Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From Atiyah-Macdonald, bottom of page 9:

"Let $f: A \to B$ be a ring homomorphism.


We can factorize $f$ as follows: $$ A \xrightarrow{p} f(A) \xrightarrow{j} B$$ where $p$ is surjective and $j$ is injective. For $p$ the situation is very simple (1.1): there is a one-to-one correspondence between ideals of $f(A)$ and of ideals of $A$ which contain $\ker(f)$, and prime ideals correspond to prime ideals. For $j$, OTOH, the general situation is very complicated. The classical example is from algebraic number theory. ..."

My questions:

1.$p = f$ and $j$ is the inclusion, no?

2.Why do we want to factor $f$?

3.Since the chapter is about contractions and extensions I assume we can use this factorisation of $f$ to find the extension and contraction of ideals in $A$ and $B$. How do we do this?

On the next page, there is an example $\mathbb Z \to \mathbb Z[i]$ for $i = \sqrt{-1}$ where some extensions are listed. But the example doesn't seem to use this factorisation. Also:

4.If the map is not explicitly given, $\mathbb Z \to \mathbb Z[i]$, is it the inclusion or any ring homomorphism?

Many thanks for your help.

share|cite|improve this question
up vote 3 down vote accepted

Here's one use of the factorization:

Consider a certain class of objects you are interested in; e.g., all commutative rings and the morphisms between them. An epimorphism in the class is a morphism $f\colon A\to B$ with the property that for all objects $C$ and all morphisms $g,h\colon B\to C$, if $gf=hf$, then $g=h$.

In the class of sets and functions between sets, for example, epimorphism is synonymous with "surjective". In other contexts, it is not. For example, in the class of rings, the map $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is an epimorphism. In the class of Hausdorff spaces with continuous functions, a continuous function $f\colon X\to Y$ is an epimorphism if and only if $f(X)$ is dense in $Y$.

The factorization in question allows us to simplify the problem: a morphism $f\colon A\to B$ is an epimorphism if and only if the inclusion $f(A)\hookrightarrow B$ is an epimorphism. This reduces the problem from studying all maps to only studying how substructures behave inside fixed structures. This leads, in many settings, to results characterizing when a morphism is onto in terms of how the image sits inside the target (like the Hausdorff space case above).

Another: In order to study all possible homomorphism from a group $G$ to groups $K$, it suffices to study all possible onto maps $G\to K$ (which in turn can be done by studying only the onto maps of the form $G\to G/N$ where $N$ is a normal subgroup of $G$). If you understand all possible onto maps of $G$, then to understand all possible maps you just consider the possible onto maps, and subgroups.

share|cite|improve this answer

1) On elements, we have $p(a)=f(a)$. But $p \neq f$ since the codomains ( = targets) differ. Many mathematicians regard the codomain as part of the data of a map. This becomes especially important in category theory and related areas. But yes, $j$ is the inclusion (not to confused with the identity; same remark about domains).

2) In some situations this becomes helpful. Sorry, but imprecise questions are rewarded with imprecise answers ;).

3) No. You have already quoted the point made by Atiyah. For surjective homomorphisms, it is easy to analyze the extended and contracted ideals. For injective homomorphisms, it is not.

3x) Well $\mathbb{Z} \hookrightarrow \mathbb{Z}[i]$ is already an inclusion.

4) For every ring $R$ (with unit, but this is the setting in Atiyah's book) there is a unique ring homomorphsm $\mathbb{Z} \to R$ (important exercise for you).

share|cite|improve this answer
Thank you, +1. It would be very nice of you if you could nonetheless write more about $2)$: I'd be pleased to see an example (a simple one, I'm an undergrad) of what we use this (seemingly not very interesting) factorisation of $f$. – Rudy the Reindeer May 26 '12 at 11:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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