Mathematics - 2012-04-16 (page 1 of 3)

Benjamin Lim

00:02

Learners drivers license!!

Eric Gregor

@MarianoSuárezAlvarez, maybe you could help me walk through something that should be routine, i guess. maybe my linear algebra is just bad

Let $\zeta$ be a primitive $16$th root of unity and consider
$\mathbb{Q}(\zeta_{16})$ over $\mathbb{Q}$.

The Galois group here is $\mathbb{Z}/4\times\mathbb{Z}/2$, with the whole group
consisting of the units in $\mathbb{Z}/16: \{1,3,5,7,9,11,13,15\}$. The elements
of order $4$ are $\{3,5,11,13\}$ and the elements of order $2$ are $\{7,9,15\}$.
Let $a=\sigma_3$ be the map sending $\zeta\to \zeta^3$ and $b=\sigma_{15}$ be the
map sending $\zeta$ to $\zeta^{15}$.

I would like to find the fixed field corresponding to $\langle \sigma_9,\sigma_{15}\rangle $.

robjohn

@BenjaminLim how old do you need to be for one there?

Mariano Suárez-Alvarez

the mathjaxy thing should be smarter... every time one edits something here, the comment gets unrendered :/

Eric Gregor

i re-edit too much

Mariano Suárez-Alvarez

why did you skip $\zeta^2$ in the list of basis elements?

Eric Gregor

00:05

oh, should it be $\zeta,\zeta^2,\dots,\zeta^8$?

Benjamin Lim

@robjohn you can get an L license when you're 16

But I'm 19 now so a bit late :D

Eric Gregor

there are only 8 basis elements

since the degree is $\phi(16)=8$

Mariano Suárez-Alvarez

I would have picked the first eight powers :D

(starting from zero!)

Nico Bellic

@Eric Gregor, If you don't mind me asking, what kind of linear algebra level is that up there? I'm a college freshman in a regular linear algebra course, and I don't think I've ever encountered such stuff.

Eric Gregor

err, right

haven't got to the linear algebra yet nico

Benjamin Lim

00:07

@NicoBellic That is not your usual linear algebra

Mariano Suárez-Alvarez

in any case, pick a basis, write down the matrices for the generators of the group

Eric Gregor

@MarianoSuárezAlvarez i just though it would be easier to use my Q-basis because you can see the structure of the group of units better

Benjamin Lim

@NicoBellic Basically you have a larger field containing a smaller field. In this case the larger field is $\Bbb{Q}(\zeta_{16})$

Mariano Suárez-Alvarez

and find their fixed spaces

then intersect

Benjamin Lim

the smaller field is $\Bbb{Q}$

robjohn

00:08

@BenjaminLim It's 15-1/2 here, but I waited until 16. not quite 3 years though :-)

Mariano Suárez-Alvarez

@EricGregor, ah, that may help: I have not tried to do the computation :)

Benjamin Lim

@robjohn haahahahahahahahahaha

@robjohn The only reason I'm getting it is for ID

When I want to buy alcohol

t.b.

While thinking about this question I came up with a tremendously slick proof that every entire function that isn't a polynomial is of the form $C \cdot e^{z}$ and managed to wonder where I went wrong... Guess that means I should go to bed :)

Benjamin Lim

@NicoBellic They are trying to calculate the dimension of $\Bbb{Q}(\zeta_{16})$ as a vector space over $\Bbb{Q}$

So right now they are trying to find a basis for $\Bbb{Q}(\zeta_{16})/\Bbb{Q}$

robjohn

@BenjaminLim 18 is the age there, correct?

It's 21 here

Nico Bellic

00:09

@Ben

Benjamin Lim

@robjohn To drink alcohol yes.

Nico Bellic

I'm still not sure what that means

Benjamin Lim

@robjohn wowowowowow

@NicoBellic Have you learned about vector spaces????

t.b.

@robjohn 16 here for beer and wine, 18 for stronger stuff.

Eric Gregor

@MarianoSuárezAlvarez when you say the matrix for the generator you just mean a 1x8 matrix with a 1 in the proper place?

