yeah, on the thoughtsis that diagram bad 
#groups-rings-fields
solar wyvern
solar wyvern
@oblique river does
[EF:k] = [E:k][F:k]/[E⋂F:k]
work generally?
also if EF/k is galois can you write its group as a product of the two galois groups quotiented by Gal(E⋂F/k) or something?
oblique river
I think as long as one of them is galois, then yes
it's actually canonically a subgroup of the product
not a quotient
If E/k and F/k are both not Galois then that degree statement isn't true
I think that if one of them is galois then it might be true. if both are galois then it's definitely true
i'd have to think about that middle case
solar wyvern
this seems like an exercise 
oblique river
I know that one of them being galois is enough to conclude that [EF : k] divides the product [E : k][F : k]
if neither is galois then you don't even get that division. one is enoguh to get the division but then asking what the quotient is might require both to be
solar wyvern
quotient meaning a quotient of groups not just of the degree numbers?
oblique river
(also if one of them isnt galois then tehre isn't even a group to take the quotient of)
but yeah now that i think more, i think the degree statement is true if only 1 is galois
solar wyvern
okay so one just nets you divisibility, other statement explicitly states the [EF:k] = [E:k][F:k]/[E⋂F:k]?
oblique river
no, i think that you only need that 1 is galois in order to get the full statement
suppose E/k is galois and that E cap F = k
we can always reduce to this case by replacing k by E cap F
then EF/F is galois, and that galois group is naturally a (sub or quotient, which one...) of Gal(E/k)
sub. there is a map Gal(EF/F) -> Gal(E/k) defined by "restrict the automorphism to E"
solar wyvern
sub?
oblique river
yeah the map Gal(EF/F) -> Gal(E/k) is injective
which is a 1-line proof
and now how to prove it's surjective...
solar wyvern
are you finishing up your dissertation?
oblique river
in some large sense, yes
i'm graduating next yera
so i'm definitely on the "wrapping up" as opposed to "getting started" side of things
(also i was just making a joke, my time isn't that valuable lol)
oblique river
nope my b lol
also actually on 3rd thought, you might need both of thema re galois
cuz i cant see how otherwise to prove that Gal(EF/F) --> Gal(E/k) is surjective (if E cap F = k)
solar wyvern
what do you do if E ⋂ F ≠ k
solar wyvern
wind steeple
Hello, I'm searching for a 6-degree simple extension over Q which has no intermediate fields
It is like an "irreductible" extension but it seems that this word doesn't exist for field extension :/
fickle brook
simple = generated by 1 element right
won't you always have an intermediate field tho
Q ⊂ Q(a^2) ⊂ Q(a)
or sth
weary terrace
Hopefully it's the right place for that..
p-adic analysis:
Let $\pi \in K$ where $K$ is an extension field of $\mathbb Q_p$ (the p-adic rationals) and $ord_p(\pi)=\frac 1 e$ where $e$ is the index of ramification.
Then any $x \in K$ can be written uniquely in the form $\pi^mu$ where $|u|_p=1$ and $m \in \mathbb Z$.
Can you please help me understand why that is right?
cloud walrusBOT
idan:
wind steeple
but I don't know how did he find it xD
nope because maybe Q(a^2) = Q(a) @fickle brook
for instance Q(e^2ipi/5) = Q(e^4ipi/5)
fickle brook
ok but let's say [Q(a) : Q] = 6
then {1, a, a^2, a^3, a^4, a^5} is Q-linearly independent right
wind steeple
no
why ?
oh yes
lel
[Q(a):Q] = d°(irr(a,X,Q))
irr(a,X,Q) is the minimal polynomial of a over Q
so yes it is Q-lineary independant
it's a basis of Q(a) over Q too
but you can have a in Q(a^2)
hollow comet
Let k be a field, Y an affine variety in X=kⁿ. Is it true that for any x in X\Y we have a polynomial f which vanishes on Y but is f(x)≠0?
hollow comet
(ping me)
oblique river
@weary terrace That's true in any discrete valuation field. Let K be a field with discrete valuation v (normalized for K). Let pi be any element with v(pi) = 1. Let x be any other nonzero element of K. Then v(x) = m for some m in Z, so v(x/pi^m) = 0, so take u = x/pi^m
The only thing missing is to show that a finite extension K of a discrete valuation field k (equipped with the extension of the discrete valuation from k) is also a discrete valuation field
which isn't hard to show, it has nothing to do with the p-adics. It's just kind of an annoying lemma that just works.
One thing that is a little annoying is that different authors use different conventions for hwo to normalize the valuations. For example sometimes you normalize the valuation on K so that v(p) = 1, so that v(pi) = 1/e
but sometimes you normalize it so that v(pi) = 1 so that v(p) = e.
but it's all the same, it's just a normalization, it doesn't actually affect the math in any meaningful way
oblique river
I was just about to ping you edelopo
hollow comet
It's much easier than that
oblique river
but your question seems easy depending on how you defined affine variety
hollow comet
If every function vanishing at the variety would vanish at that point, then the point would be in the variety
oblique river
yeah lol
hollow comet
3 hours
oblique river
that's kinda the point of varieties :)
oblique river
well i'm glad you figured it out on your own haha
@wind steeple going back to your old question while I'm here: The general idea is that such fields should come from the following setup: Let K/Q be a Galois extension with Galois group G and let H be a subgroup of G that 1) is maximal (i.e. there are no intermediate subgroups between H and G other than H and G) and 2) has index 6
then by the fundamental theorem of Galois theory, H corresponds to some subextension K^H / Q
that is degree 6 (because [G : H] = 6) and has no intermediate subfields bewteen it and Q
wind steeple
yep, I've just talked about this on another discord
oblique river
oh okay
wind steeple
How can I find such an extension ?
oblique river
yeah, my first guess for how to find such a G would be G = S_6 with H = S_5 sitting in it
that looks maximal maybe
wind steeple
oh
oblique river
so then take any S_6 extension
(of which there are plenty, you can find polys that give S_6 extensions all over the internet)
wind steeple
here in the example I found the Galois group is S6
oblique river
and then any S_6 extension will have a subextension of the form you want
yes that is good
wind steeple
mh
oblique river
that would be my answer to the problem though
just like "look in any S_6 extension, for example [this one], and use galois theory"
oblique river
no?
