Ring homomorphism and ideals Let $R$ and $R'$ be rings and let $\phi: R\mapsto R'$ be a ring homomorphism and $N$ an ideal of $R$.
Show that $\phi[N]$ is an ideal of $\phi[R]$, and give an example to show that $\phi[N]$ need not be an ideal of $R'$.
Let $N'$ be an ideal of either $\phi[R]$ or R', and show that $\phi^{-1}[N']$ is an ideal of R.
I just started learning ideals so I am having a lot of trouble with this. I know that the kernel of $\phi$ is an ideal, but I don't know how I can use this.
 A: In order to show that $\phi(N)$ is an ideal we must show that it is an additive group and closed under multiplication in $\phi(R)$

Showing $\phi(N)$ is an additive group

Identity
So as $0\in N$ and $\phi$ is a ring homomorphism we have that $\phi(0)=0\in \phi(N)$
Closure
Now take $x,y\in \phi(N)$ then by definition of $\phi(N)$ there must exists $a,b\in N$ such that $\phi(a)=x$ and $\phi(b)=y$. Now as $N$ is an ideal we have that $a+b\in N$ and so $\phi(a+b)\in \phi(N)$, then as $\phi$ is a ring homomorphism we have that $\phi(a+b)=\phi(a)+\phi(b)=x+y\in \phi(N)$.
Inverses
If we have $x\in \phi(N)$ then by definition there must be an $a\in N$ such that $\phi(a)=x$ now as $N$ is an ideal we have that $a^{-1}\in N$ and so $\phi(a^{-1})\in \phi(N)$. Now as $\phi$ is a ring homomorphism we have that $\phi(a^{-1})=\phi(a)^{-1}=x^{-1}\in \phi(N)$
So we have that $\phi(N)$ is an additive subgroup of $\phi(R)$
Now to show that it is an ideal we have to show that if we have $r\in \phi(R)$  then $r\phi(N)\subset \phi(N)$ and $\phi(N)r\subset \phi(N)$

Showing $\phi(N)$ is closed under multiplication of $\phi(R)$

Now if $r\in \phi(R)$ then there must be an $r'\in R$ such that $\phi(r')=r$
Then as $r'N\subset N$ we have that $\phi(r'N)=\phi(r')\phi(N)\subset \phi(R)$ (same argument for $r$ on the right)

Example of image of an ideal is not an ideal

We can define the inclusion map $f:\mathbb{Z}\rightarrow \mathbb{Q}$ and then take an ideal in $\mathbb{Z}$ say $3\mathbb{Z}$ then $f(3\mathbb{Z})$ is not an ideal in $\mathbb{Q}$.
Suppose that it was, then noting that $1\not\in f(3\mathbb{Z})$. But by the definition of ideal we must have, $\frac{1}{3}\in\mathbb{Q}$, $\frac{1}{3}\times 3\in f(3\mathbb{Z})$ which gives a contradiction. 
Note: as $\mathbb{Q}$ is a field then it has only the two trivial ideals so it follows directly from this.
Another example is given here: The image of an ideal under a homomorphism may not be an ideal
