Ring Homomorphism and Ideals

**Coda202** · April 3rd 2009, 09:21 AM

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'

**ThePerfectHacker** · April 3rd 2009, 11:01 AM

Originally Posted by Coda202

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)

To show $f(N)$ is an ideal basically what you need to show is that $rx,xr\in f(N)$ for all $r \in f(R)$ and $x\in f(N)$ . Now since $r\in f(R)$ it means $r = f(r_0)$ for some $r_0\in R$ and $x = f(n_0)$ for some $n_0\in N$ . Therefore, $rx=f(r_0)f(n_0) = f(r_0n_0) \in f(N)$ since $r_0n_0\in N$ for $N$ is an ideal, similarly $xr\in f(N)$ . Thus, $f(N) \triangleleft f(R)$ .

also show that f(N) doesn't need to be an ideal of R'

Consider $\phi: \mathbb{Q}[x] \to \mathbb{C}$ defined by $\phi (f(x)) = f(i)$ (evaluating polynomials at $i$ ).
Let $N = \mathbb{Q}[x]$ then $\phi(N) = \mathbb{Q}(i) = \{a+bi|a,b\in \mathbb{Q}\}$ .
This is not an ideal of $\mathbb{C}$ since the complex numbers are a field.

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