generated submodule

Definition

Given a set $X\subset M$ for an R-module $M$, the submodule of $M$ generated by $X$ (denoted $⟨X⟩$) is the smallest submodule that contains all points of $X$. More concretely, this is the submodule

$$⟨X⟩:=\left\{\sum _{x\in X}{r}_{x}x|r_x\in R\right\}$$

We say $M$ is finitely generated if $M=⟨X⟩$ for some finite set $X$.

Relation to free modules

A module $M$ is generated by $X\subset M$ if and only if there exists a surjective module homomorphism

$$F^R(X)⟶M$$

where $F^R(X)$ denotes the free R-module on $X$.

In particular, $M$ is finitely generated if and only if there is a surjective module homomorphism

$$R^{\oplus n}⟶Mn\ge 1.$$

Proof

If there exists the surjective module homomorphism $F^R(X)\to M$, then by the universal property of free R-modules, the free module is determined by the set map $X\to M$.

Then, the image of $F^R(M)\to M$ is $⟨X⟩$, so since the map is surjective $M=⟨X⟩$.

The other way is even easier. If $M$ is generated by $⟨X⟩$ then the map $F^R(X)\to M$ will get to everything in $M$ by definition, so the map is surjective.

Remark

For finitely generated, we can take $R^{\oplus n}\to M$ to be determined by sending $1_R$ to the $n$ elements $m_1,\dots ,m_n\in M$ that are the generators.

This mean that every finitely generated $R$-module is a quotient of free $R$-modules.

Submodules of finitely generated modules

Submodules of finitely generated modules do not need to be finitely generated. (This is fixed by the definition of Noetherian R-module.)

For example, consider $\mathbb{Z}[x_1,x_2,\dots ,]$. This is not a finitely generated ring, but as a free $\mathbb{Z}$ module it is finitely generated (by the element $1$). However, the submodule $(x_1,x_2,\dots )\subset \mathbb{Z}[x_1,x_2,\dots ]$ is not finitely generated.

To prove this, assume $(x_1,x_2,\dots )$ is finitely generated by generators $g_1,\dots ,g_n$. Since these are finitely generated $g_1,\dots ,g_n$ is only depends on finitely many variants $x_1,\dots ,x_N$. Since $g_1,\dots ,g_n$ are generators then

$$x_{N+1}=\sum _{j=1}^N{r}_j{g}_j$$

However, if we set $x_1=x_N=0$ in this expression then we get $x_{N+1}=0$ since $g_j$ has no constant term so they will all be 0. This is a contradiction since $x_{N+1}$ can’t be 0.