## Elementary Module Theory (I)

Modules can be likened to “vector spaces for rings”. To be specific, we shall see later that a vector space is precisely a module over a field (or in some cases, a division ring). This set of notes assumes the reader is reasonably well-acquainted with rings and groups, and has some prior experience with linear algebra, at least the computational aspects.

Definition. A (left)-module over R is an abelian group (M, +) together with a binary operation R × M → M (denoted as r·m or just rm, for $r\in R, m\in M$) such that for any $r,s\in R$ and $m,n\in M,$ we have:

1. $(r+s)m = rm + sm$;
2. $r(m+n) = rm+rn$;
3. $(rs)m = r(sm)$;
4. $1\cdot m = m$.

The map $R\times M\to M$ is often called scalar multiplication, since one should think of elements of M as “vectors” and those of R as “scalars”, following the terminology of linear algebra.

Exercise.

Find a structure which satisfies conditions 1-3 but not 4.

Let’s begin with the elementary properties of modules. For any $r\in R, m\in M$, we have:

• $0_R\cdot m = 0_M = r\cdot 0_m$, where $0_R\in R$ and $0_M\in M$;
• (-r)m = -(rm) = r(-m).

The proof is pretty straightforward:

• For the first equality, $0_R\cdot m + 0_R\cdot m =(0_R+0_R)\cdot m = 0_R\cdot m$. Adding $-(0_R\cdot m)$ to both sides gives $0_R\cdot m = 0_M.$ The other equality is left as an easy exercise.
• For the first equality, (-r)m + (rm) = ((-r) + r)m = 0R·m. Hence adding -(rm) to both sides gives us the desired equality. The other is an exercise.

Right Modules and Opposite Rings

Instead of defining a left module, we could have applied the action of $r\in R$ on the right. Thus, a right module M over R is an abelian group with a binary operation M × R → M such that for all $r, s\in R$ and $m,n\in M,$

1. m(rs) = mrms;
2. (mn)rmrnr;
3. m(rs) = (mr)s;
4. m·1 = m.

Note that if we turn the order of operation around to give R × M → M, then conditions 1, 2, 4 are exactly as above while condition 3 then becomes “(rs)ms(rm)”. Hence, if R is a commutative ring, then a left module over R can easily be turned to a right module by flipping the action as above. Conclusion: if R is non-commutative, a right R-module cannot be easily turned into a left module.

However, we can define the opposite ring:

Definition. If R is a ring, then its opposite ring Rop is defined by taking the abelian group (R, +) with the product operation *, where r*s = sr, with the RHS product taken from R.

Then a right R-module can also be described as a left module over its opposite ring Rop. Obviously, if R is commutative, then its opposite ring is identical to itself. On the other hand, even if R is non-commutative, it’s possible for $R\cong R^{op},$ in which case one can turn a right R-module into a left one via this isomorphism. For example, if H denotes the division ring of all quaternions with real coefficients, then:

$\phi:\mathbf{H}\to\mathbf{H},\quad a+ b\mathbf{i}+ c\mathbf{j}+ d\mathbf{k} \mapsto a-b\mathbf{i}-c \mathbf{j}-d\mathbf{k}$

gives $\phi(\alpha\beta) = \phi(\beta)\phi(\alpha) = \phi(\alpha) * \phi(\beta)$ where * is the product operation for the opposite ring.

Exercise.

Prove that the ring R of n × n real matrices is isomorphic to its opposite.

Finally, one notes that there are ample examples of rings which are not isomorphic to their opposite.

Examples of Modules

1. Every ring R is a module over itself.
2. Every vector space over a field K (e.g. R) is a K-module.
3. Z-module is precisely an abelian group (clearly, the underlying additive structure of a module gives us an abelian group; in the other direction, every abelian group A has a multiplication-by-n map which takes $A\to A, x\mapsto nx$ and gives us the structure of a Z-module).
4. Every ideal $I \subseteq R$ is an R-module, where scalar multiplication is given by ring product in R.
5. The n-dimensional free R-module is $R^n = R\times R\times\ldots\times R$, the Cartesian product of n copies of R. Product is given by $r(x_1, \ldots, x_n) := (rx_1, \ldots, rx_n).$
6. Consider the ring of Gaussian integers Z[i]. Now C is a Z[i]-module, where scalar multiplication is by the usual ring multiplication map in C.
7. Let $R := M_n(\mathbf{Z})$ be the ring of matrices with integer entries. Then $M :=\mathbf{Z}^n$ is an R-module, via multiplying a matrix by vector.
8. In representation theory, a representation of a group G over field K is precisely a K[G]-module.

Exercise

Let H be the ring of quaternions. Is R an H-module, if we define scalar multiplication as follows?

$\mathbf{H}\times\mathbf{R} \to \mathbf{R}, \ (a+bi + cj + dk, r) \mapsto (a+b+c+d)r.$

## Submodules

For now on, unless explicitly stated, all modules are assumed to be left modules.

