TITLE: The image of an ideal under a homomorphism may not be an ideal
QUESTION [5 upvotes]: This is an elementary question about ideals. Consider a ring homomorphism 
$$
f: \mathbb{Z} \rightarrow \mathbb{Z}[x], 
$$
and consider the ideal $\left< 2\right>$ in $\mathbb{Z}$. When why is it that $f(\left< 2\right>)$ is not an ideal? 
Some websites say that $f(\left< 2\right>)$  is not an ideal because it 
does not contain nonconstant polynomials. That still doesn't make sense on why it is not an ideal. 
Thank you all.

REPLY [4 votes]: I just want to reflect this in words, as I think the ideas are quite simple.
Your title refers to the extension of an ideal. In fact, you have extended the ring, but have not extended the ideal. As others point out there is an ideal generated by the image of your original ideal under the inclusion, and this is the proper notion of the extension of your original ideal.
So the main reason that $\left< 2\right>$ is not an ideal in the extended ring $\mathbb Z[x]$ is that it is not big enough. If you think how much you can extend a ring, your ideal can't really hope to keep up unless it is extended too.
It is not always the case that the extension will lead to a proper ideal - think about the inclusion $f: \mathbb Z \to \mathbb Q $