# Thread: Ring Homomorphism

1. ## 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.

2. 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, $r\phi^{-1}[J]\subseteq \phi^{-1}[J]$

3. 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,
$s\phi[ I ] \subseteq \phi[ I ]$

4. Thanks so much.

Can you think of an example for part three that would work?

5. 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,
$\sqrt{2} \mathbb{Z}\not \subseteq \mathbb{Z}$