Benjamin Lim

00:10

@robjohn Here they are very strict. When you drive on a P or L license your BAC must be 0.00 on the breathalyser

Nico Bellic

Yeah, fields, V.S, linear transformations, elementary ops and system of linear equations, and determinants

Eric Gregor

i can feel my question is dumb

Mariano Suárez-Alvarez

no, an 8x8 matrix

$\sigma_3$, say, is an endomorphism of $Q(\zeta)$, which is an $8$-dimensional vector space

so it has an 8x8 matrix

robjohn

@BenjaminLim sounds reasonable however.

Eric Gregor

because we have a generator $\langle \sigma_9,\sigma_{15}\rangle$

Benjamin Lim

00:11

for someone who has a full license the maximum is a BAC of 0.05 on the breathalyser

@NicoBellic Well what is the dimension of $\Bbb{C}$ as a real vector space?

Mariano Suárez-Alvarez

<strike>notice that every field automorphisms maps primitive roots to primitive roots, so, since out basis is made out of the primitive roots, the matrix will be a permutation matrix</strike>

Nico Bellic

@Ben, as real? I believe 2

Benjamin Lim

yes

so now notice that $\Bbb{C}$ is a field that contains the real numbers

in the same way whenever you have a field $E$ containing a field $F$

you can view $E$ as a vector space over $F$

Mariano Suárez-Alvarez

hmm, strike my last comment

Nico Bellic

@Ben, right.

Benjamin Lim

00:14

@NicoBellic So in this case we have that $\Bbb{Q}(\zeta_{16})$ is a field

containing $\Bbb{Q}$

To define exactly what is $\Bbb{Q}(\zeta_{16})$ we need to talk about quotient rings

you probably have not seen those.

Mariano Suárez-Alvarez

you can define it as the smallest field which contains the root...

I'd say, you must :)

Nico Bellic

Yeah. I also haven't encountered Galois groups, as well as some of the notation up there.

John Smith

hello

Benjamin Lim

@MarianoSuárezAlvarez Well ok

Eric Gregor

@MarianoSuárezAlvarez this is tedious. no wonder "ad hoc" methods are used

Benjamin Lim

00:16

But I wanted to say that $\Bbb{Q}(\zeta_{16}) \cong \Bbb{Q}[\zeta_{16}] \cong \Bbb{Q}[x]/f(x))$

Nico Bellic

@Ben, you said you're only 19. When did you start doing linear algebra.

Benjamin Lim

where $f$ is the minimal polynomial of $\zeta_{16}$ over $\Bbb{Q}$

Mariano Suárez-Alvarez

every method is ad hoc

they get embellished into "general methods' simply by making the list of hypotheses they depend on

that is why there is no difference between an example and a theorem

Benjamin Lim

@NicoBellic A while ago maybe

But it was only last year that I really started looking at things like primary decomposition, etc.

Nico Bellic

Oh wow you started pretty young. Are you in college?

Benjamin Lim

00:20

@NicoBellic Second year of university

Eric Gregor

ah i screwed this up

Benjamin Lim

@NicoBellic Right now I am looking at analysis, field/galois theory and commutative algebra

Eric Gregor

why compute these matrices and then their eigenspaces. this is going to take too long. better to just have a feel for it i guess and do this only when absolutely necessary

Mariano Suárez-Alvarez

the problem with that approach is that when it is absolutely necessary

you will not be prepared to do it

Benjamin Lim

@EricGregor I believe mathematica can compute quickly the eigenvectors/eigenvalues of a matrix

complex or not

Mariano Suárez-Alvarez

00:23

because you did not practice how to do it when you did not need to

life sucks

Benjamin Lim

@MarianoSuárezAlvarez In an assignment question I tried to show some element was in some bigger field

turns out I had to bash a $6 \times 6 $ matrix

Mariano Suárez-Alvarez

In fact, in this particular case the matrices are so simple, and the eigenspaces so easy to decribe once you see why they are, that it is worth doing the computation

Eric Gregor

@MarianoSuárezAlvarez i would normally do this, the problem is i have a test tomorrow

Mariano Suárez-Alvarez

you still don't know what the matrices look like

Eric Gregor

any method i learn will have to be doable in maybe 10 minutes

Mariano Suárez-Alvarez

00:24

that is why you think the computation will be hard :)

right now, you do not know how long it takes to do this method

Eric Gregor

if i'm doing it right it's not just a permutation matrix, there are some rows with 2 "1"s

Benjamin Lim

@EricGregor In my institution the lecturer told us explicitly that there will be no crazy computations in the exam

Mariano Suárez-Alvarez

how can that possibly be?

the map $\sigma_9$ maps each power of $\zeta$ to another power of $\zeta$

since your basis is made out of powers of $\zeta$ and $\sigma_9$ is an isomorphism, it has to permute the basis elements!

