Commutative Monoid and some properties

[Kindayr]

Hmm, I'm not sure how you would generalize all of this to algebraic numbers. But it does make sense what you're saying...

I guess I would I would have to prove a sort of analogous Fundamental Theorem of Arithmetic to the non-zero algebraics? If I can do that then I hope it all works, and I would just have to prove it to be a vector space, correct?

I never expected myself to get into doing this much for such a trivial idea I began with lol! Considering how little algebraic background I have outside of linear algebra.

 

[micromass]

I guess I would I would have to prove a sort of analogous Fundamental Theorem of Arithmetic to the non-zero algebraics? If I can do that then I hope it all works, and I would just have to prove it to be a vector space, correct?

I think such a thing would be very hard, if not impossible

I never expected myself to get into doing this much for such a trivial idea I began with lol! Considering how little algebraic background I have outside of linear algebra.

Hey, this is the perfect way to learn something more about algebra! You're doing great!

 

[Kindayr]

As a last attempt, what if we just generalize to the set

$$\mathbb{A}^{*}=\{a^{b}:a,b\in\mathbb{Q}^{*}\}$$

I would have to show this to be a field obviously.

 

[micromass]

As a last attempt, what if we just generalize to the set

$$\mathbb{A}^{*}=\{a^{b}:a,b\in\mathbb{Q}^{*}\}$$

I would have to show this to be a field obviously.

That seems plausible. Don't know if it'll be a field, though...

 

[Kindayr]

That seems plausible. Don't know if it'll be a field, though...

I agree, I don't think its closed under addition.

I guess I never really tried $\mathbb{Z}_{p}$. Or even the set $\mathbb{Q}[\mathbb{Z}_{p}]=\{a^{b}|a,b\in\mathbb{Z}_{p}\}$. Which would be closed under addition (I'd hav to prove it) I guess since every $a^{b}\equiv c<p$, so we would have $a^{b}+c^{d}\equiv e+f\equiv g$ with $e,f,g\leq p$, or something like that.

 

[Kindayr]

So working I've got some weird results, that i'll type later when you fix a $p>0$ prime, and define $\phi_{p}:\mathbb{Z}_{p}^{*}\to\mathbb{Z}_{p}^{n}$ where $1<p_{1}<p_{2}<\cdots <p_{n}<p$ are all primes less than $p$.

I can show this to be a vector space, which is good. But you get weird equivalences when you work in $\mathbb{Z}_{p}^{n}$, that are just blowing my mind a little as I write them down on paper. Which get even weirder when you move to $\mathbb{P}_{n} [\mathbb{Z}_{p}$.

I'll type it all up when I get home.

 

[micromass]

Hmm, "weird equivalences", I'm wondering about what you will type up

 

[Kindayr]

Well its trivial to show that, from the same $p$ and $n$ defined before that $\phi_{p}:\mathbb{Z}_{p}^{*}\to\mathbb{Z}^{n}_{p}$ is an isomorphism. This is seen since if $a\in\mathbb{Z}_{p}^{*}$ then $a$ has a prime factorization such that $$a=\prod^{n}_{k=1}p_{k}^{e_{k}}=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{n}^{e_{n}}.$$ So let $\phi_{p}(a)=(e_{1},\dots,e_{n})$. Its really easy to show its a vector space over field $\mathbb{Z}_{p}$

Whats interesting is when you partition the set such that for any $u,v\in\mathbb{Z}_{p}^{n}$, we have $$u\simeq v \;\; iff \;\; \phi_{p}^{-1}(u)\equiv\phi_{p}^{-1}(v).$$

If you use the definition of the isomorphism I've used before into the polynomials, with the same partition, you'll get weird things. For example, take p=7. Then n=3. So we have $$0\simeq3\simeq x+x^{2}\simeq2+2x\simeq\cdots$$

I've worked with quotient spaces in linear algebra, and I though how that collapsed a vector space was weird, but this weirds me out even more.

Its probably not weird for someone who has taken abstract"er" algebra than I have, but still weird lol

I'll be typing this all up sometime tonight or tomorrow when I have time in a paper and more rigourously, but i have other plans tongiht so I won't be able to type everything up i worked out today (like my proof that its a vector space).

I feel even that $\mathbb{Z}_{p}^{n}$ collapses $\mathbb{Q}^{n}$. Since any n-tuple's entries can be seen as a quotient a/b, but then since $\mathbb{Z}_{p}$ is a field, so this quotient exists as one of the p members of Z_p.

Maybe that's wrong.

 

[micromass]

I don't quite see why $\phi_p$ is an isomorphism. That is, I see no reason why it should be surjective. Certainly something like (p-1,p-1,...,p-1) isn't reached?

 

[Kindayr]

I don't quite see why $\phi_p$ is an isomorphism. That is, I see no reason why it should be surjective. Certainly something like (p-1,p-1,...,p-1) isn't reached?

