You need a bit stronger general algebra background, seeing tensor product, ext, Tor, direct limits helps a lot
#groups-rings-fields
next obsidian
And some basic homological algebra
There’s an appendix, but I find that trying to learn from an appendix is hard haha
blissful ice
I've heard my prof recommend it too, but never checked it out myself
blissful ice
why in specific
next obsidian
A few reasons
The first, is purely comparative
Doing A-M hurt me
The number of exercises was painful, and it felt like it dragged on
blissful ice
haha
I like atiyah but like 50% of the materials is put into exercises
next obsidian
For purely motivational reasons, doing A-M was a chore, but Matsumura is broken up into smaller sections with fewer exercises
So I get to turn pages more often, and this really helped me stay motivated
Also in comparison to Eisenbud, it’s just shorter. Eisenbud is nice in that it’s geometric (which Matsumura does not do, you’re getting pure algebra)
But it was too much for me
The other thing I like is that the arguments have this unique flavor.
Sometimes he elects to kind of dive into details, like working really hands on with a quotient of a polynomial ring
blissful ice
I don't like how eisenbud words things
for some reason he manages to confuse me on all the simple topics
next obsidian
At first this took me aback, but after working a bit I felt like I really understood how to do that much better
next obsidian
Also, there’s hints in the back for most exercises, so if you get really hard stuck you can crack it open to the back haha
I didn’t really understand commutative algebra when I first learned it, but after doing some algebraic geometry and switching to Matsumura I found I really enjoy the subject
I think the most important thing is finding a book you really enjoy and work well with, and for me, and a few ppl I know, Matsumura did it for us
blissful ice
I had a CA course last semester but I kinda stopped following towards the end (not because of difficulty, I was just busy with other things that were more important)
Now I'm just trying to relearn it and wrap it up haha
And I'm also taking algebraic geometry which does give some nice intuition, but our course has been quite classical so far, avoiding Grothendiecks langauge
I suppose everything has its time
hints do sound amazing though
next obsidian
Yeah, it’s helpful haha. I think the struggle of solving exercises is where most of the learninf happens
But if you’re disciplined and give it actual hard thought before resorting to a hint, I think the time saved outweighs what “losses in learning” you have
Plus, you still have to justify the answer or work to make the hint into a proof
The other major thing I liked was that you saw really cool theorems way early on. Things like the fact that projective over a local ring => free even without finite generation which used transfinite induction, the fact that for a finite A-module, any surjective map M -> M is actually also injective, etc
Sometimes I feel like I want to skip early chapters which do like a review of some basic ring / module theory, but I saw a lot of theorems I had no idea existed
tight otter
The degree of an element $b$ over a field $F$ is the smallest $n$ such that $b^n=1$, right?
cloud walrusBOT
panoramatopia
tight otter
silly question but i can't tell if im misremembering terminology and my textbook only provides a definition for degree of a field rather than degree of an element
next obsidian
Uhhh
I mean if you’re considering the degree of an element in F, as an element of the multiplicative group yeah
But if b lives in some field extension, which is what I think since you’re saying “over F”
Makes me think it’s the degree of its minimal polynomial over F
Aka the degree of the extension F(b)
Over F
next obsidian
soft skiff
R^2 != 0 in a Simple Ring
celest brook
units should be 1/f(x), but that won't always be defined from X->Reals
so would the units be 1/f(x) s.t. 1/f(x) in R?
multiplicative identity obviously f=1 and additive identity is g=0, given that f,g in R
i.e. take f: [0,1]->Reals f(x)=x, f's multiplicative inverse would be 1/x, but 1/x not defined on the domain, so it's not valid
unique juniper
celest brook
or am I just dumb
chilly ocean
So what requirement should we put on f?
chilly ocean
Yeah, f should be nowhere 0
celest brook
$ x \in X \mid f(x) \ne 0?$
reee
$ x \in X \text{ s.t. } f(x) \ne 0?$
matbot boke?
$2$
cloud walrusBOT
Moosey
cloud walrusBOT
Yes
proud bear
$x \in X \mid f(x) \ne 0$
cloud walrusBOT
proud bear
remove the space
cloud walrusBOT
8da
unique juniper
lol
chilly ocean
Wow, that is weird
celest brook
ohhhh
unique juniper
$ x \in X \mid f(x) \neq 0$
celest brook
ok so I'm not crazy
unique juniper
$$ x \in X \mid f(x) \neq 0$$
double $ allows a space
it seems
$$ x \in X \mid f(x) \ne 0$$
cloud walrusBOT
Yes
celest brook
ugh I feel like this proof is gonna be a page long considering all the parts, but I always see people be like 'oh you can do it in three sentences
part of me wonders how people do it
chilly ocean
Maybe this is what you are thinking, but it should be the set of f such that for all x, f(x) neq 0
celest brook
ahhh
i.e. f is an arbitary function in the set R
wait
an arbitary function in R s.t. it follows those criteria
okkk
now for the other parts. I remember someone telling me I can think of this the group under addition as a sort of vector space thing
shut halo
This was a question in my workshop last week, but I'm struggling on how to proceed.
I know that two projective varieties are birational iff their function fields are isomorphic. And from the hint I can set one from x,y,z to 1. Then I need to compute the function field, but not sure how to proceed.
Also how do I compute the function field of $C \times \mathbb{P}$?
cloud walrusBOT
snypehype
mystic tiger
How do I find the factor group (ZxZ)/H the standard from from the subgroup H={(5a+8b,10a+20b), a,b element of Z}
lethal cipher
Alright y'all I need help. My professor wants me to research from a book on evaluations and it expects a lot of knowledge about abstract algebra. So I am very lost.
