If you mean irreducible in k[X], that does not work, because X is irreducible in k[X]
#groups-rings-fields
ripe crest
in k(X)?
ripe crest
Okay - I did say that doesn't work
ripe crest
Oh
woven delta
Lol
smoky cypress
how do you prove that 1, π, π^2, π^3, ... are linearly independent over Q
@mild laurel
stone fulcrum
Do you have that π is transcendental?
smoky cypress
Sure
stone fulcrum
I mean this is equivalent to that, and good luck proving π is transcendental
woven delta
Why is this equivalent to that?
stone fulcrum
Let's say they're linearly dependent. Then finitely many are linearly independent. But then there's a minimal poly, which π can't have cuz transcendental
I think I'm missing the case where there's still infinitely many
But not every integer
woven delta
Oh hmm
stone fulcrum
For some reason I feel like that's an obvious write-off but I can't think of why
woven delta
Okay so that was transcendental implies linearly independent
That seems like the easy case
Why does linear Independence imply transcendental
stone fulcrum
That's not even the complete proof. I can't justify dropping to finitely many
woven delta
Lol you can
upper pivot
linearly independent on an infinite set for basis means theres no finite non-trivial combination that is 0
woven delta
Yeah
stone fulcrum
I'll have to look into that later, but it seems right
woven delta
But actually after thinking about it for a sec both directions seem obvious
woven delta
Yeah
upper pivot
${x^n}$ linearly dependent $\iff$ $\sum a_i x^i =0 \iff $ $x$ is algebraic
cloud walrusBOT
JohnDoeSmith:
clear obsidian
Soft question, how much background is typically recommended before I start AG? Is Galois or any commutative algebra important at all to study beforehand?
upper pivot
commutative algebra, thats what i have been told is needed for AG
bleak abyss
Commutative is generally important (depending on the treatment you may not need thaaat much), some point-set, ideally basics of category theory
mild laurel
Galois comes up but it's not super important
clear obsidian
Ok thank you, I’ll probably dive into some commutative first then
rich patrol
oblique river
haha
ripe crest
im curious how other people are doing with academics rn
like, are y'all still studying hard?
are you busy at home?
are you sick?
wrong chat sorry
chilly ocean
cool
ripe crest
I just thought this would be better suited for chill
smoky cypress
Let's take this to #discussion
ripe crest
ok
covert perch
Can someone tell what lie groups are and some source to read about them
zenith heath
Lie groups 😂
scarlet estuary
a lie group is a "continuous group", in a certain sense
formally, theyre groups that are also smooth finite dimensional real manifolds
and group multiplication (and its inverse) are smooth
i.e. the map $G\times G\to G \colon (x,y) \mapsto x^{-1}y$ is smooth
cloud walrusBOT
Namington:
scarlet estuary
as an example, consider GL_2(R)
i.e. the real matrices $\begin{pmatrix}a&b\c&d\end{pmatrix}$ such that $ad-bc \neq 0$
cloud walrusBOT
Namington:
scarlet estuary
this is a lie group
anyway, it's generally recommended to only look into lie groups after some experience with group theory and differential geometry
like you shouldn't be afraid of the phrases "smooth manifold" or "tensor ___"
Lee's Introduction to Smooth Manifolds is a good source if your diff geo background is a bit shaky/nonexistent
and it also covers lie stuff
(Lie and Lee, incidentally, are pronounced the same)
if you have some diff geo experience, the usual recommendation is something like Knapp's Lie Groups: Beyond an Introduction
there's also Fulton and Harris, which covers lie stuff from a rep theory standpoint
which is in some ways the most natural way to study lie groups (and to study representations!)
but it's a bit of a tricky text
covert perch
Thank you so much. I have done group theory but need to study diff geo properly. Will look into it!
halcyon siren
Err, why would the set of $z \in \mathfrak{g}{0,x}$ such that $(\text{ad}(Z)\vert{\mathfrak{g}{0,x}})^{\dim(\mathfrak{g}{0,x})} \neq 0$ would be open?
cloud walrusBOT
Райан:
halcyon siren
I mean, is the adjoint representation of a Lie algebra continuous or something?
vale coral
Can anyone explain subgroups of quotient groups e.g. Z/7Z. My initial thought was that ever smaller quotient group would be a subgroup of the group. However, I have found out that I misunderstood the concept. If someone could give some clarification that would be nice.
bleak abyss
So you have these "isomorphism theorems"
And one of them is this correspondence
Namely, consider the quotient map G->G/H
If you give me a subgroup of G/H, what's its preimage? That'll be a subgroup of G containing the kernel, which is H
vale coral
This is the first isomorphism theorem?
bleak abyss
First isomorphism theorem is that if you give me some homomorphism G->G', then the G/ker is isomorphic to the image
This is just a matter of, let's call pi:G->G/H the quotient map
And let's say K is a subgroup of G/H
Then pi^{-1}(K) contains pi^{-1}(0) = H
Since K contains 0
And this, I guess 3rd or 4th isomorphism theorem? Whichever one it is, it says that the map sending K\le G/H to pi^{-1}(K) is actually a bijection from the set of subgroups of G/H to the set of subgroups of G containing H
vale coral
I'll have to digest this :). Could you provide a small example as well e.g. for Z/6Z. Would be awesome!
bleak abyss
Sure
So subgroups of cyclic groups are cyclic, have you seen this fact before?
Assuming the answer is yes, we can go through the elements of Z/6Z and find out what subgroups they generate. These will be all the subgroups of Z/6Z
So we get {0}, {0,3}, {0,2,4}, and {0,1,2,3,4,5}
Note the first one is trivial, the second is isomorphic to Z/2Z, the third is isomorphic to Z/3Z, and the last to Z/6Z
Well what are the subgroups of Z which contain 6Z?
Z, 2Z, 3Z, and 6Z
So that's what's going on
smoky cypress
Didn't he say Z/7Z?
chilly ocean
Also note that there exists exactly one such subgroup of a particular order
bleak abyss
Z/7Z at first
But then "I'll have to digest this :). Could you provide a small example as well e.g. for Z/6Z. Would be awesome!"
Z/7Z is kinda trivial lol
tribal pasture
The remark about the equivalence of surjectivity of phi'. Isn't that the very definition of surjectivity? That is to say for every element in the codomain, there exists an element in the domain?
ripe crest
Haven't looked at that yet - but it's for every element in the codomain - not image
by definition, for every element y in the image, there exists an element x in the domain s.t. $x \mapsto y$
tribal pasture
Sorry yes, edited
ripe crest
Which part are you asking about? the f being in the image of phi'?
tribal pasture
The part about "the map phi' is surjective if and only if every hom..."
