I'd like to show for a field $F$ and ideal $I$, the mapping $\phi: F \rightarrow F[x]/I$ defined by $\phi(a)=a+I$ is a ring homomorphism.
Showing $\phi(a+b)=\phi(a)+\phi(b)$ is rather straightforward. We have
$$\phi(a+b) = (a+b)+I = (a + I)+(b+I) \text{ (by properties of Ideals)} $$ $$\phi(a)+\phi(b)$$
For the second part, I need to show $\phi(ab) = \phi(a)\phi(b)$. This is where I am not sure how to proceed. Probably there is a property of ideals that I am forgetting:
$$\phi(ab)=(ab)+I$$
Can I simply conclude from this that $(ab)+I=(a+I)(b+I)$, which is what I need to complete the proof?
Also, in this proof I am not really using the fact that co-domain is $F[x]/I$. Is that normal?