oblique river
this is a theorem everyone should know: the only normal subgroups of S_n for n >= 5 are A_n and {e} and S_n
wind steeple
oh yes
oblique river
that doesn't imply that S_5 is maximal in S_6 which is what we really need for this though
but I think that just some short thought will show that that is indeed the case
wind steeple
but how can I get a sub field of order 6 then ? If M = L^(S5), and [M:K] = |S6|/|S5|, that means that [L:M] = |S5| hence S5 is normal in S6
oblique river
no no no
no wait
i mean
yes yes yes
haha sorry i was confused with what you were calling M (nromally people like to use M for very big extensions, not intermediate extensions)
wind steeple
yep, K subset M subset L here
oblique river
but i'm not sure hwo you're concluding normality
wind steeple
L/K is galois
oblique river
yes
wind steeple
if we are in C
oblique river
with galois group S6
wind steeple
we have L/K normal iff [L:K] = |Gal(L/K)|
oblique river
there is a sub extension M with Gal(L/M) = S5
L is galois over M!
because it is galois over K
but M is not galois over K
wind steeple
yes
oblique river
because S5 is not normal in S6
oblique river
Aut(M/K) will end up being trivial in this example
oblique river
yeah everything is about the galosi correspondence
that is like the most important thing haha
wind steeple
yes
ok then I only have to show that S5 is maximal
and find a polynomial of Galois group S6
wind steeple
yes xD
oblique river
and even more ways to prove that such a poly must exist nonconstructively
wind steeple
but that would be nice if I could prove it
weary terrace
@oblique river so in other words, it's possible because pi has the smallest valuation in the field and hence we can multiply it by any u (with v(u)=1) to get any nonzero element of K?
wind steeple
mh btw, I want to have a simple extension constructively (I don't want to use the primitive element theorem)
oblique river
and the maximal subgroup thing I think you just do by looking at groups. consider S5 inside of S6 by permuting the first 5 objects, you should be able to argue that if you add ni any other permutation that permutes the 6th object to anywhere, you get all of S6
oh well that's gonna be harder for you then lol
@weary terrace yes
or multiply it by itself
wind steeple
if [M:K] = 6, does M = K(x) for somme root of my polynomial ?
oblique river
@wind steeple not necessarily
weary terrace
@oblique river thanks very much!
wind steeple
that's what thought
oblique river
I only say that because your poly will have 6 roots but there are more than 6 ways to embed S5 inside S6
wait no nvm
yeah if you use the obvious embedding
then yes, it will be generated by one of the roots
yeah i'm dumb my b
yeah take any poly with galois group S6
wind steeple
why ?
oblique river
and if you just add in one of hte roots then youre good
think of how S6 is acting
it acts by permuting the roots
wind steeple
yes
oblique river
so if we take the S5 that is "leave the alst one fixed"
wind steeple
yes
oblique river
then, well, the final root will be left fixed
so it will be contained in the fixed field
oblique river
but it has degree 6 so it generates the fixed field
oblique river
also i gotta go now so gl on the rest of this
chilly ocean
i had question about finding generators for a group
i kinda forgot how to
i was gonna attempt to find generators of something like a4 to get a feel for it again
stone fulcrum
It's generally not obvious. Sometimes the rules that form the group makes it clear
mild laurel
To find a minimum generating set is often hard, but you can just take all the elements and that obviously gives you a generating set
chilly ocean
Well, i was asked to find 2 generators of a5, but it has been a while since i have done any group theory. So i was gonna start with a4
see if i could find some sort of method to finding them, then hopefully apply something similar to a5
stone fulcrum
Like, a single element that generates A4?
Or, you want two elements that generate A4?
thorny slate
it's a classical result @chilly ocean
I think it's like a 3-cycle and a 5-cycle
or something
it's in galois theory books
it's a combinatorial thing
you just play around with it
stone fulcrum
Oh right, that's all you need to prove an unsolvable quintic
chilly ocean
okay
spark plank
hears quintic
thorny slate
it's kind of annoying to come up with it though
the idea is the following: try to get any 3-cycle from your initial 3-cycle
by conjugating with the other element and multiplying stuff
all 3-cycles generate A5
and then you're done
mild laurel
Yeah, showing that all the alternating groups are generated by 3 cycles might be helpful too
weary terrace
I easily proved the following for any element in $\mathbb Q_p$ but I'm struggling to do so for the rest of K:
if $K$ is a finite extension of $\mathbb Q_p$ of degree n, index of ramification $e$, and residue field degree $f$, and if $\pi$ chosen so that $ord_p\pi=\frac{1}{e}$, then every $\alpha \in K$ can be written in one and only one way as $\sum_{i=m}^\infty a_i\pi^i$ where $m=e \cdot ord_p\alpha$ and each $a_i$ satisfies $a_i^{p^f}=a_i.$
Can someone please assist?
cloud walrusBOT
idan:
little dew
I am taking abstract algebra next semester, any words of wisdom from you seasoned veterens?
stone fulcrum
t!yt Visual group theory
fossil mangoBOT
stone fulcrum
Perhaps tell me how this video series is; it looks quite good but I already know it
Otherwise, feel free to ask any questions you might have
little dew
ok thanks 😉
stone fulcrum
Hmm. Then again, it looks very long.
Anyway, group theory is fairly easy but a lot of students get caught up on how proofy it can be.
solar wyvern
@oblique river why is inverse galois hard
e: dumb q
solar wyvern
@oblique river I feel like I had this the other day but why does sum of two squares work for positive characteristic
if you can write something as an even power of a generator you're done, otherwise....
otherwise I don't know
weary terrace
Second attempt:
I easily proved the following for any element in $\mathbb Q_p$ but I'm struggling to do so for the rest of K:
if $K$ is a finite extension of $\mathbb Q_p$ of degree n, index of ramification $e$, and residue field degree $f$, and if $\pi$ chosen so that $ordp\pi=\frac{1}{e}$, then every $\alpha \in K$ can be written in one and only one way as $\sum{i=m}^\infty a_i\pi^i$ where $m=e \cdot ord_p\alpha$ and each $a_i$ satisfies $a_i^{p^f}=a_i.$
Can someone please assist?
