Proving the Ideal Property of the Preimage of a Homomorphism

[tom.young84]

"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.

 

[Hurkyl]

Then do something -- even if you don't know it will help you prove the theorem.

Definitions are pretty much always someplace to start. Trying to turn a problem into algebra is pretty much always something to try. Specific examples are often useful.

You should never have no idea how to start working on a problem.

 

[Dick]

You want to show that if x is any element of R and j is any element of f^(-1)(J) then xj is also an element of f^(-1)(J). Is that enough of a start?

 

[tom.young84]

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 0_R to 0_S and 0_S is in the ideal.

 

[Dick]

I thought you said you didn't know how to start. Why don't you try to use the properties of homomorphism to show f^(-1)(J) is closed under addition. After all, J is.

 

[tom.young84]

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)

 

[Dick]

No, no, no. You want to start with r1 and r2 in f^(-1)(J). Not r1 and r2 in R. Then prove r1+r2 is in f^(-1)(J). You want to prove f^(-1)(J) is an ideal.

 

[tom.young84]

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]

 

[Dick]

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]

You should probably get a piece of blank paper out and write down all of the definitions of everything here so you can look at them all at once. f(r1) isn't in f^(-1)(J). It's in J.

 

[tom.young84]

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
f(r1r2)f(j)=f(r1)f(r2)f(j) in J (Using the definition of homomorphism)
hence r1r2j=(r1r2)j=r1(r2j) in f^-1(J) by absorption

Yay ideal

 

[Dick]

I THINK you know what you're doing. Again, I would make your starting point, "assume r1 and r2 are in f^(-1)(J)", not "assume r1 and r2 are in R". I already said this but you want to prove f^(-1)(J) is closed, not that R is closed. You already know that. Sure, and then use properties of homomorphism and that J is an ideal of S.

 

[tom.young84]

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.

 

[Dick]

It's like at least 80% right. What I'm missing a little precision. Like EXACTLY what set something is coming from and what the properties of that set are that make your statement true. Like look at your "Inverses" part. You say (-r) in R, then you say f(-r)=(-f(r)), ok so far but then you say -f(r) in J. That's wrong. That ONLY TRUE if you start with r in f^(-1)(J) not just in R. Do you see what I'm saying??

 

[tom.young84]

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)

 

[Dick]

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)

You've got the idea. And that's true because f(r1) and f(r2) are in J. And J is closed, being an ideal and all.

 

[tom.young84]

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 for each f(r) in J, there exists a -f(r) in J (since J is an ideal).
By a few lemmas we have on negatives and homomorphisms -f(r)=f(-r).
Hence -r is in f^-1(J).

Absorption:
i is in f^-1(J)
f(ri)=f(r)f(i); we know that f(i) is in J; by using the definition of homomorphism f(ri) is in J, so ri is in f^-1(J)

Likewise for the other direction since we are assuming R and S are general rings.

 

[Dick]

You are still loosing a bit focus on what you are trying to prove. Look at absorption. You want to pick i in f^(-1)(J) and r in R and prove i*r is in f^(-1)(J), right? Do you see what's wrong with your proof??