The image of an ideal under a homomorphism may not be an ideal 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. 
 A: An ideal $I$ of a ring $R$ is, by definition, "multiplicatively absorptive"; that is, for any $s\in I$ and $r\in R$, we must have $rs\in I$.
Under the usual definition of ring homomorphism, given rings $A$ and $B$, a ring homomorphism $f:A\to B$ must satisfy $f(1_A)=1_B$, so that there is in fact a unique ring homomorphism $f:\mathbb{Z}\to R$ for any ring $R$. In the case of $R=\mathbb{Z}[x]$, this is exactly what you would expect; since $f(1)=1$, we have $f(n)=n$ for all $n\in\mathbb{Z}$.
The set $S=f(\langle 2\rangle)$ in $\mathbb{Z}[x]$ is of course just $$S=\{g\in\mathbb{Z}[x]\mid g=2k\text{ for some }k\in\mathbb{Z}\},$$ i.e. the even integers. For example, $2\in S$. But $x\in \mathbb{Z}[x]$, and $2x\notin S$, so $S$ is not an ideal of $\mathbb{Z}[x]$.
A: Strictly speaking, the extension of an ideal $I$ under a homomorphism $f:A \to B$ of rings is defined to be the ideal $f(I)B.$ As this example shows, taking the image of the ideal $I$ might not yield an ideal at all. Fortunately for algebraic geometry, the inverse image of an ideal under a homomorphism is again an ideal.
A: 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 $
A: $f : \mathbb{Z} \rightarrow \mathbb{Z}[x]$ is the $f(z) = z$. So $f(\langle 2 \rangle)$ is all the even constant polynomials.
The definition of an ideal is that for all $P \in \mathbb{Z}[x]$ and $Q \in f(\langle 2 \rangle)$, one must have $PQ \in f(\langle 2 \rangle)$. So let $P = x$ and $Q = 2 \in f(\langle 2 \rangle)$. Then $PQ = x\cdot 2 = 2x \notin f(\langle 2 \rangle)$ since $f(\langle 2 \rangle)$ consists of only even constant polynomials. So $f(\langle 2 \rangle)$ is not an ideal of $\mathbb{Z}[x]$. 
A: $I$ is an ideal of $A$ iff
$\forall i,j \in I, \forall a \in A \implies ai \in I$ and $i+j \in I$. 
$f((2))$ is not an ideal of $\mathbb{Z}[x]$, because
$i:=2 \in f((2))$
$a:=x \in \mathbb{Z}[x]$,
but
$ai=x \times 2 \notin f((2))$ because $f((2))=2\mathbb{Z}$.
You have changed of ring. The first ring is $\mathbb{Z}$. The second ring is $\mathbb{Z}[x]$.