cloud walrusBOT
idan:
oblique river
@weary terrace i think you just wanna do some kind of induction. First multiply by a suitable power of pi so that your element alpha is a unit, i.e. v(alpha) = 0
so then the expression you're looking for is a0 + a1 pi + a2 pi^2 etc
look at alpha mod pi. This is an element of the residue field and is the same as a0 mod pi
but by hensel's lemma for every element of the residue field there is a unique lift (the teichmuller lift) satisfying the polynomial x^(p^f) - x = 0
(namely every element of the res. field satisfies that poly, then apply hensel's lemma to get elements of K that satisfy that same poly and reduce to all the elements of the res. field)
so there's your a0
then subtract a0 and divide by pi, and use the saem method to find a1
then subtract a1 and divide by pi
etc
now you don't know your element can be written in the form a0 + a1 pi + a2 pi^2 at all so you can't start with that form
but you can say like "alpha - a0 is divisible by pi since alpha mod pi = a0 mod pi, so then consider (alpha - a0)/pi and repeat the process"
@solar wyvern what do you mean exactly?
i dont remember what "sum of two squares in positive characteristic" means here
solar wyvern
I think I almost have it but $$\forall a \in F:|F|<\infty, \exists b,c \in F: a = b^2+c^2$$
cloud walrusBOT
flimflam:
oblique river
I think a simple counting argument will do it
if |F| = q then there are (q-1)/2 squares, (q-1)/2 nonsquares, and then 0
take a nonsquare, call it a
subtract from a all of the (q-1)/2 squares that aren't 0
we get some subset of size (q-1)/2 of F
clearly a is not in that set because we didn't subtract 0 from a
so therefore the set must either contain 0 or one of the nonzero squares in F
in either case, we now have something that looks like a - c^2 = b^2
(where b could be 0)
so then a = b^2 + c^2
solar wyvern
subtract from a all of the (q-1)/2 squares that aren't 0
you're subtracting numbers not removing the square numbers F² from F right
solar wyvern
oblique river
the point is I wanna write a = b^2 + c^2
so i'm going to look at a - b^2
over all possible b
and by counting prove that at least one of those "a - b^2" must be a square
solar wyvern
so I was wondering if this approach works
oblique river
didn't I just demonstrate that it does? :P
solar wyvern
i don't get counting 😦
oblique river
as in you don't understand how to count or you don't understand this particular counting argument
solar wyvern
hopefully just the latter
so you're subtracting the sum of all squares from a nonsquare?
sums of squares are closed under multiplication...
and...sums
oblique river
no they aren't closed under sums
well I mean I guess they are since everythign is a sum of squares
but i dont see what you're trying to say
i'm never making any claims about sets being closed or not under addition
solar wyvern
just not sure why you'd subtract out the squares
oblique river
what? cuz that's literally what we need to do
we are given a and we want to write a = b^2 + c^2
for some b and c
so we look at {a - b^2 | b in F, b not 0}
which has size (q-1)/2
and we look at {c^2 | c in F} which has size (q+1)/2
just by the numbers, these two sets would be disjoint, which would be bad cuz we want to say they're not disjoint
but we also know that a is not in the first set (because we're not allowing b = 0)
but it's also not in the second set because we assumed it wasn't a square
which means that our two sets are both subsets of F \ {a} which has size q-1
but their sizes sum to q
so therefore they must have some intersection
i.e. there must be some b, c in F such that a - b^2 = c^2
or, a = b^2 + c^2
solar wyvern
so {a-b^2} can be thought of as the image in (F*)² of some bijection x↦-x+a, and {c^2} is simply (F)² all the squares (including zero), and by cardinality arguments, these cannot be disjoint?
oblique river
yeah but idk why you have all those fancy words
solar wyvern
image is fancy
oblique river
look at the set of elements of F of the form a - b^2 where a is fixed and b is not
and b is not zero
like if we were in the integers I could write something like {1 + 3k | k in Z}
and that's the set of things that are 1 mod 3
I don't need to say "it's the image in Z of some bijection..."
it just is what it is
and it's simpler than you are making it out to be
oblique river
you got quiet @solar wyvern how is it going
I realized that my argument can be made even simpler by just allowing b to be 0.
{a - b^2 | b in F} has size (q+1)/2 and {c^2 | c in F} also has size (q+1)/2
but they both lie in F which has size q
and therefore they intersect
this works for both cases, when a is a square and when it is not
maybe an example would be helpful?
Take F = F_7. The squares are {0, 1, 2, 4}
take a = 3
then {a - b^2 | b in F} = {3 - 0, 3 - 1, 3 - 2, 3 - 4} = {3, 2, 1, 6}
this has size 4, and {c^2} = {0, 1, 2, 4} has size 4
so therefore they must intersect since F_7 only has 7 elements
we see that they do both contain 2, for example
so 3 - 1 = 2
3 - 1^2 = 3^2
3 = 1^2 + 3^2
solar wyvern
reconsidering life options
i understand it but I wouldn't have ever been able come up with it like that. I was trying to show the nonzero squares and their translate must have nonempty intersection in F*
oblique river
isn't that what we just did?
(except we didn't restrict to nonzero squares)
we showed that the set of squares ({c^2}) and the set of their translates by a fixed number a ({a - b^2}) have nonempty intersection
solar wyvern
this is really stupid but I just don't understand why it was necessary to consider the case where either set contained 0
oblique river
it wasn't
the first argument I wrote down was more complicated than it needed to be, which is why I gave the revised argument where we didn't treat 0 as special
solar wyvern
i think I understand it but I sorta need a moment
oblique river
np, just try to write down some more explicit examples
weary terrace
@oblique river Beautiful solution, Thank you!
oblique river
@weary terrace np! glad it worked!
cloud walrusBOT
flimflam:
Compile Error! Click the 
reaction for details. (You may edit your message)
solar wyvern
just wondering if I did this okay
solar wyvern
@thorny slate
thorny slate
splits completely isn't the opposite of irreducible
solar wyvern
yikes
uncut girder
How can I most easily show the subgroup Z2 (+) Z2 in A4 given by {id, (12)(34), (13)(24), (14)(23)} is normal in A4
I dont want to write out every possible product h^-1gh
somber bramble
hey I just did that like yesterday, though I kinda handwaved it a bit
by which I mean I just looked at which “cases” there are and all the other conjugations you can get to by index permuations
like if you show that e.g. (123) (12)(34) (132) is in that group, then the same calculation also shows that (234) (23)(41) (243) is in it
that brings it down to very few different cases
because each calculation gets you 24 of the cases
actually prolly not 24 but like, a lot
I think you have to do like 2 or 3 calculations at that point
@uncut girder
uncut girder
Yeah but I didn't want to do it with variable names i j k
So I was kinda hesitant to try
woven delta
Sylow theorem
Is the easiest way
If you can show that there are exactly 4 elements of order 2^k
Then the sylow 2 subgroup is unique, and therefore normal
thorny slate
the easiest way is noticing it's a conjugation class in S4
because it's all elements of cycle structure 2,2
@uncut girder
uncut girder
Never learned about conjugation classes
somber bramble
but conjugation classes in S4 aren’t necessarily conjugation classes in A4 tho, right?
