@TedShifrin so here is my argument to show that for $L=\mathbb{Q}(\sqrt{2+\sqrt{2}})$ and $K=\mathbb{Q}$, the galois groups $Gal(L/K)$ and $Gal(L(i),K(i))$ are isomorphic. Note that if $\sigma\in Gal(L/K)$ then as $X^2+1$ is monic and irreducible with root $i$, we have by the universal property above that there exists a unique field isomorphism $\tilde{\sigma}:L(i)\rightarrow \mathbb{Q}(i)$ extending $\sigma$, and in turn fixing $\mathbb{Q}(i)$
Hence this gives us a map $\varphi: Gal(L/K)\rightarrow Gal(L(i),K(i))$ which is an isomorphism
Find a rational $\alpha\in\Bbb Q$ such that $\sqrt\alpha\notin\Bbb Q$, and such that the degree of $\sqrt[{\Large4}]\alpha$ over the rationals is $2$ (rather than the expected $4$)
@leslietownes ya it feels like all the irrational inputs for cos and sin are missing so it should be neither closed or open like the rational set but I am not sure
@copper.hat Haan I was trying to enclose in a curly bracket but I don't get how to make latex stop ignoring my curly brackets
@TedShifrin For this, $R$ should be commutative with unit element also, I think.
The question here https://math.stackexchange.com/questions/4384230/if-a-linear-map-sends-orthonormal-basis-on-orthonotmal-basis-then-it-is-an-isome made me think: Does isometry $T\in L(V)$ with inner product g remain an isometry if one changes inner product (and therefore the concept of norm for V) to g'?
@mohan10216 Sorry, the meaning of the word "degree" there is different in context. I mean the degree of the minimal polynomial of $\sqrt[4]\alpha$ (the smallest degree of a nonzero polynomial that has $\sqrt[4]\alpha$ as a root)
So the degree of, for example, $\sqrt2$ over the rationals is $2$, because the smallest polynomial (with rational coefficients) with that as a root is $x^2-2$
The degree of $\sqrt{1+\sqrt2}$ is $4$
The degree of $\sqrt2+\sqrt3$ is also $4$, but it's not obvious
(hint: write out $\{1,\sqrt2+\sqrt3,(\sqrt2+\sqrt3)^2,(\sqrt2+\sqrt3)^3,(\sqrt2+\sqrt3)^4\}$ and try to find a relation between them. The relevant tool here is linear algebra. In fact, there's another equivalent definition of degree (in this sense) in terms of the dimension of a certain vector space.)
koro: i guess i should say not necessarily. if g' is a scaled version of g then they will have the same isometries. and in finite dimensions (have not thought about the inf dim case) inner products with the same set of isometries have to be scalar multiples of one another.
koro: note that the problem you link to is about a normed space that is not an inner product space. different normed spaces can have some isometries in common. the identity is an isometry of any normed space but there can be others that distinct spaces share. e.g. permuting coordinates is an isometry in any of the ell^p norms although these normed spaces are not isometric to one another for different p.
william: hard to say without more context. the fact that the thing being plugged in changes suggests that the input of this rule is a function (of two variables a and b) and that the output is another function (again of the two variables a and b). where here x and y might be regarded as fixed for purposes of defining the rule.
If I have a symmetric matrix nxn with all positive entries i.e A_ij > 0 I know that the eigenvalues must be real, but was wondering is it always possible to obtain the sum of the eigenvalues as a trace of the matrix (tr(A))?
I know that the eigenvalues must be real and greater than zero from the characteristic equation Im just wondering if its always possible to find the sum of the eigenvalues as a trace of this type of matrix.
or wait its not guaranteed that the eigenvalues will be greater than zero. But the spectral radius should be greater than zero as it is the greatest absolute value of the eigenvalue
Consider the field extension $\mathbb{Q}(\zeta_{16})/\mathbb{Q}(i)$ . By Kummer theory we can show that there exists some $a\in \mathbb{Q}(i)$ such that $L=\mathbb{Q}(i)(\sqrt[4]{a})$
Now, I am trying to use the Lagrange resolvent to find $a$.
Note the following: If $L/K$ is a cyclic extension of...
The Q: $$
Show that } \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x \text { if }-1 \leq x \leq-\frac{1}{\sqrt{2}}
$$
I tried solving this Q a lot but I’m unable to. My answer comes different.
$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\cos \theta$
$$
\begin{array}{l}...
If interior of a set is empty, can something be said about isolated points in the set?
Is it very obvious?
Background: If S is a set without any isolated points in a complete metric space X, then S is uncountable. An attempt: Suppose on the contrary that S is countable then $S=\cup_{x\in S}\{x\}$. Since {x} is nowhere dense in X, by Baire's Category theorem $S$ should have an empty interior.
If S is a subset of X, then I define s in S as an interior point (of S) if there exists a ball with positive radius centred at s such that the ball lies completely in S.
The collection of all such s in S is defined as interior of S.
