@TedShifrin Let $R$ be a finite ring with identity. Since $R$ is closed under addition,mult. we have for any $a \in R$, $ab = 1_R$ or $ab = 0_R$ where $b$ is either a zero divisor or unit. If it weren't either, then we'd have no such element $b \in R$ such that $ab = 1$ or $ab = 0$ which implies $R$ isn't closed under multiplication (idk how to show this part)
I know $0_R$ and $1_R$ need to be the products of some other elements multiplication for some reason, like maybe existing in some equivalence relation
like $(\forall a,b\in R), (\exists ab \in R)$ and $[1_R,0_R,a \in R \implies (\exists b\in R, s.t. ab = 1_R \lor ab = 0_R)]$ I gotta prove that part
there are enough edits going on that i can't tell what you originally wrote, but if you were computing the center of M_n(R) for arbitrary n and (commutative?) ring R, i suggest first the case of the real field that we all know and love
obliv, a modified version of thorgotts observation is that all linear transformations commute with all 'scalar matrices' by definition, so the fact that the center contains all 'scalar matrices' (and not e.g. just 0 and I) is more or less immediate
there's a tiny amount of effort in proving it doesn't contain anything else, the MSE question up above collects some fairly good arguments
if you study less than full subsets of M_n(F), including e.g. subalgebras of M_n(F), that other inclusion [center subset {scalars}] may not hold at all and you can find other stuff in the center
@Jakobian I wouldn't mind eating meat if the cattle wasn't put into horrendous conditions. Like, if it were raised by someone I knew treated them well etc
Let $R$ be a commutative ring with identity and $r \in R$. Define $L_r:R\to R$ by $L_r(x) = rx$ for all $x \in R$. Prove that $L_r$ is injective iff $r$ is not a zero divisor. Prove that $L_r$ is surjective iff $r$ is a unit.
I mean there is evidence but you can choose to believe what you want. I don't think you should take anything for granted either but I find it strange that you don't believe in eternity IRL but use $\infty$ in your mathematics
1. We weren't talking about universe but about human beings (so I'm allowed some slack about what I'm saying here) 2. You are equating infinite to eternal for some reason
@Thorgott astronomy deals with massive errors so observational evidence when it comes to it doesn't matter that much. The errors are massive so the explanation "dark matter exists" may just as well be false
I believe there is a straightforward formula for the abelianization of a semi-direct product: if $G$ acts on $H$, and we form the semi-direct product of $G$ and $H$ in the usual way, and the abelianization of this semi-direct product is the product $G^{ab}\times (H^{ab})_{G}$.
(Here the subscrip...
you're shifting a lot of goalposts. the big bang theory and dark matter theory are not logically equivalent and nobody said anything like that either "must be true". physics is not that definite and variants of the big bang are still in contention, but in its general features, the big bang is accepted by physicists on the basis of multiple layers of evidence. no amount of your mental gymnastics can change that.
let $R$ and $S$ be rings. showing $R\times S$ is isomorphic to $S\times R$ does that mean I have to show $(a,b)+(c,d)=f(a,b)+f(c,d) = f(b,a)+f(d,c)=(b,a)+(d,c)$ from ring axioms
Thanks, but I already figured out the proof I needed. Right now trying to construct an isomorphism from RxS to SxR. I'm thinking if I define $f:(R\times S) \to (S\times R)$ by $f(a,b) = (c,d)$ where $a \mapsto d$ and $b \mapsto c$ that's good?
Yes. But your approach is also good. If you had $f(a, b) = (k(b), j(a))$ where $j:R\to R$ and $k:S\to S$ are isomorphism then $f$ is also an isomorphism
for me, and probably everyone here, that this property is preserved by isomorphisms is obvious and hopefully you will see it once you write it down on paper
Note that an isomorphism $f:R_1\to R_2$ is a homomorphism such that $f^{-1}:R_2\to R_1$ exists and also is a homomorphism
What's a good reference for the classification of finite-dimensional algebraic representations of $\mathrm{GL}(n,k)$ when $k$ is an arbitrary field of characteristic zero? For $k = \mathbb{C}$, Fulton does it using Young diagrams in Chapter 8 of his book Young Tableaux. Other treatments for $...
