groups-rings-fields

[chilly ocean]

But so if you can find what the coefficients of q are, then you "know" what 1/v is, and you can just multiply this to u to get u/v

Certainly $\frac{1}{a^2+3a+9}\cdot (a^2+3a+9)=1$

[cloud walrus]

Mega Euler

[chilly ocean]

Not sure where I'm going with this...

For sure, but I'm not sure what is the relevance of that equation

I'm saying we want q(a) such that (a2+3a+9)*q(a)=1

And q(a) is a 2nd degree polynomial in a

I guess I'm not sure what other element multiplied by 1/v would equal 1

Ok, so. Q(a) is a vector space over Q that has basis 1,a, a2, you agree right?

Yes

And we want to find the coefficients of u/v in this basis. An easier problem is to find the coefficients of 1/v

I agree

So we can write 1/v = b0 + b1*a + b2*a^2

(this is q(a))

Then multiply both sides by v, and solve the coefficients b0 b1 and b2

Maybe there is a better/faster way, I don't know, but at least this should work

I see

I'll look into this right now

[mild laurel]

The usual way is to use the euclidean algorithm to find polynomials f(x) and g(x) so that f(x) (x^2 + 3x + 9) + g(x) (x^3 + 2x + 1) = 1

Then, by plugging in a, you'll have that f(a) * (a^2 + 3a + 9) + g(a) (a^3 + 2a + 1) = f(a) * v = 1

[past temple]

given that f(x) is integrable

how can i prove that f(-x) is integrable

the intuition is obvious

[mild laurel]

this seems more suited for #advanced-analysis

[past temple]

oh sorry i thought i was in that channel already

my b

[chilly ocean]

Idea: merge #groups-rings-fields and #advanced-analysis for better collaboration between members of these two groups

[viscid pewter]

hnnngh

[terse crystal]

I don’t know how to type matrices or commutative diagrams yet

[unique berry]

C* algebra? 👀

[terse crystal]

And maybe differential algebra, like Liouville’s theorem or something

[cloud walrus]

Moth

Moth

Moth

Moth

Moth

[maiden ocean]

like.

its a homomorphism sure

but i cant just distribute it beacuse the x_k themselves are not necessarily in M_i

[golden pasture]

note that this sum is finite

hence ...

maybe if you rewrite in terms of like

in C

Actually maybe a more direct solution

suppose we have μ_{kN} for all k in the sum

what is μ_{iN}(x_i)

the reason why we do this is cuz we have

μ_{jk} μ_{ij} = μ_{ik} as well

[golden pasture]

whenever you see this kinda things in analysis youll think to yourself

"what if j is massive"

same thing here

[maiden ocean]

yea that was my strategy like

take a maximal j

cause its directed

[golden pasture]

ehhh

[maiden ocean]

i gues maximal is the wrong word

not like literally maximal

[golden pasture]

yea isnt the correct term here

there may not be a maximal j

[maiden ocean]

but its directed

[golden pasture]

but a sufficiently "big" j

[maiden ocean]

so theres a j greater than all the j_k

yea

[golden pasture]

yup

and you're done

[maiden ocean]

maybe i am being 4head but i dont wut we can say abt mu_iN(x_i) cause like

again its not like we can just

distribute across the sum

[golden pasture]

stare at each term of the sum

[golden pasture]

we can actually

[maiden ocean]

how tho

we dont necessarily even have that each x_k - mu_kj(x_k) is in M_i do we

just that if u sum everything all the non ith terms cancel out

[golden pasture]

the idea is that if i give you some x_i and x_j

you can find some k such that μ_{ik} and μ_{jk} makes sense

[maiden ocean]

yea but like

that doesnt mean that mu_ij(x_k) makes sense or anything

[golden pasture]

yesh you're right it doesnt

[maiden ocean]

sadge

[golden pasture]

ill jus relabel cuz letters getting annoying lol

so what you want to do is

The element $x_i-\mu_{ij}x_i\in C$ is really the element $\mu_ix_i-\mu_j\mu_{ij}x_i$

[cloud walrus]

ariana

[golden pasture]

now lets say we have some $k$ such that $\mu_{ik}$ and $\mu_{jk}$ exists

[cloud walrus]

ariana

[maiden ocean]

[golden pasture]

then consider the element $\mu_{ik}x_i$, it's injection into $M$ is $\mu_ix_i$

[cloud walrus]

ariana

[maiden ocean]

isnt mu_i x_i - mu_j mu_ij x_i zero

[golden pasture]

in M yes

[maiden ocean]

sadge

but mu_i is the projection

the restriction of the projection to M_i anyway

[golden pasture]

but you have this right

the idea is that need to get a $k$ such that you end up with $\mu_{ik}x_i-\mu_{jk}\mu_{ij}x_i$

[cloud walrus]

ariana

[golden pasture]

wait lemme unstone i rmb theres a nice way to make sense of this lol

[maiden ocean]

that part makes sense i think, i guess i just dont understand how u can go from x_i - mu_ikx_i to mu_ik x_i - mu_ik x_i

like

i get that if u could somehow make all the indices work out and slap m_jN on x_i and on mu_ik x_i theyd be equal

but i dont understand how u can do that

[golden pasture]

ah yes

in your finite sum

the M_i th component must vanish for all i

since C is jus the direct sum

so if you properly regroup the terms

should get it

[maiden ocean]

agony

[golden pasture]

agree

[mild laurel]

thought u guys were doing diff geo for a second there

[golden pasture]

am home lemme give a quick proofsketch on my com

typing on phone is pain

HAHAHA

Understandable

since we have $x_i=\sum x_j-\mu_{jk}x_j=\sum y_j$ in the direct sum, we know that $y_j=0$ if $j\neq i$ and $y_i=x_i$

$\mu_{jk}y_j=0$ for all $j\neq i$

$\mu_{ik}x_i=\mu_{ik}y_i=\sum \mu_{jk}y_j$

[cloud walrus]

ariana

[golden pasture]

andd you're done

[maiden ocean]

ok i think

i see

wait uhhalksflksdf

how do u know that mu_ik y_i = sum mu_jk y_j

[golden pasture]

cuz of the second line

it's a direct sum

so everything else must equal 0

[chilly ocean]

I don’t get the AB<G part

[mild laurel]

what exactly don't you understand?

[chilly ocean]

the part after because

Why are they equal

[scarlet estuary]

normality

[mild laurel]

try to see why $a_1 (b b_1^{-1})a_1^{-1}$ is in $B$

[cloud walrus]

Zopherus

[chilly ocean]

Because B is a normal subgroup of G I think?

And a1 and a1 inverse are in G

[scarlet estuary]

and you know why bb_1^{-1} is in B, correct?

[chilly ocean]

Yea cos B is a subgroup isn’t it?

