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.

Throughout this article, we’ll let R denote a ring, possibly non-commutative.

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.

blue-linSubmodules

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<S\right> := \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<S\right> \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<S\right>: 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<S\right>.

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<S\right> we clearly have rm\in\left<S\right> for any scalar r\in R thus giving us the inclusion Rm\subseteq \left<S\right>. On the other hand, it’s easy to check that Rm is itself a submodule of M containing m. Hence, Rm\supseteq \left<S\right> 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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