanted to clarify
shrewd ravine
thank you
rustic crown
use the map G --> G/N
by correspondence theorem, finding a normal subgroup of G containing N is equivalent to finding a normal subgroup of G/N
so how do you think we should find that normal subgroup of order 35?
shrewd ravine
the 2-sylow
rustic crown
it definitely contains the sylow 7-subgroup (which is N)!
shrewd ravine
oh
the subgroups of G/N are the groups H/N for all subgroups H of G which contain N
rustic crown
yep!! and moreover if H/N is normal in G/N if and only if H is normal in G
shrewd ravine
which we have already right
rustic crown
yee, so the problem reduces to finding a normal subgroup of G/N of order 5
can we do this?
shrewd ravine
if w5 = 1 this is true right
oh wait a subgroup of G/N
one sec
or just any 5-sylow
rustic crown
yee, it has order 40
shrewd ravine
works?
rustic crown
right, but we would need to show it's normal
else that will only give you a subgroup of order 35
we want a normal subgroup of order 35
rustic crown
yep!
shrewd ravine
okay that makes a lot of sense
shrewd ravine
yep!
last one I promise
the centre is Z(G) is {g in G: gx = xg for every x in G}
so if the subgroup is normal
it means the cosets commute
and it's also cyclic because of prime order
I don't know what to do after that
chilly ocean
cyclic is abelian
shrewd ravine
cyclic is abelian
is that enough to say it's in the centre?
next obsidian
no
chilly ocean
no
lethal dune
no
rustic crown
no
shrewd ravine
-_-
rustic crown
lethal dune
viscid pewter
it's gonna be because it's odd
shrewd ravine
so there are no 2-sylows
rustic crown
use some fun group actions maybe?
shrewd ravine
order is p1^a1 x ... x pn^an for primes >2
viscid pewter
something something conjugate??
rustic crown
so take X = subgroups conjugate to H
viscid pewter
hah
chilly ocean
conjugate is the only group action
rustic crown
G acts on X
viscid pewter
if H is not in Z(G), there exists h in H such that gh != hg, ie. ghg-1 != h...
but let h be the generator
rustic crown
oh wait, i didn't properly read the problem >.<
it's because 17 is 1 more than a power of 2
so look at H
G acts on H via conjugation
we do require H to be normal for this
This gives you a group homomorphism G --> Aut_{Grp}(H)
since H is cyclic of order 17
Aut_{Grp}(H) is isomorphic to a group of order 16
but G is odd
so image has to divide both an odd number and 16
wooden ember
nicely done
rustic crown
This shows that image of G is just the trivial automorphism on H
which means any element of G commutes with any element of H
because the action was via conjugation
next obsidian
I’m a chmonkey
rustic crown
viscid pewter
oh wow
wooden ember
i just did a past lin alg exam
the last exercise had us prove corollaries of schur's lemma
i am so confused as to why the teacher thought it would be a good idea to put rep theory into a linear algebra exam
wooden ember
it's not even hard but it's just useless
like do they expect us to come out of the exam having learnt something
no one is going to remember that one result lol
rustic crown
one of my friend had a linear algebra exam, and their prof thought it was a good idea about asking them to prove snake lemma for vector spaces
wooden ember
lmao
hidden haven
lol which prof was this
rustic crown
dunno, he was in iits
hidden haven
🥴
next obsidian
Snake lemma doesn’t need any category theory
rustic crown
last one I promise
also btw, this thing is called some N/C theorem by some people. if H is a subgroup then N(H)/C(H) can be identified with a subgroup of Aut_Grp(H)
hidden haven
I think he just means that cat theorists are snakes
wooden ember
also btw, this thing is called some N/C theorem by some people. if H is a subgro...