[scarlet estuary]

right, precisely

[chilly ocean]

Why don’t the textbooks write it like this this is so much clearer

Anyway thanks so much:)

[scarlet estuary]

so thats your value of b_2

[chilly ocean]

If H_I are normal subgroups of a group G with i=1,2,3...n, then H_1H_2H_3...H_k is a subgroup of G (prove using induction)

Can we go about the induction step like this? If the H_i’s are normal subgroups then pi product of H_i’s is just pi product gh_ig^-1 for all g in G and for all h_i in H_I which simplifies to g*(pi product h_i)g^-1. To prove this is a subgroup of G, we need to prove that this has an inverse from which it follows that it has an identity e (because G has an identity and all the H_i’s have identities e_i). Inverse of g(pi product h_i)g^-1 is just g^-1(pi product of h_i inverse where i=n,n-1,....1)*g. This shows that we have an inverse which implies an identity

Damn I rlly gotta learn latex

Associativity follows from the fact that G is a group

[chilly ocean]

no AB is defined in just one way I think

lol no, show me where AB means anything else but product of group subsets

[carmine fossil]

I think they mean a set such that elements are of form h_1 h_2 h_3 ... h_k where h_i is in H_i

[chilly ocean]

I feel like it could potentially be the smallest group containing A and B (which is the same here in this context but )

AB MEANS PRODUCT OF SUBSETS EVERHWYERE

HE DIDNT WRITE AxB

or anything

just AB I think its self explanatory

AB is the set of all xy where x is in A and y is in B

idC to read essays

pi product is defined analogously

[carmine fossil]

Hint:H is normal in G implies HK=KH for any subgroup K in G

[chilly ocean]

plz latex

fr tho i need to learn latex

the best way to learn is to just ask problems here in latex

nvm tho ill download the solution manual

[chilly ocean]

ill learn it as soon as i get a chance

[carmine fossil]

$\text{Hello world}$

[cloud walrus]

Buncho Drunk

Godel

[chilly ocean]

why b|amb and b|an proofs b|a while we don't know what are m and n

[viscid pewter]

well, it's true

there exist m and n that work

[chilly ocean]

Yes it's true, but I cannot perfectly understand it

[carmine fossil]

b|a and b|c implies b|(a+c)

[chilly ocean]

if b|anc while gcd(b,c)=1 then yes b|an

but still doesn't proof b|a

I wanna give up this is so absurdly hard:/

now got it

[toxic vessel]

What you are missing is that $b|(nc-1)$, and therefore $gcd(b, nc-1)=1$

[cloud walrus]

StellaAthena

[chilly ocean]

true 😂

Abstract algebra

[toxic vessel]

also, this convo belongs in #elementary-number-theory

[chilly ocean]

thanks a lot @chilly ocean

then why is it so hard

Bruh

when precision become limited, things look weird

[next obsidian]

Wut

How is b not in R

Also

It’s an ideal

[cloud walrus]

hungry

[next obsidian]

If a is in p, ab is always in p

You just say that if a and b are not in p

Then ab + p is not 0

So that ab not in p

The contrapositive says that if ab in p one of a,b must be in p

What

No

Wut

The thing you said is it true in general...

Is just literally always false

Haha

By definition of an ideal

If a is in an ideal I

Then for any b in R, ab is in I

[cloud walrus]

hungry

[next obsidian]

What you said is true no matter what b is

By definition of an ideal

But I don’t understand your proof

I don’t get what you mean that only one of them is in p

They could both be in p

But you aren’t contradicting the right thing

After “so a in p or b in p”

You’re done

This is a direct proof that p is prime

It doesn’t matter

Prime ideal means if ab is in p

Then one of a,b is in p

You’ve shown that already

Bruh moment

Also the line after that is wrong

You claim if a is in p and b isn’t in p

Then ab isn’t in p

I also don’t get what you wrote after that

[carmine fossil]

Are you trying to prove prime ideal => integral domain?

Or integral domain => prime ideal

[next obsidian]

The latter

Sure

But I wanted to point out what you claimed there is wrong so you don’t think it’s right then continue thinking that and get confused later

It could be both

But you only get one

Like a and b could be in p

But you’re only guaranteed one is

Which assertion above

Isn’t this the exact same thing as you wrote before

Wut

Isn’t this the same thing you wrote just above

The way we proved the reverse implication showed that if ab \in P, then a \in P or b \in P

That’s implicit in the fact you used or

But not both is encoded in the fact you said or

Wait

You meant but not both as

ab in P => ((a in P) or (b in P)) and (a,b not in P)

God no

That’s super false

Then no ideal could ever be prime

Just do 0^2 in P

But 0,0 in P

Yeh

But not one of as in both can’t be in P

You’re just only guaranteed one of them

That’s why I said ur proof is like 4x longer than it needs to be

After you show one of a,b in P you’re done

Then you do some stuff (which is false) to try and show only one of them can be

It’s false cuz it’s impossible to prove that

I mean sure that one is undgodly strong

Literally no ideal satisfies it

Any set closed under multiplication doesn’t satisfy it

Any set containing 0 doesn’t satisfy it

The one that one of a,b is in P, but in addition you don’t have a AND b in P

Eh?

Then what are you referring to here?

Like can you spell out what two assertions you’re trying to compare

And which one you say is stronger?

But those are the same thing

Saying “one of a,b in P” doesn’t rule out the possibility both do

Eat some food

Ur not u when ur hungry

9 am

For me

😎

[chilly ocean]

@chilly ocean why do you not eat

thats just sad

[next obsidian]

Eat food

If ur hungy

[golden pasture]

eating is bad

go eat

[chilly ocean]

you can eat and still be hot

[next obsidian]

Like me

[rigid cave]

Guys, if [K:F] = [F(a):F] and K is a field extension of F(a), then is F(a) = K?

[mild laurel]

yes

[rigid cave]

So say that f(X) is the minimal polynomial of a and that f(X) splits in K, then is this true?

Oh okay! Does one need to prove this or is it trivial?

If it is a field extension, then the basis must be "bigger" than the basis of the base field, right?

[mild laurel]

It's basically linear algebra, you have that F(a) is a linear subspace of K with the same dimension as K

[rigid cave]

Well I am still in highschool so I have not taking a linear algebra class...

[mild laurel]

it's not too hard to prove this

[rigid cave]

So say that B_k is a basis for K over F and B_f be a basis for F(a) over F, then B_f is a subset of B_k and since those have the same cardinality then they must be the same?

[mild laurel]

B_f isn't necessarily a subset of B_k

[rigid cave]

But K is a field extension of F(a), so why can't that be true?

[mild laurel]

I mean

Take the vector space R^2 and the subspace of the x-axis

then (1,0) is a basis for the x-axis

but (1,0) doesn't necessarily need to be in a basis for R^2

you could take (2,1), (-1,1) or something to be a basis for R^2

[rigid cave]

Ohh, a linear subspace is just a vector subspace...

