Problem: If $M$ is a module over a PID $R$, then $M$ is an injective $R$-module. Proof: Let $I$ be an ideal and $f : I \to M$ a homomorphism. Because $R$ is a PID, $I = (r_0)$ for some $r_0 \in R$. Define $\varphi : R \to M$ by $\varphi (r) = r f(r_0)$. Then $\varphi (\iota (rr_0)) = \varphi (rr_0) = rf(r_0) = f(rr_0)$, so $\varphi \iota = f$ on $I = (r_0)$.
Something about this seems off...I don't seem to use the fact that $M$ is divisible anywhere...How do I solve this problem?
if $P \in \Bbb Z[X]\subset\Bbb Q[X]$ can be factored $P = FG$ with $F,G\in \Bbb Q[X]$. Then you can write $nP$ as the product of two $F', G' \in \Bbb Z[X]$
From then on using the gauss lemma you can eventually prove that $n=1$ works
By gauss lemma, if $p = fg$ over Q then $p=(af)(bg)$ for some a,b\in Q where $af,bg\in Z[x]$
If f and g has common irreducible factor in Q[x] then f = pf', g = pg' over Q. By gauss lemma, f = (a_1 p)(b_1 f') and g = (a_2 p)(b_2 g') over Z for some $a_i,b_i\in Q$.
It seems that we can get common irreducible factor in Z[x] related to p. But I don't know how
I want to learn more about how algebras of functions on a space relate to the geometry of space. What field would this lie under? Operator theory? Algebraic geometry
@love_sodam basically if p|n, you get that p|nP, thus p divides every coefficient of P. if you look at the highest coefficient of F' and G' that are not multiples of p (assuming they exist), you see that they produce a coefficient not divisible by p in nP, which is a contradiction
Thus either p|F' or p|G' and you can divide and start over with another prime divisor of n
Another way to see that is to notice that the projection $\phi:\Bbb Z[X] \to \Bbb Z_p[X]$ sends $nP$ to $0$, So you have $\phi(F')\phi(G') = 0$, which implies $\phi(F')=0$ or $\phi(F')=0$ because $\Bbb Z_p$ is an integral domain
$d\gamma$ is analytic too and if there are infinitely many zeros on $[0,1]$, they accumulate, but then identity theorem implies it's zero, so the path is constant or sth?
If P has integer coefficients, and is the product of two polynomials with rationnal coefficient, then a multiple of P (so nP with n an integer) is the product of polynomials with integer coefficients
@Astyx Yes. If f and g has common irreducible factor in Q[x] then f = pf', g = pg' over Q. By gauss lemma, f = (a_1 p)(b_1 f') and g = (a_2 p)(b_2 g') over Z for some $a_i,b_i\in Q$.
@Tobias it just dawned on me (remember my question with $V^\infty = \bigcup V^K$ with the $K$ compact open subgroups of a group $G$ and $(\pi, V)$ a representation of $G$?) I wanted to show that $V^\infty$ was $G$-stable and your advice was that $V^\infty$ should be $G$-stable and not invariant. I can pick $v^\infty \in V^K$ for some compact open $K$.
Shouldn't $K(g) := gKg^{-1}$ work? This is compact open because conjugation is a homeomorphism and for every $g \in G$, $\pi(g)v^\infty \in V^{K(g)}$..
Oh, Now it makes sense to me. Is this generally true? I mean, if f,g has common irreducible factor in K[x] then f,g has common irreducible factor in R[x] where K is a quotient field of R
Random thought: Some person researching a new kind of property in math should name said property such that it looks like a misspelling of the name of another well-known property.
If $M$ is an $R$-module, where $R$ is a unital ring, and $f : R \to M$ is an $R$-module homomorphism, then $f$ is completely determined by $f(1_R)$, right?
I don't understand. $d\gamma$ is a map $T([0,1])\to TM$, right? I see that $d\gamma_p$ can have zeroes, but $d\gamma$ itself? Sure sometimes it can be the zero map, but then I can't just blindly apply the identity theorem or am I missing something?