oh true i forgot about this
it's pretty useful
rustic crown
i can never remember the statement lol
if it's iso or iso to subgroup
easier to reprove it
wooden ember
yeah i mean if you stop to think about it for a few seconds it makes sense
lethal dune
iso to subgroup
rustic crown
right
lethal dune
what the hell is snake lemma
wooden ember
it's all the automorphisms that are "like" inner automorphisms
rustic crown
🐍
next obsidian
what the hell is snake lemma <:monkey:651816857153699840>
lethal dune
why weren't we taught these stuffs
rustic crown
the context to learn it is homological algebra
lethal dune
ig I don't have enough algebra then
rustic crown
profs who do it in linear algebra are 
rustic crown
HK?
lethal dune
Hoffman Kunge
wooden ember
hollow knight
next obsidian
Hollow Knight
next obsidian
I wonder
For vector spaces
At least fin dim ones
Snake lemma might just fall out via a dimension counting argument or something
next obsidian
Hehehehe
True
Just pick it to be natural omegalol
Do it transfinite inductively over all diagrams to guarantee naturality
I use class sized induction
hidden haven
next obsidian
Okay actually
rustic crown
next obsidian
This might sound stupid but
I think if you can do some insane choice
Like fix a basis for all vector spaces from the start
You can do it because there’s finitely many maps in terms of matrices
Or some shit
Idk
hidden haven
Axiom of global choice 🥴
next obsidian
Ignore me
Global choice is whatever
It’s like subway, have it your way
Is that even subways slogan?
The point was supposed to be that it’s your choice
This idea is so dumb
There’s no way you can force naturality like that
hidden haven
watch me 
hidden haven
Civilisation if
next obsidian
Moldi do you know what injective hulls are
rustic crown
The dual functor would be iso to id
still the variances won't match up right
next obsidian
Do you know how maximal essential extensions are injective hulls
next obsidian
still the variances won't match up right
Oh yah
hidden haven
something about injective hulls and projective covers
lethal dune
they are speaking the language of god
hidden haven
lmao
next obsidian
To prove something I couldn’t
lethal dune
next obsidian
The point is there’s a thing called an essential extension
It’s an N>M so that for any submodule of N, that’s nontrivial, it intersects M non trivially
Then N is an essential extension of K
A maximal essential extension is injective, and they’re a minimal injective module containing M
In the sense that like for any embedding M -> I it factors through that
Every module has a maximal essential extension
But to show this, the only way I know how is to Zorn’s
But you have to already be embedded in something to do this Zorn’s
It turns out basically that for any injective I containing M, there’s a copy of every essential extension inside I
So you can do Zorn’s inside of I
Then get your result
The problem is you need to know you have enough injectives
And so I wondered if you could prove the existence of enough injectives by somehow proving there’s a maximal essential extension of M without using that there’s enough injectives
You can’t even use global choice
Since the global Zorn’s fails
Because if you try to do a union of a class sized chain, it won’t be a module because the result is too big to be a set
So maybe you can somehow set theoretically bound the size of an essential extension?
Without embedding them into an injective containing M
Or maybe you can somehow Grothendieck universe it
Anyway, that’s your hw
next obsidian
No, you can’t solve it with commutative alagrba
That’s why I gave it to you
Somehow who knows some set theory
hidden haven
lol I'll think about this
next obsidian
It’s fundamentally just a size issue
hidden haven
has size issues?
next obsidian
hidden haven
hidden haven
shrewd ravine
also btw, this thing is called some N/C theorem by some people. if H is a subgro...
that is neat
fierce perch
Remind me why Aut_K(L) -> Hom_K(L, K_{sep}) is injective for a separable extension K -> L and bijective for a galois extension?
oblique river
How is that map defined if L/K isnt separable in the first place?