Seems like just saying "the map phi' is surjective if and only if every element in Hom(D,N) aka codomain lifts to an element in Hom(D,M) aka the domain" which is basically just repeating the definition of surjectivity.
gloomy garnet
hello everyone. I'm in abstract algebra along with @cerulean siren. We're currently working with Unique Factorization Domains, Euclidean Domains, Principal Ideal Domains, and Polyomials. Good to meet you all!
ripe crest
hi!
gloomy garnet
I've got a quick question that may sound silly. Say you have an element r in a ring R and then you look at that same element in R[x], the ring of polynomials with coefficients. What do those elements look like?
Can't an element r only be in both if it's an element of R?
Like you can't say a polynomial f in R[x] is also in R if it's degree is higher than 0?
ripe crest
R injects into R[X], right?
gloomy garnet
the elements of R
ripe crest
Right, so is there a copy of R in R[X]?
gloomy garnet
yes
ripe crest
I think @mild laurel might be better to answer because what i say might be confusing. But I'll try: we can associate (1, 0, 0, 0, ...) with 1 in R but we could associate (0, 0, 1, 0, ...) with 1 in R as well. So even constant polynomials aren't "in" R. There is an isomorphism between them though. Likewise there is an isomorphism between polynomials with degree n with only 1 term and R.
Like R[X] is isomorphic to R[N] - and if you know about group rings, this is a monoid ring. We have countably infinite copies of R in R[X].
And by R[N], I mean $R[\mathbb{N}]$
cloud walrusBOT
silent flower:
ripe crest
does that help or was that just confusing ... because i can see why what i said could be confusing
tribal pasture
I've got a quick question that may sound silly. Say you have an element r in a ring R and then you look at that same element in R[x], the ring of polynomials with coefficients. What do those elements look like?
They are of the form r_0+r_1x + r_2x^2 ... +r_nx^n
So r is an element of both R and R[x]
And yes f is not an element of R if it's degree is greater than 0
@gloomy garnet
gloomy garnet
great that's what I thought
ripe crest
Strictly speaking though, f is not an element of R even if it's degree is 0
gloomy garnet
hmm
tribal pasture
How so?
cerulean siren
well, it depend how formal you want to be
I think everyone would agree 1 is a member of both Z and R
but if you see R as a polynomial ring then technically it's 1*1_R
gloomy garnet
by R do you mean the reals?
tribal pasture
Yeah but is there some criteria upon which it is correct to say that degree 0 polynomials arent elements of the parent ring?
Aside the trivial one
cerulean siren
yeah $\mathbb{Z}$ and $\mathbb{R}$
cloud walrusBOT
jynelson:
ripe crest
I guess it depends on your definition of R[X]. I learned that it was a linear combination of r_i and X_i such that a finite number of r_i were non-zero. So an element of R[X] looks like r_0 + r_1X_1 + ... + r_nX_n and we dont write the terms where r_i = 0. But formally, it's r_0 + r_1X_1 + ... + r_nX_n + 0X_n+1 + 0X_n+2 + ...
Or, (r_0, r_1, ..., r_n, 0, 0, 0, 0, ...)
So, the constant polynomials of R[X] are of form (i, 0, 0, ...)
where only the first entry is non-zero
clearly that is not an element of R
But there is an isomorphism between them
cerulean siren
I thought that was R[[x]]? i.e. polynomial rings have a finite number of terms but power series have an infinite number
gloomy garnet
yeah that's what we learned
ripe crest
R[[X]] means you can have an infinite number of non-zero coefficients
in R[X] only finite
it's the non-zero coefficients that are finite or not
cerulean siren
hmm and you're saying the terms are still there in R[x] even though the coefficients are all zero
seems a little odd but as long as it behaves the same I guess it doesn't really matter
ripe crest
yeah, it's just a page or so up
tribal pasture
Yeah but as you said formally, it is also equal to ∑r_ix^i. So if all r_i are 0 except the first one we get ∑r_ix^i = r_0 which is an element in R. No?
ripe crest
starts at: sonicollin asking their question
gloomy garnet
I'm a "he" fyi
oblique river
I didn't read all of the responses, but yes I would say that "2" is both an element of Z and of Z[x]
the image of R into R[x] is exactly the constant polynomials
ripe crest
Right - but you could also inject R into R[X] via the map r -> rX^2
oblique river
where I guess the definition of a constant polynomial is one that doesn't have any terms of positive degree
yes, although that's not a particularly nice map algebraically
it's not a ring hom
but it is an R-module hom
tribal pasture
Indeed but what if the claim is that elements of R aren't polynomials but rather (say integers?), can one say that R can be identified as the equivalence class of all such descriptions?
ripe crest
That's true
But what about:
I guess it depends on your definition of R[X]. I learned that it was a linear combination of r_i and X_i such that a finite number of r_i were non-zero. So an element of R[X] looks like r_0 + r_1X_1 + ... + r_nX_n and we dont write the terms where r_i = 0. But formally, it's r_0 + r_1X_1 + ... + r_nX_n + 0X_n+1 + 0X_n+2 + ...
Or, (r_0, r_1, ..., r_n, 0, 0, 0, 0, ...)
So, the constant polynomials of R[X] are of form (i, 0, 0, ...)
where only the first entry is non-zero
clearly that is not an element of R
But there is an isomorphism between them
oblique river
I see where you're coming from, but it's kind of like saying that 3 isn't a rational number
because it's not written like 3/1
or maybe mroe formally a rational number is an equivalence class of pairs of integers
ripe crest
Yes
I'm okay with saying 3 isn't rational in some contexts and saying it is rational in others
oblique river
In this case, for any ring R, there is a unique ring map R --> R[x]
and it's the one that takes r to r + 0x + 0x^2 + ...
tribal pasture
Ye because that's what I think the disagreement is about at the end. Because if one follows that R[x] is basically an infinite-tuple, then it doesn't follow that R sits as a subring inside R[x], where it does if we see R[x] as a polynomials
tribal pasture
fnx
ripe crest
you're right
tribal pasture
But I would argue abstracting away from all these element pov is basically what we tryna do in algebra
So long there exists an isomorphism, they're the same
ripe crest
Eh
oblique river
I mean you should be careful with that
isomorphism isn't the same as canonical isomorphism
tribal pasture
wdym?