[mild laurel]

yeah thats what I meant

[solemn rain]

hi

so i learnt today about galois fields

and how to construct finite fields from easy Z/pZ fields

i was asking if what i was doing is right when finding GF(4)

so i thought GF(4) = GF(2^2)

we look at F_2[x] and pick an irreducible polynomial

its ideal is maximal ---> F_2[x]/(p(x)) is a finite field of 4 elements

so by just 'brute forcing' all elements and trying all combinations of polynomials of degree 2

i found x^2+x+1 is the only irreducible in F_2[x]

hence F_4 = F_2[x]/(x^2+x+1)

now if i were to write F_4 as elements

without using any fancy algebra words like cosets

would i just think about remainders of polynomials quotiented by x^2+x+1?

i never thought about quotients this way

[mild laurel]

I mean, the elements are just 0,1, x and x + 1, and addition acts like how you'd expect, but multiplication works in the quotient ring basically

[solemn rain]

yes thats what i got

but is this how i should normally think about quotients from now on?

remainders of polynomials divided by polynomials of deg 2?

and i also dont know why

this quotient would yield 4 elements

like why does the polynomial have to be of degree 2

in order for this to be GF(4)

[mild laurel]

you can pick coset representatives so that they're all degree less than 2, like I did

In general, if you quotient F_2[x]/(f) where f is an irreducible polynomial of degree d

then you'll get a field with 2^d elements

Since there are 2^d polynomials in F_2[x] with degree less than d

[rigid cave]

Wait, shouldn't it be "element" + (X^2 + X +1)

[mild laurel]

I mean yes but people abuse notation a lot

[rigid cave]

Oh okay!

[mild laurel]

I could write [0], [1], [x] and [x+1] to denote the cosets I guess, but people get lazy

[rigid cave]

Yeah that's right

[solemn rain]

thank you

[thorn delta]

i have these equivalent characterizations for F being algebraically closed. I'm having trouble seeing why (v) implies any of the other though. any hints?

[sturdy marsh]

5 is saying that F is an algebraic closure of some K

try showing that algebraic closures are algebraically closed

fix any poly in F[x]. It has a root in some bigger field F'. F' is algebraic over K ...

[heavy jasper]

Hope this is the right channel for representation theory -- on my last homework, we had the question to show that given a compact group $G$ and continuous representation $(\pi, V)$ where $V$ is some Hilbert space, we can define a new inner product on $V$ such that $\pi$ is unitary with respect to the new inner product.

The way I did this was (given the existence of a Haar measure), letting $(v, w)$ denote the original inner product on $V$, define
$$\langle v,w\rangle=\int_G (\pi(g)v, \pi(g)w)d\mu$$ as our new inner product.

My question is whether this works if $V$ is infinite dimensional, since I was unable to prove if $V$ is in fact a Hilbert space with respect to this new inner product when $V$ is infinite dimensional (in the finite dimensional case it's clear since it's finite dimensional).

I went to office hours and my professor said in the infinite dimensional case, this method probably doesn't work and amended the question to assume $V$ is finite dimensional, but didn't know a counterexample, so do any of you know of one?

[thorn delta]

@sturdy marsh why would F' be algebraic over K?

[sturdy marsh]

because F' is algebraic over F

and F is algebraic over K

@thorn delta

[thorn delta]

ok hmm i didn't know algebraic-ness was transitive

[sturdy marsh]

the proof is pretty straightforward. Pick any alpha in F'. It has a min poly over F. Then K(alpha, coefficients of min poly over F) is a finite extension of K

[thorn delta]

ahh okay

okay, yeah, that makes the equivalence much more clear

[unique juniper]

how is this done?

|Z(G)| isnt 1 or p^3 so it can be either p or p^2

[sturdy marsh]

if it's p^2, then G/Z(G) is cyclic

[unique juniper]

OH

ok

i just noticed that now

ty

[next obsidian]

so what

kidding

[sturdy marsh]

implies that it is abelian

[next obsidian]

kidding!

I may be a chmonkey but I'm not a chmumbass

[latent anvil]

X

[jagged dune]

How do you prove that something is a normal subgroup of a group?

[mild laurel]

usually just by the definition?

[jagged dune]

Well, yeah. I'm just not fully understanding the process.

[mild laurel]

What example are you working on

[viscid pewter]

any context?

[jagged dune]

5Z is a normal subgroup of Z

[prisma ibex]

Construct it as the kernel of a homomorphism

[viscid pewter]

that works

[jagged dune]

Yeah I need to show that.

[prisma ibex]

Alternatively you can just show the definition directly

Also you’re in an Abelian group so every subgroup is normal anyways so there is nothing to check

[jagged dune]

So just show it's a subgroup?

[viscid pewter]

lol sure

[prisma ibex]

Yes as soon as it’s a subgroup it’s normal

[thin geyser]

stuck, would appreciate help

[mild laurel]

Do you know what Lagrange's theorem states?

[thin geyser]

ah so i've learned that for a subgroup H of group G, |H| divides |G|, and i don't think i've explicitly learned that the order of a group is precisely its cardinality

[latent anvil]

How did you define the order of a group?

[thin geyser]

ord(a) is the smallest integer m such that a^m = e

^ so, my bad, have only encountered order of an element of a group so far

[latent anvil]

That's the order of an element, yeah

The order of a group is defined to be its cardinality

[thin geyser]

makes sense and then the result in my question would follow, i'd probably need an alternative proof since i haven't encountered the order of a group yet

i actually think i have...at least what the graders would like to see

i.e. i just calculated the cardinality of the group by definition of phi(2^n) = 2^{n-1} and i know the order of any element in the group must divide that, so the result should follow

ty all!!

[thorn delta]

When can you have F = K(u1, u2, ...., un) where u1, u2, ..., un are the roots of a polynomial f but f does not split in F?

[next obsidian]

are those every root?

because then it just straight up impossible

But I see where your confusion comes from

note that under this, F already exists

We aren't defining F = K(u1,...,un) rather that F has that form

Like... the way the flow is going is that
F is a splitting field if f splits in F (meaning F has all the roots), and then F is generated over K via those roots

it says that it can't like contain any more stuff

Since you could take K-bar and then f splits in K-bar

but it's unlikely that K-bar is going to be generated by just the roots of f

The reason the other direction for like defining F to be K(u1,...,un) doesn't really make sense is that u_i need to be roots of f in F

If you want to make F using the roots of f, you need to have a place those roots live in

[thorn delta]

hmm okay i think i see. I saw a proof that every polynomial has a splitting field which inductively adjoined the roots of irreducible factors, and that had me confused or something

[next obsidian]

right right

so that's one way to get a field in which the polynomial splits

and once you get f splitting in some field you can just take the subfield generated by those roots

those roots already exist in that bigger field, and clearly f will split in the subfield generated by the roots

and well... it's generated by those roots

I think that in the case you're describing where you just adjoin roots of the irreducible factors (by adjoining x then modding out by an irreducible) polynomial should just be generated by the roots

so that this gives you a splitting field

I hope that makes sense

this really isn't too important, but it's kind of similar to how like algebraic closures aren't well-defined... but they kiiiiiiiinda are

Like you could define an algebraic closure to just be any algebraically closed field which is algebraic over your field

but most of the time you don't really think about it and you just are like "lol take k-bar" which is a similar way you think about a splitting field

[thorn delta]

Right, that makes sense. So, the context that made me revisit this definition is this:

why is it necessary to notice that each f_i splits in F for the f_i to split in E? Since E contains all the roots of the f_i, that should be enough to see that E is a splitting field?