I really need help understanding what ideals are
mint seal
do you know about normal subgroups & quotient groups? they're the ring version of that
lethal cipher
Someone tried to give me a crash course on it theother day, but that's about it
mint seal
an accessible example of an ideal in a ring is 2Z sitting inside Z
the even integers
notice that if you multiply an even integer by any integer at all, the result is even
lethal cipher
Correct
mint seal
so 2Z is closed under multiplication from the whole ring Z
lethal cipher
I can agree to that
mint seal
that's one of the defining properties of an ideal
lethal cipher
Isn't that just a property of a ring?
next obsidian
No
An ideal is closed under multiplication from anything in the entire ring
I might be properly contained in R, but for any r in R, and i in I you have ri in I
normally closure is a property internal to the thing, so it would be like for any i,j in I, you have ij in I
mint seal
right
lethal cipher
I need to talk to my professor. This is a hell of a push for knowledge in the middle of a research project. This is all a bit too much for me right now. I barely understand this
mint seal
also, nZ is an ideal of Z for any n. And all ideals of Z are of that form
lethal cipher
Please do. I am lost at the moment
next obsidian
Okay, thanks
So, there's a theorem which states that if $A$ is a Noetherian ring, and that $A_p$ is an integral domain for all $p \in $ Spec $A$ then $A \cong A/p_1 \times\dots\times A/p_n$ where $p_1,\dots,p_n$ are the minimal primes of $A$. Localization commutes with finite products, so that if we localize at any of the $p_i$ we should obtain that
$$A_{p_i} \cong (A/p_1){p_i} \times\dots\times (A/p_n){p_i}$$
Since we assumed all of these primes are minimal, it follows that
$$(A/p_j){p_i} = A{p_i}/p_jA_{p_i} = 0$$ as long as $j\neq i$, so that we should get that
$$A_{p_i}\cong (A/p_i){p_i} = A{p_i}/p_iA_{p_i}$$
but the latter is a field while the former should only be a field when $p_i = 0$
cloud walrusBOT
Chmonkey didnt get into Columbia
next obsidian
So what's going on?
mint seal
ah unfortunately I don't have any feeling for noetherian rings or localization
I don't have anything better to do tonight than try to understand that though
would that theorem be covered in Hungerford?
next obsidian
Well
I don't know
I would guess not
I called it a theorem, but really it's an exercise in Matsumura hahaha, but I saw it and thought "wait how could this be"
mint seal
ah
sturdy marsh
Chmonkey I think it can still be a field even if p_i is not zero
hmm
yeah p_iA_p_i could be zero without p_i being zero
right?
next obsidian
I mean aren't the units of A_p just the elements not in p?
sturdy marsh
they are things outside pA_p, yes
but pA_p could be zero
without p \subset R being zero
sturdy marsh
idk try Z/6Z
next obsidian
at 2?
sturdy marsh
sturdy marsh
yeah what's the localization at 2
next obsidian
I mean
if it's a field it sure isn't Z/6Z
lol
oh wait
that's not even relevant
lmao
time to think
oh right
I guess you're right
2/1 = 6/3 = 0
wtf
Okay so turns out I'm just a chmonkey
So...
Hmm
So if the hypotheses are satsified
then somehow p_iA_p_i is 0 for all p_i minimal primes
aka all A_p_i are fields
that's wack
sturdy marsh
yeah that's pretty weird
next obsidian
Someone on Twitter has a good explanation
A_p has one prime, which is equal to its nilradical
If A_p is an integral domain then the only prime is 0
So I guess for p a minimal prime, A_p is reduced iff it's an integral domain iff it's a field
cyan marten
aka all A_p_i are fields
The prime ideals of the localization A_p are those contained in p - so if p is minimal, then there is none besides (0), right?
next obsidian
Well you'd need to know that pA_p = 0
since (0) isn't always prime, but this follows since we assumed that A_p is an integral domain
xD
cyan marten
Yeah
Does localization always commute with finite products, or only if we are in a Noetherian ring?
cyan marten
that's wack
True! Are the p_i's the minimal primes associated with the zero ideal?
next obsidian
In what way do you mean associated with
This is a word which could mean a few things xD
If you mean in terms of primary decomposition
then yeah
cyan marten
If you mean in terms of primary decomposition
Yeah
next obsidian
In fact the intersection of them is 0
so you can take them to be your primary decomposition if you want
cyan marten
Makes sense
sturdy marsh
primary decomposition is for nerds
next obsidian
True
cyan marten
lol
By the way Chmonkey, can you explain how to embed Q in the double Pontryagin dual A**? I thought I understood it at first, but it turned out I was wrong.
Does it have to do with the fact that Q/Z is injective?
next obsidian
For showing that Ab has enough injectives?
cyan marten
For showing that Ab has enough injectives?
Does this mean "every abelian group can be embedded in an injective abelian group"?
next obsidian
yeah
If so, I didn't do it that way, I actually just embed any abelian group into (Q/Z)^alpha
and you can show that (Q/Z)^alpha (the direct sum) is injective for even infinite alpha
because it's divisible
Really you want to show that Q^alpha is injective, then just use that quotient of divisible is divisible
cyan marten
If so, I didn't do it that way, I actually just embed any abelian group into (Q/...
Yeah, but I don't know how to embed even Q, let alone an arbitrary abelian group lol
next obsidian
Yeah so
take A an abelian group
Wait oops it might not be Q/Z but
anyway so
Write A = Z^alpha/K
for K some subgroup of Z^alpha
you can just always do this
cyan marten
Ohh, right
next obsidian
Then from the injection of Z^alpha into Q^alpha
you just can embed A into Q^alpha/K
And it follows that Q^alpha (as a direct sum) is still injective
because it's divisible
like if you wanted to divide by n, just do it on each coordinate haha
cyan marten
That does solve the original problem
cyan marten
I was trying to embed Q
next obsidian
I did something kind of similar to extend this result to show that A-mod has enough injectives
but it's not a double dual
afaik
haha
cyan marten
So I took 0,Z,Q,Q/Z,0 (commas are arrows, this is a SES), and applied the exact functor Hom(*, Q/Z) so I got 0, (Q/Z)*, Q*, Q/Z, 0
next obsidian
A* = Hom(A,Z)?
cyan marten
but it's not a double dual
I see. There's something similar in D&F
next obsidian
I mean I think
if you want to show Ab has enough injectives you do use a double dual
I just don't know how it works
cyan marten
A* = Hom(A,Z)?
No, Hom(A, Q/Z), the Pontryagin dual
next obsidian
oh okay
Okay sure
so you have 0 -> (Q/Z)* -> Q* -> Q/Z -> 0
IIRC like
to embed A into A** you like
do some shit where you take a big enough integer or some crap
I really have no clue how it works
cyan marten
I see lol
next obsidian
¯_(ツ)_/¯
cyan marten
Well, I tried applying * again, but I got nowhere
next obsidian
I am surprised that this is how people tend to prove it
Yeah to do the embedding
you don't get it functorially afaik
You have to do some janky stuff
cyan marten
Ahaa
cyan marten
Okay, a final question. Any clue how (Q/Z)* looks like?
next obsidian
No idea lol, it looks scary
cyan marten
Quite so
next obsidian
haha
Idk why this is the standard proof, and my friends actually came to this proof by themselves e.g. Shamrock
But my first thought went to the one I did haha
and I feel it's simpler lol
cyan marten
It is cleaner, I think
cyan marten
Idk why this is the standard proof, and my friends actually came to this proof b...
