# Ring Homomorphism

• November 12th 2006, 07:21 AM
JoeDardeno23
Ring Homomorphism
Stuck on this problem.
Let f: R to S be a ring homomorphism.
-Let J-S be an ideal (the - is the triangular symbol). Prove that f^(-1) (J) (equal by definition to {r in R: f(r) in J}) is an ideal of R.
-Prove that if f is surjective and I-R is an ideal then f(I) is an ideal (where f(I)={f(i): i in I}).
-Show, by example, that if f is not surjective the assertion in the above need not hold.

I appreciate any help.
• November 12th 2006, 07:58 AM
ThePerfectHacker
Quote:

Originally Posted by JoeDardeno23
Stuck on this problem.
Let f: R to S be a ring homomorphism.
-Let J-S be an ideal (the - is the triangular symbol). Prove that f^(-1) (J) (equal by definition to {r in R: f(r) in J}) is an ideal of R.

So we have a ring homomorphism,
$\phi: R\to S$
And $J$ is an ideal in $S$.

Thus, $$ is an additive subgroup of $$. From group theory you should be know that the inverse image in a group homomorphism preserves subgroups. Thus, $\phi^{-1}[J]$ is an additive subgroup of $R$.

All we need to show is that,
$r\phi^{-1}[J],\phi^{-1}[J]r \subseteq \phi^{-1}[J], \forall r\in R$

I will show that,
$r\phi^{-1}[J]\subseteq \phi^{-1}[J]$ and leave the second anagolus part to thee to prove.
Let $x\in r\phi^{-1}[J]$ thus, $x=rj$ where $\phi(j)\in J$
Now, (by homomorphism)
$\phi(x)=\phi(rj)=\phi(r)\phi(j)$ and $\phi(j)\in J$ and $\phi(r)\in R$ but $\phi(r)\phi(j)\in J$ because $J$ is an ideal! Thus, $r\phi^{-1}[J]\subseteq \phi^{-1}[J]$
• November 12th 2006, 09:36 AM
ThePerfectHacker
Quote:

Originally Posted by JoeDardeno23
-Prove that if f is surjective and I-R is an ideal then f(I) is an ideal (where f(I)={f(i): i in I}).
-Show, by example, that if f is not surjective the assertion in the above need not hold.

In general if,
$\phi: R\to S$
With $I$ an ideal in $R$
Then,
$\phi[ I ]$ is ideal of $\phi[R]$.
But since $\phi$ is surjective $\phi[R]=R$. Thus, $\phi[ I ]$ is an ideal of $R$.
---
But it seems you never learned that theorem.
In that case, we can still show it.

First, $I$ is an ideal of $R$ that means that $I$ is an additive subgroup of $R$. Thus, $\phi[ I ]$ is an additive subgroup of $S$ because any group homomorphism preserves the images of subgroups.

Now, is it true that, $s\phi[ I ],\phi[ I ]s\subseteq \phi[ I ] \, \forall s\in S$. Again I will only show,
$s\phi[ I ]\subseteq \phi[ I ]$.
The other half is anagolous.
If, $x\in s\phi[ I ]$ then,
$x=si$ where $s \in S, i\in \phi[ I ]$
But since the map is surjective, for some,
$x=\phi(s')\phi(i')$ where $s'\in R, i'\in I$
Since this is a ring homomorphism,
$x=\phi(s'i') \in \phi[ I ]$
Because $s'i'\in I$ since it is an ideal!
Thus,
$s\phi[ I ] \subseteq \phi[ I ]$
• November 13th 2006, 04:48 AM
JoeDardeno23
Thanks so much.

Can you think of an example for part three that would work?
• November 13th 2006, 07:45 AM
ThePerfectHacker
Quote:

Originally Posted by JoeDardeno23
Can you think of an example for part three that would work?

Let,
$\phi: F\to F'$
Be the trivial homomorphism between fields.

The trivial homomorphism is defined as,
$\phi(x)=0' \, \, \forall x\in F$
Let, $\{0\}$ be the ideal of $F$.
Then, $\phi[\{0\}]=\{0'\}$
So the function is not surjective but yet we have an ideal preservation under the image of the function.

--------------
Maybe, you were asking where it would not work?
Consider,
$\phi: \mathbb{Z}\to \mathbb{R}$.
Defined as, $\phi(x)=x$
This is not a surjective map.

The set $\mathbb{Z}$ is an ideal (improper ideal).
Yet,
$\phi [\mathbb{Z}]=\mathbb{Z}$
But this image is not an ideal in $\mathbb{R}$ because,
$\sqrt{2} \mathbb{Z}\not \subseteq \mathbb{Z}$