oblique river
but in this case I really think it's fine -- there is a unique R-algebra map from R into R[x]
and so there is a canonical way that R sits inside of R[x]
I mean, if you said like "let r be an element of R and consider it as an element of R[x]"
everyone would know what you meant
ripe crest
You're right, and thanks for pointing out that it worked as R-modules
@gloomy garnet , so to answer your question, you're right - sorry for being confusing xD
oblique river
Yeah, if you're only thinking about R-modules, then R[x] is isomorphic to the countable direct sum of copies of R
gloomy garnet
no problem my guy
oblique river
but then you wouldn't really write R[x]
gloomy garnet
just getting back in the groove after two weeks of spring break, haha
oblique river
sorry I didn't really read all the discussion beforehand so I hope I was useful
ripe crest
Yup! thanks!
tribal pasture
Any idea where can I look the proof for "More precisely", especially the backward implication
oblique river
Isn't that kind of the definition of a short exact sequence?
like, 0 -> A -> B -> C -> 0 is a SES exactly when (the maps are homs, as they are here, and) there is a bijection between elements of B and pairs of elements (a,c)
given by b = (image of a) and c = (image of b)
tribal pasture
Oh the def I was given was Im ѱ' = kerφ'
tribal pasture
Oh yes I see what you mean. There is an equivalence between this and showing that there exists f,g such that g o φ is an identity and ѱ o f is an identity. Which I guess can be used here
Right? @oblique river
oblique river
sorry what do you mean "an identity"
ripe crest
i think he means an identity map
tribal pasture
ripe crest
D&F is the official text, but the prof provides good lecture notes, so I rarely look at D&F. I sometimes look at Aluffi though
tribal pasture
my prof make us write notes and I skipped that class 😦
tribal pasture
given by b = (image of a) and c = (image of b)
@oblique river Can you clarify what you mean by image of a and image of b?
oblique river
in the SES 0 --> A --> B --> C --> 0
there are maps A --> B and B --> C
I meant "image under those maps"
I just didn't give them names
here is an exercise if you haven't done it yet: 0 --> A --> B is exact if and only if the map A --> B is injective
tribal pasture
Yeah I have done that
oblique river
cool
tribal pasture
Oh smh..........
So it was literally just following the defintion
Very sorry for the silly question
oblique river
that's part of what you need to say in general that if 0 -> A -> B -> C -> 0 is exact then there is a bijection between B and AxC
tribal pasture
Yes yes got you completely
oblique river
note that i'm not saying isomorphism -- B and AxC need not be isomorphic
just that there is a correspondence
👍
tribal pasture
Yes just a set-theoretic bijection
ripe crest
What's the idea behind showing that an exact sequence doesn't split>
Like, to show it splits, we just need to find a homomorphic section
oblique river
well lots of exact sequences don't split
ripe crest
But to show that it doesn't split, we need to show that such a homomorphism is impossible
oblique river
if it splits, that means it's particularly nice
oh I see
to show something doesn't split
yeah, as a toy example, think about the SES
0 --> Z/2Z --> Z/4Z --> Z/2Z --> 0
ripe crest
Right - sorry. I meant more along the lines of, if i wanted to prove that an exact sequence didn't split, what would be the idea behind the process of showing it doesnt split
ripe crest
yup
oblique river
that is a SES which doesn't split
so see if you can find a reason "why" that you might be able to use as a template for other ones
oblique river
yeah, it's definitely a short exact sequence
bleak abyss
Math patch notes: all modules are now projective
ripe crest
which patch are we on now?
i can never keep up
yeah, it's definitely a short exact sequence
ohhhh, that's part of what you were saying above
oblique river
yeah sorry, to clarify: it is a short exact sequence, but it is not split
you're welcome to verify that it's a SES to practice
and then try to think about why it's not split
final dove
how is one supposed to read an exact sequence? I find it very unnatural to read, I need to keep reminding myself of the rule A -f-> B -g-> C means that ker(g) = im(f)
I'm studying module theory rn
oblique river
what do you mean "read"?
final dove
like
bleak abyss
Remember that doing one map and the next is 0
final dove
if I see a sequence 0 -> A -f-> B I believe that an experienced sequener can simply "read off" that f is injective
but I don't see the intuition
bleak abyss
(I'm guessing you mix up whether it's ker(g) = im(f) vs ker(f) = im(g)?)
final dove
@bleak abyss can you expand on that?
oblique river
what do you mean "intuition" here?
final dove
@bleak abyss I can resolve that by thinking about where kernel and image live. It's more that
ripe crest
How about an image to help u out
oblique river
it's just an exercise that 0 -> A -> B exact iff A -> B injective
bleak abyss
And yeah for facts like 0->A->B being injective, honestly it's more about like
bleak abyss
Doing the exercise and getting used to the fact a bunch of times
oblique river
it shows that "exactness" as a concept is a nice framework for understanding properties of maps that we carea bout
final dove
it feels more like "pattern recognition" than "insight"?
bleak abyss
Kind of is
bleak abyss
Like you just get used to it and then get quick at it
final dove
OK
bleak abyss
But it's not like there's something you should really "smell"
oblique river
short exact sequences are everywhere and recognizing when something is a SES is good
oblique river
because you want to understand B
final dove
B?
oblique river
so you squeeze it between A and C (which you nderstand better)
ripe crest
final dove
oh you mean in the sequence
bleak abyss
The nice thing is that they come up a lot and you can do computations
oblique river
and now, voila, the magic of exact sequences tells you something about B
final dove
@ripe crest indeed, I love Ghirst's book
ripe crest
(Y)
final dove
@ripe crest but I feel he skips over far too much sometimes
bleak abyss
One sneak peak of what'll come later is that you'll learn various types of "cohomology"
final dove
@bleak abyss I know De Rham cohomology
bleak abyss
And a theme is that you often get these things called long exact sequences
bleak abyss
That and also, oftentimes you can compute these groups using these long exact sequences
Are you familiar with homotopy groups?
final dove
yeah
tribal pasture
@oblique river The "theorem" that I posted asserted that I can identify middle as a direct sum only if one of the two conditions are satisfied. Is that false then?
bleak abyss
Generalizations of fiber bundles
tribal pasture
Here is it again.
oblique river
that's true
it just wasn't relevant to the question that we were discussing at the time
bleak abyss
So for example, there's a map S^3 -> S^2 given by (z,w)->[z:w] (thinking about S^3 \subset C^2 and thinking about S^2 as CP^1)
This happens to be a fibration with fiber the circle
final dove
hopf?
bleak abyss
Yup
final dove
OK
bleak abyss
So you actually get a long exact sequence of homotopy groups
tribal pasture
So for me to know that the hom seqeunce provided in the passage is exact implies that F can be set with bijection with (f,g), doesnt that first require me to establish either 1 or 2?
oblique river
no
final dove
@bleak abyss I have not studied math formally. I self-studied the math I know, so I may have huge holes
oblique river
this is what I was talking about earlier
final dove
just, like, letting you know.
