Blah @ windows.

lol

@LukasHeger Hi Lukas, can you give an algebraic explanation to why the CRT works the way it works? ie breaking the equation to many others ?

I suppose what I mean is , solving the congruence mod ( m) is same as the factors of m ( prime factors)

@Jacksoja do you know the ring-theoretic version of CRT?

I have seen something similar to this for groups

Hi @Lukas!

No I don't !

Hi @Ted

@TedShifrin Hola Ted

I know this much, Z/ pqZ is isomorphic to Z/p x Z/q

I suppose the isomorphism is the trivial one , reduction modulo each prime

@Jacksoja: The crucial thing to understand is that if $m$ and $n$ are relatively prime, then $x\cong 0\pmod{mn}$ is equivalent to $x\cong 0\pmod m$ and $\x\cong 0\pmod n$. After that you can see that what's going on is solving two simultaneous congruences and using linearity.

@Jacksoja let's just do it for two ideals for simplicity, you can generalize it inductively in a straightforward manner from there. If $R$ is a ring and $I$ and $J$ are ideals, then we always have a map $R \to R/I \times R/J$, $r \mapsto (r+I,r+J)$. The kernel of that is $I \cap J$, so we get an injective map $R/I \cap J \to R/I \times R/J$. If $I+J=R$, then this maps is surjective, so an isomorphism

and furthermore if $R$ is commutative, then $I+J=R$ implies $I \cap J=IJ$

I see

so it is just a direct consequence of those theorems

isomorphims*

so if you apply that to $R=\Bbb Z$ and $I=(n)$ and $J=(m)$ for two coprime integers $n$ and $m$, we get an isomorphism $\Bbb Z/(nm) \cong \Bbb Z/n \times \Bbb Z/m$

I like the concrete understanding before the fancy ring/module approach :P

he was asking explicitly for an algebraic explanation

Yes I see thanks ! :)

Does the homomorphism defined this way have a name? "Let $\psi$ be the function that sends each element of $A$ to the identity expect elements of the subgroup $B$, upon which it acts as the identity function."

Yes Ted haha , for something new I ask for concrete

but this CRT i used it alot before

I just was curious from where it stems algebraically

Thanks @TedShifrin @LukasHeger

@Rithaniel I highly doubt that will be a homomorphism in general

OK, then the quotient ring construction is good to understand.

Really? Where might it break down?

expect = except :P

Ah, yeah

Multiply something in $A-B$ by something in $B$?

@Rithaniel take for example $A=S_3, B=A_3$ and look what your function does to $(12)$ and $(23)$ and their product

Ah, I see.

Well, I have this claim that $\psi f\neq\phi f$ unless $\psi=\phi$ (where $\psi,\phi,f$ are homomorphisms) and I'm trying to make the claim that this means $f$ must be onto.

My thought was to explicitly construct homomorphism which don't agree on the entire domain, but do agree on $\text{Im}(f)$

@Rithaniel that's a tricky exercise

Well, I don't have to go down the explicit construction route, but it was the only option I could think of

the easiest solution to this I know probably involves some construction you haven't seen

unless you've studied algebraic topology

do you know amalgamated free products?

Did you ask Lukas about the divisibility problem?

Nope and nope on those topics. Also, I managed to get an answer I was happy with on the divisibility one, but I could post my proof to get some pointers on it

If you know amalgamated products this is easy

Oh, right, you had an alternative argument for splitting.

suppose $f:G \to H$ is a homomorphism that is not surjective, then consider $H *_{f(G)} H$ and take $\psi$ and $\phi$ be the inclusions into the two components

then by construction $\psi \circ f= \phi \circ f$, but $\psi \neq \phi$

How is the amalgamated product defined?

@LukasHeger that's a neat argument

it's the easiest to do in terms of presentations. Do you know group presentations?

Hmmm, nope :]

I can google stuff, not to worry

sorry for needing so many concepts

It's perfectly alright. You gotta learn stuff to know stuff

there might be a more elementary argument

but it's probably not as short

and I'm not seeing it right now

(Lukas is going to ask if you know what a pushout is soon)

lol

do you know what a Kan extension is?

jk

@LukasHeger ncatlab.org/nlab/show/epimorphisms+of+groups+are+surjective not really elementary, but it's a different argument

@Alessandro I like the argument with the amalgamated free product better

Agreed

Okay, so if $G$ and $H$ are two subgroups of a parent group, the amalgamated free group is effectively $\langle G, H\vert G\cap H\rangle$?

Ah, wait, I'm off

(I think)

no, not quite

the situation is this: you have a group $C$ that is simultanously a subgroup of $A$ and $B$

then if $A=\langle S_A \mid R_A\rangle$ is a presentation and the same notation for $B$, then the amalgamated free product is $\langle S_A \cup S_B \mid R_A \cup R_B \cup C\rangle$

Alright. I believe I understand