ah ... This style of question should remind you about me https://math.stackexchange.com/questions/2955488/analytic-continuation-of-a-series-raised-to-a-power-raised-to-a-power
Now to tell you what the multiplication is, I just need to tell you what happens to the basis elements, and that's easy: we say that in this ring, $g \cdot h = gh$, using the group multiplication law!
That is, two basis elements multiply to another basis element.
@MikeMiller thank you Mike, but is this book am using , this notion of group rings has been introduced without using free , modules or anything as such ^^
free $R$-module with basis $G$ just means the set of all formal finite linear combinations over $G$, or if you want an actual set-theoretic construction, the set of all functions $G \to R$ with finite support
It is just a miscommunication to say that "homology counts holes" and "g-holes torus" are supposed to mean the same thing. The latter is an easy way of saying the connected sum. Indeed, in that case, we mean the "donut holes".
In the former case we are of course counting dimension and get 2g.
anyway im supposed to submit a statement of purpose and im not quite sure what is expected for a statement of purpose for applying to conferences (in contrast to gradschool etc. where i kind of just bs about my interest in math)
i guess i should probably bs about my interest in alggeom and how i think it's going to benefit my (nonexistent) research hmm
@Mathein, are you at all familiar with how to use the inverse different to compute rings of integers? I know that you're able to sorta squeeze it in via $n\mathbb{Z}[\alpha] \subset n\mathcal{O}_K \subset \mathbb{Z}[\alpha]$ but then I'm not 100% sure how to work from there
Guys. If I know that $f' \circ j \in \mathcal{O}_X(j^{-1}(V))$ and $f' \circ j \circ f \in \mathcal{O}_X(f^{-1}(j^{-1}(V)))$ can I conclude that $f' \circ f \in \mathcal{O}_X(f^{-1}(V))$?
$V$ just an open set
$j$ just the inclusion of $V$ into $X$
and $\mathcal{O}_X$ is the sheaf of regular functions on $X$
And then the inverse different of $\mathbb{Z}[\alpha]$ being $\frac{1}{f'(\alpha)}\mathbb{Z}[\alpha]$ should have something to do with the conjugates of $\alpha$ being an integral basis of it in that case?
Okay, so you work with the discriminant to compute rings of integers. Just to be sure, the discriminant of the ring of integers in $\mathbb{Q}(\alpha)$ is gonna be the discriminant of the minimal polynomial of $\alpha$, right?
so if $O$ is any order (e.g. $\Bbb Z[\alpha]$ for $\alpha$ a primitive element) in $K/\Bbb Q$ a number field, then one has $\Delta(O)=|\mathcal O_K/O|^2\Delta(\mathcal O_K)$, so if you're lucky you can find an $\alpha$ such that the discriminant of $\Bbb Z[\alpha]$ is square-free, then you're done
then you also have the useful criterion that if the minimal polynomial for $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $\mathcal O_K/O$
so let's say you want to compute $\mathcal O_{\Bbb Q(\sqrt[3]{5})}$: you find that $\Delta(\Bbb Z[\sqrt[3]{5})=-675=-3^3 \cdot 5^2$. Now $x^3-5$ is $5$-Eisenstein, and $(x-1)^3-5$ is $3$-Eisenstein, then you're done since $\Bbb Z[\sqrt[3]{5}]=\Bbb Z[\sqrt[3]{5}+1]$
that's a bit lucky I guess...
some computations I've seen for rings of integers relied on ad-hoc methods with norms, traces and conjugates
Why must we have this \int_0^{2\pi}e^{ig(t)}e^{it}\,dt=2\,\pi\,e^{i\alpha} for some \alpha\in[0,2\,\pi) for the following question: https://math.stackexchange.com/questions/104396/continuous-function-on-the-unit-circle-must-be-c-barz