Sure it does, it corresponds to $p_{1}^{p-1}p_{2}^{p-1}\cdots p_{n}^{p-1}\equiv a$ with $0\leq a\leq p-1$.

Wait crap, that makes it non-injective. URGH.

I need to sit down with someone else and work all of this out. I need to get back to university asap. BLARGH.

 

[micromass]

Sure it does, it corresponds to $p_{1}^{p-1}p_{2}^{p-1}\cdots p_{n}^{p-1}\equiv a$ with $0\leq a\leq p-1$.

Wait crap, that makes it non-injective. URGH.

I need to sit down with someone else and work all of this out. I need to get back to university asap. BLARGH.

The problem is that you don't have a well-defined function, really.

$\mathbb{Z}_p^*$ has p-1 elements, and $\mathbb{Z}_p^n$ has $p^n$elements. So there can never be a surjection.

Do you see a way to solve this??

 

[Kindayr]

Okay, I'm meeting with my prof tomorrow, and I want to really see if this works:

Define $\mathbb{A}^{*} =\{a^{b}:a,b\in\mathbb{Q}^{*}\}$ and let $\phi :\mathbb{A}^{*}\to\mathbb{Q}^{\infty}$, where $\mathbb{Q}^{\infty}$ is the set of all $\infty$-tuples with a finite amount of non-zero entries. Note that for any $a^{b}\in\mathbb{A}^{*}$ we have a prime factorization such that $$a^{b}=\prod_{k=1}^{\infty}p_{k}^{be_{k}}.$$ So define $\phi(a^{b})=(be_{1},be_{2},\dots)=(be_{k})^{\infty}_{k=1}$. Also define, from before, $\oplus:\mathbb{Q}^{\infty}\times\mathbb{Q}^{\infty}\to \mathbb{Q}^{\infty}$ such that $$(e_{1},e_{2},\dots)\oplus (f_{1}, f_{2},\dots)=(e_{k})^{\infty}_{k=1}\oplus (f_{k})^{\infty}_{k=1}=(e_{k}+f_{k})^{\infty}_{k=1}=(e_{1}+f_{1},e_{2}+f_{2},\dots).$$ Further, $\phi$ can be shown to be an isomorphism.

We can also show $\mathbb{Q}^ {\infty}$ to be a rational vector space. Note that the scalar multiplication of any vector in $\mathbb{Q}^{\infty}$ is analogous to raising the original element of $\mathbb{A}^{*}$ by a rational exponent (the rational exponent being the scalar). That is: $$\alpha(be_{k})^{\infty}_{k=1}\overset{\phi^{-1}}{=}a^{b^{\alpha}}=a^{b\alpha}.$$ Also note that $\alpha(be_{k})^{\infty}_{k=1}=\alpha b(e_{k})_{k=1}^{\infty}=(\alpha be_{k})^{\infty}_{k=1}$.

I didn't want to type everything out, but it seems to have worked.

 

[Kindayr]

So I've met with my professor, and by the end we just were laughing at how simple and familiar it ended up working out to be.

No matter how hard I tried, I could not create a vector space, which is fine. The suddenly it came to us.

So here:

Let $(\mathbb{R}^{>0},\cdot)$ be the abelian group of positive reals and multiplication, and let $(\mathbb{R},+)$ be the abelian group of the reals over addition. Suppose there exists a function $\log:(\mathbb{R}^{>0},\cdot)\to(\mathbb{R},+)$ such that $\log(ab)=\log(a)+\log(b)$ and $\log(a^{b})=b\log(a)$. Its clear that $\log$ was just the extension of what I was trying to do before. I guess you could say its the 'analytic jump' from what I was trying to do algebraically. This has given me some new intuition into how $\log$ works. That its 'sort of' rooted in prime factorization, but more general than that.

I thought this was really interesting. And I'm glad this is all over of me attempting to find something that wasn't there, but had some sort of 'transcendental' version that explains everything.

He gave me some homework to do, the first question being one that I could probably do my self with my current skill sets, and another I'll have to wait after I take Measure Theory in the Winter Semester.

The first is that:

• 
  If $f:(\mathbb{R}^{>0},\cdot)\to(\mathbb{R},+)$ continuously with the properties that $f(ab)=f(a)+f(b)$ and $f(a^{b})=b\cdot f(a)$ and $f(1)=0$, then $f(x)=\log(x)$

• 
  There exist many non-measurable $f:(\mathbb{R}^{>0},\cdot)\to(\mathbb{R},+)$ with the properties $f(ab)=f(a)+f(b)$, $f(a^{b})=b\cdot f(a)$, and $f(1)=0$.

I hope to be able to do these sometime in the future, but we'll see what happens.

Thank you to everyone that has helped me! Especially micromass for putting up with my lack of education in abstract algebra and metric space topology. I swear I'll get better after this year when I finally get a formal introduction to everything! haha

 

[micromass]

I'm very glad you have found an answer to your (very intriguing) question. I'm also happy that you learned a bit of new mathematics this way!

I hope you enjoyed thinking about this. These are the things that math research is all about