A ring action can fail to be visible long before the whole module becomes zero. In $\mathbb{Z}/6\mathbb{Z}$, multiplication by $2$ does not kill the module, since $2\cdot 1=2$, but it does kill the element $3$, since $2\cdot 3=0$. Multiplication by $6$ kills every element. The language of annihilators records these different levels of invisibility: scalars that kill one element, scalars that kill a submodule, and scalars that kill the entire representation of a ring. This distinction matters because many algebraic constructions are controlled not by the elements we see, but by the ring elements that act as zero. Quotient rings are built by forcing an ideal to vanish, torsion modules are detected by nonzero scalars that kill elements, and dual vector spaces organise subspaces by the functionals that vanish on them. The annihilator is the common mechanism behind these phenomena. [example: First Annihilators in $\mathbb{Z}/6\mathbb{Z}$] Let $R=\mathbb{Z}$ and let $M=\mathbb{Z}/6\mathbb{Z}$ be the usual $\mathbb{Z}$-module, so $n\bar{a}=\overline{na}$ for $n,a\in \mathbb{Z}$. For the element $\bar{3}\in M$, an integer $n$ kills $\bar{3}$ exactly when $\overline{3n}=\bar{0}$ in $\mathbb{Z}/6\mathbb{Z}$: \begin{align*} n\bar{3}=\bar{0}\iff \overline{3n}=\bar{0}\iff 6\mid 3n. \end{align*} Since $6\mid 3n$ means $3n=6q$ for some $q\in\mathbb{Z}$, this is equivalent to $n=2q$, hence to $2\mid n$. Conversely, if $n=2q$, then $3n=6q$, so $6\mid 3n$. Therefore \begin{align*} \operatorname{Ann}_{\mathbb{Z}}(\bar{3})=\{n\in\mathbb{Z}:2\mid n\}=2\mathbb{Z}. \end{align*} For the whole module, an integer $n$ kills every residue class if and only if $n\bar{a}=\bar{0}$ for every $a\in\mathbb{Z}$. This condition includes $a=1$, so it forces \begin{align*} n\bar{1}=\bar{0}\iff \bar{n}=\bar{0}\iff 6\mid n. \end{align*} Conversely, if $6\mid n$, then for every $a\in\mathbb{Z}$ we have $6\mid na$, so \begin{align*} n\bar{a}=\overline{na}=\bar{0}. \end{align*} Thus \begin{align*} \operatorname{Ann}_{\mathbb{Z}}(M)=6\mathbb{Z}. \end{align*} The element $\bar{3}$ only sees the modulus $2$, while the whole module still sees the full modulus $6$, so element annihilators and module annihilators carry different information. [/example] The example also shows a recurring theme: annihilators turn module-theoretic information into ideals of the ring. Once the vanishing set is an ideal, it can be compared, quotiented out, localised, or studied through prime ideals. ## Definition The most basic question is: which scalars act as zero on a given module? If a module is a representation of a ring, its annihilator measures the kernel of that representation. This is the right definition when we want to replace the original ring by the smaller ring that acts faithfully. [definition: Annihilator of a Module] Let $R$ be a ring with identity, and let $M$ be a left $R$-module. The annihilator of $M$ in $R$ is \begin{align*} \operatorname{Ann}_R(M)=\{r \in R : rm=0 \text{ for all } m \in M\}. \end{align*} [/definition] The elementwise, faithful-action, and linear-duality variants use the same vanishing idea in more specific settings. They appear below at the points where those settings become active. ## Elementwise Annihilation and Ideals ### Element Annihilators Killing the whole module is a global condition, but many modules contain elements that behave as if they live over smaller quotient rings. To see that local structure, we need to ask which scalars kill one chosen element rather than every element. This gives a finer invariant: different elements in the same module can have different annihilators. [definition: Annihilator of an Element] Let $R$ be a ring with identity, let $M$ be a left $R$-module, and let $m \in M$. The annihilator of $m$ in $R$ is \begin{align*} \operatorname{Ann}_R(m)=\{r \in R : rm=0\}. \end{align*} [/definition] The notation is deliberately parallel, but the two objects play different roles. The module annihilator records the kernel of the entire action. The element