@TedShifrin I'd like to take some university level math courses; never done that before. I don't have any formal math education outside of high school.
@Bill Maxi nos contó un poco sobre espacios cocientes, conozco las construcciones en anillos y grupos, pero ver cosas "concretas" está bueno, como $V/S\simeq W$ si $W\oplus S=V$.
O que $V\simeq K^{(B)}$ y $V^\ast\simeq K^{B}$ si $B$ es base, en general. Esos detallecitos (nada extraños) no los tiene el libro de Sabia, Tesauri y Jeronimo. =P
@Bill OK, tomemos un compacto $K$ en $\Bbb R$. Entonces es separable (dem: tomar $1/n$-redes para $n=1,2,\ldots$). Sea $E$ tal subconjunto numerable y denso en $K$. Sea $f_i$ una familia acotada puntualmente y equicontinua.
Perdon, no.
Para cada $x_i$; $$f_j(x_i)\; j=1,2,3,\ldots$$ esta acotada.
La idea es tomar subsuc. convergentes ahora, pero de forma inteligente.
@Bill La notacion es medio engorrosa, pero aqui vamos.
$f_i(x_1)$ tiene una subsucesion convergente. Rudin la escribe $f_{1,k}$ para tal subsucesion de funciones, entonces $f_{1,k}(x_1)$ es convergente.
Prove that the coordinate ring for $V × W$ is isomorphic to $k[V ] ⊗_k k[W ]$ where $V$ is a Zariski closed set of $\mathbb{A}^n$ and $W$ is a Zariski closed set of $\mathbb{A}^m$.
Not sure how to do this some help would be nice.
This is your professor speaking. Obviously, I can't stop you from posting questions from the problem set here but I can not for the life of me understand what is the benefit to you to ask here instead of in office hours! Am I really so bad at explaining?
Also, this particular question is proved carefully in the textbook, in the very section that was assigned for reading. Your other questions (for example, about what the pullback map of coordinate rings looks like) show that you are not understanding the most basic concepts in the course. Why are you not asking the professor? The course is going to get a lot harder, and quickly! Would it be better to drop math 631 and take Math 593 instead?
How is it possible that when I compute the geometric series of a truncated power series p(x) and plug in the value x_1 I get the correct result, but If I begin by evaluating p(x) at x_1 I get a value of p(x) outside the radius of convergence for the geometric series and it fails. The expression for the geometric series of the truncated p(x) is not nice at all but it works.
So if you don't tell the geometric series that the value of the truncated power series will be outside its radius of convergence, you can fool the geometric series?
It was a quite nice answer about the connected component of a topological group. I commented that the OP seemed to be asking about the path-connected component instead, which would make the argument even simpler
The set of $n\times n$ matrices can be identified with the space $M_n(\mathbb{R})$. Let $G$ be a subgroup of $GL_n(\mathbb{R})$.
I would like to show that the set of matrices that can be joined to the identity $I$ forms a normal subgroup of G.
@cyberskull Sorry, not at all well-versed in algebraic geometry. I have no idea. The latter part seems to be an angry instructor who thinks this is a student from their class. I don't know how they could be certain, unless the asker gave their real name.
@Chris'ssis If it is anything like the others of that kind I have worked on, the diagonal, which sums to $\frac78\zeta(3)$ appears in a linear relation to the total sum.
Let $G$ be a group (finite if that makes this easier) where all elements have order dividing $2$. Can anyone think of a way to define an extra multiplication on $G$ which distributes over the original operation and such that there is a unit for each element (so the unit need not be the same for each element). And do this without using that $G$ is abelian along the way.
For example, we can take elements $e, g_1,g_2,\dots$, each time taking an element not in the subgroup generated by the previous ones, and require these to be orthogonal idempotents in the new multiplication
but I don't see a way to show that this is well-defined without using that $G$ is abelian