oblique river
there's a difference between saying that B is isomorphic to A x C
bleak abyss
... -> pi_n(S^1) -> pi_n(S^3) -> pi_n(S^2) -> pi_{n-1}(S^1) -> ...
final dove
mmhm
oblique river
and saying that there is a bijection from B to A x C
bleak abyss
But the nice thing is that we know the homotopy groups of the circle, aside from dimension 1 they're all 0
oblique river
the conclusion of the theorem is the first statement
but the original question was abotu the second.
bleak abyss
So let's say n-1 > 1, meaning n\ge 3
tribal pasture
Oh I see, but then all the exact sequence is telling me is that one map is surjective, and the other is injective. How did we go to bijectivity from that?
bleak abyss
Then our exact sequence is 0->pi_n(S^3)->pi_n(S^2)->0
tribal pasture
(sorry if it is repetitive)
bleak abyss
Which means they're isomorphic
final dove
huh
bleak abyss
In general exact sequences are often really nice for doing computations with
final dove
hmm, so you can glean facts by putting together data you know + the fact that stuff is arranged in an exact sequence?
oblique river
I'm asking you to do that as an exercise. if you have a sequence 0 -> A -> B -> C -> 0 (all maps are homs), then if the sequence is exact then there is a bijection between B and AxC
and conversely
bleak abyss
Yeah
tribal pasture
Lemme try
final dove
I see
bleak abyss
That's like a more concrete instance of "What irl info can I glean out of an exact sequence?"
final dove
right.
tribal pasture
All maps are homs mean R-mod hom. right?
final dove
I also know that you can write things such as semidirect products as an exact sequence, I think?
bleak abyss
Saying that G is a semidirect product of H and N is the same thing as saying there's a short exact sequence 0->N->G->H->0
Good exercise would be to prove it
final dove
mm
oblique river
wait, that's not true though
final dove
yeah, let me try that, actually
ripe crest
And if they're abelian groups, it's a direct product!
final dove
sounds enlightening
oblique river
yes
bleak abyss
Right
oblique river
just because you have an exact sequence of groups doesn't mean it's a semidirect product
but if it's split exact, then you have one
(and every semidirect product gives you a SES)
final dove
how do you define split for group?
bleak abyss
Same thing as for R-modules
final dove
don't we need a 'direct sum' operation?
bleak abyss
Product in the finite case
oblique river
you have to be careful
bleak abyss
Also you can define it as like, section
oblique river
in R-modules, having a section C->B is equivalent to having a retract B->A
but that's not true in groups
ripe crest
ok so, I know from class that somehow showing the intersection of the images of Z/2Z in Z/4Z have non-trivial intersection means that it doesn't split
oblique river
example: any semidirect product that's not a direct product
final dove
OK, what is a section of an algebraic object now
I know "section" from vector bundles
oblique river
a section of a surjective map B --> C is a map C --> B such that C --> B --> C is the identity
that's a general definition
for functions
ripe crest
not sure why. like I can go from 0 -> A -> B -> C -> 0 where those are abelian groups and I understand why showing A+ C = B and A \intersect C = 0 shows that B = A x B from group theory
oblique river
sorry @ripe crest I'm not sure exactly what you're trying to do here
oblique river
one way to show that it's not split is to just think about what the possible section of the map Z/4Z --> Z/2Z could be
final dove
I feel this might make it far more enlightening to me
bleak abyss
Well, you have a surjection B->C
oblique river
a SES splits if the surjective map has a section
bleak abyss
You say it's split if there's a section
ripe crest
Like, in our lecture notes, it says that showing an exact sequence does not split basically amounts to showing that A and C have non-trivial intersection in B
final dove
if there is a section of what?
bleak abyss
So for instance, if you have a semidirect product of groups
oblique river
that's kind of an odd way to put it @ripe crest -- how are you getting C as a subset of B?
bleak abyss
Of the map B->C
oblique river
@final dove sections are things that surjective maps can sometimes have. in a SES there is one surjective map, namely the one from B to C
so when someone is talking about it having a section, they mean that map
bleak abyss
Since we're not in the abelian case I'll use the multiplicative notation to be careful
final dove
OK, so in the tangent bundle case, just so I understand, the surjection is the mapping from "tangent bundle -> manifold"
bleak abyss
So let's say G is a semidirect product of N and H
oblique river
yes
ripe crest
Take any injective homomorphism C -> B. If there isn't one, we're done. If there is one, then we can identify C as a subgroup of B.
final dove
and the section is the mapping from "manifold -> tangent bundle"
oblique river
that's correct bollu
bleak abyss
Yup, so a section of the tangent bundle is the vector field
final dove
mmhm
oblique river
@ripe crest I'm still a bit uncomfortable with that, because sometimes there won't even be an injective map C --> B. also, just because there is one that has nontrivial intersection with A, that doesn't tell you anything
ripe crest
one way to show that it's not split is to just think about what the possible section of the map Z/4Z --> Z/2Z could be
Yeah, I did that. It's just 0-> 0, 1 -> 1. Or 0 -> 0 or 1 -> 3.
oblique river
0 --> Z --> Z+Z --> Z --> 0 is a good example
but those aren't homomorphisms
so those aren't sections
oblique river
then that's not a section either
ripe crest
errr
bleak abyss
So yeah that gives us a map H back into G, this is a section of the map G->H
ripe crest
you're right
oblique river
a section of B --> C means a map C --> B such that the composition C --> B --> C is the identity
tribal pasture
Is the bijection from A,C to B given by f-¹(a + gf(c)) where f:A ->B and g:B->C and f-¹ exist by being injective on the image
bleak abyss
So that says that if we have a semidirect product, we get a split exact sequence of groups
I'll let you prove that the other way holds too
ripe crest
so basically, we had 1 degree of freedom, where 1 could go, and we found taht any place it landed, we didn't get a section, hence that exact sequence doesn't split
oblique river
yep
final dove
@bleak abyss we need to prove it's a section though? I suppose it will be
oblique river
that's it
bleak abyss
Yeah I skipped that detail true, exercise show that this is the case
ripe crest
Nice
oblique river
and the "issue" is that you have an element of order 4 mapping onto an element of order 2
ripe crest
That's a good approach for small groups
bleak abyss
Unpack all the relevant maps, show that inclusion of H into G is a section
ripe crest
does it generalize?