Ohh, actually I confused you two lol
cyan marten
I see. Thanks!
next obsidian
👍
dire fractal
hi
rich ravine
@cyan marten Q/Z* is just Z^ the profinite completion of the integers
sturdy marsh
oh yeah
next obsidian
If A is an nxn matrix and V a vector in R^n, such that AV = V
latent anvil
next obsidian
is there something which says det(I - A)V_i = 0
latent anvil
next obsidian
This is for arbitrary rings tho 
latent anvil
Oh lol not \R^n
latent anvil
So (A-I)V = 0
next obsidian
Yeah
I tried toa ppeal to like
multiplicity of det
I thought about multiplying I - A with I except some V_i in one of the places
and trying to show this was singular
latent anvil
yeah this seems false
next obsidian
So like... this is what I'm drawing it from
next obsidian
latent anvil
I'm so tired lol
next obsidian
Starting at "Letting C = ..."
small bison
maybe something with the adjugate matrix?
next obsidian
blech
latent anvil
Ah yeah good idea
next obsidian
Isn't that about determinants of minors tho?
next obsidian
Uhhh
what the hell was the adjugate
was this not something you do to show inverses exist?
latent anvil
The point of the adjugate is that adj(A) A = A adj(A) = det(A) I
small bison
some magic matrix $B$ such that $AB = BA = \det{A}I$
cloud walrusBOT
young_smasher
small bison
it works in general rings
next obsidian
yeah
latent anvil
(which is polynomial in the entries of A)
next obsidian
right
latent anvil
So I think this works here
latent anvil
yes, sorry
next obsidian
so using the adjugate you know that
adj(A - I)(A - I)V = 0
but this is also det(A-I)V
so each component is 0
latent anvil
yup
next obsidian
Damn
latent anvil
clever
next obsidian
the adjugate always exists right?
latent anvil
yup
next obsidian
damn
latent anvil
It's polynomial
next obsidian
I wish I knew linear algebra
I mean wasn't it like
in each entry you take the minor
sorry
next obsidian
determinant of the minor obtained by removing those entries
next obsidian
that would make sense
small bison
you can find a formula on wikipedia
next obsidian
I'd at least belive that maybe you need a sign
latent anvil
Oh I guess it's about the order when you take the determinant ig
Anyways this works
carmine fossil
A matrix is invertible in a ring R iff its determinant is a unit,right?
delicate bloom
yep
next obsidian
It generalizes to any eigenvector right?
sorry I meant the proof of the theorem in Matsumura
See my tweet if you want to see what the proof was
Lmao
For further details refer to: (Tweet 2021)
delicate bloom
A matrix is invertible in a ring R iff its determinant is a unit,right?
the way I think about the adjugate is pretty simple, I can elaborate a bit more on this if you like
carmine fossil
I just wanted a small clarification. Thank you tho
delicate bloom
sure, I just think knowing why makes the whole thing obvious rather than some extra trivia to memorize
south temple
is there a difference between the structure theorem for finite abelian groups, and the structure theorem for finitely generated abelian groups?
chilly ocean
just a difference of Z^n right?
finitely generated abelian group = Z^n x (some finite abelian group)
south temple
when I look up just "structure theorem for finite abelian groups," the first result is for finitely generated
south temple
finitely generated abelian group = Z^n x (some finite abelian group)
ah, so finite abelian groups are a subset of finitely generated abelian groups
chilly ocean
yeah
south temple
the difference being that finitely generated abelian groups may be also a direct product of Z^n
chilly ocean
yeah
ivory cosmos
I remember seeing the structure theorem in terms of finitely generated R modules
which has both of these as a sub-case
cyan marten
If I know the theorem of finite abelian groups and the fact that torsionfree Z-modules are free, that proves the fundamental theorem, right?
I think my question is equivalent to asking whether Tor(G) is easily seen to be a direct summand of G
rich ravine
Yes, and this follows from the sequence 0 -> Tor(G) -> G/Tor(G) -> 0 being split, because G/Tor(G) is torsionfree hence free
rich ravine
So, there's a theorem which states that if $A$ is a Noetherian ring, and that $A...
$A_{p_i}$ has a unique minimal prime ideal $p_i$ ; but the intersection of the minimal prime ideals is the nilradical, which is $0$ since by hypothesis $A_{p_i}$ is integral
cloud walrusBOT
Othenor
rich ravine
So we have indeed that $p_i=(0)$ in $A_{p_i}$
cloud walrusBOT
Othenor
cyan marten
Yes, and this follows from the sequence 0 -> Tor(G) -> G/Tor(G) -> 0 being split...
Ohhh, right
cyan marten
It's still true if there is an element of finite order (regardless of the size)
rustic crown
Ohhh, right
is the proof of (fg, pid) torsion free => free any easier than proof of structure theorem?
cyan marten
This question is a bit confusing because the proof I am aware of begins by proving the first implication.
So if difficulty is monotonic, then the answer is no.
rustic crown
I know 2 proofs of structure theorem, one uses smith normal form and then uses presentations to give you a basis, other just straightaway gives you the a nice basis and is very similar to the old finite abelian group proof.
and this statement is usually a simple corollary
celest brook
ugh I just want to split this like the propostion wants
i.e. (even degrees sum)(x+1)=p(x)
but don't know how to get it to be the way I want
rustic crown
why not just look at phi_{n+1}(-1)?
celest brook
how would that help
rustic crown
what if that's 0
celest brook
...oh
how does that help me with the propostional approach though? I feel like that's cheating like 'oh you just know -1 is a root already??'
rustic crown
when you look for linear divisors... you're actually looking for roots of the polynomial
now tell me when can (x^{n+1} - 1)/(x-1) equal 0?
if the numerator is 0... then what happens?
rustic crown
now how does n being odd help you?
celest brook
difference of squares
sorry I'm just a little tired I've been staring at these problems all last night
I mean I guess what I'm asking for is some algebraic way rather than just realizing 'oh' just realize there's a linear root you can subtract -1 you can see it will be 0 at 0 so therefore must have root at -1
like the one posed in the propositon
definitely feel like I'm thinking about this too hard
rustic crown
(i don't quite understand... are we using the proposition to show that x+1 is a divisor? if you want something 'algebraic' then use difference of squares of the last line)
or maybe something like this
$\frac{x^{2m} -1}{x-1} = \frac{(x^2)^{m}-1}{(x^2)-1}\cdot \frac{x^2-1}{x-1}$
cloud walrusBOT
det
celest brook
ok...how does that help
rustic crown
the second term is just (x+1) and the first is phi_m but x replaced with x^2
celest brook
OH
wait
(2m?)
