# Modules and direct sum

Let $$A$$ be a ring and $$M$$ an $$A$$-module, given $$g\colon M \to A$$ a surjective morphism of $$A$$-module, prove that $$M \cong \ker(g) \oplus A$$.

Here's what I tought: first the following short sequence is exact $$0 \to \ker(g)\to M\xrightarrow{f} A \to 0$$ and if I show that splits I got the thesis. I define a morphism of $$A$$-modules $$f:A \to M$$ the following way $$f(a) = a\cdot1_M \space\space \forall a \in A$$ and since $$\forall a \in A$$ we have $$(g\circ f)(a) = g(a\cdot1_M ) = ag(1_M)=a\cdot1_A = a$$ We have $$gf = \operatorname{id}_A$$, the sequence splits and $$M \cong \ker(g) \oplus A$$.

What I've done is a possible solution? As last question if is right this could hold for every $$M,N$$ that are $$A$$-module and every $$h\colon M \to N$$ with $$h$$ surjective morphism and with the hypotesis that $$N$$ is finitely generated by a single element? Thank you very much.

Question: "As last question if is right this could hold for every $$M,N$$ that are $$A$$-module and every $$h\colon M \to N$$ with $$h$$ surjective morphism and with the hypotesis that $$N$$ is finitely generated by a single element? Thank you very much."

Answer: The map $$f\colon M \to A$$ has a (non-unique) section $$s: A\to M$$: pick any $$m\in M$$ with $$f(m)=1$$ and define $$s(a):=am$$. It follows $$f \circ s =\operatorname{id}_A$$ . Define $$\phi:= s \circ f: M \to M$$. It follows $$\phi^2 =\phi$$ and $$M \cong \ker(\phi) \oplus \operatorname{im}(\phi)=\ker(f)\oplus A$$.

There are isomorphisms

$$a\colon M \cong \ker(\phi) \oplus \operatorname{im}(\phi)$$

defined by $$a(m):=(m-sf(m), f(m))$$ and

$$b\colon \ker(\phi)\oplus \operatorname{im}(\phi) \to M$$

defined by $$b(u,v):=u+s(v)$$. You may verify thatt $$a\circ b=b \circ a=\operatorname{id}$$, hence $$a$$ and $$b$$ are isomorphisms of $$A$$-modules.

Note: There is no "identity element" $$1_M \in M$$. You must choose an element $$m\in M$$ mapping to $$1\in A$$ - the element $$m$$ is in general not unique.

Note: We define an $$A$$-module $$N$$ to be "projective" iff for any surjective map of $$A$$-modules $$f\colon M\to N$$ there is a section $$s\colon N \to M$$ with $$f \circ s =\operatorname{id}_N$$. There are non-projective $$A$$-modules.

• Ok now i see, I was treating M as if it was an abelian group, my bad... thank you. And for the last part, for general $M,N$ $A$-modules what i was thinking is false because as we said there's no "identity element" in modules, right? Commented Jun 4, 2021 at 10:01