thorny slate
they're bigger
so you're extra fine
@uncut girder you should, it's very intuitive
a normal group is one that's stable under conjugation
so it's precisely one you can write as a union of cojugation classes
in Sn, conjugation classes have a very simple description
conjugating by an element of Sn means changing the labels according to that permutation
for example if you have (135)(24) and you conjugate by (123), you change 1 to 2, 2 to 3, 3 to 1, and get (215)(34)
you can see this quickly by interpreting them as functions
therefore conjugating an element preserves the cycle structure
and also any two elements with the same cycle structure are conjugate in Sn (you send the labels of one to the other, in order)
so with that in mind, the conjugation class of this thing in S4 is bigger than the conjugation class in A4
so if we cojugate by elements of A4 we'll preserve the cycle structure
and we have all elements of cycle structure (2,2)
so we're fine
this came up in my math GRE lul
and I didn't remember
but it's good to keep in mind
fringe nexus
wait the GRE has questions on abstract algebra?
uncut girder
Mgre
solar wyvern
Someone help me get a handle on separable extensions
fringe nexus
Is any finite field of characteristic P a perfect field?
so every irreducible polynomial over any finite field is separable?
thorny slate
yeah
every finite field is perfect
the only finite extensions are from F(p^k) to F(p^{nk})
weary terrace
Hopefully it's the right place for that.. Can someone please explain why the part in red is correct? We're talking subsets, not subgroups, why is it right to claim h^(-1) is in H?
weary terrace
Ok, I may have got it.. seems like simple arithmetic:
$g=k_ih_i$ and $g=kh^{-1}$ $\Rightarrow k_ih_i=kh^{-1} \Rightarrow h_ih=k_i^{-1}k \in H$ because right hand side in H.
So $h^{-1}=k^{-1}g=k^{-1}k_ih_i$, hence, because $k_i^{-1}k \in H$ we get $h^{-1} \in H \Rightarrow g=kh^{-1} \in KH$.
Can you approve..? 😶
cloud walrusBOT
idan:
Compile Error! Click the reaction for details. (You may edit your message)
uncut girder
I asked before
"How can I most easily show the subgroup Z2 (+) Z2 in A4 given by {id, (12)(34), (13)(24), (14)(23)} is normal in A4"
My answer now
"This is normal in A4 because it is the subset of elements with order at most 2 and conjugation preserves order"
fickle brook
you could also look at how conjugation works in S_n
uncut girder
What
uncut girder
I already know it's a subgroup
woven delta
I know
uncut girder
Just justifying its normal
woven delta
You just didn't state that
Oh I guess you stated that earlier lol
So yeah, that should be enough
rapid tusk
Can someone help me out on this? The fact there are two inputs is really screwing with my head
its to show a homomorphism
(or that there isn't one)
fickle brook
are you asked to check whether this is a homomorphism?
bc this is certainly not one
rapid tusk
yea, I figured its not
fickle brook
not even close
fickle brook
it wouldn't be one without the +1 either.
fickle brook
i mean ok
if φ were a homomorphism then φ(0,0) would be 0, but it's not. therefore φ is not a homomorphism. QED.
or you could consider φ(1,1) vs φ(1,0) + φ(0,1)
rapid tusk
but I still don't understand how to lay out the equation with two variables
thats whats messing with me
Ф(x+a,y+a) = something I think
fickle brook
$\phi(x,y)$ is really just shorthand for $\phi\left( (x,y) \right)$
cloud walrusBOT
Ann:
fickle brook
your inputs are points of R^2
rapid tusk
Ф((x,y)+(a,b))? 
fickle brook
and to prove that your function ISN'T a homomorphism it suffices to give ONE example of the condition failing.
rapid tusk
yea I know, I just haven't seen one like this before and couldn't think how to lay it out which was the main thing bothering me
fickle brook
i'd rather write $\phi((x_1, y_1) + (x_2, y_2)) = \phi(x_1, y_1) + \phi(x_2, y_2)$ if i were to write out the condition that defines a homomorphism in this case
cloud walrusBOT
Ann:
rapid tusk
how can you add the two points though, (x1+x2,y1+y2)?
then that should equal the other side, if it is a homomorphism (which it isn't)
fickle brook
wdym "how can you add the two points though"
here R^2 comes with its usual addition operation
$(x_1, y_1) + (x_2, y_2) = (x_1+x_2,y_1+y_2)$ definitionally
cloud walrusBOT
Ann:
mild laurel
I'm currently reading commutative algebra by atiyah, mcdonald and I was wondering if people think its worth it to work through every single exercise?
I know that people say that the exercises are like a textbook in of themselves, but there are a ton of them and most of them are nontrivial so its taking me quite a while to do all of them
solar wyvern
@mild laurel I don't think they're all meant to be done chronologically at least
@mild laurel how far in are you?
weary terrace
I'm trying to figure out why it's right to claim $h^{-1} \in H$.
I thought the following solved it:
$g=k_ih_i$ and $g=kh^{-1}$ $\Rightarrow k_ih_i=kh^{-1} \Rightarrow h_ih=k_i^{-1}k \in H$ because right hand side in H.
So $h^{-1}=k^{-1}g=k^{-1}k_ih_i$, hence, because $k_i^{-1}k \in H$ we get $h^{-1} \in H \Rightarrow g=kh^{-1} \in KH$.
But then I realized there's no logic in assuming $h_ih \in H$ because, again, we're talking subsets here, not subgroups. So is it ok to assume $h_ih \in H$? if so, why?
cloud walrusBOT
idan:
solar wyvern
@oblique river @thorny slate is this true
Say E/k alg, α∈E with minimal poly f∈k[t] and E' ⊂E the subfield generated by the roots of f in E.