Not 2m+1?
i'm confused
more
oh wait
yes
it will be even
rustic crown
(m = j+1... sowwyyy >.<)
rustic crown
placid falcon
hi, how can i fix
next obsidian
I’m looking for an explanation / source which covers the following thing. I’m just quoting verbatim from Matsumura here.
Let L/K be an infinite degree Galois extension. Let K’ be the fixed field of Aut(L/K), then L is Galois over K’, and K’/K is a purely inseparable extension. If K’ ≠ K then we must have char K = p > 0
I looked at a couple things covering infinite Galois extensions but didn’t see anything about this pure inseparability
rich ravine
You should be able to reduce to the finite case, then any textbook on galois theory covering inseprability should do the trick
next obsidian
Okiedokie
sturdy marsh
hmm
yeah it shouldnt be too hard to prove
pick an element in K'
assume it's not in K
if it is the only root of it's min poly, then we're good
next obsidian
Yeah
sturdy marsh
if there's another one
rich ravine
Wait, what is your definition of galois extension ? A galois extension should be separable...
next obsidian
¯_(ツ)_/¯
Matsumura just pulls this out of nowhere
I don’t think K’ over K is assumed to be Galois
rich ravine
But your statement should hold for normal extensions
next obsidian
It’s L/K’ that’s Galois
rich ravine
I mean for Aut(L/K) to be meaningful you should maybe assume L/K normal
next obsidian
Oh yeah
sturdy marsh
probably just normal
next obsidian
L is normal over K
sturdy marsh
not galois
ok
so the min poly splits over L
so if there's another root in L
then there is an automorphism taking this root to that one
next obsidian
Right
sturdy marsh
but we assumed K' is fixed
so that should be a contradiction
something like that should work
next obsidian
I think that makes sense
And nontrivial purely inseparable extensions only exist in positive characteristic?
small bison
in characteristic 0 extensions are automatically separable right?
small bison
something something derivative
sturdy marsh
yup
rich ravine
Yes, because an irreducible polynomial will be coprime to its derivative
next obsidian
Gotcha
sturdy marsh
rip
next obsidian
I’ve come to the following conclusion
Everyone who has strong field theory also did ANT
sturdy marsh
this is true
small bison
🐜
rich ravine
The book by Pierre Samuel is a gem
sturdy marsh
neukirch
next obsidian
On field theory?
rich ravine
No on ANT
next obsidian
Ah
rich ravine
I learnt field theory from Lang
sturdy marsh
early alg NT is so dry tho 
next obsidian
Rip
Ppl have told me Lang for field theory
I know it’s way more comprehensive than the other general algebra books
But I was scared to learn anything from Lang...
rich ravine
It's actually pretty short
next obsidian
But if it’s actually good then 😭
ack how can I learn all the things at the same time I want to
Lmao
rich ravine
I mean the chapter on field theory and galois theory
next obsidian
Yeah, I was considering going through them
rich ravine
Don't rush things too much, it's not a race
next obsidian
I don’t want to rush it, I just want to learn all the cool stuff I want to learn
😔
rich ravine
I'm saying that because I tend to get anxious when thinking about all the stuff I should read/learn and then I worry instead of actually sitting down and read stuff
But you'll get there, just enjoy the ride
rich ravine
I have to sleep now but I'll think about your question tomorrow
next obsidian
Cool, thx
untold sapphire
Is computing minimal free resolutions supposed to be super hard?
I feel like I don't really understand what a syzygy module is
even though I can do some simple examples
thorn delta
I am reading something that goes like: let $T$ be an endomorphism of a finite dimensional vector space $V$ over $\mathbb F$, and I am being asked to consider the map $\mathbb F [t] \to \operatorname{End}_{\mathbb F} (V)$ where $1\mapsto \operatorname{id}_V$ and $t\mapsto T$.
cloud walrusBOT
kxrider
thorn delta
am i being dumb, or is this not technically enough information to define a ring map F[t] to End(V)?
next obsidian
You extend it linearly (I guess... polynomially?)
Over F
F[t] is the free algebra on 1 thing, so all you need to do is define where t goes
So like, for any f in F and phi in End_F(V) the expression fphi makes sense
next obsidian
you can multiply endomorphisms by an element of the field F
so then the formula 1 -> id_V and t -> T tells you for any element
a_nt^n + ... + a_1t + a_0 in F[t]
you map this to
a_nT^n + ... + a_1T + a_0id_V
thorn delta
I'm still not seeing where the action on non-identity elements of F comes from oof. is it not possible to have another ring homomorphism F to End(V) which maps 1 to id_V and induces a different ring map when t maps to T?
next obsidian
Yes it's possible
But when you define a map like that it's understood to be extending it
There's a unique F-algebra homomorphism which does that
Also I think you mean F[t] to End(V) right?
thorn delta
well, i am referencing the fact that if you have a ring map F to End(V) and you necessitate that t mapsto T, then you get a unique ring map F[t] to End(V).
next obsidian
Right, so it's only unique when you require the map to be F-linear
You could in theory have another ring map which does this
So when they say "the map given by 1 -> id_V and t -> T" they want you to extend it to be F-linear
thorn delta
i mean, yea, i understand the obvious extension. Was just curious if there was any reason why there couldn't be another extension a priori
next obsidian
Yeah, a priori there isn't a reason for that
Maybe I can cook up an example
Perhaps you can map all scalars to id?
besides 0
ah, no this won't work
it won't be additive
But I think you're right that for any map f:F -> End(V) you can extend this to a map from F[t] -> End(V)
Given by like
a_nt^n + ... + a_0 -> f(a_n)T^n + ... + f(a_0)
So if you took f to be a different ring map then you'd get a different one
since when you say "1 -> id_V" you haven't defined it on all scalars unless you extend linearly
latent anvil
Just to hammer it in, this is a unique extension to an F-algebra map
thorn delta
yea, i imagine we're just supposed to think of it as a homomorphism of algebras.
next obsidian
A_p is never 0 unless A is zero right?
like, you shouldn't be able to kill 1 unless 0 is in A\p
I mean actually you shouldn't be able to write A_p since no such p exists if A is 0
latent anvil
A_p is never 0 unless A is zero right?
yes, because A_p has a quotient which is a field
:^)
next obsidian
If you have a monic polynomial with all coefficients in A for A an integral domain, is it irreducible over A iff it's irreducible over Frac(A)?
