Monthly Archives: March 2011

Compute Hom(ZZ/(n), A) for any ZZ-module A

Let A be any (left) \mathbb{Z}-module. Fix a \in A and let n \geq 1 be a positive integer. Prove that \varphi_a : \mathbb{Z}/(n) \rightarrow A given by \overline{k} \mapsto k \cdot a is a well-defined \mathbb{Z}-module homomorphism if and only if n \cdot a = 0. Prove that \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(n), A) = \mathsf{Ann}_A((n)). (See this previous exercise for a definition of \mathsf{Ann}_A((n)).)


Suppose that \varphi_a is a well-defined \mathbb{Z}-module homomorphism. In particular, we have n \cdot a = \varphi_a(\overline{n}) = \varphi_a(\overline{0}) = 0 \cdot a = 0. So n \cdot a = 0. Now suppose n \cdot a = 0. If k \equiv \ell mod n, we have n|k - \ell. Say k - \ell = nt. Now 0 = n \cdot a, so that 0 = nt \cdot a = (k - \ell) \cdot a = k \cdot a - \ell \cdot a. Thus k \cdot a = \ell \cdot a, and so \varphi_a(\overline{k}) = \varphi_a(\overline{\ell}). So \varphi_a is well-defined. Now suppose \overline{k}, \overline{\ell} \in \mathbb{Z}/(n) and r \in \mathbb{Z}. Then \varphi_a(\overline{k} + r \cdot \overline{\ell}) = \varphi_a(\overline{k + r\ell}) = (k + r\ell) \cdot a = k \cdot a + r \cdot (\ell \cdot a) = \varphi_a(\overline{k}) + r \cdot \varphi_a(\overline{\ell}). So \varphi_a is a \mathbb{Z}-module homomorphism.

Now define \Theta : \mathsf{Ann}_A((n)) \rightarrow \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(n), A) by \Theta(a) = \varphi_a. This mapping is properly defined by the above argument. We claim that \Theta is a \mathbb{Z}-module isomorphism.

First we show that \Theta is a \mathbb{Z}-module homomorphism. To that end, let a,b \in \mathsf{Ann}_A((n)) and r \in \mathbb{Z}. Now \Theta(a + r \cdot b)(\overline{k}) = \theta_{a + r \cdot b}(\overline{k}) = k \cdot (a + r \cdot b) = k \cdot a + kr \cdot b = k \cdot a + r \cdot (k \cdot b) = \varphi_a(\overline{k}) + r \cdot \varphi_b(\overline{b}) = \Theta(a)(\overline{k}) + r \cdot \Theta(b)(\overline{k}) = (\Theta(a) + r \cdot \Theta(b))(\overline{k}). Thus we have \Theta(a + r \cdot b) = \Theta(a) + r \cdot \Theta(b), and so \Theta is a \mathbb{Z}-module homomorphism.

Now suppose a \in \mathsf{ker}\ \Theta. Then \Theta(a)(\overline{k}) = 0 for all \overline{k}; in particular, 0 = \Theta(a)(\overline{1}) = \varphi_a(\overline{1}) = 1 \cdot a = a. Thus \mathsf{ker}\ \Theta = 0, and so \Theta is injective.

Finally, let \psi \in \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(n), A), and let a = \psi(\overline{1}). Note that n \cdot a = n \cdot \psi(\overline{1}) = \psi(\overline{n}) = \psi(\overline{0}) = 0, so that a \in \mathsf{Ann}_A((n)). We claim that \psi = \Theta(a). To see this, note that for all k we have \psi(\overline{k}) = \psi(k \cdot \overline{1}) = k \cdot \psi(\overline{1}) = k \cdot a = \varphi_a(\overline{k}) = \Theta(a)(\overline{k}). Thus \psi = \Theta(a), and so \Theta is surjective.

So we have \mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(n), A) \cong \mathsf{Ann}_A((n)).

A group homomorphism which is not an R-module homomorphism

Let R be a ring with 1. Give an explicit example of a map from one R-module to another which is a group homomorphism but which is not an R-module homomorphism.


Let R be any noncommutative ring, and fix a \in R such that a is not in the center of R. Consider M = R as a left module over itself via left multiplication.

Note that the mapping \psi : M \rightarrow M given by \psi(x) = a \cdot x is a group homomorphism since \psi(x + y) = r \cdot (x+y) = r \cdot x + r \cdot y = \psi(x) + \psi(y). However, if we choose s \in R not commuting with r, then \psi(s) = r \cdot s = rs while s \cdot \psi(1) = s \cdot (r \cdot 1) = sr. These two are not equal, so that \psi is not an R-module homomorphism.

