# Show that every ideal of a ring R is the kernel of some ring homomorphism

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'?

• You don't need to show well-definedness, since you're not making a "choice" at any point. Your proof works, but it's not clear what your question is. You just need to find at least one homomorphism such that $A$ is its kernel. – lokodiz Apr 5 at 12:58
• @lokodiz I'm not quite sure that I had to use the definition with the quotient ring (what I defined). Well if you agree that the proof is correct, everything is fine :) – Ilan Aizelman WS Apr 5 at 13:10

Well, if $$A$$ is an ideal, then it is the kernel of the morphism you gave.