Why should the kernel of a ring homomorphism be an ideal?

Question

so my question is: Why should the kernel be the ideal of the field $F$? Here is the required theory. Here $(a)$ means an ideal generated by $a$.

Answer 1

To elaborate on Lord Shark's comment:

Let $\phi: R \to S$ be a ring homomorphism, and let $a \in R$, $x \in \text{ker}\phi$, so $\phi(x) = 0$.

Then $\phi(ax) = \phi(a)\phi(x) = \phi(a)\cdot 0 = 0$ (since $\phi$ is a ring hom. and by the above). So $ax \in \ker\phi$.

Showing $\ker \phi$ is an additive subgroup is equally straightforward. So the kernel is an ideal --- this is usually a proposition in a first course in abstract algebra.

Answer 2

Let $\phi: R \to S$ be a ring homomorphism. Define ker$\phi:=\{r \in R : \phi(r)=0\}$.

Let $a,b \in$ ker$\phi$ then $\phi(a-b)=\phi(a)-\phi(b)=0-0=0 \Rightarrow a - b \in$ ker$\phi$.

Similarly let $r \in R, a \in$ ker$\phi$. Then

$\phi(ra)=\phi(r)\phi(a)=\phi(r)0=0$ thus $ra \in$ ker$\phi$, by similar argument we can show $ar \in$ ker$\phi$ making it an ideal.