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.

Please help.

Thank you.

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- Dec 10th 2009, 07:41 PMkpizleExample needed please
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.

Please help.

Thank you. - Dec 11th 2009, 12:31 AMSwlabr
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 $\displaystyle \phi: R \rightarrow S$ is a surjective ring homomorphism then there exist many rings $\displaystyle S^{\prime}$ such that $\displaystyle S \subset S^{\prime}$. We can define $\displaystyle S^{\prime}$ to be pretty much anything we want, and so we can define it to be a ring where $\displaystyle S$ is not an ideal in it, and so define $\displaystyle \phi^{\prime}:R \rightarrow S^{\prime}$, $\displaystyle r\phi^{\prime} = r\phi$. As $\displaystyle S$ is not an ideal of $\displaystyle S^{\prime}$ and $\displaystyle S=im(\phi^{\prime})$ we get that the image is not always an ideal.

For instance, take $\displaystyle \phi:\mathbb{Q} \rightarrow \mathbb{R}$, $\displaystyle \frac{a}{b} \mapsto \frac{a}{b}$. As $\displaystyle im(\phi)=\mathbb{Q}$ and $\displaystyle \mathbb{Q}$ is not an ideal of $\displaystyle \mathbb{R}$ we have that $\displaystyle im(\phi)$ is not an ideal of $\displaystyle \mathbb{R}$.

Does that make sense?