“Is isomorphic to” is an equivalence relation on any set of R-modules

Let R be a ring with 1 and let \mathcal{M} be any set of (left) R-modules. Prove that the relation “is isomorphic to” is an equivalence relation on \mathcal{M}.


Recall that we say M \cong N if there exists an R-module isomorphism \theta : M \rightarrow N. We need to show that this relation is reflexive, symmetric, and transitive.

  1. (Reflexive) Let M \in \mathcal{M}. Note that the identity mapping \mathsf{id} : M \rightarrow M is an R-module isomorphism since \mathsf{id}(x + r \cdot y) = x + r \cdot y = \mathsf{id}(x) + r \cdot \mathsf{id}(y) for all x, y \in M and r \in R. Thus M \cong M, and so \cong is reflexive.
  2. (Symmetric) Suppose M,N \in \mathcal{M} such that M \cong N. That is, there exists an R-module isomorphism \theta : M \rightarrow N. Because \theta is a bijection, the converse relation \theta^{-1} is well-defined, and thus a bijective function N \rightarrow M. We claim that \theta^{-1} is in fact an R-module homomorphism. To see this, let r \in R and x,y \in N. Note that x = \theta(a) and y = \theta(b) for some a,b \in M. Now \theta^{-1}(x + r \cdot y) = \theta^{-1}(\theta(a) + r \cdot \theta(b)) = \theta^{-1}(\theta(a + r \cdot b)) = a + r \cdot b = \theta^{-1}(x) + r \cdot \theta^{-1}(y). So \theta^{-1} : N \rightarrow M is an R-module isomorphism, and we have N \cong M.
  3. (Transitive) Suppose M,N,T \in \mathcal{M} such that M \cong N and N \cong T. Then there exist R-module isomorphisms \theta : M \rightarrow N and \zeta : N \rightarrow T. Recall that \zeta \circ \theta : M \rightarrow T is a bijection; we claim that it is also an R-module homomorphism. To see this, let x,y \in M and r \in R. Then (\zeta \circ \theta)(x + r \cdot y) = \zeta(\theta(x + r \cdot y)) = \zeta(\theta(x) + r \cdot \theta(y)) = \zeta(\theta(x)) + r \cdot \zeta(\theta(y)) = (\zeta \circ \theta)(x) + r \cdot (\zeta \circ \theta)(y). So \zeta \circ \theta : M \rightarrow T is an R-module isomorphism, and we have M \cong T.

Thus \cong is an equivalence relation on \mathcal{M}.

Note also that \mathsf{id} is a unital R-module homomorphism, that if \theta is a unital R-module homomorphism then so is \theta^{-1}, and if \theta and \zeta are unital R-module homomorphisms (and the composition makes sense) then so is \zeta \circ \theta. Thus, if we replace “R-modules” by “unital R-modules”, the result still holds.

The kernels and images of module homomorphisms are submodules

Let R be a ring with 1. Let M and N be left R-modules, and let \varphi : M \rightarrow N be an R-module homomorphism. Prove that \mathsf{ker}\ \varphi and \mathsf{im}\ \varphi are R-submodules of M and N, respectively.


We use the Submodule Criterion for both sets.

Note that \varphi(0) = 0, so that \mathsf{ker}\ \varphi and \mathsf{im}\ \varphi are both nonempty.

Let x,y \in \mathsf{ker}\ \varphi and let r \in R. Note that \varphi(x + r \cdot y) = \varphi(x) + r \cdot \varphi(y) = 0, so that x + r \cdot y \in \mathsf{ker}\ \varphi. Thus \mathsf{ker}\ \varphi is an R-submodule of M.

Let x,y \in M and let r \in R. Note that \varphi(x) + r \cdot \varphi(y) = \varphi(x + r \cdot y); in particular, since \varphi(x) and \varphi(y) are arbitrary in \mathsf{im} \varphi, by the Submodule criterion we have that \mathsf{im}\ \varphi is an R-submodule of N.

Exhibit four distinct but isomorphic CC-algebra structures on CC²

Let A be the direct product ring \mathbb{C} \times \mathbb{C}. Let \tau_1 : \mathbb{C} \rightarrow \mathbb{C} denote the identity map, and let \tau_2 : \mathbb{C} \rightarrow \mathbb{C} denote complex conjugation. For any pair p,q \in \{1,2\}, define \varphi_{p,q} : \mathbb{C} \rightarrow \mathbb{C} \times \mathbb{C} by \varphi_{p,q}(z) = (\tau_p(z), \tau_q(z)).

