Recent content by DanielThrice

I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers. The question was to find all of the prime and maximal ideals of \mathbb Z_7. My answer was that because a finite integral domain is a field, the prime and maximal ideals...

This is my progress so far... Well, call the matrix M. Then you want that (Mx+B)^2=Mx+B in M2R[x][/math]. So we just have to perform the calculation and see what we get, (Mx+B)^2=MxMx + MxB + BMx+B^2 = M^2x^2+(MB + BM)x + B^2 (the x-terms act like this (commutatively) by definition)...

My professor gave us this query at the end of class, it contained two parts. 1. Show a ring is idempotent 2. Consider the degree one polynomial f(x) is an element of M2(R)[x] given by f(x) = [0 1 ______0 0]x + B (so f(x) = the matrix []x + B). For which B is an element of M2(R), if any...

Thanks Ivy, helped a lot I figured it out. I like how you reworded it.

I'm working with elementary rings, and my professor gave me about ten of these to start but it seems like a lot of work with how he managed it. I know you guys don't answer homework so I chose so I can do the others. Any help would be greatly appreciated, the groups were easy but the rings are a...

Original Query: I'm beginning to look at rings for the first time and was given this to start with: Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >. Consider the set of cosets R/I = {a + I: a is an element of R} equipped with...

Wait did I mess myself up...the original equation is % (f) = f '' - f so the homomorphism doesn't change that much, it just becomes (f + g) '' - (f + g) = f '' + g '' - f - g. Well any functions in the reals added to other functions of the reals gets you another function of the real, so % is a...

Is there an easy way to prove that these are the only three?

So the kernel would just be cos, sin, and e^x correct?

Nevermind, lol, that's what a homomorphism is. I said the kernel is just all of the constants: f ' (x) = 0 (Integrate) f (x) = C How about if we have a different function, % : G to G defined by % (f) = f '' - f ? Homomorphism? (f + g) '' + (f + g) = f '' + g '' + f + g. Well any...

Of course not, the constants are the only functions with no change in how they change over time, that's why they are constants. But how do I know to look for the zero derivatives?

Let G denote the set G = {f : R → R | f is infinitely differentiable at every point x ∈ R}. Prove that G is a group under addition. Is G a group under multiplication? Why or why not? I have proved this after much trouble, using the axioms of group theory, and I think I understand the...

Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unity in C. (That is, U∗ = {z ∈ C|zn= 1 for some n ∈ Z+}.) I don't really understand how to prove this isomorphism

H is CG(a) in this case. The operation m' is just * (written g*h or just gh). We want to show * : CG(a) x CG(a) -> CG(a) This means that for any g, h in CG(a) We need to show gh or g * h is in CG(a) g in CG(a) means what? ga = ag (commutative) similarly ha = ah How do we show gh is...

Alright, so I showed this: We have that if r+Z is an element of R/Z, and the order of r + Z = n, nr is an element of Z implies that r is an element of Q. But what about this: Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unityin C. (That is, U∗ =...