@Alessandro The point is $d\gamma$ will not be transverse to $0_M \subset M$ but wherever they intersect they intersect with finite multiplicity (by analyticity)
So the intersection locus will still be a $0$-manifold
Then you have a curve in $\Bbb R^n \times \Bbb R^n$ which basically wiggles up and down, hitting $\Bbb R^n \times 0 \subset \Bbb R^n \times \Bbb R^n$ at various times $a_1, a_2, \cdots$ as it goes, finally hitting it at $a$
There is also the curve $\gamma$ lying entirely inside $\Bbb R^n \times 0 \subset \Bbb R^n \times \Bbb R$, which completely agrees with $\gamma'$ at the points $a_1, a_2, \cdots, a$
Remember that $\gamma' = (\gamma, \gamma')$ as a curve in $TM$. It's an abuse of notation to say $\gamma'(b) = 0$, one really means $\gamma'(b) = (\gamma(b), 0)$
$TM$ keeps track of the base and the fiber, two components
Now $\gamma', \gamma$ are both analytic curves in $\Bbb R^n \times \Bbb R^n$, agreeing on a set of times $\{a_1, a_2, \cdots, a\}$ with an accumulation point. So they must agree. This means "$\gamma' = \gamma$", which means $\gamma'$ has zero second component
Problem: Show that if $f(z)$ is a nonconstant analytic function on a domain $D$, then $f$ is an open mapping. "Proof": Let $U \subseteq D$ is open, and take $w_0 f(U)$. Then there exists $z_0 \in U$ such that $f(z_0) = w_0$. If $f'(z_0) \neq 0$, then I can invoke the inverse function theorem for analytic functions to show that $w_0$ has an open nhbd contained in $f(U)$, so $f(U)$ is open....For the case of $f'(z_0) =0$, since $f'$ is analytic, it has isolated zeros, meaning I can find $r > 0$ such that $B(z_0,r) \subseteq U$ contains no zeros of $f'$ other than $z_0$. Is it possible to find…
@user193319 I am sceptical that this works. You're only using properties which also hold for real-analytic functions (inverse function theorem and having isolated zeroes), but real-analytic functions don't necessarily map open sets to open sets, e.g. consider $\mathrm{arctan}:\Bbb R \to \Bbb R$ which has the non-open range $[-\pi/2,\pi/2]$
Hello, I have a question about something in gradient descent, can you please help me? This is the link to the question: math.stackexchange.com/q/… And thank you in advance
@user193319 I'm still sceptical. You have to use some result specific to holomorphic functions which fails for real-anaytic functions on open subsets of $\Bbb R$. The approach you suggested doesn't do that
@user193319 the easiest proof of the open mapping theorem is using the inverse function theorem and Taylor's theorem to show that for every point one can always find a holomorphic chart in which the map looks like $z \mapsto z^k$ for some $k$, which is an open map by an elementary calculation (The last step fails for $\Bbb R$: for example the image of $(-1,1)$ under $x \mapsto x^2$ is $[0,1)$ which is not open)
given a small category $\mathcal{C}$, it's a fully faithful functor $\mathcal{C}\rightarrow[\mathcal{C}^{op},\mathbf{Set}]$ sending $C$ to $\operatorname{Hom}(C,-)$
now, here's an observation: $[\mathcal{C}^{op},\mathbf{Set}]$ is a cocomplete category ($\mathbf{Set}$ is cocomplete and limits of functors are pointwise)
@LukasHeger I have the following (which the book also suggest I use to handle the case $f"(z_0) = 0$): If the analytic function $f(z)$ has a zero of order $N$ at $z_0$, then $f(z) = g(z)^{N}$ for some function $g(z)$ analytic near $z_0$ and satisfying $g'(z_0) \neq 0$.
say $\mathcal{D}$ is a cocomplete category and $F\colon\mathcal{C}\rightarrow\mathcal{D}$ is a functor, you can extend this through the Yoneda embedding to a cocontinuous functor $[\mathcal{C}^{op},\mathbf{Set}]\rightarrow\mathcal{D}$
this cocontinuous functor is not unique, but unique up to isomorphism (since colimits are only unique up to iso)
But even when $f'(z_0) = 0$, I don't necessarily have that $f(z_0) = 0$, so $z_0$ is not necessarily a zero of $f$...so I don't see how to apply the above when $f'(z_0) = 0$.