(this is the kind of observation that would have been obvious by conputing the matrices of a few automorphisms; that is precisely how I know it is true)

ouch, wait

heh, you are right :)

Eric Gregor

:)

Mariano Suárez-Alvarez

well, in any case, there is no magical way to find the fixed field of an automorphism

Eric Gregor

00:28

the problem is in the case i picked we have $\langle \sigma_9, \sigma_{15}\rangle$

so it gets confusing for me to make the matrix

Mariano Suárez-Alvarez

you have to compute the matrix of $\sigma_9$

Eric Gregor

you have to consider how each element lives in the group $\mathbb{Z}/4\times\mathbb{Z}/2$

Mariano Suárez-Alvarez

no. why?

$\sigma_9$ is the autom. which maps $\zeta$ to $\zeta^9$, isn't it?

Eric Gregor

yes, but what about the pair? i'm trying to keep the galois group in mind

i'm roughly following the argument here, but trying to formalize it: mathpost.asu.edu/~zinzer/Problems7S.pdf

Mariano Suárez-Alvarez

but to find the common fixed field of the whole subgroup is the same as computing the intersection of the fixed field of $\sigma_9$ anf the fixed field of $\sigma_{15}$

because these two elements generate the subgroup.

Eric Gregor

00:31

ok, that makes sense actually

Mariano Suárez-Alvarez

I in fact don't understand what you were trying to do

Eric Gregor

so then there will surely be just one $1$ in each row

neither do i, that's why i got confused!

Mariano Suárez-Alvarez

that has nothing to do with it

Eric Gregor

just see where each element gets send after taking the 9th power

Mariano Suárez-Alvarez

for example:

$\sigma(1)=1$
and $\sigma(\zeta)=\zeta^9$ and this is $-\zeta$ because $\zeta^8+1=0$.

this tells you the two first columns of the matrix for $\sigma_9$.

Eric Gregor

00:35

you are using my automorphisms without using my Q basis, is that justified?

if you catch my drift

Mariano Suárez-Alvarez

huh?

Use $\{1,\zeta,\dots,\zeta^7\}$ as a basis, which is the most obvious choice

Eric Gregor

right, i was using a different $\mathbb{Q}$ basis

ok

Mariano Suárez-Alvarez

Don't try to be smart about the choice of the basis when you still do not know what effect that smartness will have! :D

Eric Gregor

no, my problem is understanding what automorphisms i'm thinking of

with that Q basis how do i understand the galois group

Mariano Suárez-Alvarez

the automorphism $\sigma_9$ maps $\zeta$ to $\zeta^9$

Eric Gregor

00:37

which is a group of units of $\mathbb{Z}/16$

Mariano Suárez-Alvarez

and then it maps $\zeta^j$ to $\zeta^{9j}$ for all $j$

$\sigma(\zeta^2)=\zeta^{18}$. Since this is not in the basis, we need to see what linear combination of basis elements it is. Now $\zeta^8=-1$, so $\zeta^{16}=1$ and $\zeta^{18}=\zeta^2$.

(incidentally, this means $\zeta^2$ is fixed under $\sigma_9$)

Eric Gregor

oh, maybe i understand

ok, so in this way we get a matrix and the places where there is a 1 on the diagonal correspond to the fixed elements

Mariano Suárez-Alvarez

there may be fixed elements which are not in the basis...

do the computation, do not guess

you can guess when you have sufficient information

and the information you will gather by carrying out experiments

Eric Gregor

you are a hard taskmaster, @MarianoSuárezAlvarez. ok, lemme try

just to confirm then, i am still just considering the maps $\sigma_j$ where $j$ corresponds to the group of units of $\mathbb{Z}/16$, yes?

