Sanity check: if I wanted to show that any field homomorphism is injective, I just have to show that it has trivial kernel as an additive group hom, yes?
One of my favorite elementary linear algebra proofs (that usually determined who got an A in the course) was to show that if $A$ is an $m\times n$ matrix of rank $n$, then if $\{v_1,\dots,v_k\}$ is linearly independent, so is $\{Av_1,\dots,Av_k\}$. Easy, I know, but usually predictive.
@Fargle You need the statement of the universal property of tensor product of modules. And you need to know when that universal property gives you a ring map (when the modules are actually algebras, but... I'm getting too fancy for you I think...)
I like this question but maybe I should slink back into my hole.
Here is the universal property. Let $A$ and $B$ be $k$-algebras. The "tensor product" $A \otimes_k B$ has the following universal property. If $C$ is a $k$-algebra, any ring map of $A \times B \to C$ which is bilinear over $k$ factors uniquely through $A \otimes_k B$.
@MikeMiller Okay. Let $K$ be a finite field extension of $k$---i.e., in particular, an algebra over $k$. Select a basis for $K$ over $k$. Define the map $K \times K \to k$ by $(x,y) \mapsto \pi_1(xy)$---projection onto the first coordinate. This is clearly a ring map, and clearly bilinear over $k$, so it factors uniquely through $K \otimes_k K$. But then we would have non-trivial elements of the tensor product which map to the zero element, so that in particular, the factored map is not injective
Seems like it should be. I (think that I) have constructed a ring map out of that tensor product which fails to be injective, and hence it isn't a field
Check my details if you like, I haven't checked most of them as yet
hence my uncertainty
The "non-trivial elements of the tensor product which map to zero" would be of the form $(0,x_2,\dots,x_n) \otimes (y_1,y_2,\dots,y_n)$ (where these are the coordinates w.r.t. the chosen basis of $K$ over $k$)
Wait no hang on I think I need to stop flying so dang fast and loose.
There will be things that map to zero but they won't look like that necessarily
I'm being dumb and taking the product in $K$ to be component-wise over its chosen basis, which is as they say "clearly wrong"
I feel like whatever the solution is, though, it's going to be in that vein---construct a ring map from the product, have it factor uniquely, and then show that the factored map fails to be injective
I'll need to hop off of here in order to focus on finding the right map though.
Please consider the following problem and my answer to it. The answer seems to low to me. Is my answer right?
Thanks,
Bob
Problem:
The median net worth of a certain population is $97K$. To be in the top $5\%$ you need to have a net
worth of $2387K$. Assuming the population follows the Pareto dis...
I believe the definition is An set $O \subseteq \mathbb{R}$ is open given that for all $x \in O$ there exist $\epsilon > 0 s.t (x - \epsilon, x+\epsilon) \subseteq O)$
basically epsilon neighborhood of any arbitrary $x$ in a set must be a subset of the same set
This is the real analysis definition, not to familiar with the topological definition
How do I show that $z=0$ is not a local maximum of $|p(z)|$ where $p(z)=a_0+a_1z+\cdots+a_nz^n$ if $a_i \neq 0$ for some $i>0$. We want to find $z$ in some $\epsilon$ neighbourhood such that $|p(z)| > |a_0|$. But how to construct one?
Okay, so now we have new information! Let me call this maximal element $x$.
We know that $x \in O$.
What does the definition of open set tell us now? I remember it being a condition about "for every element of $O$", and it seems we now have an element of $O$.
Ah I see! we can reach a contradiction by seeing that the epsilon neighborhood for the maximal element $x$ is $(x-\epsilon, x+\epsilon)$ where $\epsilon > 0$ so $\epsilon, x+\epsilon) \subseteq O$ however this is a contradiction as $x + \epsilon$ gives us something outside of $O$
So question, homology is a functor from the category of topological spaces to abelian groups, but homology also preserves homotopic maps, so can we view homology as a functor from the homotopy category to the category of abelian groups?
