Submodules are the internal pieces of a [module](/page/Module) that still remember the full scalar action. A module is an additive abelian group equipped with multiplication by elements of a [ring](/page/Ring); a submodule is the kind of subset on which both operations remain valid. This is the module-theoretic analogue of a [subgroup](/page/Subgroup) inside a group, a [subspace](/page/Vector%20Space) inside a vector space, and, for the regular left module of a ring over itself, a left [ideal](/page/Ideal) inside a ring. Without submodules, quotient modules, exact sequences, kernels, images, generators, and most of homological algebra would have no natural internal language. The definition is shaped by a closure problem. If $M$ is an $R$-module and $N \subset M$, then $N$ can only be treated as a module in its own right if sums, additive inverses, and scalar multiples of elements of $N$ stay inside $N$. Closure under addition alone is not enough, because the scalar action may leave the subset. Closure under scalar multiplication alone is not enough, because the additive group structure may fail. ## Definition The first definition isolates exactly the closure conditions needed for a subset to inherit the module structure from its ambient module. It points back to the parent notion of module: a submodule is not an unrelated object inside $M$, but a subset on which the same module operations restrict. [definition: Submodule] Let $R$ be a ring and let $M$ be a left $R$-module. A subset $N \subset M$ is an $R$-submodule of $M$, written $N \le M$, if: 1. $N$ is a subgroup of the additive group $(M,+)$; 2. for every $r \in R$ and every $n \in N$, one has $rn \in N$. [/definition] The notation $N \le M$ deliberately parallels subgroup notation. It records more than inclusion: it says that $N$ is closed under the relevant algebraic structure. For modules over a fixed ring $R$, the ambient module usually determines the scalar action, so authors often say simply that $N$ is a submodule of $M$. The formulae on this page use the standard algebra convention that rings have a multiplicative identity and modules are unital. This matters most for generated submodules: under this convention, $1_Rm=m$, so the set of scalar multiples of one element already contains the generator itself. [remark: Ring and Module Convention] Throughout this page, rings are assumed to have a multiplicative identity $1_R$, and left or right $R$-modules are assumed to be unital: $1_Rm=m$ for every module element $m$. [/remark] When the ring is noncommutative, the side of the scalar action becomes part of the structure. A subset stable under left multiplication need not be stable under right multiplication, so the right-handed version deserves its own explicit definition. [definition: Right Submodule] Let $R$ be a ring and let $M$ be a right $R$-module. A subset $N \subset M$ is a right $R$-submodule of $M$ if: 1. $N$ is a subgroup of the additive group $(M,+)$; 2. for every $n \in N$ and every $r \in R$, one has $nr \in N$. [/definition] This page focuses on left modules unless stated otherwise. The right-module version is obtained by moving the scalar to the right throughout the statements. Many arguments distinguish the whole module from a genuine smaller internal piece. That distinction is needed for maximal submodules, simple quotients, and induction on subobjects, so the strict version is named separately. [definition: Proper Submodule] Let $R$ be a ring, let $M$ be an $R$-module, and let $N \le M$. The submodule $N$ is a proper submodule of $M$ if $N \subsetneq M$. [/definition] Proper submodules are where maximality and simplicity enter the subject. A module with no nonzero proper submodules behaves like an indivisible object, while a module with many proper submodules has internal structure that can be studied through quotients and filtrations. Once submodules have been named, the next natural question is how all of them sit inside a fixed module. Organising them by containment makes intersections, sums, generated submodules, and quotient constructions part of one ordered framework. [definition: Submodule Lattice] Let $R$ be a ring and let $M$ be an $R$-module. The submodule lattice of $M$ is the set \begin{align*} \operatorname{Sub}_R(M) := \{N \subset M : N \le M\}, \end{align*} ordered by inclusion. [/definition] The word "lattice" is justified by the intersection and sum operations developed below. This ordered viewpoint is especially important in representation theory, where invariant subspaces and composition series are organised by containment. ## Equivalent Characterisations Checking the subgroup condition directly can be more work than necessary. Since a module already has an abelian group structure, there is a compact test using subtraction and scalar multiplication. This test is often the fastest way to verify that a candidate subset is a submodule. [quotetheorem:8497] The nonempty hypothesis ensures that the zero element is present once closure under subtraction is known. For many computations, however, even subtraction is less natural than the language of linear combinations. That motivates a second criterion closer to the way generated modules are actually built. [quotetheorem:8498] This criterion says that a submodule is precisely a subset closed under finite $R$-linear combinations of its own elements. Since arbitrary subsets usually fail to have this closure, the next construction forces closure by taking the smallest submodule containing a chosen set. [definition: Submodule Generated by a Set] Let $R$ be a ring, let $M$ be a left $R$-module, and let $S \subset M$. The submodule generated by $S$, denoted $\langle S \rangle_R$, is \begin{align*} \langle S \rangle_R := \bigcap \{N \le M : S \subset N\}. \end{align*} [/definition] This intersection definition is intrinsic: it does not depend on choosing an order for the elements of $S$. For computation, though, one wants to know which elements this intersection contains. The answer is the expected finite-linear-combination formula. [quotetheorem:8499] A single element of a module can already generate a meaningful internal piece, because all of its scalar multiples must travel together in any submodule containing it. This motivates naming the one-generator case separately. [definition: Cyclic Submodule] Let $R$ be a ring, let $M$ be a left $R$-module, and let $m \in M$. The cyclic submodule generated by $m$ is \begin{align*} Rm := \{rm : r \in R\}. \end{align*} [/definition] Cyclic submodules generalise cyclic subgroups and one-dimensional spans. In modules over a [principal ideal domain](/page/Principal%20Ideal%20Domain), understanding cyclic submodules is a major step toward structure theorems for finitely generated modules. ## Standard Examples The most familiar case occurs over a field. There, the module axioms are the vector-space axioms, and submodules are exactly the usual linear subspaces. This example shows that submodules are a broader language for linear closure. [example: Vector Subspaces as Submodules] Let $k$ be a field and let $V$ be a $k$-vector space. Since a $k$-module is the same structure as a $k$-vector space, a subset $W \subset V$ is a $k$-submodule exactly when it is an additive subgroup of $V$ and is closed under multiplication by scalars from $k$; these are precisely the closure conditions for a vector subspace. For instance, in $k^3$, let \begin{align*} W := \{(a,b,c) \in k^3 : a+b+c=0\}. \end{align*} The zero vector belongs to $W$ because \begin{align*} 0+0+0=0. \end{align*} Now take $(a,b,c),(a',b',c') \in W$. Then $a+b+c=0$ and $a'+b'+c'=0$, so their sum satisfies \begin{align*} (a+a')+(b+b')+(c+c')=(a+b+c)+(a'+b'+c')=0+0=0. \end{align*} Thus $(a+a',b+b',c+c') \in W$. The additive inverse of $(a,b,c)$ also lies in $W$, because \begin{align*} (-a)+(-b)+(-c)=-(a+b+c)=-0=0. \end{align*} Finally, if $\lambda \in k$, then \begin{align*} \lambda a+\lambda b+\lambda c=\lambda(a+b+c)=\lambda\cdot 0=0. \end{align*} Hence $\lambda(a,b,c)=(\lambda a,\lambda b,\lambda c)\in W$. Therefore $W$ is closed under the inherited addition, additive inverses, and scalar multiplication, so $W$ is a $k$-submodule of $k^3$. This shows concretely how a homogeneous linear equation cuts out a submodule. [/example] The ring itself is a module over itself, and its submodules recover one-sided ideals