Let f: R --> R' be a ring homomorphism and let N be an ideal of R.
Show that f(N) is an ideal of f(R) and also show that f(N) doesn't need to be an ideal of R'
To show $\displaystyle f(N)$ is an ideal basically what you need to show is that $\displaystyle rx,xr\in f(N)$ for all $\displaystyle r \in f(R)$ and $\displaystyle x\in f(N)$. Now since $\displaystyle r\in f(R)$ it means $\displaystyle r = f(r_0)$ for some $\displaystyle r_0\in R$ and $\displaystyle x = f(n_0)$ for some $\displaystyle n_0\in N$. Therefore, $\displaystyle rx=f(r_0)f(n_0) = f(r_0n_0) \in f(N)$ since $\displaystyle r_0n_0\in N$ for $\displaystyle N$ is an ideal, similarly $\displaystyle xr\in f(N)$. Thus, $\displaystyle f(N) \triangleleft f(R)$.
Consider $\displaystyle \phi: \mathbb{Q}[x] \to \mathbb{C}$ defined by $\displaystyle \phi (f(x)) = f(i)$ (evaluating polynomials at $\displaystyle i$).also show that f(N) doesn't need to be an ideal of R'
Let $\displaystyle N = \mathbb{Q}[x]$ then $\displaystyle \phi(N) = \mathbb{Q}(i) = \{a+bi|a,b\in \mathbb{Q}\}$.
This is not an ideal of $\displaystyle \mathbb{C}$ since the complex numbers are a field.