bleak abyss
And then show the converse
oblique river
and you can't go backward -- an element of order 2 can't map to an element of order 4
ripe crest
oh i see
final dove
now I go to bed 🙂
ripe crest
but ilke
oblique river
I'm not sure if there's a nice short "theorem" you could get out of this that would generalize
it's just an example of something you can look for
it's generally hard to prove that things aren't split
ripe crest
this sequence splits: 0 -> Z/3Z -> Z/6Z -> Z/2Z -> 0
oblique river
yes
it's a theorem that if the orders of A and C are relatively prime (and if you're working with abelian groups) then it always splits
ripe crest
and you cant map an element of order 2 to an element of order 6 still, right?
referencing: "and you can't go backward -- an element of order 2 can't map to an element of order 4"
oblique river
correct, but there's also an element of order 2 in Z/6Z which maps onto the generator of Z/2Z
so you choose that to be the section
ripe crest
But.. there is an element of order 2 in Z/4Z too that we can map onto the generator of Z/2Z
oblique river
the second part of the sentence is important
which maps onto the generator of Z/2Z
Z/4Z has an elt of order 2, but it maps to 0
ripe crest
oh interesting - right. if 0 -> Z/2Z -> Z/2^nZ -> Z/2Z -> 0 will never split by that reasoning, right
oblique river
well that's not a short exact sequence
do you maybe mean
$0 \to \bZ/2^{n-1}\bZ \to \bZ/2^{n}\bZ \to \bZ/2\bZ \to 0$
cloud walrusBOT
Buncho Bananas:
oblique river
because that is exact
as is $0 \to \bZ/2\bZ \to \bZ/2^{n}\bZ \to \bZ/2^{n-1}\bZ \to 0$
cloud walrusBOT
Buncho Bananas:
ripe crest
ah right
oblique river
here is a general way to find short exact sequences: let B --> C be any surjective map and let A be the kernel
then 0 -> A -> B -> C -> 0 is exact
(in fact every SES is of this form)
another perspective: let B be any R-module and A be any submodule. Then 0 -> A -> B -> B/A -> 0 is exact
and again every SES can be thought of this way
ripe crest
right - I can see that
Suppose G, H are abelian groups and N a subgroup of G. And also G/N = H as groups. Then we can write any element of G as: n + h where n \in N and h \in f^-1(H), where f is the isomorphism.
That seems true to me
oblique river
yeah
oblique river
yep
ripe crest
i guess we only need a section for injectivity
oblique river
i'm gonna go make dinner
gl and I'll be around more later if you have more questions
you can always ping me
👍
tribal pasture
I think I got it. We can view the exact seqeunce 0 → A →B→C→0 as 0→f(A) →B → B/ker g = B/f(A)→ 0
where the first map is the inclusion and the second is the canonical quotient map. This has a set-theoretic bijection as (a,c) to f(a)+c. Pls be correct x(
cerulean siren
I have a dumb question. I'm trying to show that in a euclidean domain, d is a unit <=> norm(d) = norm(1). I have that norm(d) <= norm(1) since cd = 1 for some c. Is there a way to show that norm(1) is the smallest possible norm in the ring?
mild laurel
use the fact that norm(cd) = norm(c) * norm(d)
cerulean siren
how do you know the norm is a homomorphism?
mild laurel
oh, you're working generally
cerulean siren
yeah for an arbitrary euclidean domain
Problem 3 (c.f. Saracino 21.7) Let D be a Euclidean domain, with v : D − {0_D } → N − {0}.
oh! It specifies that it maps to N - {0}, so 1 is automatically the smallest
there we go 🙂
just need to prove the converse now, norm(d) = norm(1) => d is a unit
mild laurel
Hm, do you know the norm of 1 is 1?
cerulean siren
... I do not 🤦
how would I do it if I knew the norm was a homomorphism? Then I'd have $\norm{d} \le \norm{cd} = \norm{c}\norm{d} = \norm{1}$ which just says $\norm{c} \ge 1$ which I already knew
ah there we go 🙂
cloud walrusBOT
jynelson:
cerulean siren
oh! maybe I could show that 1 also divides d
yes! because 1*d = d
now to prove the converse ...
mild laurel
If you know its multiplicative, then norm(1) norm(1) = norm(1), which means that norm(1) = 1
and then norm(c) norm(d) = 1, which implies that norm(c) = norm(d) = 1
cerulean siren
@ripe crest we know that the norm is in the natural numbers \ {0}
ripe crest
ah gotchu
cerulean siren
why does a^2 = a imply a = 1? intuitively that makes sense but I can't put my finger on it
oh this is still the natural numbers lol
ripe crest
xD
cerulean siren
pretty obvious there, everything else is > 1
mild laurel
Yeah, sometimes I forget that euclidean domains don't require the norm to be multiplicative
It is in almost all examples that come up
I think there's a way to do this without assuming this but its a bit harder
ripe crest
is there a way to search in just this channel?
cerulean siren
in: abstract-algebra
cerulean siren
(discord will also suggest filters if you click search on desktop)
oh! we know d divides 1
because for all a = qb +r, r = 0 or ||r|| < ||b||
wait no that doesn't work
my idea was plug in a = 1, but we don't know that there exists a q, r such that qd + r = 1
cerulean siren
I'm trying to show cd = 1, I'm going the other way now
I have ||d|| = ||1|| and that's it
and that it's a euclidean domain
ooh and I do have that $\norm{1}$ is the smallest possible norm, since if you had a c with a smaller norm you could multiply by 1 and now you have $\norm{1} \le \norm{c}$
cloud walrusBOT
jynelson:
mild laurel
Wait, you do know there exists a q,r such that qd + r = 1 with r = 0 or ||r|| < ||d||
cerulean siren
hmm yeah, q = 0 and r = 1
wait no, that gives a contradiction, ||1|| can't be less than ||1||
cerulean siren
r = 0
ripe crest
V, W are 2 finite dimensional k-vector spaces
if dim(V) = dim(W), then they are isomorphic, right
cerulean siren
that assumes that there exists a q, r such that qd + r = 1, though
ripe crest
that's just euclidean divison
mild laurel
well, assuming that d is not 0
cerulean siren
right, the problem assumed that you can't take the norm of 0
wait is that true for any element in a euclidian domain??
ripe crest
you can always divide everything in a euclidean domain by any non-zero element and have something of that form you mentioned above with |r| < |d|
mild laurel
or r = 0
ripe crest
right
cerulean siren
oh that makes sense, you can divide by any number, you might just have a remainder
and then we have d divides 1! perfect 🙂
thanks for the help!
mild laurel
np
tribal pasture
I think I got it. We can view the exact seqeunce 0 → A →B→C→0 as 0→f(A) →B → B/ker g = B/f(A)→ 0
where the first map is the inclusion and the second is the canonical quotient map. This has a set-theoretic bijection as (a,c) to f(a)+c. Pls be correct x(
@oblique river
gritty fractal
if you have a group $G$ and a G-set $X$, and $x \in X$, then if $G_{x}$ is normal then $ O_{x} $ is a group with $x$ as the identity element right?