pastel cliff
how important is chinese remainder theorem
fierce perch
you could define it by composition I guess
oblique river
what composition
There is no map L --> K_sep unless L/K is separable in the first place
lavish nexus
my dumbass be thinking 2^n-1 is prime for all n odd today
ended up proving this only for 2^n-1 prime
delicate bloom
it can only be prime when n is prime, there's a fun fact for you
lavish nexus
I hope whoever grades it have mercy
since I technically proved one case of this problem
fierce perch
True @oblique river
So why are those things true? I can see it intuitively (havent checked details) for a finite separable extension with the primitive element theorem
Then right side is basically bijective with the roots of the minimal poly which is separable hence the map is injectice. And galois means that the roots are mapped transitively
oblique river
if L/K is separable
the map is injective just by definition
postcomposition with the inclusion L --> K_sep is injective, period
the map doesnt exist if L/K is not separable
for the "L/K galois iff bijective" that comes down to your definition of galois
you can take that as the definition of galois if you would like
fierce perch
Lets say we use fixes K as definition (so it works for inf extensions)
Sorry it was ages ago that I did galois theory so this is prob straight from definitions
oblique river
its' fine haha, but yeah maybe if you review your notes you'll find a theorem like this
i cant chat for much longer now though
i gotta go
lavish nexus
ended up proving this only for 2^n-1 prime
crap it's characteristic 2
if I sum everything except 0 and 1 I don't get 0
I didn't actually prove anything big oof
delicate bloom
ended up proving this only for 2^n-1 prime
prove that if one of them is 0 then the other must be 0 first
that way we can assume both are nonzero without loss of generality
lavish nexus
I assumed both are nonzero
then (a+b)^2 = ab
there being no order 2 elements in F* I assumed a != b^(-1)
Then expanded (a+b)^p
delicate bloom
ah I wouldn't do that
once you know they're both nonzero you can divide through by b^2 and then substitute c=a/b
that way you can focus on a one variable polynomial having roots
lavish nexus
p = 2^n - 1
after expanding I found it to be the sum of all elements except 0 and 1
which... since 1 is its own additive inverse... I thought would sum to 0...
but ofc everything is its own additive inverse and I oofed
delicate bloom
once you know they're both nonzero you can divide through by b^2 and then substi...
everything you did made sense up to expanding (a+b)^p for me, I think that's an interesting way to prove a != b^-1 but what I have in mind is gonna be direct if you do this
lavish nexus
I agree
fallow plume
How I know @hidden haven is in the channel:
long obsidian
Why on earth are people interested in field extensions?
lavish nexus
field extensions aren't even in the syllabus
long obsidian
The heck they aren't
oblique river
Why on earth are people interested in field extensions?
because people care about polynomials
and roots of polynomials
long obsidian
What does this have to do with polynomials?
The only point of reference I have is extending R to C but what does that have to do with polynomials lol
uncut ridge
all* of galois theory is field extenstions
oblique river
you get C by starting with R and adding in a root of f(x) = x^2 + 1
I agree that if you only care about R then looking at field extensions is boring
since the only small field extension of R is C
but if you start with something like Q the theory is a lot richer
let f(x) be a polynomial with rational coefficients. f(x) might not have a rational root, but it will always have a root in some field extension of Q
and one can translate properties of the polynomial into properties of the extensions of Q which contain roots of that polynomial
(and vice versa)
and we prefer studying field extensions because we have lots of techniques for dealing with fields, like linear algebra for example
long obsidian
Hmmmm ok this is looking a lot like the lecture that went over my head today. Thank you lol
oblique river
i mean if you just had one lecture on field extensions it's likely that you only saw some basic definitions
and didnt really see a lot of the reasons why people end up caring about them
lavish nexus
you still working on it? There's not much more to go to finish the problem
oh no I had this on my qual today I already handed it in before posting it here lol
though it is good to know how to solve it so ty
delicate bloom
oh ok, yeah I'll help
devout crow
just to check: the map $\Phi|_U$ (the evaluation map) is not just a bijection, right? isn't it also an isomorphism?
cloud walrusBOT
∧res
devout crow
(V is finite dimensional)
oblique river
yes
south patrol
Ye phi is already linear so it'll be an isomorphism
devout crow
Ye phi is already linear so it'll be an isomorphism
thank you, mr potat from the freshers server
nice status
south patrol
aha thought i recognised you too
Well the problem was a giveaway too :)
Hope you're well
devout crow
yeah haha
south patrol
(I've swapped to maths now lol)
devout crow
yeah it threw me
south patrol
And not like "natural isomorphism" or smth
chilly ocean
And not like "natural isomorphism" or smth
Is it a category theory course? I don't see what's wrong with the latter convention
south patrol
Well saying bijection is odd when it's also a linear map imo but uh yes and no - it's not a cat theory course but the course does mention natural isomorphisms between spaces and stuff like that
well and mentioning that, say, the standard isomorphism V -> V' (once you've picked a basis) is functorial etc
chilly ocean
just to check: the map $\Phi|_U$ (the evaluation map) is not just a bijection, r...