If k_f is the splitting field of f (in algebraic closure of k), is it always true that E' = E ⋂k_f?
thorny slate
i think so
hollow peak
Is any monogenous group of cardinal n isomorph to Z/nZ?
thorny slate
what's a monogenous group?
hollow peak
a group generated by only 1 element
topaz solar
monogenerated lmao
chilly ocean
thorny slate
no one uses the word
hollow peak
actually we use this in french
thorny slate
??? this is english
hollow peak
when its generated by 1 element but not finite
thorny slate
in english you say cyclic
chilly ocean
He gets to say whatever word he wants although it might not be understood lol
thorny slate
don't shitpost in the math channels
hollow peak
in english you say cyclic even if the group isn't finite?
thorny slate
yes
chilly ocean
Its not a shit post I just think you are needlessly antagonistic over someone using a recognized synonym of a word
hollow peak
so Z is cyclic?
thorny slate
yes
hollow peak
its quiet awkward
thorny slate
I suppose so
wind steeple
🙄
uncut girder
There's an infinite cycle
wind steeple
that's a good cycle
solar wyvern
cyclic with generator ±1
uncut girder
It more like
If you ride your bicycle on the integers
You only need to start at 1
To reach every number
hollow peak
yeah but it's not like it was cyclic
solar wyvern
cyclic means it can be generated by a single element
hollow peak
Idk how to explain but in french, cyclic is for finite groups which are generated by only 1 element
i think it makes more sense to call that cyclic
chilly ocean
The word monogenous makes more sense as an umbrella term because it means descended from a single object pretty much
but in english cyclic is the word everyone uses
hollow peak
you also say non negative after all x)
solar wyvern
so this should be obvious but I'm not seeing it
Let E,F be extensions of a filed k contained in some bigger field L. We can form the ring E[F] generated by elements of F over E. Then E[F] = F[E], and EF is the quotient field of this ring
thorny slate
they're the same by definition
solar wyvern
probably just getting thrown off by "quotient field" which is actually field of fractions
thorny slate
that notation usually defines combinations in F that are free in E
solar wyvern
imagine that you can multiply things in L and E because they're both in L
thorny slate
so in order to get EF you would quotient by whatever relations E and F have
but yeah it's odd
solar wyvern
ah, but EF is the field of quotients of [elements of E[F]]
will E ⊗_k F ever be a field?
solar wyvern
I saw linearly disjoint on wiki
thorny slate
sounds about right
solar wyvern
but just wanted to be sure it wasn't like
special field notation or w/e
so if they're mutually included in some big field Ω the tensor product over k will be a field if(f?) the generators of E and F are linearly independent?
solar wyvern
this is if and only if I think, since if there's any dependencies then E⋂F won't be k and...something
thorny slate
Yeah
You'll have weird stuff
Try k[a] tensor itself
To see what goes wrong
Start with a=i and k =R for quick visualization
solar wyvern
ℂ ⊗_ℝ ℂ?
thorny slate
Yes
solar wyvern
branched covering spaces ⇔algebraic extensions of commutative rings
@oblique river could you elaborate/confirm this can be made rigorous
source is http://math.ucr.edu/home/baez/week205.html
simple agate
cloud walrusBOT
alex:
quiet cave
Yes
oblique river
@solar wyvern so that comes down to thinking about specs and ramification of primes
fringe nexus
Why are Q(sqrt(2)) and Q(sqrt(3)) not isomorphic? They are generated by the basis [ 1,sqrt(2)] and [1,sqrt(3)], so can't we map the basis to each other?
mild laurel
Isomorphic as vector spaces?
mild laurel
Yeah so for this, they're not talking about them being isomorphic as vector spaces, which they are over Q, but they're talking about them being isomorphic as fields.
fringe nexus
oh ok
for 5) i think theres only one automorphism? is it the automorphism sending the 4th root of 2 to negative?
mild laurel
Don't forget the trivial automorphism
mild laurel
Something you'll learn or maybe have is that the set of automorphisms for something like this form a group under composition
fringe nexus
I know that
mild laurel
And something you might have learned is that automorphisms must take elements to other roots of their minimal polynomial.
fringe nexus
and then Aut[k/f] is a subgroup of it which fixes F
mild laurel
So if you can find the minimal polynomial of 4th root of 2, then you can know where you can send it for it to be an automorphism
mild laurel
Not quite. Remember you're working over Q(sqrt(2)) not Q
fringe nexus
i don't know how to do p roots in latex sorry
oh in this case do i have to divide by the min poly of q(sqrt(2))?
Im not sure how to find min poly's of extensions like this
solar wyvern
$\sqrt[4]{-2}$
cloud walrusBOT
flimflam:
mild laurel
See if you can factor x^4 - 2 over Q(sqrt(2))
fringe nexus
$(x-i \sqrt(2))(x+i\sqrt(2))(x+\sqrt(2))(x-\sqrt(2))$
cloud walrusBOT
Victoria:
solar wyvern
isn't this the case where you have to be careful because you get a sqrt2 from the 8th root of unity?
fringe nexus
but you cant factor it completely over q(sqrt(2) because it doesn't include sqrt(-1) right?
im confused
is this extension the splitting field of x^2 - sqrt(2) then?
thorny slate
yeah
fringe nexus
therefore theres 2 automorphisms
the trivial one, and the one mapping the roots to one another
thorny slate
yes
fringe nexus
ok
mild laurel
Yeah you were right, but this is how you show that those are the only two
solar wyvern
If you have two extensions over k which are both contained in the algebraic closure of k along with an embedding σ between them, separability of a polynomial f is equivalent to separability of σf in the other right?
thorny slate
the embedding has to be the identity doesnt it?
hmm I guess not
only if they're normal
but yeah that should be true
lusty swallow
hi, i was wondering if this statement is true: if an element is transcendental over Q (rationals), then it is transcendental in any algebraic extension of Q properly contained in R (reals)
how would I go about proving or disproving it?
lusty swallow
@woven delta okay, i'll give it a shot. Thanks!
just grappling with the ideas of field extensions and algebraic/transcendental numbers now.
woven delta
I'm too tired to think of an actual algebraic proof of this, but if you know about definability and first order logic this is obviously true
lusty swallow
okay, thank you!
woven delta
Given a polynomial equation with coefficients in the algebraic closure of the rationals, it shouldn't be too hard to find a polynomial with coefficients in Q which has all the zeros of the first polynomial
But basically what your asking is "what is the algebraic closure of the algebraic closure of the rationals"
lusty swallow
i see
woven delta
So you should expect that the answer should be itself
lusty swallow
i'll have to work on this a little, im still a little new to the concepts
but that kinda makes sense
thanks heaps!
solar wyvern
Okay so probably big dumb but
each normal extension can be realized as a splitting field of f irreducible?