according to the side of the action. This connection is one reason module theory is central in commutative algebra: in the commutative case the distinction between left, right, and two-sided ideals disappears, while over a noncommutative ring it must be tracked explicitly. [example: Ideals as Submodules] Let $R$ be a ring and regard $R$ as a left $R$-module over itself by left multiplication. A subset $I \subset R$ is a left $R$-submodule of this module exactly when $I$ is an additive subgroup of $(R,+)$ and, for every $r \in R$ and $x \in I$, the product $rx$ lies in $I$. These are precisely the defining closure conditions for a left ideal: addition and additive inverses stay in $I$, and multiplication by arbitrary ring elements on the left stays in $I$. If $R$ is regarded as a right $R$-module over itself, the scalar action is right multiplication, so a submodule is an additive subgroup $I \subset R$ such that $xr \in I$ for every $x \in I$ and $r \in R$. Thus the right $R$-submodules of $R$ are exactly the right ideals. If $I$ is both a left and right submodule of the regular module, then for every $x \in I$ and $r \in R$ both $rx$ and $xr$ lie in $I$, so $I$ is a two-sided ideal. When $R$ is commutative, $rx=xr$, so left stability and right stability are the same condition. For example, consider the $\mathbb{Z}$-module $\mathbb{Z}$. For every integer $n \ge 0$, the subset \begin{align*} n\mathbb{Z} := \{nk : k \in \mathbb{Z}\} \end{align*} is a submodule: it contains $0=n\cdot 0$, and if $na,nb \in n\mathbb{Z}$, then \begin{align*} na-nb=n(a-b) \in n\mathbb{Z}. \end{align*} Also, for any $t \in \mathbb{Z}$, \begin{align*} t(na)=n(ta) \in n\mathbb{Z}. \end{align*} Conversely, let $N \le \mathbb{Z}$. If $N=\{0\}$, then $N=0\mathbb{Z}$. If $N\ne \{0\}$, choose the least positive integer $n \in N$. Since $N$ is closed under integer scalar multiplication, every multiple $nk$ lies in $N$, so $n\mathbb{Z}\subset N$. For the reverse inclusion, take $m \in N$. By integer division, write \begin{align*} m=qn+r \end{align*} with $q \in \mathbb{Z}$ and $0\le r<n$. Since $qn \in N$ and $N$ is closed under subtraction, \begin{align*} r=m-qn \in N. \end{align*} The choice of $n$ as the least positive element of $N$ forces $r=0$, so $m=qn \in n\mathbb{Z}$. Hence $N=n\mathbb{Z}$. The case $n=0$ is the zero submodule, and the case $n=1$ is all of $\mathbb{Z}$. [/example] Not every additive subgroup is a submodule. This failure is the reason scalar stability appears explicitly in the definition. [example: Additive Subgroup That Is Not a Submodule] Regard $\mathbb{Q}$ as a module over itself, with scalar multiplication given by ordinary multiplication of rational numbers. The subset $\mathbb{Z} \subset \mathbb{Q}$ is an additive subgroup of $(\mathbb{Q},+)$: it contains $0$, and if $a,b \in \mathbb{Z}$, then $a-b \in \mathbb{Z}$, so it is closed under subtraction. However, $\mathbb{Z}$ is not a $\mathbb{Q}$-submodule of $\mathbb{Q}$. The submodule condition would require $rq \in \mathbb{Z}$ for every scalar $r \in \mathbb{Q}$ and every element $q \in \mathbb{Z}$. Taking $q=1 \in \mathbb{Z}$ and $r=\frac{1}{2} \in \mathbb{Q}$ gives \begin{align*} \frac{1}{2}\cdot 1=\frac{1}{2}. \end{align*} Since $\frac{1}{2}\notin \mathbb{Z}$, the scalar-closure condition fails. The same subset $\mathbb{Z}$ is a submodule of the $\mathbb{Z}$-module $\mathbb{Q}$. If $n \in \mathbb{Z}$ is a scalar and $m \in \mathbb{Z}$ is an element of the subset, then the scalar action is ordinary integer multiplication: \begin{align*} n\cdot m=nm. \end{align*} Because the product of two integers is an integer, $nm \in \mathbb{Z}$. Thus the obstruction above comes entirely from enlarging the scalar ring from $\mathbb{Z}$ to $\mathbb{Q}$; changing the scalar ring changes the submodule question. [/example] Representations require a version of submodule that records invariance under the acting algebra. For a [Lie algebra](/page/Lie%20Algebra) action, the right internal pieces are vector subspaces that are preserved by every element of the Lie algebra. [definition: Lie Algebra Submodule] Let $\mathfrak{g}$ be a Lie algebra over a field $k$, and let $V$ be a $\mathfrak{g}$-module with action map $\mathfrak{g} \times V \to V$ given by $(x,v) \mapsto xv$. A $k$-subspace $W \subset V$ is a $\mathfrak{g}$-submodule of $V$ if for every $x \in \mathfrak{g}$ and every $w \in W$, \begin{align*} xw \in W. \end{align*} [/definition] Here the additive and $k$-linear structure is supplied by the vector space $V$, while the extra requirement is invariance under the Lie algebra action. The following example shows how scalar stability becomes stability under operators. [example: A Lie Algebra Submodule] Let $\mathfrak{g}$ be the Lie algebra of diagonal linear operators on $V=k^2$, and let \begin{align*} W := \{(a,0) : a \in k\} \subset k^2. \end{align*} First $W$ is a $k$-subspace of $k^2$: it contains $(0,0)$ because $(0,0)=(a,0)$ with $a=0$; if $(a,0),(b,0)\in W$, then \begin{align*} (a,0)+(b,0)=(a+b,0)\in W; \end{align*} and if $c\in k$, then \begin{align*} c(a,0)=(ca,0)\in W. \end{align*} Now take a diagonal operator $x\in\mathfrak{g}$. Since $x$ is diagonal, there are scalars $\lambda,\mu\in k$ such that \begin{align*} x(u,v)=(\lambda u,\mu v) \end{align*} for every $(u,v)\in k^2$. For $w=(a,0)\in W$, this gives \begin{align*} xw=x(a,0)=(\lambda a,\mu\cdot 0)=(\lambda a,0). \end{align*} Because $\lambda a\in k$, the vector $(\lambda a,0)$ lies in $W$. Thus $W$ is stable under every element of $\mathfrak{g}$, so $W$ is a $\mathfrak{g}$-submodule. By contrast, let \begin{align*} L:=\{(a,a):a\in k\}. \end{align*} Consider the diagonal operator $y$ defined by \begin{align*} y(u,v)=(u,0). \end{align*} The vector $(1,1)$ lies in $L$, but \begin{align*} y(1,1)=(1,0). \end{align*} The vector $(1,0)$ is not in $L$, because membership in $L$ would require $(1,0)=(a,a)$ for some $a\in k$, forcing $a=1$ from the first coordinate and $a=0$ from the second. Thus $L$ is not stable under all diagonal operators. The example shows that a Lie algebra submodule is not just a linear subspace; it must also be preserved by the operators in the action. [/example] ## Properties Submodules are stable under intersection. This matters because it guarantees that the submodule generated by a set exists and that common solution spaces to compatible linear conditions remain submodules. [quotetheorem:8500] The theorem is a closure principle: any property expressed by membership in a family of submodules can be imposed simultaneously. Unions behave differently, and seeing this failure motivates the separate operation of adding submodules. [example: Union of Submodules Can Fail] Let $k$ be a field and view $k^2$ as a $k$-module. Define \begin{align*} N_1 := \{(a,0):a\in k\} \end{align*} and \begin{align*} N_2 := \{(0,b):b\in k\}. \end{align*} The set $N_1$ is a submodule: it contains $(0,0)$ because $(0,0)=(a,0)$ with $a=0$; if $(a,0),(a',0)\in N_1$, then \begin{align*} (a,0)-(a',0)=(a-a',0)\in N_1; \end{align*} and if $\lambda\in k$, then \begin{align*} \lambda(a,0)=(\lambda a,0)\in N_1. \end{align*} The same verification gives $N_2\le k^2$: it contains $(0,0)$, and for $(0,b),(0,b')\in N_2$, \begin{align*} (0,b)-(0,b')=(0,b-b')\in N_2, \end{align*} while for $\lambda\in k$, \begin{align*} \lambda(0,b)=(0,\lambda b)\in N_2. \end{align*} Their union $N_1\cup N_2$ is not a submodule. The vector $(1,0)$ lies in $N_1$, and the vector $(0,1)$ lies in $N_2$, so both lie in $N_1\cup N_2$. But their sum is \begin{align*} (1,0)+(0,1)=(1,1). \end{align*} If $(1,1)\in N_1$, then $(1,1)=(a,0)$ for some $a\in k$, forcing the second coordinate to satisfy $1=0$, impossible in a field. If $(1,1)\in N_2$, then $(1,1)=(0,b)$ for some $b\in k$, forcing the first coordinate to satisfy $1=0$, again impossible. Hence $(1,1)\notin N_1\cup N_2$, so the union fails closure under addition. The submodule containing both axes is their sum. Indeed, every vector $(x,y)\in k^2$ decomposes as \begin{align*} (x,y)=(x,0)+(0,y), \end{align*} with $(x,0)\in N_1$ and $(0,y)\in N_2$, so $N_1+N_2=k^2$. [/example] Because unions do not preserve submodules, one needs a replacement that contains several submodules and remains closed under addition and scalar multiplication. The sum of submodules is designed to be exactly that replacement. [definition: Sum of