Its about this no? This sounds more like category theory than algebra when you talk about double duals and functorials
but then again i've read very little and listened to even less cat theory
south patrol
well this is our like second year lin alg and it seems to be pushing more in the direction of sort of higher/category-level stuff
it's cool stuff
chilly ocean
nice that they're talking about dual spaces, my school never touched on that stuff and i still get confused about it, kinda aggravated by it but if i learn it'll come i suppose
I think the arguement needed there looks more like a bijection since you're talking about equivalent subsets, still awkward to have used that wording earlier in the problem and call it something different
south patrol
ye
south patrol
nice that they're talking about dual spaces, my school never touched on that stu...
aw sure ye
when i first did dual spaces i didn't realise there were also dual maps and stuff, that stuff's cool lol
lethal dune
if A,A' or B,B' are different then aren't the Morphism already different (different domain/range) so why do we need an extra axiom asserting it?
next obsidian
Because like
Oh
Mor isn’t a set of maps
A and B don’t have to be things with underlying sets
Mor(A,B) could be {🥺}
lethal dune
next obsidian
It’s just stating here that all the morphisms are distinct, there’s absolutely no way to say that a morphism between distinct objects are the same
lethal dune
Mor(A,B) aren't maps?
next obsidian
No
We tend to think of them as maps
But they don’t have to be
Take any poset
And make Mor(x,y) = {*} if x <= y
This is a kosher category
The existence of a morphism isn’t a map, it just signals that x <= y
lethal dune
ou okay 
chilly ocean
This is a kosher category
can you find a source on this cause I want to see more of it, but google doesn't give me anything good, just stuff about kosher food LOL
oblique river
he just meant that it's a valid category
there's no such thing as a "kosher category"
chilly ocean
oh lmao
gritty sparrow
can you find a source on this cause I want to see more of it, but google doesn't...
Kosher just means acceptable to jews
chilly ocean
but what if the category is boiled in it's mother's milk then what
fossil shuttle
i am a logician and i have an orthodox jewish friend who is also a logician
he used to talk to me a lot about logic in the hebrew bible
and deducing what the right way to behave is in a given situation
the talmud contains essentially like, a self contained logical system for deriving new commands from the given ones
fossil shuttle
Also, the French Jewish philosopher / Talmudist who we call Gersonides is arguably the inventor/discoverer of mathematical induction, although of course such a concept is hard to pin down exactly
see
fossil shuttle
if A,A' or B,B' are different then aren't the Morphism already different (differ...
it depends on your underlying set-theoretic formalism. You could define a category C by fixing
- a set, called Obj(C), and then
- a set of ordered triples {(c,d,Mor(c,d))} which is a function whose domain is Obj(C) x Obj(C)
...
but then it would be conceivably possible for the function Mor to send the pair (c,d) to the same set as it sends (c', d'), so we are asking that they are distinct
hidden haven
if A,A' or B,B' are different then aren't the Morphism already different (differ...