A point t in S but not an interior point of S is isolated point of S.
well, you could take irrational numbers say, take some homeomorphism to the Baire space, extract a dense countable subset, then go back to the homeomorphism, you get a dense countable subset of irrationals, then it's also dense in R
ig
then you'd get some kind of dense countable subset in terms of continued fractions
Like, you could take those irrational numbers whose continued fractions are eventually constant, those form a countable set and are dense in $\mathbb{R}$.
say $(x,y,\theta)$ corresponds to infinite copies of the $x-y$ plane where real number (angle) $\theta$ indexes the planes and $0 \ge \theta \lt 2pi.$ For a given function $f$ on $(x,y,\theta)$ for every $\theta,$ the union of all such $f$ gives a surface of revolution. If the domain of all the $f$ can be maximally extended to a region of $\Bbb C,$ then can the union of all the maximally extended $f$ give a complex surface?
geo: you probably don't mean a function "on (x,y,theta)" there (i.e. a function defined on the entire plane). you might mean, you have the graph of a function of one variable function in that plane, and you regard that as the slice of a surface of revolution corresponding to the angle theta. (i.e. if the curve in the theta plane is t -> (t,f(t),theta) then you look at (t, f(t) cos(theta), f(t) sin(theta))? without something like that there's no obvious way of getting a surface from that data
you'll also need to assume a whole lot of stuff about f to extend it anywhere. if this is a generalization of something specific, it might help to just ask the specific thing
I don't know names for lots of these results. I just need to know the theorem :P Koro was saying Cauchy proved that the Cesaro averages converge to the limit of a convergent sequence; I had never heard a name on that one, either.
@TedShifrin Yes. A local friend has a court appearance via Zoom and I got him set up last week and wanted to make some suggestions about lighting before his appearance.
@TedShifrin He is in his mother's room and all there was on the wall was her painted plate collection, and no, there weren't any risque plates (she's in her 90s)
The attempted solution (which I am not sure can be completed or not): Suppose on the contrary that X is countable. Noting that {x} is nowhere dense in X so $\cup_{x\in X}\{x\}=X$ has empty interior. Question: Can we get a contradiction from here?
@Jakobian Suppose on the contrary that X is countable. So X can be indexed as $x_1,...x_n,...$. Noting that $X_i=X\setminus \{x_i\}$ is dense in X. We must have $\cap X_i$ dense in X. But clearly the intersection is empty, which contradicts Baire Category theorem. :)
You need to carefully read the definition of perfect set you are using. The definition in Wikipedia is "a subset of a topological space is perfect if it is closed and has no isolated points", which I think is standard. The empty set is required to be closed in all topological spaces and has no ...
yes. All points of S are limit points implies S has no isolated points (but not that S is perfect).
$f(x)=\begin{cases}a_n+\sin \pi x; x\in [2n,2n+1], n\in \mathbb Z\\ b_n+\cos \pi x; x\in (2n-1, 2n), n\in \mathbb Z\end{cases}$ is given to be continuous on R and it is asked to find $a_n, b_n$ Does this question make sense?
My confusion: what is the meaning of $a_n$ for negative $n$?
It is seen by continuity at 2n, where n is any positive integer that: $a_n=b_n+1$ and by continuity at 2n+1, we get: $a_n=b_{n+1}-1$. Solving the recursion we get: $a_{n+1}=2n+a_1, b_{n+1}=1+2(n-1)+a_1$
Let $\Phi$ be a root system for a reflection group $W$, let $\Pi \subset \Phi$ be a positive system, and let $\Delta$ be a simple system in $\Pi$. I know that if $\alpha \in \Delta$, then $s_{\alpha}(\Pi \setminus \{\alpha \}) = \Pi \setminus \{\alpha \}$. My question is, is it also true that $s_{\alpha}(\Delta \setminus \{\alpha\}) = \Delta \setminus \{\alpha \}$?
I feel like it should be true; maybe it has to do with the fact that positive systems contain unique simple systems...
can i get a prize for not solving collatz? because i did that already
i actually mentioned the collatz conjecture in my thesis. i was near finished and my advisor was pushing me to include more examples of something. he suggested a setting for some interesting examples. i said no. he said why not. i said, because you can encode the collatz conjecture in that setting. he said, maybe a good time to stop.
i'm not the guy to ask, but my vague impression is that a lot of that stuff requires the 'metric' to give vectors negative or zero values, so my guess would be no
@robjohn I was talking about perfect spaces more than perfect sets
@Koro of course it makes sense. A sequence is just a function $a:\mathbb{N}\to \mathbb{R}$, here all you're doing is replacing your domain with $\mathbb{Z}$.
@copper.hat No, you're not... the dates on things are no longer relevant. You got one recently. It is the date of the question that is given in the table. That is stupid.
this is the new and improved, more responsive interface (that sucks)
@copper.hat I mean, sometimes it is, and people complain about their cheese being moved, but, in this case, I think that there are a lot of legitimate gripes. SE seems to be very narrowly focused on mobile, and appears to be abandoning the desktop site. :/
You need to carefully read the definition of perfect set you are using. The definition in Wikipedia is "a subset of a topological space is perfect if it is closed and has no isolated points", which I think is standard. The empty set is required to be closed in all topological spaces and has no ...
