My school is using Pinter's "A Book of Abstract Algebra" for both semesters of Modern Algebra. For a class assignment a couple weeks ago, regarding rings, I was tasked with the following problem (chapter 18, problem I.3):
Let $A$ and $B$ be rings. Prove that if $f:A\to B$ is a homomorphism from $A$ onto B with kernel $K$, and $J$ is an ideal of $A$ such that $K\subseteq J$, then $f(J)$ is an ideal of $B$.
Just looking at the raw information, I couldn't find a proof easily, so I started drawing a picture. The black is what is given, and the blue ($f(b\eta)$) is what is wanted. I hope it's legible:
Represented from the information given:
$f$ is surjective from A to B$
$J$ is an ideal (for $a\in A, \phi\in J\implies a\phi\in J$)
My question is: does anyone know if the graphic is enough to gain any insight into the proof? That is, what is the graphic "saying" that isn't quite clear just from words, if anything?
I can't really read your handwriting, but surjectivity is the important thing here and I think your picture suggests that. An additive subgroup $J'$ of $B$ is an ideal if for each $b\in B$ and $j\in J'$ we have $bj\in J'$. $f(J)$ is an additive subgroup of $B$, so we need to know that for each $b\in B$ and $f(a)\in f(J)$ we have that $bf(a)\in f(J)$. We know that there exists (possibly more than one) $b'\in f^{-1}(\{b\})$ due to surjectivity. Hence $bf(a)=f(b')f(a)=f(b'a)$, and since $b'a\in J$ we have that $f(b'a)\in f(J)$.