Let \(f\colon A\to B\) be a ring homomorphism. Then \(\ker f\) is an ideal of \(A\), \(\im f\) is a subring of \(B\), and \(f\) induces a ring isomorphism \[ \bar f\colon A/\ker f \to \im f,\qquad a+\ker f\mapsto f(a). \] In particular, if \(f\) is surjective, then \(A/\ker f \cong B\). As abelian groups, \(0 \to \ker f \to A \mapsvia{f} \im f \to 0\) is exact.
If \(x,y\in\ker f\) and \(a\in A\), then \(f(x-y)=0\) and \(f(ax)=f(a)f(x)=0=f(x)f(a)=f(xa)\), so \(\ker f\) is an ideal. The image contains \(1_B=f(1_A)\) and is closed under subtraction and multiplication, so it is a subring. The map \(\bar f\) is well defined and injective because \(f(a)=f(a')\) if and only if \(a-a'\in\ker f\); it is surjective onto \(\im f\) by construction, and it is a ring homomorphism because \(f\) is.