1
$\begingroup$

Let $R, S$ be rings.

Suppose $\phi: R \rightarrow S$ is a ring homomorphism. Clearly we have a map $R \rightarrow S/J$ defined by the canonical map.

However, for any ideal $I \subset R,$ can I form a ring homomorphism $R/I \rightarrow S/J?$

This feels wrong for some reason as $I, J$ are unrelated.

However, it seems like I can just send $a + I \mapsto \phi(a) + I.$ I don't see anything wrong with this. $\phi(a + I) + \phi(b + I) = \phi(a) + I + \phi(b) + I = \phi(a + b) + I,$ and so on...

I feel silly for asking such a basic question but it feels quite strange that this is possible as, again, there is no relation between these two ideals.

$\endgroup$
1
$\begingroup$

Such a definition would be meaningful (i.e. would lead to a ring homomorphism) in case the ideal $I, J$ and $\phi$ satisfy the condition $\phi(I)\subset J$. Otherwise the value will not be well-defined. Taking a different coset representative will give a different value making it NOT a functiton.

$\endgroup$
  • $\begingroup$ oh brain fart! I forgot to check that it is a well defined function $\endgroup$ – 伽罗瓦 Mar 1 at 7:22
  • 1
    $\begingroup$ Using a function to to define another function on a quotient object will always bring the issue of well-definedness. Once it passes this hurdle 'most likely' it will be a good function. $\endgroup$ – P Vanchinathan Mar 1 at 7:27
2
$\begingroup$

A homomorphism $\phi\colon R\to S$ induces a homomorphism $R/I\to S$ iff $I\subseteq \ker \phi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.