# Abstract Nonsense

## Submodules

Point of Post: In this post we discuss the notion of submodules and prove a few basic theorems.

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Motivation As always, once we have defined a new structure we’d like to discuss the substructures inherit in the definition. Luckily for us, thinking of substructures shouldn’t be a big stretch since they are just generalizations of subspaces for vector spaces. Indeed, the subsets of a module $M$ for which the restriction of $R$ action is a module are precisely those subgroups of $(M,+)$ which are stable under the $R$-multiplication. Moreover, just as was the case for rings the submodules of a given module form a lattice. So, let’s get started!

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Let $R$ be a ring and let $M$ be a left $R$-module. We say that the additive subgroup $N$ of $M$ is a left submodule if the restriction of the $R$-action to $N$ is still a left $R$-module, we denote this by $N\leqslant M$. To be more explicit, suppose our ring action is $\phi:R\to\text{End}(M)$, then we say that $N\subseteq M$ is a submodule if $N$ is an additive subgroup of $M$ and the map $\psi:R\to\text{End}(N)$ given by $\psi(r)=\phi(r)_{\mid N}$ produces a well-defined module structure on $N$. It’s pretty easy to see that this is equivalent to saying that $N$ is stable under $R$-multiplication in the sense that $\phi_r(N)\subseteq N$ for every $r\in R$ since the only real “issue” that could arise with this mapping $\psi$ to be well-defined is that $\phi(r)_{\mid N}$ is not in $\text{End}(N)$ which is equivalent to having $\phi_r(N)\not\subseteq N$. The definition of a right submodule of a right $R$-module is defined similarly. Since it should be clear from context which kind of submodule we are using we shall often just speak of submodules.

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So, when is $N$ an additive subgroup which is also stable under $R$-multiplication. Well, when $R$ and $M$ is unital there is a nice way to think about this:

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Theorem: Let $R$ be a ring and $M$ a left $R$-module. Then, $N\leqslant M$ if and only if $N\ne\varnothing$ and for every $x,y\in N$ and $r\in R$ one has $x+ry\in N$.

Proof: To see that $N$ is an additive subgroup of $M$ we merely note that if $x,y\in N$ we then have by assumption that $x+(-1)y=x-y\in N$, and so $N$ is a subgroup as claimed. Moreover, $N$ is stable under $R$-multiplication since given any $x\in N$ and $r\in R$ one has that $0+rx=rx\in N$. The converse is clear.$\blacksquare$

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Ok, so let’s get some examples of submodules. As we mentioned earlier one of the prime examples of modules are rings thought of as modules over themselves (or any subring of themselves). Indeed, suppose we are given some ring $R$ and let’s consider it as a left $R$-module over itself. What are the submodules? Well, they are precisely the subgroups $I$ of $R$ with the property that $rx\in I$ for every $x\in I$–sound familiar? You guessed it! The submodules of a ring thought of as a left module overthemselves are precisely the left ideals of the ring. What about the other important class of modules, abelian groups? Indeed, suppose that $A$ is an abelian group considered as a left $\mathbb{Z}$-module in the usual way. Note then if $B$ is a subgroup of $A$ then automatically we know $nx\in B$ for every $n\in\mathbb{Z}$ and $x\in B$ since this is one of the defining characteristics of subgroups. Thus, every subgroup of $A$ of $B$ is a submodule, and clearly the converse is true so that submodules of abelian groups are just the subgroups.

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Now, let’s look at a more exotic sort of example. Suppose that are given an arbitrary ring $R$ and we look at the product ring $R^\mathbb{N}$. Clearly then $R^\mathbb{N}$ naturally has the natural structure of a left $R$-module by defining scalar multiplication coordinatewise (i.e. if $r\in R$ and $(r_n)\in R^\mathbb{N}$ then $r(r_n)=(rr_n)$). We claim that the direct product $R^{\oplus\mathbb{N}}$ of all elements of $R^\mathbb{N}$ with finite support (i.e. only finitely many non-zero coordinates) is a submodule of $R^\mathbb{N}$. Indeed, it’s easy to see that if we define $\text{supp} (a_n)$ to be the set of nonzero coordinates of $(a_n)$ then $\text{supp }(ra_n)\subseteq \text{supp }(a_n)$ and $\text{supp }((a_n)-(b_n))\subseteq \text{supp }(a_n)\cup\text{supp }(b_n)$ and so evidently both $\text{supp }r(a_n)$ and $\text{supp }((a_n)-(b_n))$ are finite if $\text{supp }(a_n),\text{supp }(b_n)$ are so that $R^{\oplus \mathbb{N}}$ is a submodule as desired.