They are not already different, your argument is that their domain is different implies that they are different, but domain is not a well defined function if the Mor sets are allowed to intersect, so that becomes a circular argument
There's a #category-theory channel too btw
lethal dune
There's a <#497227343337881640> channel too btw
I guess I've never scrolled that low
or don't wanna
summer cradle
if A,A' or B,B' are different then aren't the Morphism already different (differ...
for the existence of such set
we need the axiom
like
the mor(a,b)
summer cradle
Im sorry man okay
chilly ocean
if A,A' or B,B' are different then aren't the Morphism already different (differ...
we need the axiom
untold cloud
Hi, guys, for a grading vector space $V=\bigoplus V_i$, if the union of basis for $V_i$ is a basis for V?
cloud walrusBOT
cuiyuze0728
oblique river
yes
chilly ocean
yeah most likely
oblique river
this is just a fact about direct sums
chilly ocean
sniped 
untold cloud
cheers!
summer cradle
we need the axiom
I meant a logically equivalent axiom pdsojvopdsfojvspodjfv
spice whale
what's the significance of tropical algebra
rustic crown
it's cute. and you can use it like in some shortest distance graph algorithms
rustic minnow
If I am asked to find all irreducible polynomials in the ring of polynomials $\mathbb{Z}_2[x]$ over the field $\mathbb{Z}_2$ how do I know up to what degree to look for?
I figured since the only elements are 0 and 1 I can exclude all even number of non zero since plugging 1 into them would return zero, i.e $1^2 + 1 = 2 \equiv 0 \mathbb{Z}_2$.
cloud walrusBOT
Fredrikpiano
rustic minnow
The other similar exercise I solved had some "up to degree" condition for the irreducible polynomial , so thats why I got confused
rustic crown
there are infinitely many such polynomials >.<
you can reuse the proof of infinitely many primes in Z, if you wish.
(also slightly harder to show, but there is at least one such polynomial of each degree)
rustic minnow
So maybe a "up to degree 2" was omitted or maybe I am missing something
rustic crown
yep quite possible
notice that a polynomial of degree 2 or 3 is irreducible if and only if it has no roots
so your strategy will work, all you would need to do is check 0 and 1 are not roots.
rustic minnow
yes, there is a theorem relating this fact if I remember correctly in the textbook Im using
like checking x^2 +1, x^2 +x , x+1 and x^2 for up to degree 2 if thats all of them?
rustic crown
x^2 + x has a root
rustic minnow
oh missed that one.
rustic crown
x and x+1 are linear irred
rustic crown
rustic minnow
but what about x +1 it also has even terms?
rustic crown
ah, but it's linear you see
for degree 2 and 3 have a root is bad, because it means it has a linear factor
rustic minnow
oh I see. Thanks
pastel cliff
i wanna prove that whenever m is not 0, gcd(0, m) =|m|.
my proof is just showing that |m| divides both m and 0, and then pointing out that m can only have divisors in the range [-m,m] and |m| is the largest possible one of these - is this valid?
rustic crown
yep!
pastel cliff
pog
next obsidian
you could do a really stupid thing and like...
use Bezout's lemma
this is sorta the exact same thing, but you can just note that 0k + mn is minimized at |m| for positive integers
idk
this is stupid to even note
rustic crown
make sure that you know that this is the place where we used m is not 0
m can only have divisors in the range [-m,m]
next obsidian
Bezout's lemma says that gcd(m,n) is the smallest positive integer of the form km + jn for k,j integers
pastel cliff
yes yes i have a verbose version
next obsidian
It's probably most often seen to just say that if m,n are coprime then there's a linear combination which equals 1
which is very very useful
next obsidian
This statement shows up a lot in finite group theory
There’s a lot of stuff about “if things have coprime order then…”
And as a corollary, if either of m or n is prime, then gcd(m,n) is always 1
So it gives you a lot of stuff to work with
rustic crown
primes
next obsidian
Probably the most important basic number theory result to know for basic group theory
next obsidian
Rings first?
pastel cliff
yeah
next obsidian
Ah okay
rustic crown
all of rings first or Z first?
pastel cliff
i know a group is like
next obsidian
Group is just an associative operation, with identity and inverses
next obsidian
Think of the units of a field under multiplication
pastel cliff
but im still just trying to internalize rings to begin with so
pastel cliff
GL?
