Recent content by TopCat

I don't see which contour would be best to use in this situation.

That's why I multiplied by 1/2. But now that I think about it, that wouldn't work since the function isn't symmetric about the y-axis, right?

Sure. My contour is the upper half of a circle of radius r, the segment from -r to -a-e, the upper half of a circle of radius e around -a, and the segment from -a+e to r. I took the limits r -> infinity and e -> 0. The integral over the arc goes to zero, so I end up with the integral over the...

Homework Statement I need to compute \int_0^\infty \frac{dx}{x^3+a^3}.Homework Equations If f = g/h, then Res(f, a) = \frac{g(a)}{h'(a)}. The Attempt at a Solution In the first I've used a semicircular contour in the upper plane that is semi-circular around the pole at -a. So I calculate the...

Since G is finite abelian, G \cong \sum \textbf{Z}_{p_{i}^{n_i}}. If |x| has maximal order, then isn't A just removing one of the direct summands, so that, say, A \cong \sum <e> \oplus \textbf{Z}_{p_{i}^{n_i}}, with i \geq 2, which is a subgroup?

Homework Statement If G is a finite abelian group, and x is an element of maximal order, then <x> is a direct summand of G. Homework Equations The Attempt at a Solution I claim that the hypothesis implies that A = G\<x> \bigcup {e} is a subgroup of G. If so, then since G = < <x> \bigcup A>...

Thanks for the help, morphism. So then I have [K(u,v):K] = [K(u,v):K(v)][K(v):K] = [K(u,v):K(u)][K(u):K] = n[K(u):K]. Since v trans|K, [K(v):K] is inf, and by equality [K(u):K] must be inf. But this means that u is trans|K, so there is an isomorphism from K(u) → K(v) mapping u → v. But this...

Homework Statement Let F|K be a field extension. If v e F is algebraic over K(u) for some u e F and v is transcendental over K, then u is algebraic over K(v).Homework Equations v transcendental over K implies K(v) iso to K(x). Know also that there exists f e K(u)[x] with f(v) = 0.The Attempt at...

Hungerford says that But if we take K = ℝ and K(x_{1}) = ℝ(i) = ℂ, we have that i is not in ℝ yet is algebraic over ℝ. Guess I'm missing something here. Is it that this need not be true for simple extensions if the primitive element is algebraic over the field?

To see that the red statement is true, observe that \angleADO = \angleCDB > \angleBOD by Euclid I.32 (book I proposition 32). In fact, that statement is superfluous. Once you know that \angleADO > \angleBDO = \angleADC = \angleAOD + \angleOAD, you know that \angleADO > \angleOAD, and by...

I strongly suggest you try and use the forums tags to typeset anything that looks that nasty, because it is really confusing to try and parse it. It'll be a worthwhile skill to have in the long run and you'll get more responses. LaTeX help What you need here is how to evaluate the limit of a...

Ah, thanks for the tip. I needed to prove the lemma m|h, n|h imply [m,n]|h. And my apologies. I'll be sure to use the template next time.

Let G be an abelian group containing elements a and b of order m and n, respectively. Show that G contains an element of order [m,n] (the LCM of m and n). This is true when (m,n)=1, because mn(a+b) = e, and if |a+b|=h, then h|mn. Now, hm(a+b) →m|h and similarly I find that n|h. But (m,n)=1...

D'oh! Thanks, mathwonk.

So I just proved that Q (the rationals) is not free abelian since it must have an empty basis. But Q is isomorphic to Z^2 = ƩZ which is equivalent to Q having a nonempty basis. I must be wrong about the isomorphism but I don't see why.