Morphism between quotients of a ring by ideals.

Question

Let $I, J$ be ideals of a ring $R$ and $I \subset J$. Is there a homomorphism from $R/I$ to $R/J$? I check that $x + I \mapsto x+J$ satisfies the conditions of a homomorphism. I think that $R/J \to R/I$, $x+J \mapsto x+I$ does not define a homomorphism. Is this correct? Thank you very much.

Answer 1

Well, in general there are several such homomorphisms. However, if we let $\eta_I:R\to R/I$ and $\eta_J:R\to R/I$ be the respective quotient morphisms, then there is a unique homomorphism $\kappa:R/I\to R/J$ such that $$\eta_J=\kappa\circ\eta_I$$ In fact we can say something more general. Let $\eta:R\to R/I$ be the quotient map and let $f:R\to S$ be any other morphism such that $I\subseteq \text{ker}(f)$. Then there is a unique lift $\tilde f:R/I\to S$ such that $f=\tilde f\circ\eta$.

To prove this assertion, observe first that $\eta$ is surjective, so for each $q\in R/I$ there is at least one $r\in R$ such that $q=\eta(r)$. So $\tilde f(q)$ must be $f(r)$. It follows that $\tilde f$ is unique where it exists.

It remains to show that there is in fact such $\tilde f$. Let $\chi$ return for each element of $R/I$ an element of $R$ such that $\eta(\chi(q))=q$. Note that $\chi$ is a function of the underlying sets, as opposed to a homomorphism of rings. The existence of such $\chi$ is guaranteed by the surjectivity of $\eta$ and AoC. We claim that $\tilde f:q\mapsto f(\chi(q))$ defines a ring morphism.

Firstly, $\chi(0)\in \text{ker}(\eta)\iff \chi(0)\in I\implies \chi(0)\in \text{ker}(f)$, so $\tilde f$ preserves the additive identity. Likewise, $\chi(1)-1\in \text{ker}(\eta)\implies \chi(1)-1\in \text{ker}(f)$, so $\tilde f$ preserves the multiplicative identity as well. Furthermore, for any two $q_1,q_2\in R/I$, $$\chi(q_1+q_2)-\left(\chi(q_1)+\chi(q_2)\right)\in \text{ker}(\eta)\implies \chi(q_1+q_2)-\left(\chi(q_1)+\chi(q_2)\right)\in \text{ker}(f)$$ And $$\chi(q_1q_2)-\chi(q_1)\chi(q_2)\in \text{ker}(\eta)\implies \chi(q_1q_2)-\left(\chi(q_1)\chi(q_2)\right)\in \text{ker}(f)$$ So $\tilde f$ is additive, multiplicative, and thus a ring homomorphism $\blacksquare$

Answer 2

There always is a homomorphism $R/I\to R/J$, when $I\subset J$. Indeed, $I$ is contained in the kernel of the canonical projection $R\to R/J$, so the homomorphism theorems provide a homomorphism $R/I\to R/J$ (which is exactly $x+I\mapsto x+J$).

There can be homomorphisms $R/J\to R/I$.

Example: let $F$ be a field and take

• $R=F[x,y]$
• $I=(y)$
• $J=(x,y)$

Then $R/J\cong F$, $R/I\cong F[x]$ and there is a homomorphism $F\to F[x]$ (it is not $x+J\mapsto x+I$, of course).

In other cases this doesn't happen. If $R=\mathbb{Z}$, $I=4\mathbb{Z}$, $J=2\mathbb{Z}$. Then there is no ring homomorphism $R/J\to R/I$.

This happens in more generally in $\mathbb{Z}$: if $I=n\mathbb{Z}$ and $J=m\mathbb{Z}$, then there exists a homomorphism $\mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ if and only if $n\mid m$.