next obsidian
Invertible matrices
pastel cliff
ah
next obsidian
The issue with R\{0}
Is that the operation is commutative
In general this isn’t true
So with invertible matrices you can multiply, you have inverses, an identity
And it’s associative
Anyway, you’ll get around to them later lol
next obsidian
Oh hahaha
pastel cliff
yeah i can read
ok perhaps a bit silly, but if i send the problems from the section on rings can you point out which you think might be the most useful
give me homework
rustic crown
do all problems in the start >.<
pastel cliff
there's 20 questions and doing them all will take me like 2 days of work which i dont have unfortunately
i am slow with problems
5 looks cool
pastel cliff
oh rly
next obsidian
Yah
next obsidian
Yeah, it follows pretty immediately from Bezout’s lemma
c. follows form a and b
Then d. is also useful
pastel cliff
that lemma only comes up till 200 pages from now
so i'll assume i can prove it without bezout's
next obsidian
Hurb
Bezout’s lemma is definitely in the box of “assumed for an abstract algebra book”
I mean 5a and 5b essentially proved a weaker version of Bezout’s
And Bezout’s lemma isn’t hard to prove either
You can get the parts you need just from noting that lcm(a,b) = ab/gcd(a,b)
pastel cliff
actually damn it barely even mentions it
pastel cliff
time to go learn it then
pastel cliff
Bezout's lemma says that gcd(m,n) is the smallest positive integer of the form k...
yeet
rustic crown
there is also another cute argument which you can do but i kinda like bezout more
pastel cliff
to prove 5a?
rustic crown
yee
pastel cliff
well all i did try and rewrite stuff formulaically
rustic crown
you can use the fact that if (a, m) = 1 then ab = ac in Z_m implies b = c in Z_m
pastel cliff
do you mean gcd(a,m)?
pastel cliff
mmmmm yes inner product of a and m
next obsidian
Isn’t inner product usually <,>
pastel cliff
yeah my prof was just a fuckin weirdo and did ()
rustic crown
gcd, point in 2d space, ideals, column vectors, ...
pastel cliff
he also put matrices in big parentheses instead of []
is there anything else in said box of shit that is assumed tho
rustic crown
you can use the fact that if (a, m) = 1 then ab = ac in Z_m implies b = c in Z_...
(the bad part about this is that, to prove this you need some sort of Bezout's lemma, or unique prime factorization. but it does give like a different flavor of argument, which is used later as well, to show finite integral domains are fields and in the exact same way replacing finite with "finite dimensional k-algebra")
pastel cliff
i wonder if it's a bad idea to study out of two books
next obsidian
No
pastel cliff
im considering going through Artin as well
next obsidian
Not at all
pastel cliff
even though i only have two weeks of self study left
next obsidian
You should probably skim stuff you already have seen before, but new perspectives can help and if one’s explanation is hard to understand the other’s might be better
pastel cliff
the only thing stopping me is the different order
the book i have only does groups until ch6 but artin starts with that
at the same time, that's kinda why i wanna go through artin, but i also dont wanna spread too thin over too little time
also i only have a pdf of artin which kinda sucks
rustic crown
i kinda like the order your book is following >.<
rings are something people are more comfortable with because there are lots of cute examples like Z, Q, R, C but groups can appear very weird abstract objects if you haven't encountered enough examples to justify the abstraction
pastel cliff
the only thing is that like... shouldnt abstract algebra be abstract
though imi admittedly even having a hard time caring about rings
not for lack of examples, probably just time spent with the material
rustic crown
the only thing is that like... shouldnt abstract algebra be abstract
well, i think it should be more like "oh these properties are appearing quite often, let's abstract out all these examples and study in a single framework and get theorems for each of the examples at once!"
studying some object without knowing why i should care about it is kinda sad
pastel cliff
it's why im getting bored
and thus looking for a bit of a change
but it could also be that im not interacting w the material enough
i have a tendency to give up before i should
rustic crown
i have a tendency to give up before i should
i like to take things slow, because i've noticed whenever i try to speed run through something, it often doesn't make me learn it any better and in the end i'm just a little sadder than at the start. speed running is only nice if you've already read something properly once.
there are lots of reasons to care about rings btw...
