# Recent content by UD1

1. ### A (probably) simple combinatorial problem

Thanks for your reply, Dick. I don't see how you can cover a circle of radius 4 by 3 circles radius 3 each, but 4 such circles is definitely enough. For instance, putting the center of the coordinate system at the center of the circle, so that it has equation x^2+y^2=4^2, the four small...
2. ### Isomorphism with z sub prime#

Let f be an automorphism of Z_p, and x a generator for Z_p, so that <x>=Z_p. Explain why f is determined by what it does to x, i.e. why knowing f(x) suffices to to know where f sends any other element of the group. Now think about whether (x is a generator) => (f(x) is a generator) is true...
3. ### A (probably) simple combinatorial problem

Homework Statement In a circle city of radius 4 we have 18 cell phone power stations. Each station covers the area at distance within 6 from itself. Show that there are at least two stations that can transmit to at least five other stations. Homework Equations The Attempt at a...
4. ### Kernel of GL(n,F) acting on F^n

Homework Statement Suppose GL(n,F) acts on F^n in the usual way. Consider the induced action on the set of all k-dimensional subspaces of F^n. What's the kernel of this action? Is it faithful The Attempt at a Solution Well, I anticipate that the kernel of this action consists of scalar...
5. ### GF(q); x^p-x+a has no root => irreducible [PROOF]

So, we know that, assuming that b is a root in some extension of F, then f(x)=\prod_{j=0}^{p-1}(x-(b+j)). Now suppose that f is reducible, that is f=gh for some polynomials g and h whose coefficients are in GF(q). There must exists a proper subset I of {0,...,p-1} such that...
6. ### Partial derivatives and chain rule

I apologize. The solution is correct though.
7. ### GF(q); x^p-x+a has no root => irreducible [PROOF]

So, how does that help?
8. ### GF(q); x^p-x+a has no root => irreducible [PROOF]

OK. It looks that there is some notation misunderstanding. I'm not that well familiar with finite fields so I might have used the type of notation not usually used. By GF(q) I mean a finite field of order q (with q elements) where q is a power of prime p, and that p is a characteristic of...
9. ### Partial derivatives and chain rule

d/(dv)[du/ds]= d/(dv)[du(s,t(s,v))/ds]= d/(dv)[du/ds + du/dt * dt/ds]= d/(dv)[du/ds+du(s,t(s,v))/dt * dt(s,v)/ds]= d^2u/dsdt * dt/dv + d^2u/dt^2 * dt/dv * dt/ds + du/dt* d^t/dsdv Write it down in usual partial derivative notation and you'll see where it all comes from.
10. ### GF(q); x^p-x+a has no root => irreducible [PROOF]

Homework Statement Let q=p^e, where p is a prime and e is a positive integer. Let a be in GF(q). Show that f(x)=x^p-x+a is irreducible over GF(q) if and only if f(x) has no root in GF(q) Homework Equations The Attempt at a Solution One of the directions seems obvious. Namely, if...
11. ### Char(F)=p; poly NOT irreduc. => must have a root?

Why are you referring to a root of x^p-a as THE root? How do you know there is only one root? What if they behave like roots in the complex field?
12. ### Char(F)=p; poly NOT irreduc. => must have a root?

How about these arguments. All the roots of x^p-a are p-th roots of a. Suppose that x^p-2 is not irreducible and factorize in into irreducibles. All of the will be of degree >=2. Consider any of them. Look at the constant term of that factor. It MUST (BUT WHY?!) look like the p-th root of a to...
13. ### Char(F)=p; poly NOT irreduc. => must have a root?

First of all, it is not true that x^p-2 can be represented as a product of only two irreducible polynomials.
14. ### Char(F)=p; poly NOT irreduc. => must have a root?

If it has a root everything is clear. What is not clear is why the fact that x^p-a is NOT irreducible implies that it has a root.
15. ### Char(F)=p; poly NOT irreduc. => must have a root?

Because it is indeed correct? In a field F of characteristic p for any two x and y in F we have (x+y)^p=x^p+y^p. This is to see using the mentioned binomial expansion and convincing yourself that p divides every binomial coefficient except the frist and the last. Thus, each term in the expansion...