Mariano Suárez-Alvarez

(in the particular cases of $\sigma_9$ and $\sigma_{15}$ the matrices are so simple that there is not much to do; it is up to you to see why that happens!)

Eric Gregor

00:44

basically the odd guys between $1$ and $16$

Mariano Suárez-Alvarez

yes

how does that correspondence work? an imnvertible element $j\in\mathbb Z/16$ corresponds to the unique automorphism which maps $\zeta$ to $\zeta^j$

call it $\sigma_j$

it then maps $\zeta^i$ to $\zeta^{ij}$ for all $i$, because it is a map of rings

Eric Gregor

interesting matrix

if you shift the basis it's a diagonal matrix

Mariano Suárez-Alvarez

what matrix?

Eric Gregor

it's a shift matrix

corresponding to $\sigma_9$

i don't know if shift is the current terminology

Mariano Suárez-Alvarez

the matrix corresponding to $\sigma_9$ is a diagonal matrix with $1$s and $-1$s along the diagonal, no?

Eric Gregor

00:50

wait, i screwed up

arg

oh, ok

so a 1 in the first entry, then we have it alternating

1,-1,1,-1 along the diagonal

i am slow

Mariano Suárez-Alvarez

for $\sigma_9$?

I got $\left(
\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1
\end{array}
\right)$

Eric Gregor

yes, right

Mariano Suárez-Alvarez

since the matrix is diagonal, it is obvious what the invariant subspace for it is

$\sigma_{15}$ is more interesting

$\left(
\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 & 0 & 0
\end{array}
\right)$

Eric Gregor

dumb question: why is it obvious what the invariant subspace is? you mean just the guys with 1 in the diagonal are invariant, right?

if so i agree it's obvious

Mariano Suárez-Alvarez

the subspace generated by those basis elements

Eric Gregor

00:57

by those basis elements with 1s in the diagonal, right?

Mariano Suárez-Alvarez

the invariant subspace is the eigenspace for the eigenvalue $1$

Eric Gregor

ok, good

my linear algebra is awful

need to work on that this summer!

so with $\sigma_{15}$ do we try to diagonalize this guy? is that useful?

Mariano Suárez-Alvarez

no

write down the equations which express that a vector is invariant under $\sigma_{15}$ and solve them

Eric Gregor

um, ok

Mariano Suárez-Alvarez

that 'um' comes from an overestimation of the work involved

Eric Gregor

01:02

so i want to solve $a_0+a_1\zeta+\cdots+a_7 \zeta^7=$ "itself under the transformation of coordinates?

Mariano Suárez-Alvarez

well, that is one way to put it

you can also try to solve the linear system Ax=x with A the matrix of $\sigma_{15}$ and $x$ an element of $\mathbb Q^{8}$

David Wheeler

hello, folks, what are we doing today?

it looks like cyclotomic extensions

Eric Gregor

01:21

@MarianoSuárezAlvarez, just to be clear the fixed field of $\sigma_9$ then is $\mathbb{Q}(\zeta^2+\zeta^4+\zeta^6)$?

Mariano Suárez-Alvarez

no sums

Eric Gregor

commas?

Mariano Suárez-Alvarez

as a vector space, it is generated by $1, \zeta^2, \zeta^4$ and $\zeta^6$

as a field, it is generated over Q by $\zeta^2$

Eric Gregor

i'm having trouble getting mathematica to solve the equation to find the eigenspace of $\sigma_{15}$

Mariano Suárez-Alvarez

solve it by hand!

David Wheeler

01:24

if an automorphism fixes $\zeta^2$ it surely fixes any power of $\zeta^2$, no?

Mariano Suárez-Alvarez

the system of equations has 8 equations, each of the form $\pm x_i=x_j$

you surely can solve it by hand...

David Wheeler

it looks like it's (the $E_1$-eigenspace of $\sigma_{15}$) 4-dimensional

Eric Gregor

er ok

you are right

Mariano Suárez-Alvarez

@DavidWheeler, let him do the computation :D

David Wheeler

well, i just did it myself by hand....

Mariano Suárez-Alvarez

01:27

one of the most difficult things in explaining something is often not doing something :)

Eric Gregor

so all of the "even terms" have to be zero

and the other terms can be anything

Mariano Suárez-Alvarez

for $\sigma_{15}$?