one of my favorites is showing that if a prime number is 1 mod 4, then it can be written as sum of two squares. this is a fact you can state by staying entirely in the ring Z. but that's totally not the natural place to ask such a question. To truly understand this, you have to look at Z[i].
as soon as we define an object, we also get to hop from one object to the other, and this hopping usually tells us a lot about these objects themselves!!
subtle ivy
are field isomorphisms between ordered fields also order isomorphisms? 
rustic crown
nope
subtle ivy
i would think not
rustic crown
but you can use the isomorphism to define order on the other field
simplest example would be to look at the isomorphism Q(sqrt(2)) and Q(-sqrt(2))
i mean sending a + b * sqrt(2) to a - b * sqrt(2)
next obsidian
If (K, <) is an ordered field
If you just took the opposite order
Doesn’t that make it an ordered field still?
subtle ivy
um no
next obsidian
Uh oh, spaghetti-os
next obsidian
Wdym
rustic crown
squares have to be positive
next obsidian
True
I was worried about multiplication
But I just stopped thinking about it
Cuz I’m lying down with my cat
rustic crown
next obsidian
Ordered fields r weird
next obsidian
sturdy marsh
is that a cat or a dog
sturdy marsh
that one
rustic crown
my laptop has 6% battery and i'm too lazy to go find the charger >.<
i guess i'll end the day
sturdy marsh
just buy a new laptop
rustic crown
(but i'm too attached to my current one >.<)
next obsidian
It’s annoying buying a new one every 12 hours
sturdy marsh
12 hours lmao
next obsidian
Yeah
subtle ivy
yeah as if 12 hours
sturdy marsh
my laptop drains in like 4 hours
subtle ivy
i get like 4
sturdy marsh
next obsidian
Lmfao
rustic crown
lol
sturdy marsh
4 gang
subtle ivy
but also
why r laptops like this
it is 2022. it has been like this for a decade or more
rustic crown
my laptop is being so optimistic >.<
subtle ivy
ooh linuckz
next obsidian
Mine used to actually survive like a half hour at that level
Now it says “you are low, try to plug in”
And proceeds to die a minute later
sturdy marsh
next obsidian
Like if it gives that message and I have to dig my charger out of my bag, it’ll probably die before I can charge it
subtle ivy
i have hibernation turned on so it will just turn off and keep all my programs open n stuff
rustic crown
Uptime: 4 days, 23 hours, 36 mins
sturdy marsh
5:17:19:00
hidden haven
Yal living the good life, I unplug my laptop for 2 minutes and it dies, and that is not even an exaggeration
prisma ibex
next obsidian
Hmmm
Non injectivity means that f(x) - a has a multiple root (if a is a repeated output of x) and this is measured via the derivative test
The derivative of f(x) and f(x) - a are the same, while the roots of f(x) - a are shifted by a
Maybe this is something you can try to use?
So I guess you can describe it concretely by saying f’(x) and f(x) - a never share a root, but if f’ has a root at a we can just shift f so that f(x) - f(a) has a root at a
So is it precisely when f’ doesn’t have a root?
@delicate bloom
oblique river
lol
that was literally my approach the other day
hahahaha
but the problem is that i dont think it's true
next obsidian
Hurb
oblique river
how do you know that non-injective implise that f(x) - a has a multiple root for some a
oblique river
like i said that was literally my approach too
next obsidian
It’s when f(x) - a has more than one root
delicate bloom
yeah, double roots aren't really the problem
delicate bloom
I was confused about that too lol
oblique river
i think that something like that might be true still because it's true over fields like R
next obsidian
Yeah, idk why, it just felt true in the moment
delicate bloom
very simple sounding problem isn't it
oblique river
it's not that doubel roots are the problem
it's that double roots are easily detectible
next obsidian
Ugh
I haven’t thought about polynomial injectivity
Isn’t this isn’t about this problem, but isn’t like every injective polynomial over R[x] or C[x] also surjective or something?