  1. Prove that each \varphi_{p,q} is an injective ring homomorphism and that the \varphi_{p,q} all agree on the subring \mathbb{R} \subseteq \mathbb{C}. Deduce that A has four distinct \mathbb{C}-algebra structures, and explicity show the action z \cdot (u,v) in each case. Let A_{p,q} denote the \mathbb{C}-algebra structure induced by \varphi_{p,q}.
  2. Prove that if (p,q) \neq (s,t) then the identity map on A is not a \mathbb{C}-algebra homomorphism A_{p,q} \rightarrow A_{s,t}.
  3. Prove that for any pair (p,q) there is a ring isomomorphism \theta_{p,q} : A \rightarrow A such that \theta_{p,q} : A_{p,q} \rightarrow A_{1,1} is a \mathbb{C}-algebra isomorphism.

We begin with a lemma.

Lemma: Let R be a ring and let \{\varphi_i : R \rightarrow S_i\}_{i \in I} be a family of ring homomorphisms. Define \theta : R \rightarrow \prod_I S_i by \theta(r)_i = \varphi_i(r). Then \theta is a ring homomorphism. Moreover, if some \varphi_i is injective, then \theta is injective. Moreover, if each \varphi_i is unital, then \theta is unital. Proof: For all r,s \in R and i \in I, we have \theta(r+s)_i = \varphi_i(r+s) = \varphi_i(r) + \varphi_i(s) = \theta(r)_i + \theta(s)_i = (\theta(r) + \theta(s))_i and \theta(rs)_i = \varphi_i(rs) = \varphi_i(r) \varphi_i(s) = \theta(r)_i \theta(s)_i = (\theta(r) \theta(s))_i. So \theta(r+s) = \theta(r) + \theta(s) and \theta(rs) = \theta(r)\theta(s), and hence \theta is a ring homomorphism. Now suppose some \theta_k is injective. If \theta(r) = \theta(s), then in particular \varphi_k(r) = \theta(r)_k = \theta(s)_k = \varphi_k(s). So r = s, and \theta is injective. Now suppose each \varphi_i is unital; then \theta(1)_i = \varphi_i(1) = 1, and thus \theta(1) = 1. \square

Using the lemma, since \tau_1 and \tau_2 are both injective unital ring homomorphisms (in fact isomorphisms) \mathbb{C} \rightarrow \mathbb{C}, the mappings \varphi_{p,q} are all injective unital ring homomorphisms \mathbb{C} \rightarrow \mathbb{C} \times \mathbb{C}. Moreover, because A is commutative, \mathsf{im}\ \varphi_{p,q} \subseteq Z(A).

Thus each \varphi_{p,q} yields a \mathbb{C}-algebra structure on A, whose induced module action is given by z \cdot_{p,q} (u,v) = \varphi_{p,q}(r)(u,v) = (\tau_p(r)u, \tau_q(r)v). Specifically, we have (letting bars denote complex conjugation) z \cdot_{1,1} (u,v) = (zu, zv), z \cdot_{1,2} (u,v) = (zu, \overline{z}v), z \cdot_{2,1} (u,v) = (\overline{z}u, zv), and z \cdot_{2,2} (u,v) = (\overline{z}u, \overline{z}v).

We can see that these \mathbb{C}-algebra structures are different:

  • i \cdot_{1,1} (1,1) = (i,i)
  • i \cdot_{1,2} (1,1) = (i,-i)
  • i \cdot_{2,1} (1,1) = (-i,i)
  • i \cdot_{2,2} (1,1) = (-i,-i)

We let A_{p,q} denote the \mathbb{C}-algebra structure induced by \varphi_{p,q}. Suppose now that (p,q) \neq (s,t). If the identity mapping \mathsf{id} : A \rightarrow A is in fact a \mathbb{C}-algebra homomorphism, then we have $latex i \cdot_{s,t} (1,1) = i \cdot_{s,t} \mathsf{id}(1,1) = \mathsf{id}(i \cdot_{p,q} (1,1)) = i \cdot_{p,q} (1,1). However, we showed above that this is not the case for (p,q) \neq (s,t); hence the identity mapping is not a \mathbb{C}-algebra homomorphism A_{p,q} \rightarrow A_{s,t}.