Now submodules are simply subsets of a module M which can inherit the module structure from M.

Definition. A submodule of M is a subset $N\subseteq M$ satisfying the following:

• $0\in N$;
• if $m, n\in N$ then $m+n \in N$;
• if $r\in R, n\in N$, then $rn\in N$.

Immediately, the second and third conditions give us: for any $m,n\in N,$ mnm+(-1)n also lies in N. Thus, we can replace the first condition with $N\ne\emptyset$ since we can then pick $n\in N$ and infer that $0 =n-n\in N$ from the second and third conditions.

Given a collection of submodules of M, there’re at least two ways to create another submodule:

Proposition. If $\{N_i\}$ is a collection of submodules of M, then $N = \cap_i N_i$ is a submodule of M;and

$N' = \sum_i N_i := \{n_1 + n_2 + \ldots + n_k : n_j \in N_{i_j} \text{ for } j =1,\ldots,k\}$

is also a submodule of M. Thus $N'=\sum_i N_i$ is the set of all sums of finitely many terms from $\cup_i N_i.$

Proof.

• For N, clearly $0\in N$. Next, if $m,n\in N$, then $m,n\in N_i$ for every i. Hence, $m-n\in N_i$ for every i, which gives $m-n\in \cap_i N_i.$ Finally, suppose $n\in N$ and $r\in R.$ Then $n\in N_i$ for each i so $rn\in N_i$ for each i. Hence $rn\in \cap_i N_i.$
• For N’, again $0\in N'$ and clearly N’ is closed under addition since if $x,y\in N'$ are sums of finitely many terms from $\cup_i N_i$, then x+y is obtained by concatenating the two sums. Finally $r(n_1 + n_2 + \ldots + n_k) = (rn_1) + (rn_2) + \ldots + (rn_k)$ where each $rn_j \in N_{i_j}.$ ♦

Generated Submodules

Let $S\subseteq M$ be any subset and consider:

$\Sigma := \{ N : N\supseteq S, N \text{ is a submodule of } M\},$

i.e. the collection of all submodules of M which contain S. This collection is non-empty since it at least contains M. Let’s take the intersection of all these modules:

$\left := \cap_{N\in\Sigma} N.$

Being an intersection of submodules of M, <S> is also a submodule.

Definition. The resulting <S> is called the submodule generated by S.

Observe that <S> satisfies the following two critical properties:

• $\left \supseteq S$: indeed, it’s an intersection of N‘s, each of which contains S.
• if N’ is a submodule of M such that $N'\supseteq S$, then $N'\supseteq \left$: indeed since N’ contains S, it’s found in the collection ∑. Since <S> is defined to be the intersection of N’ and some other submodules, we have $N'\supseteq \left.$

It is for this reason that we often say:

The generated submodule <S> is the smallest submodule containing S.

The case where S = {m} is a singleton set is of special interest: we claim that <S> is the product $Rm = \{rm : r\in R\}.$ Indeed, since $m\in \left$ we clearly have $rm\in\left$ for any scalar $r\in R$ thus giving us the inclusion $Rm\subseteq \left.$ On the other hand, it’s easy to check that Rm is itself a submodule of M containing m. Hence, $Rm\supseteq \left$ and the two sets are equal.

Examples

1. Every module M has two obvious submodules: {0} and M. Any other submodule is called a proper submodule.
2. Consider ring R as a module over itself. Its submodules are called left ideals (compare this with the normal, two-sided ideals). Unwinding the definition, a left ideal of R is a subset $I\subseteq R$ such that
1. $0\in I$;
2. if $r,s\in I$, then $r+s\in I$;
3. if $r\in R, s\in I$, then $rs\in I$.
3. Consider C as a Q-module. Then the set S = {1+i, 3-2i, √2} generates the submodule comprising of all abic√2, for all $a,b,c\in \mathbf{Q}.$
4. If I is a left ideal of R and M is a module, then IM is a submodule of M. Here IM is defined as the set of all finite sums $x_1 y_1 + x_2 y_2 + \ldots + x_k y_k,$ where $x_i\in I$ and $y_i \in M.$
5. Consider the ring $R = M_n(\mathbf{Z})$ of all n × n integer matrices, with module $M = \mathbf{Z}^n.$ Then $N = 2\mathbf{z}^n = \{2\mathbf{v} : \mathbf{v}\in\mathbf{Z}^n\}$ is a proper submodule of M.
6. On the other hand, for the ring $R = M_n(\mathbf{R})$ of n × n real matrices and module $M =\mathbf{R}^n,$ one sees that there’s no proper submodule. In other words, any single non-zero $\mathbf{v}\in M$ generates the whole module M. Such modules are called simple modules and we’ll encounter them again later.
7. Let RR × R and consider MR as a module over itself. Then N := {0} × R is a left R-module since N is an ideal of M, and in particular a left-ideal.

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