[chilly ocean]

I think what chmonkey was saying was that ui are all the roots of the fi in F, so eg if fi did not split in F then some fi could have a root that is not one of the ui

[mild laurel]

think about the correspondence theorem. The ideals of R/I are in one-to-one correspondence with the ideals in R that contain I

correct

[chilly ocean]

(a+I)(b+I) = 1 + I

[mild laurel]

yes thats right

[chilly ocean]

yes

R is a field iff the only ideals of R are (0) and R

with ease

[delicate bloom]

start by supposing you have a nonzero ideal and play around

[mild laurel]

its basically the reverse of the proof you just did

[chilly ocean]

wait a minute

isn't that the question you posted yesterday where arturo was rude

well this is just a special case of that one

haha

the contrapositive is redundant

if R is a field, then forall a ≠ 0, exists b such that ab = 1, thus (a) contains 1 and therefore all of R

[final pasture]

(another way of saying it, for the converse: if R is not a field, then there exists a nonzero element a without an inverse, and then aR is a non trivial ideal)

[delicate bloom]

I do it directly, for a nonzero ideal you can then take for arbitrary r in R and i in I and since it's a field i^-1 exists so r = (r*i^-1) * i is in I, meaning I=R

[chilly ocean]

g is bijective

and i guess also that g is additive

if two groups are isomorphic, then their underlying sets are equinumerous

yes

but for rings, bijective homomorphisms are isomorphisms

it works for all

[final pasture]

You already know that:

• the map phi is bijective (because the thm is know to be true in the case of a group)
• the map is additive (because the thm is know to be true in the case of a group, again, and that a ring homomorphism is also a group homomorphism for the underlying additive group)

so now it remains to show that this actually generalizes to rings, so that the map also preserves multiplication and that Ker(phi) is actually an ideal, to be able to speak about R/Ker(phi) as an ideal

can you state the thm in the case of a group ?

Indeed. Now if you have two rings (R, +, x) (R', +', x') and a ring homo f: R -> R'

is f a group homo between (R, +) and (R', +') ?

so R/Ker(f) iso to Im(f) (in the context of groups), right ?

oh, sorry I confused f and phi

[cloud walrus]

mirza

[chilly ocean]

ol

Why do you call it an extension

Are there any extra elements?

i think they are really the same underlying map anyway, set theoretically

yeah

that was what mirza was confused about

since g is an isomorphism of groups, it must be a set bijection, and it suffices to show that is a ring homomorphism for it to be a ring isomorphism

lol

yes

lol homo

when 2 people of same gender marry. they get a ring homo

[final pasture]

yh

[shut halo]

I'm having trouble understanding the solution to the following problem (taken from the book "Algebraic Geometry: A Problem Solving Approach"):

Let $P(x,y,z)$ be a homogenous polynomial and $l = ax + by + cz$ be a linear polynomial. Now suppose $(x_0 : y_0 : z_0) \in V(P) \cap V(l)$. Show that $(x_0 : z_0)$ is a root of the homogeneous polynomial $P(x, ax + cz,z)$ and that $y_0 = ax_0 + cz_0$.

Now, working over $\mathbb{CP}^2$ I can assume that $b = -1$ and we know that $P(x_0, y_0, z_0) = 0$ and $ax_0 + by_0 + cz_0 =0$. So from here we deduce that:
$$ax_0 - y_0 + cz_0 = 0 \implies y_0 = ax_0 + cz_0 $$.
Then substituting for $y_0$ we also obtain that $P(x_0, ax_0 + cz_0, z_0) = 0$.
However, I don't quite understand how one can show that $(x_0 : z_0)$ is a root of this polynomial.

[cloud walrus]

snypehype

[golden pasture]

uh root of polynomial -> polynomial evaluates to 0?

[shut halo]

@golden pasture oh so $P(x_0, ax_0 + cz_0, z_0)$ vanishes on $(x_0, z_0)$ by definition but this doesn't imply that $P(x_0, z_0) = 0$ right?

[cloud walrus]

snypehype

[golden pasture]

notice that P(x_0,z_0) doesnt actually make sende

the problem is saying like

consider the new polynomial P(x,ax+cz,z) in x and z

show that (x_0,z_0) is a root

[shut halo]

Right so that is just true because above I showed exactly that. Thanks

[carmine fossil]

For every element in R/I_1 x R/I_2,if you can find an element in R that maps to said slement,the map is surjective

[cloud walrus]

morza horizon 4

[carmine fossil]

a+I_1 is also r_1+I_1 and a+I_2 is also r_2+I_2

That's what we need

[cloud walrus]

morza horizon 4

[chilly ocean]

yes

[cloud walrus]

morza horizon 4

[chilly ocean]

I mean those congruences show that for r1 and r2 in r+I_1 and r+I_2 there is corresponding element r + I_1 * I_2 under the standard mapping

r = r1 mod I_1 and r =r2 mod I_2 means r \in R/I_1 * I_2 no?

Hmm, how I understand it is from definition r in R/cap I_i if it satisfies such congruences. If we have such r as in the problem, then it gets mapped r +I_1 I_2 \mapsto (r_1 + I_1, r_2 + I_2). If we show for any r1 r2 we can find such r such that it maps to (r_1 + I_1, r_2 + I_2), then we show surjectivity right?

Can you paste how you define R/(\cap I_i)?

If you think about it as remainders then its pretty much analogous to the normal division with remainders

So the problem is you're not sure that any element in R/(I_1\cap I_2) can be written as such congruence relations right?

[cloud walrus]

morza horizon 4

[chilly ocean]

I think this is just definition though? elements of R/I are r+ I for r in R

[cloud walrus]

morza horizon 4

morza horizon 4

morza horizon 4

morza horizon 4

[carmine fossil]

Map r+(I_1 inter I_2) to (r_1+I_1,r_2+I_2)

R to R/I_1 x R/I_2 will be a surjective hom

And then you quotient by kernel to get isomorphism

Just read the 2nd half of the proof

[chilly ocean]

just wanted to say that this proof sucks and theres a cooler one: just need to show (1+I,0,0,...,0) is in the image and rest follows

[carmine fossil]

This would mean r + I_1=r_1 +I_1

[chilly ocean]

what book is this?

[carmine fossil]

and that's it

[chilly ocean]

read introductory ant by kenneth williams

[carmine fossil]

Your map takes in some r and outputs (r+I_1,r+I_2)

You want that output to be (r_1+I_1,r_2+I_2)

So we want r+I_1=r_1+I_1

To show map is surjective,we need a element r which maps to any given element in range. The element in range here is (r_1+I_1,r_2+I_2)

[chilly ocean]

are they distinct?

[carmine fossil]

Why should they be same?

We are taking an arbitrary element of R/I_1 x R/I_2

[cloud walrus]

morza horizon 4

[thin geyser]

a bit stuck, i can apply euler's totient thm but i'm unsure what to do with e

[chilly ocean]

just write it out what it means for e to be equal 1 mod phi(m)

also a^e = a (mod m) implies (a^(e-1) = 1 mod (m))

[thin geyser]

ahh thank you

[cloud walrus]

bilyan

[chilly ocean]

just list them out

[terse crystal]

The multiplication group of units of Z/mZ has order φ(m).and the order of any element divides the order of the group...

