Recent content by kimberu

I'd say no, it can't, because p*p can't divide something that doesn't have at least p as a factor. But what about if p does divide x^2 - a?

Homework Statement Let p = a prime. Show {x}^{2} ≡ a (mod {p}^{2}[/tex]) has 0 solutions if {x}^{2} ≡ a (mod p) has 0 solutions, or 2 solutions if {x}^{2} ≡ a (mod p) has 2. The Attempt at a Solution OK, my mistake, I don't think this has anything to do with the phi function. But I don't...

using the formula on (1,2,3) in S5, I got that it has 10 conjugates, which is wrong -- which N and R should I use for this example if not 5 and 3 (or am I calculating wrong)?

Homework Statement The question is, "How many conjugates does (1,2,3,4) have in S7? Another similar one -- how many does (1,2,3) have in S5? The Attempt at a Solution I know that the conjugates are all the elements with the same cycle structure, so for (123) I found there are 20...

Homework Statement Show that a field of order p2 exists for every prime p. The Attempt at a Solution In an earlier problem I found that there were p2 monic quadratics in Zp[x], but I don't know if that's useful. Any ideas or theorems would be super helpful, thanks!

Homework Statement using the following definition: ln x = \int dx/x, give best possible upper and lower estimates of ln n, n a positive integer. then, use this result to show that the limit of the function as n approaches infinity is infinity. Homework Equations -- The Attempt...

Thank you so much for walking me through! :)

So from here...can I say that s must equal 0 and x = aq for all x, since otherwise it's a contradiction because a is the smallest element?

Would it be (m+n)R?

You mean, say that n = qd + r, or for m? Sorry, I'm totally lost on this problem.

Homework Statement If R = Zn, show that every ideal of R has the form mR for some integer m. Homework Equations -- The Attempt at a Solution Well, by a previous problem I showed mR is the principal ideal of the ring, but I don't know if that's relevant. I was given the hint to try...

Okay, so I have: For any element x, |x|=p^r, x^p^r = 1 = (((x^p)^(r-1))^p) by the properties of exponents. Then x^p^(r-1) equals an element of order p. Thank you so much!

(x^(p^(r-1)))^p is equal to x^(p^r)? So, is (x^(p^(r-1))) always of order p?

Using the example of the cyclic group of order 9, I found that (x^(p^(r-1)))^p is 1 as well. Is that correct?

Sorry, I don't even know which power of x would yield order p in the case of p^2! Is it x^2?