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Of course, just as the case with rings we can form new submodules of a given ring out of certain subsets, just by intersecting all the submodules containing them. This will end up giving us a lattice structure just as it did for rings. Indeed, this is all based off of the following observation:

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Theorem: Let $R$ be a ring and $M$ a left $R$-module. Then, if $\left(M_\alpha\right)_{\alpha\in\mathcal{A}}$ is a set of submodules of $M$ then $\displaystyle \bigcap_{\alpha\in\mathcal{A}}M_\alpha\leqslant M$.

Proof: We know the intersection of subgroups of $M$ is a subgroup, and so it suffices to show that the intersection is stable under the $R$-multiplication. But, this is clear since if $x$ is in the intersection then $x\in M_\alpha$ for every $\alpha\in\mathcal{A}$ and since $M_\alpha\leqslant M$ we have that $rx\in M_\alpha$ and so $rx\in M_\alpha$ for every $\alpha\in\mathcal{A}$ and so $rx$ is in the intersection. $\blacksquare$

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Just as in the case of  rings we can define for a subset $X\subseteq M$ of a module $M$ the submodule generated by $X$, denoted $\langle X\rangle$ to be the intersection of all submodules of $M$ containing $X$. If $X=\{x_1,\cdots,x_n\}$ we are liable to write $\langle x_1,\cdots,x_n\rangle$ opposed to $\langle X\rangle$. If $M=\langle x_1,\cdots,x_n\rangle$ for some $x_1,\cdots,x_n\in M$ we say that $M$ is finitely generated and if $M=\langle x\rangle$ then $M$ is said to be cyclic (note that this jives with our previous group theoretic definition of cyclic groups).

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Now, note that in linear algebra there is a notion of something which looks fishly close to the above definition, the span of a set of vectors,denoted $RX$ (or just $Rx$ if $X=\{x\}$) or $\text{span}_R(X)$, for a set $X\subseteq M$, which was defined to be the set of linear combinations of elements of $X$ (linear combination here has the same meaning as it did in linear algebra). To be more formal if $M$ is a left $R$-module and $X\subseteq R$ we define $\text{span}_R(X)$ to be the following set $\left\{r_1 x_1+\cdots+r_n x_n:r_k\in R\text{ and }x_k\in X\right\}$. The first thing to note is that $\text{span}_R(X)$ is always a submodule of $M$ trivially. That said, the important thing to note is that in general $\text{span}_R(X)$ is not equal to $\langle X\rangle$. To see this consider $\mathbb{Z}$ as a module over $2\mathbb{Z}$, note then that $2\mathbb{Z}\{1\}=2\mathbb{Z}$ and so does not even contain $1$! The thing to note is that if $M$ is not a unital there is no reason for $Rx$ to even contain $x$ (for example). That said, if $M$ is unital then $\langle X\rangle$ and $\text{span}_R(X)$ do coincide (as is consistent with our previous knowledge of linear algebra over fields):

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Theorem: Let $R$ be a unital ring and $M$ a unital left $R$-module, then for any $X\subseteq M$ one has $\langle X\rangle=\text{span}_R(X)$.

Proof: Clearly now $x\in\text{span}_R(x)$ for every $x\in X$ since $x=1x\in\text{span}_R(X)$. Thus, $\text{span}_R(X)$ contains $x$ and is a submodule of $M$ from where we may conclude that $\langle X\rangle\subseteq\text{span}_R(X)$. The converse is always true since $\langle X\rangle$ being a submodule of $M$ containing $X$ implies that every linear combination in $X$ is in $\langle X\rangle$ so that $\text{span}_R(X)\subseteq\langle X\rangle$. The conclusion follows. $\blacksquare$

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Now, we make the observation that just like the case for rings these ideas allow us to define a lattice structure on the set $\mathfrak{L}(M)$ of submodules of the module $M$. Indeed, if $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}\subseteq\mathfrak{L}(R)$ we define the supremum of $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ to be the submodule generated by the union over all the submodules in $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ usually denoted $\sup\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$. Similarly, we define the infimum of $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ , denoted $\inf\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$, to be the intersection over the submodules in $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$. Now, it’s clear from the previous theorem that if $M$ is unital then $\sup\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ is equal to the sum $\displaystyle \sum_{\alpha\in\mathcal{A}}M_\alpha$ defined to be the set of all finite sums of things in $\displaystyle \bigcup_{\alpha\in\mathcal{A}}M_\alpha$ (note this agrees with our previous knowledge of the lattice of ideals in a ring). It is not difficult to check that if $M$ is unital that $\mathfrak{L}(M)$ is a modular complete lattice (cf. the discussion of this issue on the lattice of ideals for a ring).

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.