[lavish pike]

So for the rational root test of this $f(x)=-x^{4}+x^{3}+x^{2}+x+2$ in Q, i get the r/s thingy and u just plug those in and would end up with f(-1)=0, f(2)=0. So then f(x)=(-1)(x+1)(x-2) but then in the book it also lists $(x^{2}+1)$ as one but why is that?

[cloud walrus]

NocuousNick

[mild laurel]

why can't it?

The rational root test only tells you where the rational roots are, not where other roots could potentially be

also your original polynomial has degree 4 so you have 4 total (complex) roots

[lavish pike]

so then like how do i figure out that x^2+1 is also a root

[mild laurel]

Do you know synthetic division?

In this case, you can synthetically divide f(x) by (x+1) and (x-2) to get that

[lavish pike]

but if u do synthetic division on f(x) wouldnt u get zero?

[mild laurel]

No?

You look for f(x)/(x+1)

[lavish pike]

huhhhh? am i missing something? sorry for the sloppiness lol

is that not synthetic division for (x+1)?

[mild laurel]

Yeah that's right

So you get that f(x)/(x+1) = -x³ + 2x² - x + 2

[lavish pike]

I haven't done synthetic division in so long i forgot about that

[vestal snow]

Am I supposed to think of these as Q-algebras/C-algebras or Z-algebras/R-algebras?

[thorn delta]

why algebras and not just Z-modules and R-modules?

[terse crystal]

There are R algebra when the tensor product is taken over R......

[steady belfry]

how to do 1b

[terse crystal]

Kerψ={z from T such that z^2=1}={1,-1}

[steady belfry]

how does that work

[latent anvil]

For the isomorphism bit or the tensor product bit?

Presumably they mean isomorphic as Z algebras in (a) and R algebras in (b)

But to take the tensor you need to consider them as algebras over whatever you're tensoring over

[terse crystal]

What do you mean by saying work?

[steady belfry]

like how do you calculate that lol

[terse crystal]

Observation

[terse crystal]

Like the second equation doesn’t hold because the former is not an integral domain but the latter is.

[vestal snow]

I meant isomorphisms

Thanks

[latent anvil]

👍

Well, they aren't even isomorphic as rings

So it's a bit of a moot point

[terse crystal]

$(i \otimes 1+ 1\otimes 1)(i \otimes 1-1 \otimes 1)=0$

[vestal snow]

For b), I used a dimension argument

C (x)_R C is a 4 dimensional R vector space

and C (x)_C C = C is a 2 dimensional vector space

Does this work?

I think it does

[terse crystal]

It works but how to type tensor product symbol in latex😂😂

[vestal snow]

C (x) _C C = C

C (x)_R C is not C

[cloud walrus]

Cogwheels of the mind

[vestal snow]

over R

tensoring over R

$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$

[cloud walrus]

Have a Banana, Bitch

Cogwheels of the mind

[vestal snow]

Yes

But C is a two dimensional vector space over R

[latent anvil]

Actually banana I might have been wrong

They might mean as Q or C vector spaces

I wasn't thinking when I wrote my first reply

but it's not much harder to show that they're not isomorphic as rings

[latent anvil]

Like, how did you compute this?

Oh I guess probably just by general facts about the dimension of a tensor product

So maybe it is trickier

The C dimensions are still different so w/e

[vestal snow]

The dimension thing proves that they are not isomorphic as R-modules or R-algebras right?

[latent anvil]

Yup

[terse crystal]

This one is correct not the one I wrote later btw...

[latent anvil]

But I realized I might be wrong about the isomorphism being over R

[vestal snow]

They still might be isomorphic as rings

[latent anvil]

And not C

[latent anvil]

Yes, they might be a priori. But they aren't

[vestal snow]

But I don't think the question is asking that

[latent anvil]

yeah I just meant I'm not sure if it's over R or C

When earlier I said it was over R

[vestal snow]

Gotcha

[old hollow]

this is prob a dumb question

but how do I do this

[thorn delta]

you have k[x] is iso to k[x] (by the identity). Map y to mx + b. This induces a ring hom k[x,y] to k[x]

not sure if that is enough direction. Does that help at all? @old hollow

[old hollow]

I'm having trouble intuiting what k[ℓ] would look like

uhhh

[long beacon]

it's just polynomials of x and y

but instead of y, you have mx+b

[old hollow]

oh wait

[thorn delta]

my advice dodges working with elements of k[ell]. You can find a surjective map k[x,y] to k[x] and quotient out by the kernel to get an iso (if you've seen first isomorphism theorem)

[old hollow]

that sounds similar to how k[x]/(x^2 + 1) is polynomials of x, but x^2 is replaced with -1

oh god

this is killing me

why is this so confusing rn

[thorn delta]

yea, its exactly like that. Reducing modulo a polynomial is like reducing modulo an integer. Just.... with polynomials lol. The elements of k[ell] are equivalence classes of functions which agree on y = mx + b. Anyway, I don't think that helps much with the exercise

[chilly ocean]

So a quick question - I'm getting ahead of class and starting on ring theory - the textbook i'm working from doesn't include the need for a multiplicative identity for a ring but I decided to because many other sources do and it makes more sense to me. However I've gotten up to defining the kernel of a homomorphism as an ideal but this could not hold since the kernel can't contain 1. How would I get around this? Woudl I just have to make an exception in the proposition (hope i'm not interupting)

[long beacon]

note that if in k[x,y] you replace all your y's with mx+b you don't really get anything new polynomials

[old hollow]

yeah that sorta makes sense

[long beacon]

so intuitively, doing that doesn't really make it any different from k[x]

[thorn delta]

are you familiar with the theorem that says if you have a ring homomorphism R to S, and you let an indeterminate x map to an element of S, you get a ring homomorphism R[x] to S? That is what im talking about here

you have k[x] is iso to k[x] (by the identity). Map y to mx + b.

[long beacon]

it's helpful to think of what happens in other cases-- take for instance k[x,y]/(x^3-y^2)

[old hollow]

with the other example of R = k[x]/(x^2 + 1), the fact that x^2 is replaced with constant makes all the elements of R have degree less than 2

[long beacon]

so i mean, now you can't just replace all your y's with something involving x.... you need at least a y^2 before it turns into an x^3

[old hollow]

o lemme see

[long beacon]

so xy is genuinely a new addition to k[x] in that above ring

and hence you don't have that particular isomorphism

[long beacon]

? what's the problem? an ideal containing 1 becomes the entire ring by definition

[chilly ocean]

if we require rings to have multiplicative identities, then all ideals must contain the multiplicative identity since they are subrings, right

[thorn delta]

ideals need not be subrings (under the convention that rings have 1)

[chilly ocean]

right. so when defining an ideal, even if we consider rings to require 1, you still just define it as being a subgroup under addition and absorbing products from R? we don't require 1 to be in the ideal also?

that would make sense

so then kernels can be ideals

is every ideal the kernel of some homomorphism?