Or maybe it goes the other way
I thought I remember hearing something like that idk
oblique river
delicate bloom
something weird I'm playing with at the moment is, $|f^{-1}({c})|=1$ means injective and $|f^{-1}({c})|=\deg f$ when the field is algebraically closed, it sort of feels like we can leverage this niceness of the algebraic closure to say something about it
oblique river
i think that's what you're thinking of chm
cloud walrusBOT
Merosity
next obsidian
Oh boy lmao
oblique river
interesting
next obsidian
I guess you could try to do some base change to k-bar somehow on A^1 or something???
But I have no idea how this would work
oblique river
so a galois group acts on the set f^{-1}({c})
next obsidian
I’m not even sure if that’s how it works
oblique river
namely the galois group offf(x) - c
next obsidian
i think that's what you're thinking of chm
Yeah it was, thx
delicate bloom
someone left a nice comment which didn't cross my mind but good to know, if f(x) in K[x] and f(x) is injective in the extension field K(alpha)[x] then, it's injective in K[x], seems handy
next obsidian
Ah okay
delicate bloom
I should have called it L or something
idk is there a relationship between the roots of polynomials and the roots you use to extend the field by?
next obsidian
Could you maybe like
No, this seems not true
I was thinking if maybe you can get an iff for injectivity after base change to some nicer field
But I doubt it
delicate bloom
every time you extend the field, you cut out polynomials
next obsidian
I was thinking maybe to a splitting field of f or something
delicate bloom
until eventually you add everything and are left with just linear polynomials
I like buncho's idea of doing some sort of galois group acting on the preimage
next obsidian
This seems like it has to have been considered by someone before
next obsidian
Which either means like
It’s solved
Or it’s very hard
IMO
Hahahaha
I’m honestly more curious about integer polynomials now
delicate bloom
well I think it's fun and managed to make some progress on it last night there lol
next obsidian
That seems like there would be some serious number theory involved controlling stuff
delicate bloom
if you want we can focus on any special cases, it's not too serious
specific cases can help inform the bigger picture as a good stepping stone
yeah that's a good point
what are good restrictions on fields that would make this seem more doable?
next obsidian
Well okay for R I think there’s stuff you can say
It has to have derivative strictly positive or negative
delicate bloom
yeah the case for R is solved actually I found a question
next obsidian
By like IVT
This suggests to me maybe the derivative controls stuff, but perhaps that’s just bad intuition because you can do analysis on R
delicate bloom
yeah, the problem is it's too specific to R yeah, like even though we have formal derivative on polynomial rings they're not really good for more that I can tell past checking for multiple roots
next obsidian
Maybe you can examine p-adic fields? I don’t know shit about them but you at least have the tools of analysis?
delicate bloom
I have thought about it but the topology is too different to directly take anything
next obsidian
Hmmm yeah idk much else at the moment
I’ll put this in the file of stuff to think about randomly
delicate bloom
yeah same
personally I think my approach there can be completed to generate all polynomials
by some extra extension of how you generate them
like adding in intermediary non polynomial, injective functions possibly
next obsidian
Hmm
delicate bloom
which can be composed to make polynomials
next obsidian
Nothing strikes me as a particularly natural thing
You have to restrict which other functions to add to get anything useful
delicate bloom
for instance $f(x)=sign(x) |x|^\sqrt{3}$ makes $f(f(x))=x^3$ and is injective
cloud walrusBOT
Merosity
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
delicate bloom
well, we already have x^3 but trying to focus on easier tasks like creating x^3+x in R[x] first to try it out to see if it's viable
delicate bloom
hah no pressure I don't think anyone's gonna come up with any insight on this problem that fast, but we've sorta laid out a handful of good ideas so we'll see what happens
maybe this is specific enough to say something, if $f(x) \in K[x]$ is injective and $L=K(y)/(g(y))$ we have that $f(x)$ is not injective in $L(x)$, then can we say something like $gcd(f(x)-c,g(x)) \ne 1$ for some $c$? not really thinking this through to know if that's right
cloud walrusBOT
Merosity