cloud walrusBOT
StevenDoesStuffs:
grim edge
1+1=2 👀
chilly ocean
I think it is the orbit
If G_x is the stabilizer of x in G then it looks like he's trying to state the definition of a transitive group action
G_x usually isn't normal, unless the action is transitive iirc
mild laurel
Stabilizer subgroups can be normal without the action being transitive
To answer your question, yes, you could put a group structure on O_x by making it isomorphic to the group G/G_x @gritty fractal
gritty fractal
thanks, i had a hunch from accidentally assuming that the relation used to prove the orbit stabilizer theorem was a homomorphism, cuz i forgot to chekc that the operation was well defined
solemn rain
can any1 help me with this
if N is normal in G and H is a hall subgroup of G
prove that H intersects N is a hall subgroup of N
very hard for me xd
mild laurel
What's your definition of hall subgroup
solemn rain
a subgroup of G ,H, is called hall iff (|G|,[G:H]) =1
mild laurel
you mean, (|H|, [G:H]) right
wind steeple
that implies this since |G| = [G:H] * |H|
mild laurel
the one that says that $\frac{N}{N \cap H} \cong \frac{NH}{H}$
cloud walrusBOT
Zopherus:
mild laurel
note that NH is a subgroup since N is normal
somber bramble
isn’t tat the second one?
third one was the “cancellation” one iirc
(G/K)/(H/K) = G/H
brisk granite
I don't think there is a standard way to number them.
cerulean siren
are euclidean domains necessarily commutative?
oblique river
euclidean domains are domains which are usually commutative by hypothesis, yes
cerulean siren
hmm it's not given in this problem
I'm trying to show that for a Euclidean domain D and a norm that's a constant function, D is a field. I have that it's a division ring, I just need to show that it's commutative
oh and the norm is non-zero if that helps:
Let D be a Euclidean domain, with v : D − {0 D } → N − {0}.
oblique river
no, what i'm saying is that a euclidean domain is, by definition, an integral domain, and integral domains are, by definition, commutative
cerulean siren
oh then I just don't understand the definitions haha, why are euclidean domains an integral domain?
is that just by definition?
oblique river
I mean it has "domain" in the name
gloomy garnet
if I have a given function from a domain to the natural numbers, and I'm trying to prove that the domain is Euclidean, how would I go about proving the division part of the Euclidean definition?
mild laurel
This uh, is really dependent on your domain and function
Usually maybe some sort of descent or well ordering argument is used but
gloomy garnet
sorry by domain I mean integral domain, but I figured it out
ripe crest
@cerulean siren , this is a common picture that's in algebra texts: https://www.math3ma.com/blog/the-integral-domain-hierarchy-part-1
lime skiff
Hey! I'm tasked with showing x⁴ + 1 is reducible but I don't really see what to do
all the roots are complex
over R, my bad
I know all polynomials of degree >= 3 are reducible but I don't really know why
especially in this context, considering there's no real root I can factorize
wind steeple
bc it has a complex root a
and necessarly its conjucate b is a root
so you can factor it by (x-a)(x-b) which is real
lime skiff
I don't follow, those are polynomials over C, not R right?
how is this real?
especially here since there are four roots over C
lime skiff
oh I see
wind steeple
ab is real for the same reason
lime skiff
but we're still missing something right?
wind steeple
?
lime skiff
x⁴+1 is of degree 4
so we can't have it equal to (x-a)(x-b)
we still need a polynomial of degree 2
how do I know that polynomial is real
wind steeple
it's equal to (x-a)(x-b)(x-a')(x-b')
wind steeple
yes
lime skiff
thanks a lot!
upbeat burrow
According to Wiki, every irreducible polynomial over the reals is either degree 1 or degree 2
cloud walrusBOT
iamtim:
woven delta
🐶
mild laurel
Its not irreducible
Sage factors it as (x^2 - 1.53208888623796x + 1.00000000000000) * (x^2 - 0.347296355333861x + 1.00000000000000) * (x^2 + 1.87938524157182*x + 1.00000000000000) for me
And if you read the discussion, you can probably find a nicer form for this
median aspen
We can talk about bounded above only in a total order set ...right?Or in a subset of a poset which has total order right?
brisk granite
I don't understand why splitting fields and normal extensions aren't always the same? Clearly, splitting fields are normal extensions. And, if we wanted to construct the normal extension of some set of polynomials, then that's just the splitting field of the product of those polynomials. So, splitting field = normal extension?
stone fulcrum
The splitting field is a field
A normal extension is a field extension.
These are not the same object, but you're correct that they're very related
The splitting field of a polynomial generates a normal extension from the field its coefficients are in
@brisk granite
brisk granite
"The splitting field of a polynomial generates a normal extension from the field its coefficients are in". I don't really understand what this means
wdym by generate
stone fulcrum
Like,
x² - 2
You'd agree it has coefficients in Q.
Its splitting field is Q[√2]
The extension Q → Q[√2] is normal
The coefficient field
When extended to the splitting field
Is always a normal extension
brisk granite
ok, so, what I'm tryin to say is that the larger field in any normal extension is a splitting field
because we just need to multiply the polynomials
Anyway, I see that my question is kinda pointless now
Actually here is my question: what is the point of calling an extension normal when we could just say the larger field is a splitting of field
stone fulcrum
I think you're correct, a normal extension is an extension to a splitting field
Less words?
brisk granite
I see
here is what my book says about it
my question came because when he said "closely related concept", I thought it mean't different
ACtually, he does say infinite collection of polynomials
and in that case we can't just multiply them
and say that larger field in the normal extension is the splitting field of the product
stone fulcrum
I don't think this class is ready to handle an infinite product
brisk granite
wdym?
stone fulcrum
Well, maybe yours is, how would I know?
brisk granite
not in a class (self study)
chilly ocean
The Sun
brisk granite
is a deadly lazer
chilly ocean
@brisk granite
brisk granite
yeah?
stone fulcrum
Why isn't your pfp Gabe from the office?
brisk granite
It was
stone fulcrum
Yis
brisk granite
I'll probably change my name soon tho
so, just to clarify. This is wrong right? an extension normal when we could just say the larger field is a splitting of field
stone fulcrum
What's more interesting, to me, are extensions that aren't normal. Q → Q[³√2] is all fucky
brisk granite
yeah, my book does called normal extensions well behaved
what's wrong with it tho?
