Theorem – Show that every ideal of a ring R is the kernel of some homomorphism R to some other ring.
My proof: Let A be an ideal of $R$
Define: $f : R \rightarrow R/A$
$f(r) = r + A , \forall r \in R$
To show: f is ring homom. (and well defined) and then to show $ker(f) = A$
$f$ is well defined: $x=y$ then $x + A = y + A$, then $f(x) = f(y)$.
$f(x+y) = (x+y) + A = (x + A) + (y + A) = f(x) + f(y)$
$f(xy) = (xy) + A = (x+A)(y+A) = f(x)f(y)$
To show: $ker(f) = A$
Proof: $kerf = \{x \in R | f(x) = 0 + A\} = \{x \in R | x + A = 0 + A\} = \{x \in R | x + A = A\} = \{ x \in R | x \in A \}$. Thus, $ker(f) = A$.
My question: I'm not sure the proof is true, does $f : R \rightarrow R/A$ generalizes to 'any ring homomorphism'?
Best Answer
Well, if $A$ is an ideal, then it is the kernel of the morphism you gave.
On the other hand, the kernel of a ring homomorphism is an ideal.
I'm not sure the answer you expect for your question.