Recent content by tom.young84

So given this identity: [V,W] = \nablaVW-\nablaWV ^^I got the above identity from O'Neil 5.1 #9. From this I'm not sure how to make the jump with vector functions, or if it is even possible to apply that definition to a vector function [xu,xv].

For the first question, I understand your proof, but I don't understand why that answers the question. Your contradiction is that it violates the definition of a prime ideal. I don't understand why this demonstrates that A\P is closed under multiplication.

Outside of the case with {0}, I'm not sure why this is a multiplicative set.

1) R is an integral domain and P is a prime ideal. Show R\P (R complement P or R-P) is a multiplicative set. -Well since R is an integral domain it contains 1. -{0} would be a prime ideal, and that was removed (is this too much to assume) I'm not sure how to show multiplication is...

I just want to formally prove this so as to make a good answer hopefully available. Nonempty: f(0) is in J (since its an ideal) Hence 0 is in f-1(J) Addition is closed: Suppose an r1, r2 in f-1(J). f(r1+r2)=f(r1)+f(r2) which is in J. Hence r1+r2 is in f-1(J) Inverses: We know that...

I'm starting to see what you are saying. I'm reworking having addition closed. r1, r2 in f^(-1)(J) f(r1+r2)=f(r1)+f(r2) which is in J and hence r1+r2 is in f^(-1)(J)

So my proofs should go something like r1, r2 in f-1(J) => Definition of homomorphism in J => operation is closed in f-1(J) This problem is giving me a hard time. Last semester I butched a similar problem that dealt with the preimage of a subgroup.

These is the chunk of the proof i was having difficulty with, I think I corrected it. Closed under addition r1, r2 in R and j in J f(r1+r2)=f(r1)+f(r2) in J (definition of homomorphism) hence, r1+r2 f-1(J). Inverses -r in R f(-r)=-f(r) in J hence -r in f-1(J) Absorption...

I keep wanting to pull the idea of homomorphisms into this so we have this r1, r2 in f-1[J] f(r1+r2)=f(r1)+f(r2) in f-1[J] This shows r1+r2 is in f-1[J] Thus r1+r2 is in f-1[J]

Is this allowed? r1, r2 in R f(r1+r2)=f(r1)+f(r2) in f^-1[J] or would I have to toss in a f(j) such that f(r1+r2)f(j)=f(r1)f(j)+f(r2)f(j) in f^-1[J] (this makes use of both absorption and the property of homomorphisms)

I get how to show something is in f-1[J] through absorption. Suppose an r in R and j in J f(rj)=f(r)f(j) which is in J I'm stuck on show that f-1[J] is an ideal by showing that addition is closed and that there are inverses. I know 0 is in it since homomorphism carries 0R to 0S and 0S...

"Preimage" of a homomorphism f:R-->S is a homomorphism of rings and suppose J is an ideal of S. Prove that f^-1[J]={r in R: f(r) in J} is an ideal of R. I'm more concerned about how to even start this proof as I am lost.

I've tried using solver...I get something that isn't the answer. Is there a good tutorial?

Introduction to Operations Research, Hiller, 3.4-12 I believe 1 2 3 4 5 New Alloy tin 60 25 45 20 50 =40 zinc 10 15 45 50 45 =35 lead 30 60 10 30 10...

The larger ring is C (the complex number), we just have to show that this subset is in fact a subring. I thought a subring needed to suffice 4 axioms: closure under addition, additive inverse, additive identity, and closure under multiplication Additionally, I forgot to answer this. It's...