Submodules] Let $R$ be a ring, let $M$ be a left $R$-module, and let $(N_i)_{i \in I}$ be a family of submodules of $M$. The sum of the family is \begin{align*} \sum_{i \in I} N_i := \left\{\sum_{j=1}^{n} x_j : n \in \mathbb{N}\cup\{0\},\ i_j \in I,\ x_j \in N_{i_j}\text{ for }1 \le j \le n\right\}, \end{align*} where the case $n=0$ denotes the empty sum $0 \in M$. [/definition] The definition uses finite sums even when the index set is infinite because submodules are closed only under finite addition. Thus the construction has to solve a precise containment problem: it must include every $N_i$, remain a submodule, and avoid adding any elements not forced by those requirements. [quotetheorem:8501] The two constructions have an ordered meaning: intersections give the largest common submodule below a family, while sums give the smallest common submodule above it. This is the promised reason for calling $\operatorname{Sub}_R(M)$ a lattice. [quotetheorem:8108] Here the relevant closure operator is the operation that sends a subset of $M$ to the submodule it generates. Its closed sets are exactly the submodules of $M$: applying the operator again adds nothing precisely when the subset was already closed under addition, additive inverses, and scalar multiplication. The complete-lattice theorem explains why the earlier intersection and sum constructions are not isolated tricks. Meets are intersections, joins are generated sums, and arbitrary families of submodules can therefore be organized by containment without separately reproving each lattice operation. The next major construction uses a submodule in a different way: instead of combining submodules, it collapses one to zero. This shift from ordering submodules to forming new modules is the point of quotients. [definition: Quotient Module] Let $R$ be a ring, let $M$ be a left $R$-module, and let $N \le M$. The quotient module $M/N$ is the abelian [quotient group](/theorems/790) of $M$ by $N$, equipped with scalar multiplication $R \times (M/N) \to M/N$ defined by $(r,m+N) \mapsto rm+N$. [/definition] A quotient module should be an honest module, not merely a set of additive cosets. The obstruction is well-defined scalar multiplication: if $m+N=m'+N$, then one must have $rm+N=rm'+N$ for every scalar $r$, and this is precisely where closure of $N$ under scalar multiplication is used. [quotetheorem:8502] Quotient modules are the foundation for exact sequences and isomorphism theorems. To connect quotients with maps, one first needs the basic fact that the natural subobjects attached to a homomorphism are submodules. [quotetheorem:8333] This theorem explains why submodules appear throughout homological algebra. Exactness of a sequence is the assertion that one submodule, an image, equals another submodule, a kernel. The [first isomorphism theorem](/theorems/791) is the basic bridge between quotients and homomorphisms. It identifies the quotient by the kernel as the actual image of a map. [quotetheorem:862] The theorem is often the first place where submodules show their power: the obstruction to injectivity is not just a set of elements killed by the map, but a submodule that can be quotiented out. ## Relationship to Other Concepts Submodules specialise many familiar constructions. For a vector space over a field, they are linear subspaces. For the regular module $R$ over itself, they are left ideals, and in the commutative case they are ordinary ideals. For abelian groups, viewed as $\mathbb{Z}$-modules, submodules are exactly subgroups. [quotetheorem:8503] This result is a useful calibration point. Module theory simultaneously generalises linear algebra and abelian group theory. Once submodules are the allowed internal pieces of a module, the most indecomposable case is a nonzero module with no proper nonzero internal pieces at all. [definition: Simple Module] Let $R$ be a ring and let $M$ be a nonzero left $R$-module. The module $M$ is simple if its only submodules are $0$ and $M$. [/definition] To construct simple modules from larger modules, one needs a submodule that cannot be enlarged without becoming the whole ambient module. This is the module-theoretic maximality