annihilator records the relations satisfied by a cyclic submodule generated by one element. ### Ideals from Vanishing The first structural fact is that annihilators are not arbitrary subsets. They are closed under exactly the operations needed to become ideals, with a small asymmetry in the noncommutative case. This matters because it lets us pass from module elements to quotient modules over quotient rings. [quotetheorem:7850] This theorem is the reason annihilators are algebraically useful rather than just descriptive. Once a vanishing set is an ideal, it can be used as the kernel of a quotient construction. ### Cyclic Submodules The elementwise version is especially important because every element generates a cyclic submodule. A cyclic submodule is the smallest place where the action of $R$ on one element can be studied without interference from the rest of $M$. The next result says that the annihilator is exactly the ideal of relations in that cyclic presentation. [quotetheorem:7851] The theorem says that an element is not just an element: it is a quotient of the base ring. Its annihilator is the ideal of relations that must be imposed to obtain the cyclic module it generates. [example: Cyclic Submodules of $\mathbb{Z}/12\mathbb{Z}$] Let $M=\mathbb{Z}/12\mathbb{Z}$ as a $\mathbb{Z}$-module, with $n\bar{a}=\overline{na}$. For the element $\bar{4}$, the multiples are determined by the residue of $n$ modulo $3$: \begin{align*} 0\bar{4}=\bar{0},\quad 1\bar{4}=\bar{4},\quad 2\bar{4}=\bar{8},\quad 3\bar{4}=\overline{12}=\bar{0}. \end{align*} If $n=3q+r$ with $r\in\{0,1,2\}$, then \begin{align*} n\bar{4}=(3q+r)\bar{4}=3q\bar{4}+r\bar{4}=q\overline{12}+r\bar{4}=r\bar{4}. \end{align*} Therefore \begin{align*} \mathbb{Z}\bar{4}=\{\bar{0},\bar{4},\bar{8}\}. \end{align*} Now compute the annihilator. An integer $n$ kills $\bar{4}$ exactly when \begin{align*} n\bar{4}=\bar{0}\iff \overline{4n}=\bar{0}\iff 12\mid 4n. \end{align*} If $12\mid 4n$, then $4n=12q$ for some $q\in\mathbb{Z}$, so $n=3q$ and hence $3\mid n$. Conversely, if $3\mid n$, say $n=3q$, then $4n=12q$, so $12\mid 4n$. Thus \begin{align*} \operatorname{Ann}_{\mathbb{Z}}(\bar{4})=\{n\in\mathbb{Z}:3\mid n\}=3\mathbb{Z}. \end{align*} The map $\mathbb{Z}/3\mathbb{Z}\to \mathbb{Z}\bar{4}$ defined by $\bar{n}\mapsto n\bar{4}$ is well-defined because changing $n$ by a multiple of $3$ changes $n\bar{4}$ by a multiple of $3\bar{4}=\overline{12}=\bar{0}$. It is surjective by the definition of $\mathbb{Z}\bar{4}$, and it is injective because $n\bar{4}=\bar{0}$ holds exactly when $n\in 3\mathbb{Z}$. Hence \begin{align*} \mathbb{Z}\bar{4}\cong \mathbb{Z}/3\mathbb{Z}. \end{align*} The annihilator records the exact modulus seen by the element $\bar{4}$, not the modulus of the whole module. [/example] The computation is small, but the pattern is general. To understand an element of a module, study the ideal of scalars that kill it; to understand a finitely generated module, study how the annihilators of generators interact. ## Faithful Actions and Quotients If a module has a nonzero annihilator, then the original ring contains information the module cannot see. The natural response is to divide out by the invisible ideal. The quotient ring then acts without changing the module action. [definition: Faithful Module] Let $R$ be a ring with identity, and let $M$ be a left $R$-module. The module $M$ is faithful if \begin{align*} \operatorname{Ann}_R(M)=\{0\}. \end{align*} [/definition] Faithfulness says that the action map remembers the ring. If $M$ is not faithful, the obstruction is not in the module but in the ring action: two scalars that differ by an element of $\operatorname{Ann}_R(M)$ act in exactly the same way on every element of $M$. The question is whether identifying precisely those indistinguishable scalars loses any module information, or whether it simply removes the part of the ring that was already invisible. Quotienting by $\operatorname{Ann}_R(M)$ is the construction that answers this