Recent content by UD1

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...

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...

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...

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...

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...

I apologize. The solution is correct though.

So, how does that help?

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...

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.

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...

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?

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...

First of all, it is not true that x^p-2 can be represented as a product of only two irreducible polynomials.

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.

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...