1
$\begingroup$

I have the following question :

Find homomorphism between $T=\frac{\mathbb{R[x]}}{J}$ and $S=\frac{\mathbb{R[x]}}{I}$ that is surjective.

I have read Finding all homomorphisms between two groups - couple of questions yet the problem is when speaking about cyclic group and finite and problem is relatively easy, But how to handle with infinite groups and also homomorphism between groups that are not the "same"?

The solution is less important for me, I'd like to know how to approach these kind of question is there a method to find surjective homomorphisms? Is it about practice? I can't even find a homomorphism which is not surjective between those groups, how do you start?

Thank you in advance.

$\endgroup$
1
$\begingroup$

Here is a general fact: given a ring homomorphism $\phi:R\to S$, and an ideal $I\subseteq\ker\phi$, there exists a unique map $\bar{\phi}:R/I\to S$ making the diagram: $$\require{AMScd} \begin{CD} R @>{\pi}>> R/I\\ @V{\phi}VV @VV{\bar{\phi}}V\\ S @>>{id_S}> S \end{CD}$$ commute. This is usually called the universal property of quotients. The identity is redundant but as far as I know there is no way to make diagonal arrows. Note that since $\phi = \bar{\phi}\circ\pi$, the map $\bar{\phi}$ is surjective if and only if $\phi$ is.

Here is your case. You have rings $R$ and $S$, and ideals $I\subseteq R$ and $J\subseteq S$. If you can find a surjective homomorphism $\phi:R\to S$ satisfying $I\subseteq\ker\phi$, then you will get a unique surjective map $\overline{\phi\circ\pi_S}:R/I\to S/J$ (since $\phi\circ\pi_S$ is surjective and $I\subseteq\ker\phi\circ\pi_S$) making the diagram: $$\require{AMScd} \begin{CD} R @>{\pi_R}>> R/I\\ @V{\phi}VV @VV{\bar{\phi}}V\\ S/J @>>{id_{S/J}}> S/J \end{CD}$$ commute. This is usually how you construct maps out of a quotient.

$\endgroup$

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.