@AkivaWeinberger lol I did not even knew it existed. As far I knew, the only thing I get is a blank screen with the youtube logo, which returns to normal after a refresh
I understand that bilinear map only problem is why $\beta_s$ ($1 \otimes x_i$, $1 \otimes x_j$, ) is invertible? Also how to do this
argumment for projective case?
Problem: Let $B = \{(1,2),(2,3)\}$ and $C = \{(-1,1),(2,1)\}$ be bases for $\Bbb{R}^2$. If $[v]_{C} = (1,2)$ what is $[v]_{B}$?...The problem doesn't specify what $v$, so I'm assuming it is some abstract vector. To find $[v]_B$, do I just compute $[I_{\Bbb{R}^2}]_{C}^{B}[v]_{C}$, where $I_{\Bbb{R}^2}(x) = I_2x = x$?
I think this is right, because $[I_{\Bbb{R}^2}]_{C}^{B}$ is the change of coordinate matrix...but I'm not entirely certain...
$[v]_{B}$ is the coordinate vector of $v$ under the basis $B$. Since the vector space is only $\Bbb{R}^2$ you can let v = (x,y) and then compute that by matrix multiplication. Alternately, you can derive the change of coordinate matrix by trying to express the vectors in $C$ as linear combination of $B$, then the $[I_{\Bbb{R}^2}]_{C}^{B}$ matrix should only act on the components of said vectors (as under the basis C, all column vectors should behave like e.g. (1,0, and (0,1), thus does not contribute)
Let's say I have a great math book, and I am not a math student so I have all of the time in the world to study in as much depth as I wish. Do I take it slow and solve every problem before getting into another, or rush, and skip whatever I can't solve in a set amout of time, say a day, or a week?
(The book isn't all encompassing, and I'd really like to get into more refined parts of the domain)
I am asked to prove that the cylinder and the punctured plane are homeomorphic.
I understand that I need to find a function that maps every point in the plane to a point on the cylinder.
I can represent every point on the plane in polar coordinates and now I just need to find a mapping from e...
@LeylaAlkan Yes. Alternately, you can also deduce they are clopen by noting that each of those open intervals are open in the subspace topology, and then the relative complement of these wrt X are thus closed, hence concluding they are all clopen
@Secret Yeah I tried that but I got an error such as "MovieWriter ffmpeg unavailable" even though I did install ffmpeg ... I'm looking for someone who has done it in the past because all documentation I found was not very helpfull unfortunately
I recall there is something relate to adding the path for ffmpeg to your main path environment variable. That's how one tells python to refer to any packages or something
`Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/Users/Eric/anaconda3/lib/python3.6/site-packages/matplotlib/animation.py", line 1200, in save
writer.grab_frame(**savefig_kwargs)
File "/Users/Eric/anaconda3/lib/python3.6/contextlib.py", line 99, in __exit__
self.gen.throw(type, value, traceback)
File "/Users/Eric/anaconda3/lib/python3.6/site-packages/matplotlib/animation.py", line 241, in saving
self.finish()
File "/Users/Eric/anaconda3/lib/python3.6/site-packages/matplotlib/animation.py", line 367, in finish
Is it true that: "$S_n$ contains all groups of order $\le n$"? As, $S_n$ contains the group of order $n$ by Cayley theorem and as $S_n$ contains all elements of $S_{n-1}, S_{n-2},S_{n-3},\cdots, S_2$, I think the answer is true. Can anyone tell me I am correct or wrong?
@KenOno your reasoning is correct, though a more careful (possibly pedantic) formulation would be "each group of order $\leq n$ is isomorphic to a subgroup of $S_n$"
@MatheinBoulomenos Have you seen the following problem before? I posed it in here yesterday; it was an old qual problem. Leaky and I have different solutions. I like both.
Let $K/k$ be a finite nontrivial field extension. Then $K \otimes_k K$ is not an integral domain.