I'm gonna stick to the direct product because I'm dumb and can't think about anything in generality. for $f:(R\times R)\to R$ to define an isomorphism we need $(a,b)\cdot(c,d)=f(a,b)\cdot f(c,d)$
$f(a,b) = g$ and $f(c,d) = h$ for some $g,h \in R$ then $(ac,bd) = gh$?
Balarka's approach is interesting though I think $\Lambda$ should be $S^1$. I don't have a deep understanding about modular group and I think number theoriest (who uses hyperbolic geometry) knows much better
I'd never heard of that theorem and I thought it's a theorem in analysis or number theory but weirdly graph theory came out and I asked if I searched correctly
well it says algebraic independence and cantor set
@onepotatotwopotato No, Mycielski's theorem about equivalence relations on Polish spaces (even though you need a slightly more general version here, the one presented in Kechris's book for example)
@onepotatotwopotato $\Lambda$ above is the orbit of $0$ under $\mathrm{PSL}_2(\Bbb Z)$; I agree the limit set of $\mathrm{PSL}_2(\Bbb Z)$ is the entire circle.
In any semigroup, D-class is defined to be there is c such that a L c and c L b If we permutate all the possible a,b in S, we get abstractly, a structure consists of all possible pairs of (aLc, cLb) which is almost like a automorphism between pairs of elements. Question ia, is this always have the be the same as the automorphism semigroup of a given semigroup?
@Secret if $S$ is the zero semigroup i.e. $xy = 0$ for all $x, y$ then $x D y$ iff $x = y$. Then your map sends pairs $(a, a)$ to themselves. But if $f$ is any bijection from $S$ to itself with $f(0) = 0$ then $f$ is an automorphism of $S$
@Balarka sanity check: does every pair $(X,A)$ have the weak homotopy type of a CW-pair? its standard to build a weak homotopy equivalence $A'\rightarrow A$ with $A'$ a CW-complex cell-by-cell and then extend the composite $A'\rightarrow A\subseteq X$ to a weak homotopy equivalence $X'\rightarrow X$ cell-by-cell s.t. $(X',A')$ is a relative CW-complex. now, the cell attachments are out-of-order, but we can use cellular approximation to homotope the attaching map of, say, a $(k+1)$-cell of $X$ into the $k$-skeleton of $A$ and doing this inductively over the skeleta of the relative CW-complex…
That's true, but for advanced topics like quantum mechanics and whatnot I don't think it's possible to grasp it as easily as elementary scientific ideas
@Thorgott What's the easiest proof that an infinite cyclic cover of any compact connected oriented $3$-manifold $M$ is obtained from cutting $M$ along a compact surface $S \subset M$, and pasting infinitely many copies of these end-to-end?
$3$ is not essential above, you can replace surface by a codimension $1$ submanifold. The proof I have in mind is that such covers are classified by $\mathrm{Hom}(\pi_1(M), \Bbb Z) = [M, S^1]$, one can take a regular value of $M\to S^1$ and its preimage is $S$.
Surely there's a by-hand proof where you look at the infinite cyclic cover $\widetilde{M}$, consider a fundamental domain, and modify the action slightly so that the fundamental domain is a nice manifold with corners?
Halmos on the standard convention regarding "sets with structure"
My question is a bit irrelevant, but when he says that the second coordinate (i.e. just the order relation) encodes all of the information in the ordered pair I am not sure I agree
Surely $dom \, \leq \subset X$ and it can be a proper subset in general (the order is partial)
Did someone hear a mention of a classification of subgroups of Q² ? I can't seem to figure out whether it has a neat classification like subgroups of Q do, or if the set of subgroups has a much bigger cardinality than the dim 1 case.
@Tapi For starters, $f$ is differentiable everywhere in $S$ and $\nabla f(a,b)\cdot (h,k)>0$ for all $(h,k)\in S$. But I don't think the question quite makes sense. Why is $(a+th,b+tk)\in S$ for all $(h,k)\in S$ and all $-\delta<t<0$?
The classification is that such subgroups have the form $\{r, v_p(r) \geq n_p\}$, where $n_p$ is a sequence in $\mathbb{Z} \cup \{-\infty\}$ indexed by primes.
I've been looking at triplets $(a,b,c) \in \mathbb N^3$ such that $\gcd(a,b,c)=1$ and no two elements of the triplet are pairwise co-prime. For all the triplets I've noticed I can always find a common prime divisor $p$ of any two elements (WLOG say $a$ and $b$ so that $p | a,b$) such that we have $\pmod p: \gcd(c,a) \neq \gcd(c,b)$. I'm not sure if this always holds though... can someone help me find counterexamples or hint for a proof??