Gotcha

[old hollow]

wait is this 1st isomorphism theorem

nvm

what am I doing

[chilly ocean]

so just to make sure I have it completely clear, if we require rings to have 1, then every subring also has 1, but ideals are then not necessarily all subrings

[long beacon]

any ideal I is the kernel of a map R -> R/I

[old hollow]

yeah

[chilly ocean]

awesome thank you :D

[old hollow]

that seems super similar to the kernel / normal subgroup thing for groups

oh

didn't see that lol

haha

ah

the conjugacy requirement in normal subgroups seems kind of random

where did "they" come up with that

[long beacon]

quotients don't make sense without it

[thorn delta]

when H is normal, you can do the following manipulation: (aH)(bH) = a(Hb)H = a(bH)H = abHH = abH.
This is how lang proves the quotient group operation is well-defined

[long beacon]

if you have a subgroup H then with a "G/H" you want aHbH = abH

only choice is for Hb = bH

[old hollow]

ohhh I see

[old hollow]

ah

[latent anvil]

There's only finitely many mirza

If you have a finite set of sets (like these ideals) it automatically has a minimal element

So like

Take one of them

Is it minimal?

If not, take something smaller

Keep doing this at most r+s times

Yup, a finite poset has a minimal element

Formally you could prove it like

Do induction on the size

Take some element

If it's minimal we're done

Otherwise look at the poset of strictly smaller elements

This has size at most n-1

Right?

And a minimal element there will be minimal overall

Reduce both sides of the equation mod p1

The left hand side becomes zero

Right?

And the right hand side will be a product of nonzero elements

Ah, probably a typo

Yeah I think the argument goes through if you change i to 1

[prisma ibex]

Yea keep in mind that the proof of uniqueness here is identical to the proof of uniqueness of prime factorizations in Z

And yea that is a typo

[latent anvil]

I don't think I saw that before the general case

Never took like a number theory course

[prisma ibex]

[latent anvil]

Where's the contradiction coming from?

Okay yes lol but I'm asking you to justify why the quotient isn't a field

I don't think I agree

[prisma ibex]

The quotient is going to be a field though

Think of the situation with Z

[latent anvil]

How do you know the q_i don't all become (1)?

Also ng if you are answering me I know why the quotient isn't a field

Mirza dont read ||if you reduce the Πp = Πq equality mod p1 you get that a product of nonzero elements is zero||

[prisma ibex]

The contradiction isn’t coming from this; rather, the contradiction is that if all the q_i were nonzero mod p_1 then we could not have p_1...p_n=q_1...q_n

[latent anvil]

Or nvm lol

[prisma ibex]

Since the left hand side would reduce to 0 mod p_1 but the right side wouldn’t

[latent anvil]

I think this is a really weird way to argue this though

like, I would say that p1 divides the left hand side so it divides the right, and so by primality it divides some qi

[prisma ibex]

Is it weird? I would guess that is what the screenshot is referring to as the contradiction

[latent anvil]

By irreducibility of qi we have qi = u p1 for some unit u

[latent anvil]

Oh I agree that's what the screenshot is doing

I just don't see why you need to do stuff with minimality

Or really reduce mod p1 (although this is just the stuff I'm saying with divisibility)

Yup

But also

[latent anvil]

Did this make sense?

I think the screenshot is sort of obfuscated

Yup

I've never seen the minimality proof

Also as @carmine fossil pointed out all of these ideals are automatically minimal

by irreducibility

Yeah they deleted it

Idk why

I sort of steamrolled to explain the screenshot's logic

Okay time for comm alg

Or jjk final epsiode

:himb:

Oh yeah shoot

so I think the usual way to prove this is by proving it's a euclidean domain

have you seen polynomial long division (like in a high school algebra class or whatever)?

but I can also think of a meme proof rn

yup, that works

[carmine fossil]

Do you know what an Euclidean domain is?

So,Define norm(p)=deg(p)

[latent anvil]

This is how you prove it in a euclidean domain, right?

taking an element of minimal norm

[carmine fossil]

By polynomial division,given f and g!=0 ,you can always find an element r such that f=gq+r , where deg(r)<deg(q) or r=0

So,F[x] is a ED

and your result follows

anyway,Well ordering is very cool

I is (f) right?

You have f=$\sum_{k=0}^{n}{c_k x^k} \implies f(\alpha)=0 \implies \newline \sum_{k=0}^{n} {c_k \alpha^k}=0$

[cloud walrus]

Buncho Drunk

[carmine fossil]

So you can rewrite \alpha ^n in terms of the other terms

How is that set LI?

is f like the polynomial with minimal degree,such that f(alpha)=0?

[vestal snow]

Any ideas how to prove the if direction for the second part?

I know that F (x) F = F[x]/(f(x)) where f is the minimal polynomial of \alpha over k

I only need to somehow show that all roots of f are in F

[rustic crown]

use CRT

[vestal snow]

Chinese remainder theorem or something else?

[rustic crown]

yea

[vestal snow]

Ah got it

[rustic crown]

so if direction is <= right?

[vestal snow]

I think I got it

Let me try it first

I don't think I got it

I tried doing something like assume f(x)=(x-a_1)...(x-a_n)g_1(x)g_2(x)...g_m(x) where g_i(x) is irreducible in F

and g_i have degree greater than or equal to 2

Any ideas?

[rustic crown]

count the number of idempotents

x^2 = x in the ring

[vestal snow]

I see

Thanks

[rustic crown]

[next obsidian]

Holy shit I did this problem in Aluffi

I totally forgot about it

I should do it again

[rustic crown]

it took me 2 days to understand the question lol

[next obsidian]

Hahaha

[rustic crown]

i was seeing tensor products like for the first time

[next obsidian]

right

I’m pretty sure I asked Shamrock for help maybe for one direction? I don’t even remember anymore

[latent anvil]

Yup

Also, I think I have it

Say we mod out by g+h

I think g will multiply to 0 with any non unit element

In the quotient

Because xg = yg = zg = tg = 0 in the quotient by g+h

and so ann(g) contains (x, y, z, t)

Is that what you did @next obsidian?

So showing this ring has depth 0

[next obsidian]

Yup!

[latent anvil]

Nice

[next obsidian]

Like you know g,h have no constants

[latent anvil]

Yup

[next obsidian]

So when you want to kill x or y

Pretend g is -h

Or whatever

[latent anvil]

It was obvious after I thought about the accumulator

Yeah exactly

*annihilator

[next obsidian]

Yup so it’s like silly

Because it’s really dumb

I found it silly at least

[latent anvil]

I think I'm going to go to bed

[next obsidian]

Yee

[vestal snow]

@rustic crown I had totally forgot that idempotents were even a thing lol

[rustic crown]

yea i felt like aluffi should have said something about products of rings and idempotents

[vestal snow]

How did you think of looking at idempotents?

[rustic crown]

our ring theory prof proved that a ring is a product of two rings if and only if there is a non-trivial idempotent

[vestal snow]

I guess that makes sense

[rustic crown]

so it was just a thing i tried and it worked

[vestal snow]

Hold on

I remember that proof

Is it something like R = Re x R(1-e)?