stone fulcrum
I think it's correct, at least from what I've read. You need a bit more effort to prove it though
stone fulcrum
Np! Feel free to ask if you have anything else
mild laurel
splitting fields are the same as normal extensions in the finite case
They're the same in the infinite case as well, if you allow splitting fields over infinite families of polynomials
@brisk granite
mild laurel
Eh, I mean, yeah usually
But for finitely many polynomials, you can just take consecutive splitting fields to get a splitting field for all of the finitely many polynomials
mild laurel
nighty night
daring sage
need a bit of help proving that Z[sqrt(10)] is not a unique factorization domain
the hint i got is try to work with 6 and the norm (a^2 - 10b^2)
no idea where to go with this
oblique river
have you seen the proof that Z[sqrt(-5)] is not a UFD
daring sage
i have not
oblique river
ah ok, I only ask because that's usually given as the "standard" example of a non-UFD
daring sage
gotcha
oblique river
The goal is to factor 6 in two different ways
daring sage
right
oblique river
I presume you already know one of those ways
daring sage
2,3
oblique river
great
now how can we use the norm to help? suppose 6 = x*y. Then N(6) = N(x)N(y) where N is norm
for 6 = 2*3, we have N(6) = N(2)*N(3), which is 36 = 4*9, which checks out
daring sage
ahhhh
oblique river
now we want to find an honestly different factorization. So instead of trying to factor 6 in Z[sqrt(10)], let's take a hint by factoring N(6) in Z
daring sage
so do i solve some diophantine equations?
daring sage
12, 3
oblique river
there's one more that you're missing :P
daring sage
obviously {6,6}
oblique river
and it's the important one haha
yes
so let's just say for a second that we can find some x and y with N(x) = N(y) = 6 and x*y = 6
(which we will do in a minute)
how do we know that we have really violated unique factorization?
daring sage
hm
oblique river
after all, we can write 36 = 4*9 = 6*6 but that doesn't violate UF because we know that we can break down 4, 6, and 9 into smaller parts
daring sage
so we need to make sure its a prime factorization
oblique river
so how do we know we can't break down 2, 3, x, and y down into smaller parts? again we can use the norm for this.
daring sage
oh
oblique river
yep, exactly (although the word "prime" has a technical meaning that doesn't exactly apply in this case)
daring sage
irreducible?
oblique river
yep
oblique river
yeah so there are two steps now
finding numbers with norm 6 (and trying to get them to multiply together to get 6)
and then explaining why that violates UF
oblique river
ok!
gritty fractal
if i have a subgroup H < G and some homomorphism from G called f, does |f(H)| divide |H|?
mild laurel
Yes
gritty fractal
where can i find a proof?
ive tried googling but im not too good at googling math 😦
mild laurel
just restrict the homomorphism f onto the subgroup H
gloomy garnet
if I have a Principal Ideal Domain, what does that tell me about the elements of that domain?
I was trying to work with a problem about gcds in a PID, but I couldn't get anywhere because I don't think I fully understand PIDs
mild laurel
All PID's are UFD's and so you can do unique factorization and stuff like you'd want
toxic zephyr
Is an element of a group is mapped to the identity element of another space by an isomorphism, can we say that the element must have been the identity in the original space?
smoky cypress
Yes
woven delta
Yeah
Or even just by an injection
Well actually what you said is equivalent to your homomorphism being injective
toxic zephyr
Okay cool
I'm working on a proof now and I'm not sure exactly how to show the result. It's to prove that if a mapping from G to G is an automorphism, then it is abelian.
$\Phi : G \to G. \Phi(x)=x^{-1}$
cloud walrusBOT
nix:
toxic zephyr
So what would it look like exactly to say the isomorphism is an automorphism? Like a statement to get to where I can say: "See, it's an automorphism"
mild laurel
Uh, it seems like the way you've written it, you're assuming that the mapping is an automorphism
and using that to show that the group is abelian
median aspen
So phi is an auto
toxic zephyr
What I wrote was exactly what was given
mild laurel
Then yeah, you're assuming the mapping is an automorphism
So you don't need to show that its an automorphism
toxic zephyr
Because an automorphism is just an isomorphism to the same space?
mild laurel
Yes
toxic zephyr
Okay that is what was tripping me up. I wasn't sure exactly what they wanted to prove that lmfao. It just is by definition.
Right that part of the proof makes sense to me
toxic zephyr
Alternatively I could have done xy=Phi(xy inv)=Phi(y inv x inv)= Phi(y inv)Phi(x inv)=yx, right?
median aspen
xy=(Yinv xinv)inv=phi(xy inv)..yeah..it does
toxic zephyr
Alright thanks for the help everyone :)
gilded trellis
Hi there ! so whilst reading Artin's Algebra I've gotten a bit confused here :
I tried reading on but that didn't help..
what does the set T refer to here
(the set that phi sends to)
mild laurel
any set
gilded trellis
oh
heh, I'm a still a bit confused about the whole introduction of the phi application. I think I'll go watch some lectures and hope the lecturer covers it.
anyways, thanks ! 🙂
mild laurel
i can try to help too if you'd want
gilded trellis
ah
thanks !
so, why was the phi application introduced here ?
like, erm, I don't know exactly what I don't understand/where I'm having an issue in understanding
it's just that the introduction of the phi application felt really sudden .. heh
mild laurel
Can you show me the stuff right before this picture?
mild laurel
It's fine
Hm
Maybe its best to just say that its an important example of how you can get equivalence relations on a set
But well, they give an example of this
When they talk about Z -> {(even), (odd)} this is exactly what they're doing
the preimage of (even) are exactly the even numbers in Z
and the preimage of (odd) are exactly the odd numbers in Z
so you get your equivalence relation on Z in this way
gilded trellis
(ah,sorry had to go for a bit... irl stuff...)
Ah
I see
Yeah I'm less confused now, thanks !!!!! 🙂 !
rancid narwhal
Started groups this semester, was wondering when is the image of a homomorphism say f: G -> H equal to H? I've read when the homomorphism is surjective however I don't see how that makes makes the image(f) = H. This is in regards to the first isomorphism theorem where G/ker(f) is isomorphic to image(f), however I've seen it extended to G/ker(f) isomorphic to H.
chilly ocean
surjective means exactly that im g = H
rancid narwhal
I think im getting confused then, couldn't the image have less elements then H? For example if G maps to the same elements of H .
chilly ocean
well but surjectivity says that for each element of H, there is an element in G that maps to that element
scarlet estuary
yeah this is exactly the deifnition of surjective
what do you think surjective means?
rancid narwhal
Ah, its each element of H, I see. I was getting confused thinking only the mapped elements of G have a corresponding g. Never mind then. I must of gotten confused when doing a previous surjective statement showing that im(f') = im(f).