condition. [definition: Maximal Submodule] Let $R$ be a ring and let $M$ be a left $R$-module. A proper submodule $N \subsetneq M$ is a maximal submodule if there is no submodule $L \le M$ such that \begin{align*} N \subsetneq L \subsetneq M. \end{align*} [/definition] The reason to name maximal submodules is that quotienting by a larger submodule collapses more of $M$. This raises the natural test for when a quotient module has become indivisible: its submodules should correspond exactly to submodules of $M$ lying between $N$ and $M$. Thus simplicity of $M/N$ should be equivalent to the absence of any genuine intermediate submodule. [quotetheorem:8504] This theorem connects the internal ordering of submodules with the external structure of quotients. Another major way to understand a module is to ask when it breaks into complementary submodules rather than merely having quotients. [definition: Internal Direct Sum of Submodules] Let $R$ be a ring, let $M$ be a left $R$-module, and let $A,B \le M$. The module $M$ is the internal direct sum of $A$ and $B$, written \begin{align*} M = A \oplus B, \end{align*} if $M=A+B$ and $A \cap B=\{0\}$. [/definition] This definition is the internal version of the external direct sum. The two conditions play different roles: $M=A+B$ guarantees existence of a decomposition $m=a+b$, while $A\cap B=\{0\}$ is the obstruction-free condition that prevents two different decompositions of the same element. [quotetheorem:8505] Direct-sum decompositions are central in semisimple module theory. A module that splits into simple submodules is much easier to understand than one whose submodules are entangled through nonsplit extensions. In commutative algebra, submodules of free modules are used to encode systems of equations, syzygies, and presentations. If $R$ is a commutative ring, a submodule of $R^n$ can be viewed as a family of algebraic relations among $n$ generators. This is the setting of Noetherian modules, finite generation, and many structure theorems. In representation theory, a representation is often treated as a module over a group algebra, Lie algebra, or associative algebra. Submodules are then invariant subspaces. A representation is irreducible precisely when it has no submodules other than $0$ and itself. ## Common Pitfalls The first common mistake is to confuse subset inclusion with submodule inclusion. A subset $N \subset M$ may contain the right-looking elements but fail closure under scalar multiplication or subtraction. The notation $N \le M$ should only be used after the closure conditions have been verified. The second common mistake is to ignore the scalar ring. The same underlying subset can be a submodule for one scalar ring and fail to be a submodule for another. The example $\mathbb{Z} \subset \mathbb{Q}$ distinguishes the $\mathbb{Z}$-module structure from the $\mathbb{Q}$-module structure. The third common mistake is to assume that unions preserve submodules. Intersections preserve submodules, and sums provide the correct replacement for unions. The operation $N_1+N_2$ is built precisely to include sums of elements from both pieces. ## Beyond and Connections Submodules are the entry point to many later constructions in algebra. Quotient modules lead directly to exact sequences, kernels, cokernels, and the isomorphism theorems; these tools measure how modules differ by controlled collapse rather than by elementwise comparison. Direct sums and internal decompositions turn submodules into a language for splitting a module into independent pieces. The lattice of submodules also connects module theory with order-theoretic and representation-theoretic questions. Simple modules are the atoms of this structure, maximal submodules detect simple quotients, and composition series study how a module can be assembled from simple layers. In commutative algebra, submodules of free modules generalize ideals and support constructions such as presentations, syzygies, and localizations. ## References [Module](/page/Module). [Ring](/page/Ring). [Ideal](/page/Ideal). Dummit and Foote, *Abstract Algebra* (2004). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Lang, *Algebra* (2002).