somber bramble
wasn’t a normal extension just “splitting field of a separable polynomial with coefficients in the ground field”
thorny slate
but maybe
somber bramble
or is that my teacher using nonstandard definitions
thorny slate
that's galois
solar wyvern
if it has any roots in the base field, then it's not normal i think
somber bramble
I know he’s used at least one
thorny slate
it doesn't need to be separable
solar wyvern
yeah, this is just about normality
somber bramble
I’m p sure that’s what we called normal, and then showed it to be equivalent to galois
thorny slate
in char 0
somber bramble
galois being “base field is the fixed field of the automorphism group”
solar wyvern
imagine working in characteristic zero
somber bramble
I know my prof used a nonstandard defintion for a word somewhere, it was probably this
thorny slate
anyway a normal extension is definitely given by splitting field of polynomials
solar wyvern
it's not a standard def, it's a restricted case @somber bramble
somber bramble
but we definitely defined normal extension to be “splitting field of separable polynomial”
thorny slate
but the irreducible part is suspect
solar wyvern
if it's not irreducible I think it won't be normal?
solar wyvern
wait what
thorny slate
yeah you're just adding a bunch of roots
solar wyvern
oooh
i think i get it
but if you do choose a irreducible factor, its roots will all be in the splitting field
so it should be normal...
so other dumb q is
what does it mean for an embedding to induce an automorphism
i thought an embedding into itself was always auto
thorny slate
it will be normal
the hard direction is the other one
normal -> splitting of irreducible poly
solar wyvern
ah, this is an embedding of K into the closure kbar all over k
e: i think I can figure this out
I don't imagine there's any canonical embedding of an arbitrary field into its closure is there
solar wyvern
okay im just gonna work through the proof in lang
welp I'm dumb
for σ: K_f →Kª an embedding from a splitting field of some poly in K into the closure, im σ = K_f so you can define σ' ∈Aut K_f such that σ’(x)=σ(x)
e: sorry for janky notation
feel as though this satisfies some universal property
tempted to say σ'=Coker σ, but p sure that's not quite right
hollow peak
Let (G,.) be a group whose order is p^2 where p is a prime show that G is abelian
I considered Z(G) which is a subgroup of G, then used Lagrange's theorem:
-If the order is p^2 then Z(G)=G so G is abelian
-if the order is p, then Z(G) is cyclic so it exists a in G such as Z(G)=<a>. Then we consider G/Z(G) which order is also p so it is cyclic of generator b, and we have G=<a,b>. Knowing that a is in Z(G) we also have G abelian
-But if |Z(G)|=1 I have no idea how to start the exercise
I have almost no tools to do it
only lagrange's theorem
I can show that G=<a,b> like I did when |Z(G)|=p
but then how to show that a and b comute?
hollow peak
It uses too much notions that I didn't see
thorny slate
ok let me put it this way
consider the conjugation action
the elements of Z(G) are exactly the fixed elements
so the orbits of size 1 of this action
hollow peak
what's an action?
thorny slate
therefore |G| = |Z(G)| + U {bigger orbits}
woven delta
Welp
thorny slate
oh
hollow peak
I'm new to general algebra
hollow peak
okay
thorny slate
one can prove the size of each orbit of the action must divide the size of the group
and from this and my previous comments
|G| = |Z(G)| + U {bigger orbits}
and you know that p divides both |G| and the right side
so it must divide |Z(G)|
that's the idea
you should read up on group actions
hollow peak
It's a bit strange because it's one of my exercises, and we didn't do group actions
hollow peak
its maybe on purpose because its the last one
woven delta
Using lagranges theorem you can show that if G is nonabelian then every non identity element must have order p
hollow peak
yes
woven delta
Maybe you can use that
hollow peak
even if its abelian in fact
woven delta
What do you mean
hollow peak
because <x> is a subgroup of G
thorny slate
I dunno man the usual way to solve this is to use the lemma that I used
hollow peak
and its different from {e}
thorny slate
it's weird that you weren't given actions
woven delta
I don't think undergrads even learn actions in a first algebra course
thorny slate
anyway ok
woven delta
My first course focused on rings
hollow peak
yeah we're focusing on rings and groups
thorny slate
woven delta
Oh that's cute
hollow peak
actually we didn't finish the course. Maybe we'll see actions at the end
bleak finch
Forgot about the class equation
fringe nexus
Aut(K/F) can be automorphisms that fix F, but also fix some E such that $F \subset E$ right?
cloud walrusBOT
Victoria:
fickle brook
sure ofc
fringe nexus
do you have an example
fringe nexus
Idk what im trying to ask here tbh
fickle brook
but like idk take any two trivially-intersecting extensions of something
like idk Q(sqrt(2)) and Q(sqrt(3))
fringe nexus
But theres a collary that says Aut|K/F| = [K:F] iff F is the fixed field of Aut(K/F)
fickle brook
fixed field of Aut(K/F) = biggest field fixed by all the autos
fringe nexus
o ok
fickle brook
some may fix a bigger field that's fine
wind steeple
Cube root x)
worthy kindle
$\sqrt[3]2$
cloud walrusBOT
Tuong:
fringe nexus
Oh
wind steeple
Too fast
worthy kindle
Hehe
wind steeple
But the two other roots are imaginary not real
Also we have |Aut(K/F)| = [K:F] iff K/F is separable
Nope
Mb
wind steeple
|Aut(K/F)|=[K:F] iff K/F is separable and normal
fringe nexus
if so then thats true i think
wind steeple
We have |Hom_F(K,omega)|=[K:F] iff K/F is separable, and Hom_F(K,omega)=Aut(K/F) iff K/F is normal
Hom_F(K,omega) is the set of homomorphisms from K to omega that fix F.
And omega is an algebraic closure of K
fickle brook
$\bbQ(\sqrt[3]{2}) \subset \bbR?$
cloud walrusBOT
Ann:
wind steeple
Yes
fickle brook
o ye ofc
wind steeple
$$\mathbb{Q}(\sqrt[3]2) = {a+b\sqrt[3]2+c\sqrt[3]2^2,(a,b,c)\in \mathbb{Q}^3}$$
cloud walrusBOT
Zak:
patent viper
Yes, but careful - its splitting field is not in R, so it cannot be normal
solar wyvern
Q: if you have two alg, sep extensions k(a), k(b) with minimal polys f,g respectively, will k(a,b) always be iso to Frac(k[X]/(fg))?
wind steeple
k[X]/(fg) isn't an integral domain
cl(f)cl(g) = cl(fg) = 0
you can't take the fraction field
@solar wyvern
wary sphinx
As a quick sanity check, the power of a unit in a ring is always going to be a unit, right? I have a homework problem that looks like it's supposed to be a lot more complicated than that.
thorny slate
write an inverse for it
wary sphinx
I mean
if xy=1, then x^ny^n=(xxx...x)(xy)(yyy....y), and it just cascades into one-ness
thorny slate
yes
wary sphinx
It just seemed like I had to be missing something, but I guess not
solar wyvern
@oblique river when you say a prime P ramifies over L/K, that means O_L/(P) has nilpotent elements? (P) the O_L-submodule generated by elements of that prime P from O_K.
also even if you generalize beyond algebraic extensions, any Q-algebra A will be separable right?
