Mathematics - 2023-11-28 (page 2 of 2)

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Jakobian

23:03

Huh

$x^2-1 = (x-1)(x+1)$ always

@oscarmetalbreak what kind of isomorphism

I call this theorem Fubini for poor people

Idk. It might be strong for gauge integrals

Though I don't think Bartle was aiming high with this one

oscarmetal break

Just the very beginning kind of isomorphism that we saw in abstract algebra. Proving the map is both bijective and homomorphism.

Jakobian

What?

What kind of isomorphism...

oscarmetal break

I only know one kind of isomorphism

Jakobian

And what kind of isomorphism is it

Of what objects

oscarmetal break

From $F[x]/(x^2+1)$ to $F\times F$ where $F$ has prime number of element and is equaled to $4k+1$.

This kind

Jakobian

23:17

You're not answering my question

oscarmetal break

I don't understand...

I don't understand what "kind" do you mean?

Jakobian

Isomorphism of vector spaces, rings, groups, idk. There are isomorphisms of sets too

Its such a vague question

oscarmetal break

I see.

Jakobian

In abstract algebra alone you have so much of them

So telling me you are considering abstract algebra one tells me nothing at all

mick

i "found" a weird function

Q: About $f(x) = \frac{\sum_{n=1}^{\infty} \sin^2(x/n)}{x}$

Consider for $x>0$ $$f(x) = \frac{\sum_{n=1}^{\infty} \sin^2(x/n)}{x}$$ I was fascinated by the behaviour of this function. It is easy to show that $$\lim_{x \to +0} \frac{f(x)}{x} = \zeta(2) = \frac{\pi^2}{6}$$ It seems we get for all $x>1$ $$f(1) \leq f(x) \leq \zeta(2)$$ and for all sufficient...

oscarmetal break

23:22

In this case, since $F[x]$ is a polynomial ring and $(x^2+1)$ is a principle ideal and $F\times F$ is a vector space, so it will be the quotient ring $F[x]/(x^2+1)$ to the vector space $F\times F$.

PM 2Ring

@oscarmetalbreak Primes not of the form 4k+3 can always be written as the sum of two squares. And that is related to the fact that $x^2+1\equiv 0 \pmod p$ always has a solution for such primes. See en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares

oscarmetal break

Sorry I didn't realize that premise that I omit in my question

Jakobian

@oscarmetalbreak so are you considering them as F-vector spaces

oscarmetal break

Yes

That is why I was looking if $x^2+1$ is reducible or not so that I can write a direct sum

@PM2Ring I don't think I need the solution for $x^2+1\equiv 0\pmod p$.

Jakobian

I think its pretty clear that it has dimension at most $2$

PM 2Ring

23:28