David Wheeler

"even terms"?

Eric Gregor

oh, sorry

David Wheeler

i used the 8x8 matrix Mariano wrote down above

Eric Gregor

01:31

ok, so we get $a_1$ is free, $q_2=-q_8$, $q_3=-q_7$, $q_4=-q_6$, $q_5=-q_5$, etc

so $q_5$ is zero

Mariano Suárez-Alvarez

yes

Eric Gregor

and then we get 3 pairs that are related

Mariano Suárez-Alvarez

that means the subspace is spanned by $1$, $\zeta-\zeta^7$, $\zeta^2-\zeta^6$ and $\zeta^3-\zeta^5$

Eric Gregor

so it is 4-dim

Mariano Suárez-Alvarez

it would be nice to have generators as a field, though

you can check that $\zeta-\zeta^7$ generates it

David Wheeler

01:34

i see how he got it: $\sigma_{15}(\zeta) = \zeta^{15} = -\zeta^7$

Eric Gregor

am i right that the intersection of $\sigma_{15}$ and $\sigma_9$ will then be generated by $\zeta^2-\zeta^6$?

Mariano Suárez-Alvarez

well, a element of the intersection is a linear combination of even power, by the first computation

Eric Gregor

yes, that was what i was reasoning

Mariano Suárez-Alvarez

and at the same time a linear combination of $1, ζ−ζ^7, ζ^2−ζ^6 and ζ^3−ζ^5$

that is only possible for elements which are linear combinations of $1$, and $\zeta^2-\zeta^6$.

Eric Gregor

ok

@MarianoSuárezAlvarez, thank you for your remarkable patience

and thanks for making me do the damn thing

i really do appreciate it

i think i may have actually learned a thing or two

Mariano Suárez-Alvarez

01:40

Notice that the square of $\zeta^2-\zeta^6$ is $2$.

so it indeed generates a field of degree 2

Eric Gregor

so the field is really $Q(\sqrt{2})$

nice

@MarianoSuárezAlvarez do you learn anything when you help out here?

Mariano Suárez-Alvarez

haha

sure

David Wheeler

so one would expect $\sqrt{2}$ to figure in the "normal" complex number form of an eighth root of unity

Mariano Suárez-Alvarez

$\cos(2\pi/8)$ is $1/\sqrt2$.

David Wheeler

$1/\sqrt{2}$ and $\sqrt{2}$ generate the same field over $\Bbb{Q}$

Mariano Suárez-Alvarez

01:56

Notice that for example $\cos(2\pi/16)$ is $\frac{1}{2} \sqrt{2+\sqrt{2}}$

(we had started with $16$th roots of unity)

so $(2\cos(2\pi/16))^2-2$ is $\sqrt2$

and that cosine is $(\zeta_{16}+\zeta_{16}^{-1})/2$

There are other subgroups of index 2 in the galois group of $Q(\zeta_{16})$

David Wheeler

which suggests there is a real extension of $\Bbb{Q}$ contained in $\Bbb{Q}(\zeta)$ of degree 4

Mariano Suárez-Alvarez

so there are other square roots in the field

a nice problem is to find them

that is not hard to check

being real means that it is fixed under complex conjugation, which is an element of order 2 in the galois group, which is $Z/4\times Z/2$

that shows there is one real subfield of degree 4

but does not tell us if its galois group is cyclic or not :)

which is it?

David Wheeler

isn't conjugation $\sigma_{15}$?

Mariano Suárez-Alvarez

yes, because it maps $\zeta$ to $\zeta^{15}$ which is $\zeta^{-1}$ which is $\bar\zeta$

and which element in $Z/4\times Z/2$ is it?

David Wheeler

02:11

thinking...it's been a loooong time since i did this stuff

ok, $\sigma_3$ is an element of order 4

David K.

So, I'm working on this problem and I am having a lot of trouble. Can someone tell me if I am supposed to assume that $K/F$ is Galois?

David Wheeler

but $\sigma_{15}$ isn't in $\langle \sigma_3 \rangle$, so i conclude the galois group of the real subfield of order 4 isn't cyclic.