question. [quotetheorem:7852] This result explains the conceptual meaning of the annihilator of a module: it is the largest part of the ring that can be removed without changing how the ring acts on the module. [example: Removing Invisible Scalars] Let $R=k[x]$ and let $M=k[x]/(x^2)$, viewed as an $R$-module by multiplication. Write $\bar{f}$ for the class of a polynomial $f\in k[x]$ modulo $(x^2)$. If $h(x)\in k[x]$, then \begin{align*} (h(x)x^2)\bar{f}=\overline{h(x)x^2f(x)}=\bar{0}, \end{align*} because $h(x)x^2f(x)\in (x^2)$. Thus every element of the ideal $(x^2)$ kills every element of $M$. Conversely, suppose $g(x)\in k[x]$ kills all of $M$. In particular, it kills the class $\bar{1}$, so \begin{align*} g(x)\bar{1}=\bar{0}. \end{align*} Since multiplication by $\bar{1}$ does not change the residue class, \begin{align*} g(x)\bar{1}=\overline{g(x)}. \end{align*} Therefore $\overline{g(x)}=\bar{0}$ in $k[x]/(x^2)$, which means exactly that $g(x)\in (x^2)$. Hence \begin{align*} \operatorname{Ann}_{k[x]}(M)=(x^2). \end{align*} Now let $k[x]/(x^2)$ act on the same module by \begin{align*} \bar{g}\cdot \bar{f}=\overline{g(x)f(x)}. \end{align*} If $\bar{g}$ kills every element of $M$, then it kills $\bar{1}$, and \begin{align*} \bar{g}\cdot \bar{1}=\bar{g}. \end{align*} So $\bar{g}=\bar{0}$. Thus the quotient ring has removed exactly the invisible scalars, and the resulting action is faithful. [/example] This example is a model for algebraic geometry and commutative algebra. A module over a large coordinate ring may only live on a smaller closed subscheme; the annihilator is the ideal cutting out the part of the ring that acts as zero. A faithful module lets us study ring elements by their action. A nonfaithful module forces us to distinguish equality in the ring from equality as operators on the module. [remark: Annihilator as a Kernel] The action of $R$ on $M$ gives a ring homomorphism from $R$ to the endomorphism ring of the underlying abelian group of $M$. Its kernel is $\operatorname{Ann}_R(M)$. In this sense, the annihilator is the kernel of the representation of $R$ on $M$. [/remark] The kernel viewpoint also explains why quotienting by the annihilator is not an auxiliary trick. It is the standard passage from a representation to a faithful representation. ## Annihilators in Exact Sequences ### Short Exact Sequences Modules are often studied through short exact sequences. If $B$ is assembled from a submodule $A$ and a quotient $C$, then a scalar that kills both pieces should kill $B$ after multiplying twice in the right way. This produces useful containment relations rather than a single equality in full generality. [quotetheorem:7853] The right containment says that if a scalar kills $B$, it kills every submodule and quotient. The left containment says that killing the quotient moves an element into the submodule, and then killing the submodule finishes the job. [example: Why Equality Need Not Hold] Consider the short exact sequence of $\mathbb{Z}$-modules \begin{align*}0 \to 2\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0.\end{align*} Here $2\mathbb{Z}/4\mathbb{Z}=\{\bar{0},\bar{2}\}\subset \mathbb{Z}/4\mathbb{Z}$, and the quotient map sends $\bar{a}\in \mathbb{Z}/4\mathbb{Z}$ to the class of $a$ modulo $2$; its kernel is exactly $\{\bar{0},\bar{2}\}$. For the submodule $2\mathbb{Z}/4\mathbb{Z}$, an integer $n$ kills every element if and only if it kills $\bar{2}$, since $\bar{0}$ is already killed by every integer. We have \begin{align*}n\bar{2}=\bar{0}\text{ in }\mathbb{Z}/4\mathbb{Z}\iff \overline{2n}=\bar{0}\iff 4\mid 2n\iff 2\mid n.\end{align*} Thus \begin{align*}\operatorname{Ann}_{\mathbb{Z}}(2\mathbb{Z}/4\mathbb{Z})=2\mathbb{Z}.\end{align*} For the quotient $\mathbb{Z}/2\mathbb{Z}$, an integer $n$ kills the whole module if and only if it kills $\bar{1}$: \begin{align*}n\bar{1}=\bar{0}\text{ in }\mathbb{Z}/2\mathbb{Z}\iff \bar{n}=\bar{0}\iff 2\mid n.\end{align*} Therefore \begin{align*}\operatorname{Ann}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z})=2\mathbb{Z}.