As in for any other Q alg B, A ⨂_Q B will be reduced (nilpotent free).
chilly ocean
Hi, can someone explain why isomorphism has a trivial kernel?
wind parrot
If it didn't have trivial kernel, there would be at least two different elements in it, which both map to the identity. so the homomorphism wouldn't be injective, so not bijective either
chilly ocean
thanks
simple valley
isos in concrete categories are all bijective
all split monics are injective, all split epis are surjective
ergo qed
oblique river
@solar wyvern both of those are correct
(the first one is definitely correct, and I believe the second one is as well)
solar wyvern
@oblique river thanks, I wasn't sure if I was messing up on the terminology for ramification stuff 😅
this seem okay @thorny slate
For $f = t^n -x = \sum_{k = 0}^n \beta_k t^k \in F(x)[t], (x) \in Spec F[x]$,
$$
\beta_k \in (x) \iff k<n, \beta_0 \nin (x)^2=(x^2) \implies (f) \in Spec F(x)[t]
$$
cloud walrusBOT
flimflam:
thorny slate
what
why do you write it as betas?
b0 = -x, bn = 1, the rest are 0
am I being pranked
solar wyvern
just making it clear (hopefully) that it's standard application of eisenstein
thorny slate
you sohuld have just said "we can apply eisenstein" instead of doing the buildup
the formalism usually confuses
solar wyvern
(i did)
thorny slate
more than anything
solar wyvern
rip
For $x$ trancendental, we can apply eisenstein criteria using $F[x]$ as our domain. Consider the prime ideal $(x) \subset F[x]$: $x$ divides every nonleading coefficient but not the leading coefficient and $x^2$ does not divide the constant term.
cloud walrusBOT
flimflam:
solar wyvern
maybe better to just say that tho
solar wyvern
yeah
thorny slate
don't use (f) in Spec F(x)[t] to say it is irreducible
solar wyvern
Is this correct?
An extension $K/F$ is separable if for each $\alpha \in K$,
$$[K:F(\alpha)]_s = [K:F(\alpha)]$$
that is the inseparability degree of each extension of a simple extension equals the degree of the extension.
@thorny slate
cloud walrusBOT
flimflam:
thorny slate
dunno
oh, you're allowed to pick a in F
then yeah sure
since [K:F] = [K:F]_s
but then this is useless
just pick a = 1
somber bramble
but it says “for each a”, not “for some a”
solar wyvern
for each, yeah
thorny slate
yeah it's useless
it's equivalent to [K:F] = [K:F]_s
which is trivially equivalent to separability
solar wyvern
okay, was checking reality
thorny slate
you sure you aren't doing it on purpose
it's v suspicious
all your questions contain extraneous variables
solar wyvern
which is extraneous
thorny slate
free variables that don't do anything except muddy the question
solar wyvern
simple extensions are easy to work with
solar wyvern
u mean the id map?
thorny slate
yeah
solar wyvern
okay I see what you mean
solar wyvern
If you have a poz char field extension $E/F$ and $S \subset E$ linearly independent over $F$, and $S^p$ the image under frobenius you'll have $F(S^p) \cong F(S)$ right?
cloud walrusBOT
flimflam:
thorny slate
finte fields are perfect but the others might fail
take F = Fp(x^p), E = Fp(x),
assuming by the frobenius you mean x -> x^p
solar wyvern
yeah
actually just need to show that S^p is linearly independent over the base field
solar wyvern
does that work for infinite prime char fields?
thorny slate
what
solar wyvern
so E/F is of finite degree, but they're not necessarily finite fields
thorny slate
yeah
it fails if they're infinite
as I showed
for finite ones it should work
maybe
actually no
not even
when you take S to be linearly independent like that
you need S to be a set of minimal generators
or something
then it's more tricky
solar wyvern
sorta not getting what "linearly independent roots" means
well
the significance or how to say anything about it
e: wait my brain cell activated e2:
solar wyvern
@thorny slate I may be totally wrong about this but the idea is:
Embeddings are objects and the extensions between embeddings are morphisms. Then you have a certain sort of "distinguished category" which means
(1) E/k ⇔ E/F and F/k
(2) E/k, F/k ⇒EF/F, EF/E
(3) E/k, F/k ⇒EF/k
and sep extensions form such a distinguished cat, as do the algebraic and finite extensions.
[technically you also need a common field for E,F, but unless you're dealing with some really big cat you'll always have a terminal object like Kbar(X1,...)]
[E:k]_s \leq [E:k] is the statement that when your extension is separable, the faithful functor including stuff from SepAlg_k into Alg_k is full. Degree multiplicity is hopefully encoded into the functor which changes your (distinguished) category over base field k to a (distinguished) category over another field (of same char).
thorny slate
this is just the category of fields
and you can distinguish separable embeddings yeah
a field extension is just a nonzero field map
solar wyvern
not really useful but you can replace "field" with "k-algebra"
what am I doing with my life
solar wyvern
If you have some prime char field K, E = K(t_1,...,t_n), F = K(t_1^p,...,t_n^p) then for D the formal derivative on E, F=ker D and [E:F] = p corresponding to E/ker D = {f: leading coefficient after modding has exponent in k+pℤ}
@thorny slate is this wrong
solar wyvern
I just don't know anymore
solar wyvern
wait n=2
still confuse
next part is that there are infinite int extensions so I imagine this is an exercise where you show primitive element thm fails if you drop separability
still not sure why [K(t_1,t_2):K(t_1^p,t_2^p)] = p
@oblique river
oblique river
sry just got here
is K a field of characteristic p? or any field?
