Recent content by winter85

Good day, I just need someone to tell me if this is correct. If L and K are subfields of M, their composite LK is the smallest subfield of M that contains both L and K. is this correct \alpha \in LK if and only if there are positive integers n and m, polynomials f(x_1,x_2,...,x_n) \in...

Good day, I am reading Stewart's Galois Theory (3rd ed). I'm up to chapter 8 where he starts tackling the issue of solubility by radicals. The author considers independent complex variables t_1,t_2,...,t_n and forms the polynomial F(t) = (t-t_1)(t-t_2)...(t-t_n) which he calls the general...

I've thought of that, but any such interval maps to the whole codomain [-1,1], which is open in the relative topology, so not really sure how that helps?

Homework Statement This problem is from Schaum's Outline, chapter 7 #38 i believe. Let f: (0, inf) -> [-1,1] be given as f(x) = sin(1/x), where R is given the usual euclidean metric topology and (0,inf) and [-1,1] are given the relative subspace topology. Show that f is not an open map...

oh i see, Thanks HallsofIvy. I know now where my confusion comes from. The definition in the book I'm using (Halmos) says: the tensor product of two spaces U and V is the dual space of the space of all bilinear forms on UxV. the tensor product of vectors u in U and v in V is z in U(x)V defined...

Homework Statement Let P(n,m) be the space of all polynomials z with complex coefficients, in two variables s and t, such that either z = 0 or else the degree of z(s, t) is <= m - 1 for each fixed s and <= n - 1 for each fixed t. Prove that there exists an isomorphism between Pn (x)...

you're welcome. it helps to draw a diagram to get a feel for what's going on :)

if you let q = (b*c), the you have a*q = e.. what is q called in this case, with respect to a? what are its properties?

hi ForMyThunder, first let's go over some basic definitions and facts, 1) A collection of continuous functions \{ f_{a} : a \in A \} on a topological space X is said to separate points from closed sets in X iff for every closed set B \subset X , and every x \notin B , \exists a \in A...

Hi jeff1evesque, You can't assume the group is abelian when the question doesn't say it is abelian. Actually, the result is true even if the group is not abelian. The associative law is helpful. can you state in plain english what a*(b*c) = e means?

ur welcome :)

ok consider this... fixing y = y0, the function u(x,y0) is a real function of one real variable which is x. so you can treat it as a normal function R -> R. its derivitaive with respect to x is 0, so as x varies, u(x,y0) doesn't change. but holding y0 constant and varying x means walking in...

well, du/dx = 0 and du/dy = 0. now if you fix a value for y, say y = y0, consider the function g(x) = u(x,y0). dg/dx = 0, so g is a constant. similarly, fix a value for x, say x = x0, consider the function h(y) = u(x0,y). dh/dy = 0, so h is a constant. imagine you start at a point z0, first...

why did you assume u(x,y) = x? that's not true in general. you only know v(x,y) = 0.

ok so for z = x + iy , we have f(z) = u(x,y) + i v(x,y) . since f is real valued, v(x,y) = 0 for all z = x+ iy. substitute that in the CR equations, what do you get? then pick two points in the square, join them with a special line, and try to figure out from the CR equations you got how the...