Thanks for clarification
latent anvil
If you have a submodule M of a finitely generated free module F, and M is free, is rank M <= rank F?
oblique river
I think you need the ring to be commutative for rank to make sense
latent anvil
Sorry, I was assuming commutativity
latent anvil
It's true if there's always a maximal ideal m such that Tor1(F/M, A/m) = 0
By tensoring with A/m and using linear algebra (A is the ring)
oblique river
if the ring is a domain then it's true
although I don't think my proof is very efficient
latent anvil
By going to the field of fractions?
oblique river
oh that wasn't my argument but yeah that should work
oblique river
yes localization is exact
so tensoring with field of fractions is, too
yeah that's definitely easier than my proof which was much more "hands-on"
well, it requires more machinery
latent anvil
What was your proof?
oblique river
basically submodules of R^n have to be products of ideals because you can multiply by, for example, (1,0,...,0)
and the rank of an ideal is no more than 1
latent anvil
That's not true. Take n = 2 and the submodule { (x, y) in R^2 : x = y }
latent anvil
I think it's something like linear relationships between ideals?
oblique river
let me figure out where I went wrong
latent anvil
I did this for division rings
oblique river
oh I'm silly, I was thinking of "R^n-submodules of R^n"
latent anvil
Oh lol
oblique river
not "R-submodules of R^n"
latent anvil
Yeah I see
oblique river
and yes it's true that ideals in a product ring are products of ideals
latent anvil
Yeah
oblique river
hmm, yeah tensoring with the field of fractions is slick
latent anvil
So you can prove stuff like A^n ≈ A^m implies n = m by tensoring with A/m, where m is a maximal ideal
oblique river
you might be able to do it directly as well -- this question is equivalent to "can a n-by-(n+1) matrix have rank n+1"
latent anvil
But A/m is not flat in general
oblique river
A/m is flat if and only if m = m^2 I think
latent anvil
Yeah but I don't really need that it's flat
Just that Tor1(A/m, F/M) = 0 for some m
oblique river
that seems hard to show
what is the context of this? do you just need the answer or is it homework or something?
I only ask because you can google and find some proofs, and depending on what you need that could suffice
latent anvil
No I'm just curious
I thought it was interesting
I needed to show that a certain direct summand of a free module would have to have rank 2, and I thought I could show this, but realized that it's much easier because it's a direct summand
If A = B (+) C and A, B are free of rank n, m, then if C is free, it must have rank n - m
oblique river
ah, yeah
latent anvil
Even more specifically, I was showing that the module of continuous vector fields on S^2 is projective over the ring of continuous functions on S^2, but is not free
oblique river
oh nice
that's fun
there's an "algebraic" version of that as well that just uses polynomials
you take A to be R[x,y,z]/(x^2 + y^2 + z^2 - 1)
so this is the functions on the sphere x^2 + y^2 + z^2 = 1
and the module to be
M = {(f,g,h) in A^3 | (f,g,h).(x,y,z) = 0 in A}
where by . I mean "dot product of vectors"
so something in A^3 is a triple of functions from S^2 to R which you can think of as a vector field on S^2
and then the dot product condition is just asking that it's actually tangent to the surface
bleak abyss
Why, that's just Serre Swan 🙃
oblique river
idk what that is
bleak abyss
Oh it's just a generalization of this lol
It says that if you give me any smooth vector bundle over a smooth manifold, the space of sections is a projective C^{\infty}(M)-module
And vice versa
latent anvil
Yeah, that's why I thought of it in the first place
oblique river
oh that's cool
latent anvil
Well this is the continuous version
bleak abyss
Yeah there are like, 3 versions
latent anvil
I want a concrete example of a projective module for my algebra students
bleak abyss
One is where everything is smooth but you don't need compactness
One is continuous stuff over a compact Hausdorff space
And finally there's one about varieties
oblique river
oh yeah in that case I think this is a good example
bleak abyss
Oh I guess "Vice versa" might've been strong it might be some weird equivalence of categories something something. But yeah I've been meaning to learn that theorem at some point
oblique river
depending on the context, "any non-principal ideal in a dedekind domain" could also work
latent anvil
Yeah, unfortunately they don't know what a dedekind domain is
bleak abyss
You can change that 🙃
oblique river
haha
latent anvil
I'd like to learn serre swan at some point
bleak abyss
Tbh there are a lot of things I'd like to do regarding breadth before I hit the research phase of things
bleak abyss
Or I could just pull a Benson Farb
latent anvil
I'm taking a bundles course next winter but it doesn't say we cover serre swan
latent anvil
oblique river
(on the previous topic, this seems to be a good writeup about the projective thing https://kconrad.math.uconn.edu/blurbs/linmultialg/stablyfree.pdf)
latent anvil
Using an unpublished book by Lee
latent anvil
Yeah I'm definitely more excited for bundles
But apparently the riemannian geometry stuff will show up there
So it's necessary
bleak abyss
Oh snap nice
scarlet estuary
tribal pasture
Say I have to prove given two basis of a free module {a_i}_n, {b_j}_m, then n =m. Would it be correct to do the following
+^n R_i a_i ≊ F ≊ +^m R_j b_j
And then just prove R^n ≊R^m implies n =m?
cerulean siren
So I recently got the following question on a homework:
Let $\mathbb{Z}[\zeta_3]:={a+b\zeta_3 \mid a,b \in \mathbb{Z}}$ be
the subring of $\mathbb{C}$ where $\zeta_3$ is a complex root of $x^2+x+1$.
Show that $7$ is not irreducible in $\mathbb{Z}[\zeta_3]$.
I've already turned in the homework, but I have no idea how I would approach this. Does anyone have ideas?
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latent anvil
@tribal pasture yes
But your notation is a little wonky. What do you mean by R_i a_i?
gentle pendant
Probably just Ra_i
latent anvil
Yeah that was my assumption
mild laurel
@cerulean siren You would just want to write it out, here I'll let zeta be z. To show its not irreducible, you want to have that 7 = (a + bz)(c + dz) where neither a + bz nor c + dz are units.
just multiply out the right hand side and compare coefficients
latent anvil
You get some nasty equations though. ac - bd = 7 and ad + bc - bd = 0
Idk, maybe there's a trick that makes that easy
Like you get ad + bc = ac - 7
Maybe you just guess?
cerulean siren
guessing did not help
the systems of equations idea is good, I didn't think of that
cerulean siren
I've been really lost in class lately
mild laurel
oh maybe you can use norm ideas, what's the norm function on this
cerulean siren
I have no idea
latent anvil
I thought about that too
cerulean siren
🤷