If A was a UFD this is just Gauss's lemma
next obsidian
I think this is false in general
celest brook
ugh...what would be a way to slip past proving all the possible combinations? I mean the 1 step subgroup sets says I have to prove for ALL a,b in a subgroup H of G that ab^{-1} in H
next obsidian
You could do it in a very slick way
Let G be the group of symmetries of the square
make a group homomorphsim G -> GL(2,R)
prove the image is exactly that set
the image of any group homomorphism is a subgroup
next obsidian
Let A be an integral domain, and K its field of fractions. Let alpha be an element in B > A another integral domain. Considering alpha as an element of Frac B, let f be alpha's minimal polynomial over K.
If it happens that alpha's minimal polynomial (monic) over K has all coefficients in A, is it true that f divides g(x) in A[x] if and only if f divides g(x) in K[x]?
Is this something about the euclidean algorithm working the same way?
Oh it totally is, sick
I sure am glad I now understand the Euclidean algorithm
latent anvil
If you have a monic polynomial with all coefficients in A for A an integral doma...
Any non normal ring A provides a counterexample, I think
Eg A = k[t^2,t^3]
The field of fractions is k(t)
t satisfies the polynomial x^2 - t^2
Which is monic and irreducible over A but not over k(t)
next obsidian
normal in what sense?
all localizations at primes are integrally closed domains?
Also I think you're right, there's a problem in the book which says that this is true if A is an integrally closed domain
Oh wait, lol I assumed it's a domain so you just meant an integrally closed domain for normal
latent anvil
Normal means integral domain which is integrally closed in its field of fractions in my head
next obsidian
I think it's common to have that be the definition
latent anvil
That's what stacks defines it as
next obsidian
the other definition I gave is like due to Serre and in EGA I think
Which for integral domains is actually equivalent
latent anvil
Probably
next obsidian
since A is integrally closed iff A_m iff A_p is for all m, p
latent anvil
I don't know what it means to be integrally closed if you're not a domain
next obsidian
yeah so that's the thing
For A to be normal under this more general thing
You need A_p to be integrally closed and an integral domain
latent anvil
Ah, gotcha
next obsidian
So that's why it's kinda weird
latent anvil
Anyways you see my point
next obsidian
yeah
latent anvil
Take something in the fof which is integral over it
Take like a minimal irreducible divisor of the polynomial
(that feels a little sketchy)
that will be a counterexample
But it definitely works for the ring I gave
next obsidian
Yeah
latent anvil
k[x, y]/(x^3-y^2)
candid pier
Hello , Can someone here help me with these 2 questions =
mild laurel
what have u tried
candid pier
so i think for first question it might be ** 5 elements ** but i am not sure .
mild laurel
and why do you think that?
candid pier
because ** injection ** , 1 to 1 .
mint seal
If it were isomorphic to a subgroup of a permutation with 5 elements, it would be a permutation group with 5 elements
By lagrange's theorem
But that's not good
Not only that, there is no (full) permutation group with only 5 elements
candid pier
so maybe ** 25 elements ** is correct answer ?
mint seal
Caley's theorem says any group is isomorphic to a subgroup of S_n for some n
The order of S_n is n!
mint seal
yeah
wraith obsidian
imo that question is just complicated because of misleading wording
You always hear the phrase „permutation group over n elements“ or „permutation group over a n-element set“ to denote n
…but what they write here is „a permutation group with n elements“
It's just confusing the cardinality of the group with the cardinality of the set it's acting on by ambiguous grammar
candid pier
imo that question is just complicated because of misleading wording
You always h...
What do you think of Question 2 ?
rich ravine
I’m looking for an explanation / source which covers the following thing. I’m ju...
Let $L/K$ be any normal (algebraic) extension, not necessarily finite. Denote $G=\mathrm{Aut}(L/K)$ and $F=L^G$.
cloud walrusBOT
Othenor
rich ravine
- L/F is Galois : we will use that L/F Galois iff $L^{\mathrm{Aut}(L/F)}=F$. But we have $\mathrm{Aut}(L/F)=\mathrm{Aut}(L/K)=:G$ since any automorphism of L/K fixes F, by definition of F. Thus the claim follows from the equality $L^G=F$
cloud walrusBOT
Othenor
wraith obsidian
What do you think of Question 2 ?
i think this is clearly worded in contrast to the first assignment 🙃
rich ravine
- F/K is purely inseparable. Fix an algebraic closure $\bar{L}$ of $L$. We want to show that any morphism $\sigma/K\to \bar{L}$ extends uniquely to $F$. Let $\tau,\tau'$ be any two extensions of $\sigma$ to $F$, and extend them further to $\bar{L}$. Then $\tau^{-1}\tau'$ fixes $K$ so by normality of $L$, $(\tau^{-1}\tau')_{|L}$ is a well-defined automorphisms of $L$ that fixes $K$ hence it fixes $L$. This show that $\tau_L=\tau'_L$
Now being purely inseparable can be checked on elements, i.e. on finite subextensions of the base field, where it is by definition "the separable degree is 1" i.e. "morphisms to an algebraically closed field can be uniquely extended", so you see that in the infinite case we indeed have (purely inseparable) iff (morphisms to an algebraically closed field can be uniquely extended)
cloud walrusBOT
Othenor
candid pier
cool , thanks
paper flint
Also, is this a quiz/test? 
candid pier
nah , just a small mock/ practice test .
paper flint
pine spoke
Hello! I just started going through a book on abstract algebra and have a basic definition question. (sorry if this is more appropriate for one af the questions channels). A relation p is transitive if a p b and b p c implies a p c. What if i have a relation where there does not exist any a, b, c where both a p b and b p c can't ever be simultaneously true? Would the relation not be transitive in that case?
hot lake
it would be transitive
pine spoke
Okay, thanks!
celest brook
You could do it in a very slick way
Sadly I haven't learned that 'slick' way, though my professor advised me something like this. Idk how it would help.