[rustic crown]

yea

but the operation on Re looks disgusting

it was something like define ae*be = (ab)e

it makes sense and all but when i saw it, it felt very very awkward

like the first time i saw in group theory (aH)*(bH) = abH

[latent anvil]

I mean

It's the same as the multiplication in R

Since e^2 = e

[rustic crown]

i mean i was sad that identity is now e

[latent anvil]

sure haha

[rustic crown]

lol this is true why did i not think aboout this lol

[vestal snow]

for 3, can i use other universal properties or do I have to explicitly construct a homomorphism?

@rustic crown did you do this one?

[rustic crown]

maybe use the universal property of tensor to get the map, then show its actually an R-Alg hom

[vestal snow]

That doesn't work

[latent anvil]

Why not?

[vestal snow]

To do that, you would need to map S, T into S x T first

and the way to do that is to send s to (s,0) and t to (0,t)

[latent anvil]

I'm not sure what you mean

[rustic crown]

no i directly look at the map S x T --> U given by (s, t) |--> fs(s)*ft(t)

this is bi-linear so factors through the tensor

now you got the map, we still need to check that it preserves multiplication

for that we need U to be commutative

[vestal snow]

Yeah I messed up

You're right

[rustic crown]

oh were you like trying to get the map S x T --> U also by universal properties?

[vestal snow]

Yes

[rustic crown]

ughhh, the arrows don't go the way we want

[rustic crown]

lol i would be just saying this but in a complicated way, like use the multiplication map, U x U --> U and so on

[unique juniper]

how is part b done?

[rustic crown]

from part (a) we get a hom G --> S_n

what can you say about the kernel of this map?

[unique juniper]

its normal in G

[rustic crown]

is it non-trivial?

[unique juniper]

idk tbh

[rustic crown]

okie so lets check the case when kernel is trivial

in this case the map G --> S_n is injective!

[unique juniper]

hm yes

[rustic crown]

so this map identifies G with a subgroup of S_n

[unique juniper]

yeah

[rustic crown]

but then |G| divides |S_n| = n!

[unique juniper]

what did you mean by this

[rustic crown]

G is isomorphic to the image of the map which is a subgroup of S_n

[unique juniper]

yes

i see

thank you

[rustic crown]

[rigid cave]

Okay, let's say that $F \subseteq L \subseteq K$ and let $\psi: Gal(K/F) \rightarrow Gal(L/F)$ be defined as $\psi: \varphi \mapsto \varphi|_L$. I know that this is a homomorphism, but is it surjective?

[cloud walrus]

older sister

[mild laurel]

yes

[vestal snow]

I think you need to use the splitting field characterization to prove this

[rustic crown]

how do you know phi restricted to L maps from L to L and not K

[rigid cave]

Because Galois fundamental theorem

[mild laurel]

I mean, I assume you're assuming that both L and K are galois over F

because that's necessary

[rustic crown]

(yea, wanted to confirm that... some people define Gal(K/F) as just the auto group without caring if the extension is galois or not)

[rigid cave]

It turns out that if L is a normal intermediate field extension then the "fixed field" is a normal group and $\varphi(L) \subseteq L$

[cloud walrus]

older sister

[rigid cave]

Yes

[vestal snow]

That is basically the definition for galois

Separability is inherited from L

[rigid cave]

Okay but how does one prove that it is surjective?

[rustic crown]

just use the thing on phi^-1

to get the other inclusion?

[rigid cave]

Wait, what thing?

[unique juniper]

Let G be a group which has a subgroup of index 6. Prove that G has a normal
subgroup whose index is a divisor of $720$.

[cloud walrus]

Yes

[unique juniper]

just wanna check my answer because im not completely sure

[rustic crown]

oh sorry i read it wrong, i thought you were asking why the restriction of phi on L is surjective to L

[rigid cave]

Oh, okay! I meant to ask why psi is surjective. Is it maybe because Gal(L/F) < Gal(K/F)?

But you assume that H is normal and then prove that H is normal, or am I completely lost?

[rustic crown]

what does < mean?

that's no way a subgroup

[rigid cave]

Ooh, I meant less then

So the cardinality (order) is smaller

[rustic crown]

oh okie

[mild laurel]

No it's not that easy. You want to start by looking at f such that K is the splitting field of f over F

[unique juniper]

oh yeah, i did it wrong

[rustic crown]

yea you need to use that you can lift an isomorphism to isomorphism of splitting fields

for a more precise statement,
if F1 --> F2 is an isomorphism, and it takes the polynomial f1 to f2 then the you can lift it to an isomorphism K1 --> K2 where K1 is splitting field of f1 over F1 and K2 is splitting field of f2 over F2

[rigid cave]

Oh yeah, that's right

[rustic crown]

in our case, F1 = F2 = L and K1 = K2 = K and f1 = f2 = f in F[x]

[rigid cave]

Do you mean that f1 = f2 is in L?

[rustic crown]

so K is splitting field of some (separable) polynomial f in F[x] over F. Hence K continues to stay the splitting field of f in F[x] contained in L[x] over L.

K     K
|     |
L --> L
 \   /
   F

K is splitting field of f over F, then we can also so that K is the splitting field of f over L

now since a map in Gal(L/F) fixes F, it will send the polynomial f to f

so by the theorem, you can lift the middle arrow

[rigid cave]

But how can this be true?

[rustic crown]

what else would the splitting field of f over L be?

K = F(r1, r2, ..., rn) where r_i are the roots of f

but then K = L(r1, r2,..., r_n)

[rigid cave]

Oh, yeah, that's true.

But then if I lift the arrow, how do you proceed?

[rustic crown]

phi': K --> K
      |     |
phi:  L --> L
       \   /
         F

notice that under the map psi, phi' maps to phi

which shows its surjective

cause the phi we started with was an arbitrary element of Gal(L/F)

i'll change the names cuz i realized that you called your map psi

[rigid cave]

Ohhh, okay! Thank you so much for taking your time, seriously. I love your "diagrams" as well!

[rustic crown]

[rigid cave]

Or like this: Let $[G:H] = 2$. Then $G = H \cup aH = H \cup Ha$ which gives that $aH = G \setminus H = Ha$

[cloud walrus]

older sister

[rustic crown]

from the part (a) of the theorem, we get a map G --> S_6, now i claim kernel of this map is the required normal subgroup. indeed, you get the map G/N --> S_6, and now this map is injective, so like last time |G/N| divides |S6| = 720

[rigid cave]

Never mind

[rustic crown]

(i'm referencing this)

[hot lake]

did you assume that every subgroup is normal

when you talk as if G/H is a group

what's the exponent of G/H

you literally used the induction hypothesis on G/H

[final pasture]

you're speaking about an exponent for G/H

[hot lake]

so you are using that G/H is a group

"n is an exponent of G/H and so the order of G/H divides a power of n by the induction hypothesis"

oh wait you said G is a finite abelian group

I am blind, nevermind

delete everything i said

xD

[rustic crown]

if G is finite abelian then you have subgroup corresponding to every divisor and the whole world is normal... so H is a subgruop of index 6, then 6 divides 720 lol

[rustic crown]

it was this

[hot lake]

can we just pick H=G ?

