# Unique homomorphism between quotients

I am working on an exercise I found rather entertaining, albeit I found myself struggling at how to attack this problem as I don't know from which angle to approch it and tips or tricks would be appricaited.

Let $\varphi:R_1\to R_2$ be a ring homomorphism such that $\varphi(I_1)\subseteq I_2$ where $I_1$ is an ideal of $R_1$ and $I_2$ of $R_2$. Show that there is a unique homomorphism $\phi:R_1/I_1\to R_2/I_2$ such that the following diagram commutes

$\require{AMScd} $$\begin{CD} R_1 @>\varphi>> R_2\\ @VV{\eta_1}V @VV{\eta_2}V\\ R_1/I_1 @>\phi>> R_2/I_2 \end{CD}$$$

Where $\eta$ are the natural homomorphisms between rings and their quotient rings.

I am unfamiliar somewhat with tackling how to show that morphisms are unique which is my main issue, I don't know how to demonstrate that.

• try the map $x+I_1\mapsto\varphi(x)+I_2$ – janmarqz Jun 2 '15 at 15:23
• That did strike me but I don't see how that helps – Zelos Malum Jun 2 '15 at 15:24

This theorem relies on a very important fact about homomorphisms.

If $\alpha\colon R\to S$ is a ring homomorphism, $I$ is an ideal of $R$ such that $I\subseteq \ker\alpha$ and $\eta\colon R\to R/I$ is the canonical projection, there exists a unique ring homomorphism $\beta\colon R/I\to S$ such that $\alpha=\beta\circ\eta$.

(Uniqueness) Suppose $\beta$ exists. Then, for $r+I\in R/I$, we have $$\beta(r+I)=\beta(\eta(r))=\beta\circ\eta(r)=\alpha(r)$$ so the action of $\beta$ is determined by $\alpha$ and this settles uniqueness.

(Existence) We want to show that, if $r+I=r'+I$, then $\alpha(r)=\alpha(r')$, so that setting $\beta(r+I)=\alpha(r)$ does not depend on the particular representative of the coset. This is true because $r+I=r'+I$ is equivalent to $r-r'\in I$, which implies $r-r'\in\ker\alpha$ and therefore $\alpha(r-r')=0$: thus $\alpha(r)=\alpha(r')$ as desired.

Therefore the position $\beta(r+I)=\alpha(r)$ defines a map $R/I\to S$; checking it's a ring homomorphism is easy.

Now that we have the general theorem, we can apply it to our present situation. Let $S=R/I_2$ and $\alpha=\eta_2\circ\varphi$.

If we prove that $I_1\subseteq\ker\alpha$, we can apply the theorem and get a unique ring homomorphism $\psi\colon R_1/I_1\to R_2/I_2$ such that $$\psi\circ\eta_1=\alpha=\eta_2\circ\varphi$$ (I use $\psi$ instead of $\beta$ as in the theorem to comply with your notation; $\eta_1\colon R_1\to R_1/I_1$ and $\eta_2\colon R_2\to R/I_2$ are the canonical projection).

All we need is to show that $I_1\subseteq \ker\alpha=\ker(\eta_2\circ\varphi)$, that is, for $x\in I_1$, $\eta_2\circ\varphi(x)=0+I_2$. But, by assumption $\varphi(x)\in I_2$, so $\varphi(x)\in\ker\eta_2$ and therefore $$\eta_2\circ\varphi(x)=\eta_2(\varphi(x))=0+I_2$$ as desired.

• Could you please rewrite it somewhat? I see you speak of $R$ and $I$ here but I cannot deduce which of the two rings and ideals we're talking about, I am somewhat getting the idea but not quite there, maybe that detail is needed for me. – Zelos Malum Jun 2 '15 at 15:28
• @ZelosMalum I have more clearly separated the two parts. – egreg Jun 2 '15 at 15:38
• Thanks, it helped a bunch :) Now I understand it. – Zelos Malum Jun 2 '15 at 15:43