I mean I know it's true
I really really don't wanna draw out the multiplication table for both of them
sturdy marsh
writing out the mult table shouldnt be hard
celest brook
it's just a bit tedious
and I don't even know what I would do after I write up the table
just draw a bunch of arrows and list all of them and show an injective+surjective mapping between all 64?
sturdy marsh
anyway, you just need to show R and J satisfy the relations that you want them to
and then write down the isomorphism
celest brook
I already did prove they formed a subgroup of GL2
using determinant properties, and the table itself
celest brook
...generators?
celest brook
would I need to make D_2 in the form of matrices as well?
sturdy marsh
no
celest brook
how can I define such a mapping then?
sturdy marsh
wait what is your definition of D_2
celest brook
symmetries of a square
sturdy marsh
I mean do you have a definition of the form <r, s | ...>
where the "..." are the relations
celest brook
not that I recall
sturdy marsh
how exactly did you define "symmetries of a square"
if you have a very precise definition for what it means to be a symmetry
you could "draw" a square in R^2
and look at what the matrices do to it
celest brook
yes, that's how the book presented it.
I know we did it for symmetries of a triangle,
D_3
sturdy marsh
R rotates by 90
sturdy marsh
J flips
celest brook
wait
sturdy marsh
along an axis
celest brook
hmm
sturdy marsh
so draw a square centered at (0,0)
celest brook
ok
sturdy marsh
and look at what R and J do to it
celest brook
ahhhh
what would I do after?
would I need to look at all combinations of RJ and see what actions they coresspond to in D_3?
sturdy marsh
show that that group is "the symmetries of the square"
yeah show that you get all of D_2
celest brook
i.e. just show that the 8 elements also correspond to 8 elements in D_3?
sturdy marsh
yes
celest brook
yeah D_2*
getting used to notation
wait
why D_2
@sturdy marsh
do you mean D_4?
o
weird
mild laurel
It's cause the group of symmetries of an n-gon has 2n elements, so the algebraic people denote it based on the number of elements and the geometric people denote it on what its acting on
obsidian sleet
hello
is the converse true as well
i feel like it should but
maybe i am missing something
i cant really see how the converse could be false
mild laurel
whats the converse
cursive temple
if G_1 x H_1 is a subgroup in G x H then G_1 and H_1 are subgroups of G and H respectively
probably
viscid pewter
think so
mild laurel
Another example is if you take a group G and consider G x G, and then consider all the elements of the form (g,g) for some g in G. This is a subgroup of G x G, but isn't of the form you want
@obsidian sleet
viscid pewter
??
mild laurel
I mean yeah it's abstractly isomorphic to G, but as a set its not equal to a direct product of two subsets of G
Although I guess I misread what the converse is
cursive temple
i was just riffing with the "converse" so i could be wrong
viscid pewter
what?
no but like
well actually the converse is practically a tautology
<(g, g)> isn't a product of two subgroups
ok so statement is (A -> B)
then converse is (B -> A)
yeah
i mean wait
i might have screwed it up
ok do-over
if G_1 x H_1 is a subgroup in G x H then G_1 and H_1 are subgroups of G and H respectively
so if we have this
then G_1 and H_1 might not have to be subgroups
they could potentially be sets
they wouldn't be, but it might be non-trivial
cursive temple
Ouch
viscid pewter
yeah it's very easy
cartesian products are nice bc they have basically every reasonable nice property you can think of right
so relaxing
well it's just like
cursive temple
I mean G and H dont really interact with each other in any meaningful way in G x H
viscid pewter
GxH is a group, G and H are isomorphic to subgroups of GxH, those subgroups are normal, the quotient groups are isomorphic to H and G... etc.
yeah
it's peaceful difficulty
cursive temple
:D
celest brook
wouldn't I still need to check that the function between them satisifies f(xy)=f(x)f(y) for x,y in (J R?)
hmm
I mean I've proven the one to one correspondence and surjectivity
so would I still need to check painstakingly for all of them that this holds? I mean there could be some shortcuts since JR=R^3j
obsidian sleet
if G_1 x H_1 is a subgroup in G x H then G_1 and H_1 are subgroups of G and H re...
yes this is the converse i mean
i don’t understand zoph’s example i think
but i only asked because i wanted to try to use the converse somewhere in a proof but i think i can try to let leverage someplace else
viscid pewter
it's a subgroup but it's not a cartesian product of a subset in G and a subset in H
so it's not related
just a random subgroup
thorn delta
does anyone have any idea on how to approach part b? F[t] is acting on the space by multiplication
(context is developing RCF, JCF, SNF)
chilly ocean
if you follow your nose i think h(t)p(t) = lambda p(t) comes out to (t-alpha)^m divides (h(t)-lambda)p(t), so i guess this means h(t)-lambda should be a multiple of (a power of) t-alpha. not sure whether this is always possible
well, h(t) divided by t-alpha must have a scalar remainder, so that's 1 eigenvalue, at least
thorn delta
hmm, i mean, it seems like there should be a way to use part a. I'm trying to find the eigenvectors of t^n right now.
chilly ocean
so i think (t-alpha)^(m-1) is always an eigenvector of h(t). (and in a trivial sense there are infinitely more eigenvectors, any multiple of (t-alpha)^(m-1) by a polynomial without a t-alpha factor). but other eigenvectors, it's not as clear to me (maybe this has to do with generalized eigenvectors, lambda can be chosen to make h(t)-lambda have a factor of t-alpha, then (h(t)-lambda)^n p(t) = 0 for some n)
obsidian sleet
hello
so in a separate problem i was working on
i found a group which had a multiple of phi(d) elements of order d
i am correct in interpreting the group as not being cyclic?
er i guess i am asking if i can interpret the theorem as an iff
ooga
mild laurel
isn't that just the contrapositive of this statement
obsidian sleet
right that's what im gonna use when i write it down
oh
i don't even need to interpret it as iff
bruh
🤦♂️ ok ty
mild laurel
This statement is iff though. Since you have varphi(n) >= 1 elements of order n, that implies that there's a generator so the group is cyclic
obsidian sleet
ah alright
delicate bloom
does anyone have any idea on how to approach part b? F[t] is acting on the space...
http://mathb.in/51193 I typed up an answer, ask me anything if you have any problems with it
might as well paste it but I figured it'd be kinda long:
cloud walrusBOT
Merosity
delicate bloom
oh and in case it isn't clear, just as an example to show $h(t)=t^n= (\alpha+ (t-\alpha))^n = \alpha^n + (t-\alpha)g(t)$ and so $t^n$ has the eigenvalue $\alpha^n$ like you expect
cloud walrusBOT
Merosity
chilly ocean
admittedly i didn't read through the careful index and coefficient manipulations. but you are not saying that (t-alpha)^(m-1) is the only eigenvector right?
