Recent content by ircdan

very nice, thanks

"[every element of an extension of prime degree p has degree 1 or p]" i wasn't thinking of it that way, but yes i agree with your statement "in fact you omit the only non trivial part of your solution, checking that it cannot be 1" well i guess i made the mistake and assumed it to be trivial...

let b = (3^(1/5))^4 + 29*(3^(1/5))^2 + 17*3^(1/5) - 16. Q(b) is contained in Q(3^(1/5)) and [Q(3^(1/5):Q] = 5, so the degree of Q(b) over Q divides 5, so it's 1 or 5, but it can't be 1 so must be 5

when they say solvable in M(or N) they mean "has a solution in". So you are to show that for any y in N and a in D, there is x in N s.t. ax = y if and only if there is z in M s.t. az = y. So suppose N is a direct summand of M, so M = N + K (here "+" = direct sum) for some submodule K of M...

your priority should be on solving problems from past quals if the exams are available, don't neglect these! I think most schools try to keep their quals each year similar, so doing problems from old exams helps a ton. I would of course as already mentioned, ask at your school. everyone worries...

well suppose L(A) = L(C), so AB = CB, then ...some stuff... implies A = C, so L is 1-1, you can fill in the missing step your way, suppose L(A) = 0, so AB = 0, so ...some stuff... implies A = 0, so kerL = {0} so it's 1-1, again you can fill in the missing stepnote the missing step is the same...

yup that's right, think about what halls just said though and it answers your question, you know (g^-1xg)(g^-1x^-1g) = e. This tells you what?

for continuity at an irrational point x_0 you can use epsilon-delta so let e > 0, you need a d such that |x-x_0| < d => |f(x)-f(x_0)| < e If x is irrational it's trivial. Now for x rational, choose M s.t. 1/M < e. What can you say about all the numbers x = m/n which are at most some...

i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))some observations, assuming a_n > 0 always I take it if sum(a_n) converges, then they both converge by comparison since |a_n/(1 +...

i don't understand your argument for bounding a_n, this doesn't mean it's wrong though everything else makes sense to me i did this, sum(a_k/(1 + a_k)) converges, so lim a_k/(1 + a_k) = 0, so there is an N s.t. for all k > N, a_k/(1 + a_k) < 1/2, so a_k < 1/2(1 + a_k) = 1/2 + a_k/2, so a_k/2...

well then spivak is too easy

there are so many books, it's hard to say because it depends what lvl you are at i'm assuming you have access to a school library, so i'd recommend checking out rudin, bucks, and bartle and sherberts and maybe some others you see that look readable i do think rudin is worth owning though, but...

ahh typo, no wonder! recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e i'll be extra terse on purpose because this is a good problem, made me think a_k > 0 by hypothesis(this is used in the proof of course) claim...

I get the feeling you are trying to start one of those long posts over something silly like (0/0 = 1?). Best of luck in your efforts. And don't worry, I'm done here:)

that function is not injective and your question is not clear the function you just typed does show that the statement f(A n B) = f(A) n f(B) for all sets A, B, and any function f isn't necessarily true, and as you already know, it's true if f is injective