cuz that extension should have degree p^2
what is true about that extension (assuming K has characteristic p) is that every element of K(t_1, t_2) has degree 1 or p over K(t_1^p, t_2^p)
which shows that the extension doesn't satisfy the primitive element theorem and thus isn't separable
@solar wyvern
what is alpha
solar wyvern
$[K(α):K] = p, α∈ K(t_1,t_2) \ K(t_1^p,t_2^p)$
cloud walrusBOT
flimflam:
oblique river
why do you delete messages :(
solar wyvern
because I'm a waste of space
solar wyvern
now it shouldn't matter whether there are 2 or n indeterminants I think
oblique river
correct
solar wyvern
do trancendental elements correspond to adjoining free generators to your field and taking inverses?
solar wyvern
so I've been wondering this but there are fields which aren't quotients of K[X_1,...] right?
oblique river
K(X) isn't
solar wyvern
algebraic fields extensions I mean
oblique river
Q(sqrt(p) | p prime)
isn't a quotient of a finite Q[x_1, ..., x_n]
if that's what you're looking for
solar wyvern
is it a quotient of the polynomial ring in countable indeterminates?
oblique river
yes, quotient by x_i^2 - p_i where p_i is the i-th prime
in general if you want to consider K(alpha_i) where the alpha_i are algebraic over K (maybe infinitely many)
then you can just adjoin that many variables to K
K[x_i]
and then consider the (surjective) map K[x_i] --> K(alpha_i) via x_i -> alpha_i
as long as the alpha_i are algebraic that map is surjective
solar wyvern
so this sorta dumb and I don't know how to phrase it, but for a simple trancendental extension K(X) why do you know this can't be an extension of K(Y,Z) for Y,Z also trancendental over K
oblique river
^this
solar wyvern
I mean intuitively you'd know it's true because transcendental elements are the ones which don't have relations between them
nothing makes sense aaaaaaaaa
thorny slate
solar wyvern
this is actually readable 🤔
solar wyvern
@oblique river you still there?
oblique river
kinda
solar wyvern
still fuzzy on failure of the iso between the
{zero set of a set of polys} ↭ {extensions of embeddings wherethey vanish}
--and separability
if an (alg) extension is inseparable a finite subextension must also fail to be separable?
oblique river
the second is true, and im having trouble parsing what the first question is
also sry but i have to head to bed now, im reaching "trouble parsing questions" phase of tired
i'll get back to you tomorrow tho
thorny slate
seems unintelligible to me too
solar wyvern
same
oblique river
also @solar wyvern I just woke up and I'm reading your questions again. I still can't understand the first one but I realize now that I misread the second one
The second one is false actually
for example consider F_p(x)(x^(1/2p))
this is not separable but it has F_p(x)(x^(1/2)) as a subextension which is separable over F_p(x)
sry about that
oblique river
ahh okay, yes if they meant "purely inseparable" then it is true
solar wyvern
purely inseparable means everything in the subfield is some prime power of something in the field?
oblique river
the prime has to be the same as the characteristic
but yes that is one way to think about it
do you know about "separability degree" and "inseparability degree"?
solar wyvern
yes, but probably not as well as I should
oblique river
its ok i also had to look up the definition again just now lol
the point though is that purely inseparable is the same as "separable degree 1"
solar wyvern
oh yeah
oblique river
which is just saying that nothing in the field is separable over K
solar wyvern
[E:F]_s=1
oblique river
yes
solar wyvern
i am also waking up hopefully
solar wyvern
bye
chilly ocean
Maybe someone here can answer this. I am looking into taking a grad class in lie groups and lie algebra. What prerequisites are usually good to know for a basic and elementary intro to lie groups and algebras? I found a PDF of a book by Brain Hall and he says in the preface linear algebra is largely used in the first part of the book
somber bramble
from what I can tell, the usual prereqs are basically some group theory and some differential geometry - how much will depend heavily on the lecturer. it’s a pretty deep topic and the more prereqs it assumes the less time it’ll have to spend on them, but you could just as well develop all the group theory you need in the first few classes…
at my school, Lie Groups is typically taken as a followup to Differential Geometry I & II
uncut girder
At my school Grad Alg 1, Grad Top 1 and Grad Top 2 are prereqs.
Top 2 is smooth manifolds
full blaze
im gonna take Grad Alg 1 in the fall
chilly ocean
Cool. Thanks peoples
full blaze
woot
hollow peak
if a and n are coprime, then a is in (Z/nZ)*. Knowing that |(Z/nZ)*| =phi(n) then by Lagrange's theorem we have |<a>| divides phi(n) so a^(phi(n))=1
is that right?
hollow peak
ah oui corrigé
onyx mirage
You guys might know
f(x) can be seen as a function f, acting on a tuple (x) right? How is the binary operator * in f*(x) called and does f*(x) = (x)*f
solar wyvern
You can identify R[x] with a subring of the endomorphism ring Endo R[x] by identifying f with the function which sends x to f
I think that's how it works
@left lily
@thorny slate Writing Kª for the algebraic closure of a field K, will there be a bijection between the irreducible elements of K[X] and Kª iff K=Kª?
thorny slate
write K for the algebraic closure of a field k
solar wyvern
thorny slate
and yes
irreucible elements are exactly linear polynomials of the form ax + b with a nonzero
wait
that's not what you asked
solar wyvern
generally do you have any ring map from k[X] to K tho?
thorny slate
i misread
well
no, you don't have a bijection between irreducibles of k[X] and elements of K
wait I didn't misread
what the fuck
my brain is mush
so yeah what I said before applies
send ax + b to b/a
so when k = K you have a bijection
otherwise you don't
you need more data
and the data is like
solar wyvern
seems like there's no obvious reason for there to be a multiplicative homomorphism. although it'd be nice if for f,g: (f) = ker ev_a, (g) = ker ev_b, fg mapped to ab, but no reason that'd actually work
thorny slate
irreducible f in k[x] <-> K/k galois orbit of a root of f in K
solar wyvern
in general your orbits won't be the same size? eg for C/R you have a 2-1 map for nonreal values of C and a 1-1 for stuff already in R
well, rather than in general, unless your field is algebraically closed I guess
thorny slate
yeah
solar wyvern
depending on the base field k, K/k won't necessarily be separable right? actually not sure
thorny slate
yeah
the algebraic closure includes finite inseparable extensions
so actually my statement is true for the separable closure
solar wyvern
which one
thorny slate
irreducible f in k[x] <-> K/k galois orbit of a root of f in K
actually it still works I think
probably
solar wyvern
I mean the simplest case it could fail would be like F_p(x) and its alg closure