@LukasHeger Okay, without loss of generality $f(z_0) = 0$. Then apply the theorem to write $f(z) = g(z)^N$ with $g'(z_0) \neq 0$, and then apply the inverse function theorem to $g(z)$, and use the fact that $g(z)$ and $z \mapsto z^k$ are open to conclude that $f$ is open?
in general, if you have two functors $F\colon\mathcal{A}\rightarrow\mathcal{B}$, $G\colon\mathcal{B}\rightarrow\mathcal{A}$ and an isomorphism $\operatorname{Hom}(GB,A)\cong\operatorname{Hom}(B,FA)$, which is natural in both variables, this is an adjunction
So, a 2-category (strict 2-category to be precise, but they're nicer to talk about) is a category enriched over the category of small categories. This means that each $\operatorname{Hom}$-set is itself given the structure of a category in such a way that composition is functorial. The classical example is the 2-category of small categories, where each Hom-set is of course given the structure of a category as the functor category.
Now, there's one natural way to define what an adjunction of 2-categories should be: it should be the same thing as what I said above, except these isomorphisms of sets should now be isomorphisms of categories
E.g. in the example with the Yoneda embedding, a functor C->D correspond to an isomorphism class of cocontinuous functors [C^op,Set]->D
This says the hom-categories are equivalent
And even when just doing 1-categories, you may perhaps notice that equivalence is sometimes a better notion of sameness for categories than isomorphism
Equivalence exhibits a pattern that is very pervasive through lots of stuff in 2-category theory, namely that the classical notion of a commutative diagram in a 1-category theoretical setting (think of two functors being inverse as example), we instead just require that diagram to be commutative up to a 2-morphisms (a natural transformation in the case of functors, thus giving equivalence)
stuff like lax functors are relaxations of functors in the same way
instead of requiring your functor preserves composition and identities, you require it preserves them up to a choice of 2-morphism (if you require those to be invertible, you get a pseudo-functor, I think) and that these collections of 2-morphisms themselves adhere to a couple necessary coherence conditions
another relevant notion is that of a pseudonatural transformation between 2-categories, which is the same as a natural transformation, except you require that all the squares only commute only up to a choice of invertible 2-morphism satisfying some coherence conditions
the reason I mention this is cause the equivalence of categories for "adjunction" given by the Yoneda embedding turns out to be pseudo-natural in both variables
(except those work even more generally between non-strict 2-categories and non-strict 2-functors, but that gets very messy)
so, the point I'm trying to make is that the Yoneda embedding is a natural thing and it embeds a small category into a cocomplete category
and this embedding has a universal property, except it turns out that this property is not about existence of unique morphisms making a diagram commute, but only about the existence of unique up to 2-morphism morphisms making a diagram commute
and when we try to express that in the language of adjunctions, we naturally arrive at the 2-categorical weakening, the biadjunction
it's called the "free cocompletion", because it's an embedding into a cocomplete category that's universal in this precise sense
@user193319 just for reference, here's the lemma that I would use to prove the open mapping theorem (which is also intersting in it's own right). Suppose that $f:X \to Y$ is a holomorphic map between Riemann surfaces. Then for all points $x \in X$, there exist charts of $X$ and $Y$ in which $f$ looks like $z \mapsto z^k$ for a unique $k \in \Bbb N_0$. Proof: Using Taylor's theorem, we can expand $f$ in a chart around $0$ as a power series $f=\sum a_n z^n$. Now let $k$ be the smallest $n$ such that $a_n \neq 0$. Then $f=a_k \cdot z^k (1+\sum_{n \geq 1} b_n z^n)$ with $a_k\neq 1$. Rescale the…
i am writing $F = \int dF$, then observing by symmetry that while doing the integral (which is integral of a vector valued function so integrate componentwise), the x and y coordinates vanish by some "symmetry argument" (think of integrating normal vector of a hemispherical bowl)
so I am writing $F = \int dF = \int dF \cdot e_3$
LOL
i feel like the grader deserves this for being a physicist