delicate bloom
I reason out that if f(t) is an eigenvector with some coefficient that's nonzero, it implies that all the coefficients of h(t) are 0 except the constant term
and so it's just multiplying by a scalar
the induction argument forces this cancellation on higher terms, so doesn't apply when you look at an eigenvector which has the highest term nonzero (in this t-alpha basis)
so this is proving the uniqueness too yeah
it might be much clearer if you just look at my formula when n=0 and try to work out the next case and see why it works
I kinda had a lot of messy scratch work on cases where I realized this was gonna be how it'd go
so that's why it comes across as a bit mysterious with the whole indices lol
chilly ocean
maybe this is just a technical issue, but eg (t-alpha)^(m-2) is also an eigenvector, for h(t) = (t-alpha)^2 + lambda, for any lambda
i think your induction is missing a basecase
thorn delta
^ i also noticed that.
delicate bloom
the base case is taken care of with the part involving "Then, $b_0c_n = \lambda c_n$ and so $b_0 = \lambda$."
cloud walrusBOT
Merosity
delicate bloom
by showing b_0=lambda that satisfies the base case and then we use this to show b_j=0 for j>0 since that was the j=0 case
it's a bit funky cause it's the only nonzero term while the higher terms are all 0 is what we're showing
@chilly ocean sorry for the delay on the response I was skyping my brother, figured I'd ping you so you see my response
thorn delta
anyway, thanks merosity! im going to take a close look at this a little later and ill let you know if i have any questions.
delicate bloom
oh was this your question originally whoops tagged wrong person xD
you're welcome, if there's a nicer proof that avoids the induction argument let me know lol
oh also realized a simple taylor expansion could have made the last comment explicit,
$h(t) = h(\alpha) + (t-\alpha) g(t)$, as in the eigenvalue is always $h(\alpha)$
cloud walrusBOT
Merosity
chilly ocean
ah ok, so i see that h(t) must be of the form lambda + (t-alpha)^(m-n)g(t)
delicate bloom
just a single (t-alpha) needed, g is completely arbitrary
there's an entire equivalence class of h(t) that have the same eigenvalue, mod (t-alpha)
chilly ocean
i agree what those two statements, but i think there is some care to be taken in identifying what the eigenvectors of h are
eg if we subsume all but 1 power of t-alpha into g(t), then it's hard to say what is the smallest power of t-alpha that is an eigenvector of h(t)
delicate bloom
no, that's what my proof shows, the only eigenvector is (up to scalar multiples) (t-alpha)^{m-1}
chilly ocean
maybe this is just a technical issue, but eg (t-alpha)^(m-2) is also an eigenvec...
but here is a counterexample
delicate bloom
I do this by assuming there's a nonzero c_n with n<m-1 and this ends up forcing that condition on h(t)
oh then my proof is completely wrong
well, probably it's salvageable but I'm not gonna try right now lol
chilly ocean
no, i agree that h(t) = lambda + (t-alpha)^(m-n) g(t), where g(t) has no factors t-alpha. but there is some annoyingness of turning this into identifying the eigenvectors of h, without already knowing it
yeah, algorithmically i think it could be something like: given h(t), divide it by (t-alpha)^(m-i) for i=1,2,3,.... if you get a scalar remainder lambda, then all eigenvectors are (t-alpha)^j for j=i,i+1,...,m-1.
delicate bloom
I'm thinking I messed up my induction argument at the end and in the step of showing all higher terms are 0 I actually could have shown that there's a ceiling where the last term happens which is nonzero
and I didn't think about the cancellation there
which might be enough to fix it, since that induction step entirely could work in just plain F[X]
I didn't use F[X]/<X-alpha>, so that's a red flag
rich ravine
Your indices for coefficients are wrong, it should be $i+j\equiv k[m]$
cloud walrusBOT
Othenor
rich ravine
Nah that's wrong, sorry I got confused
rich ravine
Ok so we can translate by $\alpha$ to reduce to the case where $\alpha=0$ (this is the same as writing things in the basis $(t-\alpha)^i$), and we can suppose without loss of generality that the degree of $h$ is less than $m-1$. Then we can write explicitly (exercise : do it) the matrix of the action of $h=\sum a_i t^i$ in $(1,t,..,t^{m-1})$. Observe that is is a lower triangular matrix with $a_0$ on the diagonal. Hence $a_0$ is the only eigenvalue, and the eigenvectors are the elements in $\mathrm{ker}(h-a_0\mathrm{Id})$. But since the matrix of $h$ is lower triangular with $a_0$ on the diagonal, the matrix of $\mathrm{ker}(h-a_{0}\mathrm{Id})$ is lower triangular with zeros on the diagonal ; in particular it is column-reduced. The last column is zero, corresponding to the eigenvector $t^{m-1}$ with eigenvalue $a_{0}$, and the i-th column starting from the right will be zero if and only if $a_{1}=\ldots=a_{i}=0$, in which case all the columns to its right will also be zero. Thus let $i$ be such that $a_j=0$ for $1\leq j \leq i$ and $a_{i+1}\neq 0$ ; then we read on the matrix that a basis of eigenvectors for the unique eigenvalue $a_0$ is $t^i,...,t^{m-1}$.
cloud walrusBOT
Othenor
vestal snow
Suppose $P$ is a place of a function field $F$ over $k$ where $k$ is algebraically closed. Is it true that the if there is some holomorphic differential $\omega$ with order $n$ at $P$, then $n$ is/isn't a gap number at $P$?
cloud walrusBOT
Have a Banana, Bitch
vestal snow
I'm trying to prove something and either is/isn't would work
@prisma ibex So you can see
terse orchid
how complicated is the Galois thereom? I would like to do my extended essay on it, but I don't know if I'll be able to get to the bottom of it without reaching math i cannot do
thanks!
mild laurel
Which theorem are you talking about?