[rustic crown]

abelian yep

[unique juniper]

guys

if we wanted to do it with group actions

$\varphi : G \rightarrow$ Sym($[G : H]$)

[cloud walrus]

Yes

[unique juniper]

this is a homomorphism

but

im not completely sure, how the image is subgroup of S6

[hot lake]

well if phi : G1 -> G2 is a group morphism, the image of phi is a subgroup of G2

[unique juniper]

sure

[hot lake]

so I'm not sure where the problem is

[unique juniper]

but um

[hot lake]

maybe you're annoyed that we are identifying the group of permutations of G/H with the group of permutations of {1;2;3;4;5;6}

[unique juniper]

how can it be isomorphic?

umm

[rustic crown]

lol

[final pasture]

I don't understand why it's not just showing that e(G) = ord(G) when G is an abelian finite group though. That wouldn't be much longer

[hot lake]

I don't think that exponent(G) = order(G) is true for G = (Z/2Z)²

[final pasture]

well i'm fairly sure it is true

wait

[carmine fossil]

Exponent(G) is 2

Ord(G) is 4

You can say exponent divides order of group

[final pasture]

wait

yeah

i'm totally drunk

[carmine fossil]

Ok,That might not be true

[hot lake]

but each element has its order that is a divisor of the order of G

so the order of G is a common multiple of all of them

[rustic crown]

can we not say exponent is the smallest n such that g^n = e for any g in G?

oh so we can only conclude n <= |G|

divide is stronger

[hot lake]

and thus, the least common multiple of the orders of the individual elements can't be larger than the order of G

because the order of G is a common multiple

[rustic crown]

oh we can conclude n | |G|

[carmine fossil]

LCM of things divide a common multiple

[summer geyser]

So I have f=X^4-4X^2+5 in Q[X] and I found the Gal(f) to be D_8 and so I constructed the following diagram of subgroups,

Then labelling the roots of f, x_1=sqrt(2+i), x_2=-sqrt(2-i), x_3=-sqrt(2+i), and x_4=sqrt(2-i) with tau=(1,3) and sigma=(1,2,3,4)

I tried to construct the corresponding diagram of intermediate fields and I got this far,

where zeta=x_1+x_2

Problem is I'm stuck on trying to find what field corresponds with sigma (I think sigma^2 follows from finding sigma)

I thought it would be Q(i) but I find that x_1^2-2=i and 2-x_2^2=i, so if I let sigma act on i I get -i and hence it can't be in this field, so I'm kinda stuck on what else to try. Thanks in advance.

[hot lake]

so maybe one of the Q(sqrt5) or Q(sqrt(-5)) is wrong ?

and it should be Q(i) in place of one of them

[summer geyser]

hm

I have sqrt(5)=x_1x_4 and so is invariant under tau sigma and sigma^2

so I think that's fine, so probably Q(sqrt(-5)) is wrong then?

[hot lake]

probably

[summer geyser]

okay yeah so Q(i) fits in where Q(sqrt(-5)) was, so what about be in place of sigma?

Q(sqrt(-5)) I guess? but I'm not sure how to reason this

[hot lake]

well the fixed field of sigma² is going to be Q(i, sqrt5)

[summer geyser]

yeah

[hot lake]

and there is only one other field that you can put between that and Q

[summer geyser]

ah I see alright thanks!

[hot lake]

you can also try to get sqrt(-5) as an expression in the roots

though if you already got one for i and one for sqrt5 you can just multiply them

then you can check how the group act on it

[summer geyser]

alright I see, thanks again

[hot lake]

and yeah it makes much more sense for Q(x1) = Q(sqrt(2+i)) to be a degree 2 extension of Q(i) instead of Q(sqrt-(5))

[summer geyser]

yeah haha I basically tried to follow an example in the book, and I guess not everything is the same

[chilly ocean]

yo Im not sure if my argument works: I want to show that [Q(sqrt2,sqrt3) : Q] = 4 - so like is saying sqrt2 and sqrt3 are linearly independent over Q enough? Therefore the basis of this extension would be 1, sqrt2, sqrt3, sqrt2sqrt3?

cuz I wasnt sure about Qsqrt3 over Q sqrt2 - would the minimal polynomial be the same? (x^2 -2)

I guess this is the same thing, need to show you cant get sqrt2 in Q(sqrt3)

[rigid cave]

Do not trust me because I am still a noob but I like to think about this like this: Q(sqrt2, sqrt3) = Q(sqrt2)(sqrt(3)= {a+bsqrt3 | ab in Q(sqrt(2)}. You could also use the "tower law", if $F \subseteq L \subseteq K$ then $[K:F] = [K:L][L:F]$. Like I said, I am still a beginner so I am not sure if your argument is correct. Let's hope someone else answers

[cloud walrus]

older sister

[chilly ocean]

I mean I think thats exactly what I wrote xD but yeah Im just kinda looking for the fastest proof for this and if my reasonings correct

[rigid cave]

Oops lol

[chilly ocean]

I showed it in a different way, by showing q(sqrt2,sqrt3) = q(sqrt2 + sqrt3) and then finding minimal polynomial for sqrt2 + sqrt3, but it wasnt that fast and obviously not always works

[rigid cave]

Yeah that also works. A faster way would (maybe) be to use the tower law or to think about q(2, 3) as q(2)(3)

[unique juniper]

$|G| = 28$ prove $G$ has a normal subgroup of order 7

[cloud walrus]

Yes

[unique juniper]

so by cauchy there exists an element of order 7

then we have a subgroup of order 7, H

now if we consider G acting on the set of left cosets of H in G

the corresponding permutation representation would be x : G -> sym(S)

now the kernal of x is of order something from {1,2,4,7,14}

im stuck here

:/

[prisma ibex]

Use the Sylow theorems

[unique juniper]

i cant use sylow

[prisma ibex]

Hmmm

[unique juniper]

well its not interesting with sylow

[prisma ibex]

Yea it’s too easy with Sylow

[celest brook]

lAgrAnGe thEoReM

[unique juniper]

hm

[unique juniper]

how so

like

[celest brook]

i.e. 7 divides 28

[unique juniper]

errr

[prisma ibex]

I mean if you know that Sylow will make it easy why not mimic the proof of Sylow here?

[unique juniper]

im pretty sure this is meant to be done without sylow tho

[viscid pewter]

yeah, that was what i was thinking, just do a hack job

[prisma ibex]

Let H be this order 7 subgroup, let S be the set of order 7 subgroups of G. Let H act on S by conjugation. Then the orbit of H has size 1, and all the other orbits have to have size divisible by 7 (check this). Then by the class equation |S|=1+7x so |S|=1 mod 7

[unique juniper]

i dont think we need this tho

[final pasture]

How to use Sylow without saying the word "Sylow", a tutorial

[prisma ibex]

Yea

[final pasture]

I love your profile picture btw nGroupoid

[prisma ibex]

Ahh here’s a clever way to do this

[unique juniper]

waiittttt

|kernal of x| is either 2 7 or 14