@MikeMiller wlog it suffices to treat the separable and the purely inseparable case. (since everthing over a field is flat) in the separable case by the primitive element theorem, we have $K=k[x]/(f)$, so that we get $K\otimes_k K \cong K[x]/(f)$ which is not an integral domain as $f$ has a root in $K$ For the purely inseparable case, we have that there is some $a \in K\setminus k$ such that $a^p \in k$, where $p$ is the characteristic, then we get that $a \otimes 1-1\otimes a$ is nilpotent in $K\otimes_k K$
I was given an exercise with a hint, which I solved, but for the life of me I cannot see how to use the hint. Do others have these kinds of mathematical blindspots?
Sometimes I talk to people about how to prove something and I just have no idea what they're trying to say to me, even though they seem like they know what they're talking about
And I just go home and prove it myself some other way
I remember at the start of last year I was given the task of finding the Galois group of $P(X) = X^5 - 7$ and was given the hint "If $\alpha_1, \alpha_2$ are roots, what is the minimal polynomial of $\alpha_1^2\alpha_2^3$"
Consider the action of $\mathbb{R}$ on the unit circle $S^1=\{z\in \mathbb{C}~|~~|z|=1\}$ by $t\cdot z=e^{2\pi it}\cdot z$. I want to find the $Stab(1)$, $Stab(1)=\{t~|~~e^{2\pi i t}\cdot 1=1\}=\{t~|~~(e^{2\pi i})^t=1^t=1\}=\mathbb{R}$. I am actually not sure about the last equality. Am I correct that $1^{irrational}=1$?
I would have said that's true by Euler's equation $e^{2\pi i t} = \cos(2 \pi t) + i \sin(2 \pi t)$, and you want the $\sin$ term to vanish and the $\cos$ term to be one.
@KenOno Okay, I think this is maybe not productive. I will try something else. The problem is that everything you want to try involves abstract generalities, and you eventually need to use something about the specific group.
@dalbouvet My question is: $\pi(x)$ is the prime counting function. This function is very important in number theory.
$\Phi(x)=Ad \int_2^xc^{Ab}dx,$ where $A=1/\ln(x),$ where $d,c,b$ are constants. Is there any combination of these constants that will give an asymptotic bound on $\pi(x)?$ I've tried using $c=5/4,$ and $d,b=1,$ but I'm curious if there is some optimized combination.
Quick question, will $\int_C z^{-1+i}dz$ where $C:z(t)=e^{it}$ for $t\in [0,2\pi]$ equal 0? Because that's the result I'm getting, and I have a sneaking suspicion that's wrong.
@Astyx Let $ x,y \in \mathbb{R} $, $ x<y $. If, there exists a $ r \in \mathbb{Q} $ such that $ x<r<y $, we say $ \mathbb{Q} $ is dense in $ \mathbb{R} $.
The relevant point is that $\Bbb Q$ is "divisible". Given any element $a \in \Bbb Q$, and any integer $n > 0$, there is an element $b \in \Bbb Q$ so that $nb = a$. In fact, $b$ is unique.
Either use that to show $\Bbb Q$ has no finite index subgroups or use that to show that $\Bbb Q$ acts trivially on every finite set.
So your function is $e^{(-1+i) \ln z}$ on the domain of the branch of $\ln$.
Now, observe that to define that integral, you really need to take the limit of the integrals along $[\epsilon, 2\pi - \epsilon]$, because your function is undefined at the endpoints.
I understand that bilinear map only problem is why $\beta_s$ ($1 \otimes x_i$, $1 \otimes x_j$, ) is invertible? Also how to do this
argumment for projective case?
I am asked to prove that the cylinder and the punctured plane are homeomorphic.
I understand that I need to find a function that maps every point in the plane to a point on the cylinder.
I can represent every point on the plane in polar coordinates and now I just need to find a mapping from e...