David Wheeler

02:39

or in your terms, conjugation is the element (0,1) in $\Bbb{Z}_4 \times \Bbb{Z}_2$ (15 is not a square modulo 16)

@DavidK i don't see where K has to be galois (for example we might have $K = \Bbb{Q}(\sqrt[3]{2})$

John Smith

is there a fast way to determine the number of ways to write 1/n as a sum of 2 unit fractions?

David K.

@DavidWheeler I didn't think so. In that case, I have no idea how to proceed with this problem.

David Wheeler

are you working just on the part (d) posted?

David K.

@DavidWheeler Yes. And I can't make heads or tails of the answer that is there, perhaps because the notation is a bit confusing.

David Wheeler

do you understand the notation $H^K$ for two subgroups H,K of a group G?

hmm...it looks like he's just using the superscript to indicate the fixed field.

David K.

02:57

@DavidWheeler No. If $K$ were a finite group of automorphisms of a field $H$, then I would see $H^{K}$ as the fixed field of $K$.

David Wheeler

it appears to me as if the answer essentially depends on the fact that a subgroup of a subgroup just refines the coset partition of the larger subgroup

i'm willing to try to talk about it, but it's been at least 2 decades since i last took galois theory

David K.

@DavidWheeler I appreciate it, but maybe you can just help me to see how the OP arrives at the conclusion that $[H':H]=n/d$. I've been trying to figure that out, as I feel that that is the direction I should be working in.

David Wheeler

03:14

in any group, [G:H] = [G:H'][H':H]

that's what i mean by "refinement"

if you break G into H-sized pieces, that is the same as breaking G into H'-sized pieces, and then breaking each H'-sized piece into H-sized pieces

David K.

@DavidWheeler Right, and in this case $G=\operatorname{Gal}(L/F)$ ?

David Wheeler

the galois group of the galois extension L, yes

John Smith

can anyone help me with my unit fraction problem? or do they know a good way to calculate tau(n)?

David K.

Ok, so I see how (using your notation) $[G:H]=n$, by Galois Correspondence. What I don't quite see is how to get that $[G:H']=d$. I have already established that $d$ divides $n$.

David Wheeler

for example, we can take L to be the splitting field of $m_\alpha$

David K.

03:22

Oh wait. I see now. @DavidWheeler

David Wheeler

right, because we know that H' corresponds to $F(\alpha)$

David K.

@DavidWheeler Yes. Zactly. Ok so I now see how $[H':H]=n/d$. Not quite sure how that implies the final result that I need. But, its closer.

David Wheeler

one question...what are the $\sigma$?

David K.

@DavidWheeler $\sigma$ is a coset representative for $H$ in $\operatorname{Gal}(L/F)$. So an automorphism...

David Wheeler

so we're talking about a traversal of H in Gal(L/F)

David K.

03:38

@DavidWheeler I'm not really familiar with that term...

David Wheeler

meaning we pick one element of Gal(L/F) for each coset of H, for gH, we pick "g"

David K.

@DavidWheeler Yes.

So aren't there just $n$ such embeddings?

I mean there are $[G:H]$ cosets right?

David Wheeler

the thing is, we don't know that because K is not galois

there could be subfields isomorphic to K that include F in the algebraic closure

David K.

@DavidWheeler But by hypothesis $n$ is the degree of the extension $K/F$. And then the Galois Correspondence theorem says that $[L:K]=\operatorname{Ord}(H)$ and $[K:F]=\operatorname{Index}(G:H)$.

David Wheeler

i see what you're saying, yes there are n cosets

David K.

03:50

So, $d<n$, and therefore there should be $n/d$ 'extra' terms in the product yes?

David Wheeler

what we can do is factor that product through H'

David K.

How would that work?

David Wheeler

that is first form the product of H'-coset representatives, and then break every such product into H-coset representatives

so we're splitting n into (n/d) and d

do you see why we want to do this?

David K.

So there are $n/d$ coset representatives of $H'$, say $\tau$, and there are $d$ coset representatives of $H$, say $\sigma$. Then $$\operatorname{N}_{K/F}(\alpha)=\left(\prod_{\tau}\tau(\alpha)\right)\left( \prod_{ \sigma }\sigma(\alpha)\right)$$

David Wheeler

fix your latex

Transcript for

$Mathematics$ Mathematics