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'?

  • $\begingroup$ 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. $\endgroup$ – lokodiz Apr 5 at 12:58
  • $\begingroup$ @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 :) $\endgroup$ – Ilan Aizelman WS Apr 5 at 13:10

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.


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