\end{align*} For the middle module $\mathbb{Z}/4\mathbb{Z}$, the same generator test gives \begin{align*}n\bar{1}=\bar{0}\text{ in }\mathbb{Z}/4\mathbb{Z}\iff \bar{n}=\bar{0}\iff 4\mid n,\end{align*} so \begin{align*}\operatorname{Ann}_{\mathbb{Z}}(\mathbb{Z}/4\mathbb{Z})=4\mathbb{Z}.\end{align*} Finally, the product ideal $(2\mathbb{Z})(2\mathbb{Z})$ is $4\mathbb{Z}$: every product $(2a)(2b)$ equals $4ab$, and every element $4q$ equals $(2)(2q)$. Hence in this exact sequence the containment from [Annihilators in a Short Exact Sequence](/theorems/1547) reads \begin{align*}(2\mathbb{Z})(2\mathbb{Z})=4\mathbb{Z}\subset 4\mathbb{Z}\subset 2\mathbb{Z}.\end{align*} The middle term remembers an extension that is not detected by looking only at the two outer modules separately. [/example] This failure of equality is one reason annihilators are sensitive to extension data. They behave well enough to give bounds, but not so rigidly that they erase the difference between a direct sum and a nonsplit extension. ### Direct Sums A direct sum has no hidden extension data: an element is a finite tuple of elements from independent components. Therefore a scalar kills the whole direct sum precisely when it kills every summand. This gives a computation rule that turns decompositions into intersections of annihilators. [quotetheorem:7854] The intersection reflects a simple principle: a scalar kills a direct sum exactly when it kills every component. This is often the fastest way to compute annihilators of decomposed modules. ## Linear Annihilators and Duality ### Linear Vanishing Module annihilators are about scalars that vanish by acting on vectors. Linear algebra has a dual version, where the actors are functionals and the objects being killed are vectors in a subspace. This version is needed because subspaces are often described not by their spanning vectors, but by the linear equations they satisfy. [definition: Linear Annihilator] Let $k$ be a field, let $V$ be a $k$-vector space, and let $U \subset V$ be a linear subspace. The annihilator of $U$ in $V^*=\operatorname{Hom}_k(V,k)$ is \begin{align*} U^0=\{f \in V^* : f(u)=0 \text{ for all } u \in U\}. \end{align*} [/definition] This is the dual-space version of the same idea: a subspace is studied by all linear measurements that vanish on it. The notation $U^0$ is common in linear algebra, while $\operatorname{Ann}$ is common in module theory. ### Order Reversal In vector spaces, annihilators turn inclusions around. A larger subspace imposes more vanishing conditions, so fewer functionals survive. This order reversal is the algebraic core of many duality arguments. [quotetheorem:7855] The statement matches the intuition: vanishing on more vectors is a stronger condition. In finite dimensions, the remaining question is quantitative. A subspace $U\subseteq V$ can be described by linear equations, and its annihilator $U^0$ records exactly the equations that vanish on $U$. The obstruction to guessing its size directly is that different equations can be linearly dependent; the dimension formula below counts only the independent ones. [quotetheorem:7856] This theorem says that the number of independent equations cutting out a subspace is its codimension. The annihilator is therefore the dual home of the linear equations defining $U$. [example: A Plane in $\mathbb{R}^3$] Let $V=\mathbb{R}^3$ and let \begin{align*} U=\{(x_1,x_2,x_3)\in \mathbb{R}^3 : x_1+x_2+x_3=0\}. \end{align*} Every linear functional $f\in V^*$ has the form \begin{align*} f(x_1,x_2,x_3)=ax_1+bx_2+cx_3 \end{align*} for unique real numbers $a,b,c$. If $f\in U^0$, then $f$ vanishes on every vector of $U$. Since $(1,-1,0)\in U$ and $(1,0,-1)\in U$, we get \begin{align*} 0=f(1,-1,0)=a-b \end{align*} and \begin{align*} 0=f(1,0,-1)=a-c. \end{align*} Thus $a=b$ and $a=c$, so $a=b=c$. Writing this common value as $\lambda$, we have \begin{align*} f(x_1,x_2,x_3)=\lambda(x_1+x_2+x_3). \end{align*} Conversely, if $f(x_1,x_2,x_3)=\lambda(x_1+x_2+x_3)$ and $(x_1,x_2,x_3)\in U$, then $x_1+x_2+x_3=0$, so \begin{align*} f(x_1,x_2,x_3)=\lambda\cdot 0=0. \end{align*} Therefore \begin{align*} U^0=\operatorname{span}_{\mathbb{R}}\{(x_1,x_2,x_3)\mapsto x_1+x_2+x_3\}. \end{align*} Hence $\dim_{\mathbb{R}}U^0=1$. The subspace $U$ itself is two-dimensional: if $(x_1,x_2,x_3)\in U$, then $x_3=-x_1-x_2$, so \begin{align*} (x_1,x_2,x_3)=x_1(1,0,-1)+x_2(0,1,-1). \end{align*} The two spanning vectors are linearly independent, because \begin{align*} \alpha(1,0,-1)+\beta(0,1,-1)=(0,0,0) \end{align*} forces $\alpha=0$ from the first coordinate and $\beta=0$ from the second coordinate. Thus $\dim_{\mathbb{R}}U=2$. The annihilator is exactly the one-dimensional space of linear equations defining the plane. [/example] ### Double Annihilators The plane example suggests that a subspace should be recoverable from all linear equations vanishing on it. To make this precise, we must take the annihilator inside the double dual. In finite dimensions, evaluation identifies each vector with a functional on $V^*$, and the annihilator operation then returns to the original subspace. [quotetheorem:7857] The finite-dimensional hypothesis prevents pathologies. In infinite-dimensional spaces, the double annihilator depends on which dual and which topology are being used, leading naturally into functional analysis and weak topologies. ## Ideals, Supports, and Commutative Algebra ### Quotients and Closed Sets For commutative rings, annihilators connect module theory with the geometry of prime ideals. A module may not live over the whole spectrum of a ring; its annihilator cuts out a closed set containing the primes where the module can still be seen. Before relating annihilators to prime ideals, it is useful to isolate the special case where the module is the ring itself modulo an ideal. This case answers the reverse question: which ideals arise as annihilators? The answer shows that quotient rings are the basic source of module annihilators. [quotetheorem:7858] So every ideal of a commutative ring appears as the annihilator of some module, namely the quotient module it defines. But ideals are also modules in their own right, and their internal vanishing can reveal nilpotents and zero divisors. That motivates a separate name for the annihilator of an ideal. [definition: Annihilator of an Ideal] Let $R$ be a commutative ring, and let $I \trianglelefteq R$ be an ideal. The annihilator of $I$ in $R$ is \begin{align*} \operatorname{Ann}_R(I)=\{r\in R : ra=0 \text{ for all } a\in I\}. \end{align*} [/definition] This definition treats an ideal as an $R$-module. It is useful when an ideal is a submodule whose internal vanishing reveals nilpotents, zero divisors, or decompositions of the ring. [example: Nilpotents in a Cubic Polynomial Quotient] Let $R=k[x]/(x^3)$, and write $\bar{x}$ for the class of $x$. We compute the annihilator of the ideal $(\bar{x})\subset R$ directly from the definition. Since $(\bar{x})$ is generated by $\bar{x}$ as an $R$-module, an element $r\in R$ annihilates $(\bar{x})$ if and only if $r\bar{x}=0$. Every element of $R$ has a unique form $r=a+b\bar{x}+c\bar{x}^2$ with $a,b,c\in k$, because $\bar{x}^3=0$. Multiplying by the generator gives \begin{align*} (a+b\bar{x}+c\bar{x}^2)\bar{x}=a\bar{x}+b\bar{x}^2+c\bar{x}^3=a\bar{x}+b\bar{x}^2. \end{align*} The residue classes $1,\bar{x},\bar{x}^2$ form a $k$-basis of $R$, so $a\bar{x}+b\bar{x}^2=0$ holds exactly when $a=0$ and $b=0$. Thus the annihilating elements are precisely the multiples $c\bar{x}^2$, and therefore \begin{align*} \operatorname{Ann}_R((\bar{x}))=(\bar{x}^2). \end{align*} This ideal is nonzero because $\bar{x}^2\ne 0$ in $R$, but it squares to zero after one more multiplication by $\bar{x}$. The annihilator detects the last nonzero layer in the nilpotent filtration $R\supset (\bar{x})\supset(\bar{x}^2)\supset 0$. [/example] ### Torsion and Support Annihilators also measure torsion. Over an integral domain, the phrase "torsion element" means that a nonzero scalar kills the element. Annihilator language sharpens this by recording the full ideal of scalars that kill it. [definition: Torsion Element over an Integral Domain] Let $R$ be an integral domain, and let $M$ be an $R$-module. An element $m\in M$ is a torsion element if there exists $r\in R\setminus\{0\}$ such that \begin{align*} rm=0. \end{align*} [/definition] With annihilator language, this definition says that $m$ is torsion precisely when $\operatorname{Ann}_R(m)$ contains a nonzero element. The annihilator records not merely that torsion exists, but also its exact algebraic order. [example: Torsion over a Polynomial Ring] Let $R=k[x]$ and let $M=k[x]/(x^2)$. We determine the annihilators of two elements of $M$ by reducing products modulo the ideal $(x^2)$. For the class $\bar{x}\in M$, a polynomial $f(x)\in k[x]$ kills $\bar{x}$ exactly when $f(x)x$ is divisible by $x^2$, which is equivalent to $f(x)$ being divisible by $x$. Hence \begin{align*} \operatorname{Ann}_{k[x]}(\bar{x})=(x). \end{align*} For the class $\bar{1}\in M$, the equality $f(x)\bar{1}=0$ means exactly that $f(x)\in (x^2)$. Therefore \begin{align*} \operatorname{Ann}_{k[x]}(\bar{1})=(x^2). \end{align*} Both $\bar{1}$ and $\bar{x}$ are torsion elements, but their annihilators are different. The element $\bar{x}$ has smaller order because multiplication by $x$ already kills it, while $\bar{1}$ requires multiplication by $x^2$. The annihilator records this exact algebraic order rather than merely recording that torsion is present. [/example] The geometric slogan is that an annihilator defines where a module is forced to vanish. A large annihilator means the module is supported on a smaller algebraic set; a small annihilator means the module sees more of the ambient ring. [remark: Radical of the Annihilator] In commutative algebra, the radical $\sqrt{\operatorname{Ann}_R(M)}$ often controls the closed set associated to $M$ in $\operatorname{Spec}(R)$. The annihilator gives the scheme-theoretic vanishing, while its radical gives the underlying set-theoretic vanishing. [/remark] This distinction between an ideal and its radical is central when nilpotents are present. The annihilator can remember infinitesimal thickness that the radical forgets. ## Beyond and Connected Topics Annihilators sit at the boundary between algebraic structure and representation. In [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules), they appear naturally when modules over rings are introduced and when quotient modules are used to encode relations. In linear algebra, the annihilator of a subspace is part of the broader duality story developed in [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). There it should be read alongside dual bases, quotient spaces, rank-nullity, and the canonical map from a finite-dimensional vector space to its double dual. In homological algebra, annihilators interact with exact sequences, resolutions, and derived functors. The exact-sequence estimates above are an entry point toward the methods in [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions), where modules are studied through chains of maps rather than individual presentations. In functional analysis, linear annihilators become topological annihilators in dual Banach spaces. The finite-dimensional identity $(U^0)^0=U$ is replaced by closure statements involving weak and weak* topologies, making [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis) the natural continuation for infinite-dimensional duality. In commutative algebra and algebraic geometry, annihilators help define support, associated primes, and scheme-theoretic structure. A module over $R$ can be viewed as a module over $R/\operatorname{Ann}_R(M)$, so its annihilator is the first approximation to the part of $\operatorname{Spec}(R)$ on which the module lives. ## References Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Lang, *Algebra* (2002). Dummit and Foote, *Abstract Algebra* (2004).