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.
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.
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...