@arctictern I honestly don't understand how this helps me. I don't know what a tensor product is. My idea was to use the fact that $R=\mathbb{M}_n(\mathbb{C})$ is simple so the only algebra homomorphisms $R\to M$ are either injective or the zero map... then to show a homomorphism to $M=\mathbb{C}^n$. But this seems stronger...
@TobiasKildetoft I want to show that since $R$ is simple as a $\mathbb{C}$-algebra, then any left $R$-module has dimension at least $n$ over $\mathbb{C}$.
@BalarkaSen Ah I see. For the normal derivative, we would have to apply Leibniz's rule for $\partial_\mu(V^\nu e_\nu)$. But the covariant derivative has the $\partial_\mu e_\nu$ term "baked in", so it gets $C^\infty$-linear, which means $D_\mu (V^\nu e_\nu)=(D_\mu v^\nu)e_\nu$. Right?
@Semiclassical maybe at one point in time we can travel between universes. That doesnt say that there isnt a universe of universes. (say if its too philosophical)
looks good @meow, could be written more clearly with something like if $u\in I$ then $u^{-1}\in I$ so $1\in I$ and $r1=r\in I$ for every $r\in R$ but it's correct
@arctictern If you can't help, that's okay as well. For a prime $\mathfrak{p}$ of a number field $K$ and a prime $\mathfrak{q}$ above $\mathfrak{p}$ in a number field $L$ we look at $K_{\mathfrak{p}}$ and $L_{\mathfrak{q}}$.
@Semiclassical to prove something you can use a truthtable, which is a test for the statement. (i.e. something is equal to whatever) this is a practical test.
@arctictern And then I want to show that $L_{\mathfrak{q}}$ is a cyclotomic extension of $K_{\mathfrak{p}}$, if $\mathfrak{q}$ is unramified. I have shown that the Galois group is cyclic and I need some hints for the cyclotomic part
you define an equivalence relation on $R$ given by $x\sim y$ iff $x-y\in S$ (where $S$ is the ideal you're taking the quotient by), $R/S$ is the quotient set $R/\sim$ with operations given by $[a]+[b]=[a+b]$ and $[a][b]=[ab]$, where $[a]=a+S$ so the elements look like $a+S$, not $aS$ @meow
what if you simulated a universe you made a ball in the universe behind you then you turned around and see the same ball... makes sense in nested determenistic simulations
the only objection i would have is if true it implies quantum immortality which is inevetiablly tested and can be performed right now by anyone depressed enough
To put it a little differently: To the extent that I find religion/theology interesting, it is how such questions weigh upon our actions, i.e. how we live our lives and such.
The speculative aspect of it I find almost entirely uninteresting.
@Alessandro so the elements are of the form $r + I$ for quotient ring $R/I$, and addition is given by $(r + s) + I$ and multiplication is given by $(rs)+I$?
@arctictern Revision: I think I understand that the sum decomposition would show the dimension is at least $n$. But I want to show that using the fact that $R$ is simple... where does simplicity come into the argument?
Hm, I'm not sure @meow, if you take the quotient of an euclidean domain by the ideal generated by an irreducible element you'll get a field, but I don't know how much can be said in the other direction
technically if you wrote everything down in symbolic logic you could still explain it without them speaking english you just cant guarentee mutual understanding of the symbols
@Alessandro the thing is since $R/I$ is a field, for every $u + I \in R/I$, there exists a $v + I \in R/I$ such that $(u + I)(v + I) = (1 + I)$. Therefore, for every $u \in R$, wouldn't there be the same $v$ such that $uv = 1$, by definition of multiplication on quotient rings
no, take as an example $\mathbb{Z}/3\mathbb{Z}$, in this field $2\times2=4$, but $4=1$ since they are in the same equivalence class, while this does not happen in $\mathbb{Z}$ @meow
look at technology it makes everyones lives better not just the people who made them. math research doesnt only help the person who publishes but everyone who reads something from the publications and is inspired we have progress not stalled as it would be in a zero sum game (game theory)
so $2$ (let's omit the brackets for the sake of commodity) is its own multiplicative inverse in $\mathbb{Z}/3\mathbb{Z}$, but that doesn't imply that the same is true in $\mathbb{Z}$ @meow
Hmmm... I still don't understand the proof, and I am searching an elementary proof! Like just using some of the basic theorems and things like that... If anyone could help... Here is the problem: Show that there is an infinity of $a$'s so that if $a$ divides $(x+y)^5-x^5-y^5$ then it must divide $(x+y)^7-x^7-y^7$... Anyone?
i think a good assumtion is that: "a is an element of a infinite field"
I know this is unrelated to math, but since there's like no one in tex room, any idea how to center an image (which is a graph) in the middle of a page in sharelatex? I tried every code, but nothing seems to work lol..
@AndyMiles Well, not really. All I meant was that you can write $D_\mu(\sum_i V^i e_i)$ as $\sum D_\mu(V^i e_i)$. They are simply denoting $V^i e_i$ by $V^i$, using the canonical identification with $\Bbb R^n$ by sending $e_i$ to the basis vectors, in the equation (3) you quoted.
Look at the comments below the answer in the source you linked for (3). That'd clarify.
@BalarkaSen maybe you can help me with something that I am seeing correctly.
Let G be a group of order 12 with following presentation $<a,b : a^3 = b^4 = e, bab^{-1}a = e>$. Find the center Z(G). So I proved that any $c \in G$ is of the form of $a^ib^j$.
@shaihorowitz just as a mathematical approach: would it make sense to say "everyone is a winner" in respect to the whole "hell-heaven"-thing? (just say if it is too broad)
