• Dec 10th 2009, 07:41 PM
kpizle
Hi there.

Hammering away at this abstract algebra business.

Need to drive a few ideas home and am having a hard time thinking of an example that justifies why the image of a ring homomorphism doesn't have to be an ideal.

Thank you.
• Dec 11th 2009, 12:31 AM
Swlabr
Quote:

Originally Posted by kpizle
Hi there.

Hammering away at this abstract algebra business.

Need to drive a few ideas home and am having a hard time thinking of an example that justifies why the image of a ring homomorphism doesn't have to be an ideal.
The reason an image does not have to be an ideal is because it is not dependent on where where you are mapping to. What I mean is that if $\phi: R \rightarrow S$ is a surjective ring homomorphism then there exist many rings $S^{\prime}$ such that $S \subset S^{\prime}$. We can define $S^{\prime}$ to be pretty much anything we want, and so we can define it to be a ring where $S$ is not an ideal in it, and so define $\phi^{\prime}:R \rightarrow S^{\prime}$, $r\phi^{\prime} = r\phi$. As $S$ is not an ideal of $S^{\prime}$ and $S=im(\phi^{\prime})$ we get that the image is not always an ideal.
For instance, take $\phi:\mathbb{Q} \rightarrow \mathbb{R}$, $\frac{a}{b} \mapsto \frac{a}{b}$. As $im(\phi)=\mathbb{Q}$ and $\mathbb{Q}$ is not an ideal of $\mathbb{R}$ we have that $im(\phi)$ is not an ideal of $\mathbb{R}$.