Factoring a ring homomorphism

Question

From Atiyah-Macdonald, bottom of page 9:

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

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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. ..."

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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.

Answer 1

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).

Answer 2

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.