I think regardless, it's pretty beyond the reach of a high school student, depending on your background
terse orchid
symmetry groups and their relation to quadratic - cubic - quartic and quintic formulas
trying to show why a quintic formula cannot be possible
it is supposed to be something that challenges you but of course I won't try it if I can't
mild laurel
Well, do you know any abstract algebra?
terse orchid
do you know anything about IB, the school program?
mild laurel
I graduated with an IB diploma yeah
mild laurel
oh huh, I didn't know that existed
terse orchid
terse orchid
formula sheet to get some understanding
mild laurel
So you need some basic group theory knowledge, probably like half a semester of a college course's worth
Technically there are no prereqs for group theory, but abstract linear algebra helps a lot
Also group theory is more proof oriented and it can be pretty hard to adapt
terse orchid
yeah hmm
i picked math for my EE
idk if it was a good move tho
i don't need to prove the theorem
only explain it
mild laurel
even like explaining the theorem and where it comes from is pretty difficult
terse orchid
hm
would something like
group theory real world applications be too broad?
maybe I could look at one specific application
mild laurel
I don't think so, but that's mostly cryptography
And that's a pretty reasonable EE
one of my friends explained RSA for his EE
cursive temple
Arent ruler and compass constructions a slightly easier topic
terse orchid
hehe
mild laurel
you could just explain the basics of group theory and talk about some geometric notions of groups like groups acting on the rubix cube or something
terse orchid
i guess RSA would go well since i also take cs
cursive temple
I remember there being a somewhat nice mathologer video on the topic
terse orchid
the reason why i thought of group theory was becayse of the 3blue1brown video
thorn delta
yea, its not too bad. You just need to understand what the degree of a field extension is.
mild laurel
what would you even talk about for ruler and compass constructions idk
i mean defining a field extension, and talking about its degree already requires abstract linear algebra
cursive temple
Sure
mild laurel
not to mention all the notions of groups and fields
mild laurel
Kinda along the same lines of the quintic impossibility, you can use galois theory to show that certain compass and straightedge constructions are impossible
cursive temple
terse orchid
that looks interesting
thorn delta
i feel like it might be possible to write a paper on. idk it depends on how technical you have to be
its definitely possible to understand
terse orchid
well i'll take a note of all of these and check with my teacher tomorrow
i think the doubling cubes would be nice tho
is there any other interesting question that comes to ur mind
along any lines of symmetry
ive still got tons of research to do
mild laurel
Symmetry is super broad haha, groups show up literally everywhere in math
thorn delta
moonshine 
mild laurel
haha, I gave a talk about it in #events a couple weeks ago, but its pretty complicated
chilly ocean
I want to "research" a "moonshine" 
mild laurel
I mean, monstrous moonshine relates the two seemingly separate fields of finite simple group theory/representation theory and the theory of modular forms so understanding both of these would be pretty hard
chilly ocean
what was your extended essay about, zoph
mild laurel
If you're talking about Gannon's book, that's meant for grad students
uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
I wrote about P vs NP I think
chilly ocean
i see
mild laurel
It's mostly a CS problem about algorithms
terse orchid
scarlet estuary
fun fact: gannon taught me intro analysis
mild laurel
oh cool
terse orchid
what u get
thorn delta
i still like the ruler/compass constructions / why you can't trisect angles, double cubes, square cirlces, etc...
It seems like there would be a lot to write about other than the actual field theory.
terse orchid
yes ive moved that up to first on my proposed RQ
mild laurel
I mean I guess if you need the points to get the diploma
but that wasn't really an issue for me
terse orchid
what score
scarlet estuary
i think gannon's moved to berkeley since then
terse orchid
for diploma
scarlet estuary
he was a pretty good prof
though its hard to judge someone as a prof based on an intro course
especially since i skipped half the classes
mild laurel
don't you only need 24 points for the diploma or something
chilly ocean
Gannon sounds like a pretty cool name, almost like ganondorf
terse orchid
irrelevant but i was just asking what u got overall in ib
mild laurel
oh
uh
37 I think?
I don't remember what I got on the tok points
but I got 7 on HL math, 6 on HL English, History and Physics, and 5 on SL Spanish and CS
mild laurel
idk this was like 6 years ago
mild laurel
yeah, I can talk more about moonshine too if you want to hear more about that
There's a lot of "symmetry" happening there too
terse orchid
yeah as soon as i can ill watch ur talk
mild laurel
I don't know how much of it you'll be able to understand, I gave it at a slightly higher level
viscid pewter
it's not
thorn delta
what are R and A supposed to be here? Subrings or ideals?
next obsidian
It’s just like, take r in R, a in A, then look at the set of ra. Now close this under addition
next obsidian
It’s pretty useful to be able to have a notion of being able to take the product of two subobjects or something
This is kind of like “we want all possible products, but it’s not closed under sums, so we just throw in all sums of the products”
unique juniper
ok thanks, i think i understand.
thorny mountain
need help
unique juniper
i cant do this
my initial approach was, x,y in R such that xy =0 then x or y is in P but got stuck there
thorn delta
P is prime means: if x and y are not in P then xy is not in P
unique juniper
think we have different definition?
thorn delta
well, its probably the contrapositive of your definition
xy in P => x or y is in P
(for a commutative ring)
x or y is in P
unique juniper
nvm
chilly ocean
this is a confusing wording though. when it says "P contains no zero divisors", does this mean P is an integral domain itself, or that none of the zero divisors of R are in P?
(i think it must be the latter, but it is weird)
thorn delta
im 90% sure it means the latter lol
chilly ocean
yeah, but it is ambiguous
thorn delta
i think its pretty common to think of stuff like this when working with rings of fractions (whether your multiplicative subset contains zero divisors)
unique juniper
ok
mild laurel
Is there a difference between those two?
unique juniper
i think yes, cant be sure that P has identity ?
mild laurel
I mean ok if we require that integral domains have identity then the first is never true yes
Anyways
You're on the right track
if xy=0, then you know that x or y is in P like you said
But you also know that x and y are both zero divisors right
thorn delta
i think there is a typo here. Does anyone know the correct statement of Smith normal form? I can interpret this as phi is an injection which has the matrix representation of the transpose of the matrix above in some basis, or i can interpret phi as a surjection which has the matrix representation of the above matrix in some basis. Which is it?
chilly ocean
what are you reading? these look nice
thorn delta
ah, my professor wrote this.
latent anvil
New tterra pfp(!)
chilly ocean
ah, my professor wrote this.
do you mind sharing? my algebra class has been touching on this stuff recently
latent anvil
Yeah kxrider this seems sus
chilly ocean
it's okay if not, e.g. dox
latent anvil
Ah wait
Yeah I think they just need the transpose of what they wrote
I like aluffi's section on Smith normal form
I think I've only ever heard it stated for maps Z^n -> Z^n
thorn delta
yea terra, i can dm it. its like a self-guided worksheet on canonical forms and stuff
thorn delta
ok, so the conclusion ive come to is that the way its stated is contradictory, but its probably easier to assume phi is injective to apply this theorem
that way the image of phi is isomorphic to a subgroup of Z^n.
next obsidian
Now being purely inseparable can be checked on elements, i.e. on finite subexten...
I forgot to say thanks for this! It helps a lot
unique juniper
anyone know any good resources/books on rings?
chilly ocean
d&f
unique juniper
besides that?
nova plank
Rotman