@Sahaj doesn't this just follow by assumption $(a,b)\neq 1, (b,c)\neq 1, (a,c) \neq 1$ so there exists $p_1 \mid a,b$ and $p_2\mid b,c$ and $p_3 \mid a,c$
for the "no two elements of the triplet are pairwise co-prime" part
$p_1,p_2,p_3$ are gcds greater than 1 idk why I labeled as if they were prime
and it follows if $p_1 \mid a,b$ then $p_1 \nmid c$ and $p_2 \mid b,c \implies p_2 \nmid a$ etc
so why isn't $R\times R$ isomorphic to $R$ if $R$ is an integral domain? last night I tried to argue injectivity fails but @TedShifrin said it's the homomorphism properties
let $f:(R\times R)\to R$ and $(x,y) \in (R\times R)$ then for $f$ to define a hom. we need $(x_1,y_1)+(x_2,y_2) = f(x_1,y_1)+f(x_2,y_2)$ and $(x_1,y_1)\cdot (x_2,y_2)=f(x_1,y_1)\cdot f(x_2,y_2)$
@Obliv so with that reasoning we have that $p \nmid \gcd(c,b)$ and $p\nmid \gcd(c,a)$. But its not clear to me how that means $p \nmid \gcd(c,b) - \gcd(c,a)$.. For example $3\nmid 5$ and $3 \nmid 2$ but $3\mid (5-2)$...
i said the geometry is hard for me, not impossible :P
the more general question i should raise is the following
given only a two-sided codimension $1$ surface $S$ in a manifold $M$ and letting $\tilde{M}$ denote the cover obtained by cutting $M$ along $S$ and gluing infinitely many copies, how do we determine the subgroup of $\pi_1(M)$ corresponding to $\tilde{M}$?
i dont have a good answer
in particular, the corollary question is how to tell if two surfaces determine the same cover
it's only easy for knot complements cause they only have one subgroup that is co-$\mathbb{Z}$
I assume $S$ is not closed, otherwise $M \setminus S$ is disconnected. Let $\mu$ be a circle transverse to $S$, cutting it at exactly one point. Then $\pi_1(\tilde{M})$ is amalgamated product of infinitely many copies of $\pi_1(M \setminus S)$ where you identify a loop $\lambda_{+}$ given by a positive pushoff of a loop $\lambda$ in $S$ with $\mu \lambda_{-} \mu^{-1}$ where $\lambda_{-}$ is the negative pushoff
@Thorgott This is not difficult, the surface is related to the cover by the map $M \to S^1$. Any two regular values determine the same cover, so the answer is that the surfaces must be related by an ambient (cyclic) cobordism.
also, your description sounds reasonable (though idk why exactly $\mu$ exists), but the issue is that $\pi_1(M\setminus S)$ is not at all easy to understand. it's not straightforwardly clear how to realize any co-\mathbb{Z}$ subgroup of $\pi_1(M)$ by this process.
@Obliv it isn’t necessarily the case. For example take $(6,10,15)$. Note that $\bmod 5: gcd(6,10)\neq \gcd(6,15)$. Whereas $\gcd(6,10)=2$ and $\gcd(6,15)=3$
$[(S, \partial S)]$ gives a second homology class in $H_2(M, \partial M; \Bbb Z) = H^1(M; \Bbb Z) = H_1(M; \Bbb Z)$, pick $\mu$ to be a loop representing the element in the last group
The point is that $R$ has no zero divisors and $R\times R$ has many. An isomorphism must preserve products (and sends $0$ to $0$), and so an isomorphism — or its inverse, depending on the direction you use — would produce zero divisors in $R$.
still, what if two regular values of $M\rightarrow S^1$ don't lie in the same component of the regular values in $S^1$? how do you get a cobordism then?
@psie That's basically the generalization of the fundamental theorem of calculus you find in most graduate real analysis books, I think. I no longer have any of the books, but check Folland or papa Rudin.
What's a good reference for the classification of finite-dimensional algebraic representations of $\mathrm{GL}(n,k)$ when $k$ is an arbitrary field of characteristic zero? For $k = \mathbb{C}$, Fulton does it using Young diagrams in Chapter 8 of his book Young Tableaux. Other treatments for $...
