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Jakobian

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
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Jakobian
00:06
So Euclidean algorithm (in usual sense as anank pointed out), does not work in $R[x]$ if $R$ is not a field.
Jakobian
00:05
And $F$ is a field iff $F[x]$ is a PID, in which case it's an Euclidean domain
Jakobian
00:03
Being an Euclidean domain is a stronger property than being a PID
Jakobian
00:00
Because all non-zero elements of F are units
Jakobian
yst 23:58
@sku the ideal generated by 2 and x should be non-principal
Jakobian
yst 21:55
I was referring to my thoughts
Jakobian
yst 21:52
no problem
Jakobian
yst 21:40
then you'll have $\mathcal{A} = \mathcal{A}_Y$
Jakobian
yst 21:40
given a completely regular algebra $\mathcal{A}$, you want to take the compactification associated with $\mathcal{A}$
Jakobian
yst 21:39
part c) does show surjectivity
Jakobian
yst 21:39
oh okay no
Jakobian
yst 21:38
part b) shows injectivity
Jakobian
yst 21:37
part a) shows that the map is into what we want it to be into
Jakobian
yst 21:37
@psie there is no part that shows surjectivity as far as I can see
Jakobian
yst 21:27
you then show injectivity, surjectivity
Jakobian
yst 21:24
First step would be to show this map is well-defined so that if $Y, Y'$ are equivalent compactifications, then $\mathcal{A}_Y = \mathcal{A}_{Y'}$
Jakobian
yst 21:24
@psie $[Y]\mapsto \mathcal{A}_Y$
Jakobian
yst 17:21
replace it with $c_1...c_n$ for arbitrary placing of brackets if you want non-associative
Jakobian
yst 17:02
That's how you do it with any algebraic structure
Jakobian
yst 17:01
You need to first ask yourself what needs to be in such subalgebra, and then show that if you take all those things that need to be in there, it will already be a subalgebra
Jakobian
yst 17:00
Yes. But that doesn't get us closer to what it is
Jakobian
yst 16:58
subalgebras, and "contain the algebra" is uh. I don't know what you mean
Jakobian
yst 16:56
of course I'm assuming your algebras are non-unital since your algebras were non-unital in other theorems (this has bearing on what we mean by subalgebra)
Jakobian
yst 16:55
In this context, the subalgebra generated by $\mathcal{C}$ will consist of linear combinations of finite products $c_1...c_n$ where $c_k\in \mathcal{C}$. You should prove it
Jakobian
yst 16:53
there is two ways in which we can think of those subspaces
Jakobian
yst 16:53
you should recall when we were talking about vector subspace generated by a set
Jakobian
yst 16:53
equivalently you can give it another characterization
Jakobian
yst 16:52
this will be the intersection of all subalgebras $\mathcal{A}_0\subseteq \mathcal{A}$ such that $\mathcal{C}\subseteq \mathcal{A}_0$
Jakobian
yst 16:52
If $\mathcal{A}$ is an algebra and $\mathcal{C}\subseteq \mathcal{A}$ then we can consider subalgebra generated by $\mathcal{C}$
Jakobian
yst 16:50
@psie because in the context I was referring to this specific example
Jakobian
yst 16:47
You are misquoting me
Jakobian
yst 15:36
the fact that the two results together are equivalent to AC is an odd curiosity
Jakobian
yst 15:35
But Boolean prime ideal theorem is strictly weaker than AC
Jakobian
yst 15:11
this is a fundamental tool in our reasoning
Jakobian
yst 15:11
the more advanced stuff - can be too
Jakobian
yst 15:10
the basic stuff is proven using induction, that's what it's for
Jakobian
yst 15:09
its maybe worth keeping in mind that all these schema for proving things allow us to reason formally about things and are the greatest tool
Jakobian
yst 15:06
:68059980 I just explained it. What?
Jakobian
yst 15:06
It allows you to continue to reason by induction even past the natural numbers
Jakobian
yst 15:05
Or the class of all ordinals
Jakobian
yst 15:04
@pie It's induction in case instead of natural numbers $\omega$ you have an arbitrary ordinal
Jakobian
yst 15:03
another one is transfinite induction
Jakobian
yst 15:03
One that is quite straightforward is "induction on reals"
Jakobian
yst 15:02
There are generalizations of induction, yes
Jakobian
yst 15:01
"for countably many $P$" - makes no sense to me
Jakobian
yst 15:01
@pie what do you mean? Induction says that if $P_0$ is true, and for all $n\in \omega$ the implication $P_n\implies P_{n+1}$ is true, then for all $n\in \omega$ the statement $P_n$ is true
Jakobian
yst 14:57
@pie what's a case in this context
Jakobian
yst 14:30
contrary to its name, the Van Douwen line is $2$-dimensional
Jakobian
yst 02:12
26
Q: Using higher-order Bring radicals to solve arbitrary polynomials

Daniel MillerIt is well known that there is no general formula for the solution of the quintic. Of course, what this really means is that there is no general formula that only involves addition, subtraction, multiplication, division, and the extraction of $n$-th roots. Indeed, if one is allowed to use the Bri...

polynomials galois-theory big-picture
Jakobian
yst 02:11
so I'd expect we'd have to have Bring radicals of different kinds as well