Note that both \tau_1 and \tau_2 are ring isomorphisms. Given p,q \in \{1,2\}, define \psi_{p,q}(u,v) = (\tau_p^{-1}(u), \tau_q^{-1}(v)). We claim that \psi_{p,q} : A_{p,q} \rightarrow A_{1,1} is a \mathbb{C}-algebra isomorphism. Certainly \psi_{p,q} respects addition, and moreover is a bijection. Thus it suffices to show that it respects scalar multiplication. To that end, let z \in \mathbb{C} and let (u,v) \in A_{p,q}. Then \psi_{p,q}(z \cdot_{p,q} (u,v)) = \psi_{p,q}(\tau_p(z)u, \tau_q(z)v) = (\tau_p^{-1}(\tau_p(z)u), \tau_q^{-1}(\tau_q(z)v)) = (z \tau_p^{-1}(u), z \tau_q^{-1}(v)) = z \cdot_{1,1} (\tau_p^{-1}(u), \tau_q^{-1}(v)) = z \cdot_{1,1} \psi_{p,q}(u,v), as desired. Thus A_{p,q} and A_{1,1} are isomorphic as \mathbb{C}-modules.

An R-module which is a ring satisfying an extra property is naturally an R-algebra

Let R be a ring with 1 and let A be a ring. Suppose A is a unital left R-module which satisfies f \cdot (ab) = (r \cdot a)b = a(r \cdot b) for all r \in R and a,b \in A. Prove that the mapping \varphi : R \rightarrow A given by r \mapsto r \cdot 1_A is a unital ring homomorphism and that \mathsf{im}\ \varphi \subseteq Z(A). Conclude that A is a left R-algebra and that the natural R-module structure induced by this R-algebra structure coincides with the original R-module structure on A.


First, let r,s \in R. Note that \varphi(r+s) = (r+s) \cdot 1_A = r \cdot 1_A + s \cdot 1_A = \varphi(r) + \varphi(s) and that \varphi(rs) = rs \cdot 1_A = r \cdot (s \cdot 1_A) = r \cdot \varphi(s) = (r \cdot 1_A)\varphi(s) = \varphi(r) \varphi(s), so that \varphi is a ring homomorphism. Moreover, \varphi(1_R) = 1_R \cdot 1_A = 1_A, since A is a unital module. Finally, let r \in R and a \in A. Then \varphi(r)a = (r \cdot 1_A)a = r \cdot a = a(r \cdot 1_A) = a \varphi(r), so that \mathsf{im}\ \varphi \subseteq Z(A).

Now A is a left R-algebra via the mapping \varphi. This map induces a natural left R-module structure on A given by r \star a = \varphi(r)a. Note that r \star a = \varphi(r)a = (r \cdot 1_A)a = r \cdot a, so that the induced module structure coincides with the original structure.

Exhibit a left submodule which is not a right submodule

Let n > 1 be a natural number, let F be a field, and let R = \mathsf{Mat}_n(F) be the ring of n \times n matrices over F. Let N = \{ [a_{i,j}] \in M \ |\ a_{i,j} = 0\ \mathrm{if}\ j \neq 1 \}. Show that if we consider M = R as a left module over itself via left multiplication, then N \subseteq M is an R-submodule, but if we consider M = R as a right module over itself via right multiplication, then it is not an R-submodule.


First, let M = R be a left module over itself via left multiplication. Note that N \neq \emptyset since in particular 0 \in N. Now let A = [a_{i,j}], B = [b_{i,j}] \in N and let T = [r_{i,j}] \in R. Note that A + T \cdot B = [a_{i,j} + \sum_{k=1}^n r_{i,k}b_{k,j}]. Now if j \neq 1, then a_{i,j} = 0 and b_{k,j} = 0, so that a_{i,j} + \sum_{k=1}^n r_{i,k}b_{k,j} = 0. Thus A + T \cdot B \in N, and by the submodule criterion N \subseteq M is a submodule.

Now let M = R be a right module over itself via right multiplication. Let A = [a_{i,j}] where a_{i,j} = 1 if (i,j) = (1,1) and 0 otherwise; note that A \in N. Now let T = [r_{i,j}] where r_{i,j} = 1 if (i,j) = (1,2) and 0 otherwise. Consider A \cdot T = [\sum_{k=1}^n a_{i,k} r_{k,j}]. Suppose (i,j) = (1,2). Then r_{k,j} = 0 unless k = 1, and a_{1,1} = 1. Thus the (1,2) entry of A \cdot T is 1, and so A \cdot T \notin N. In particular, N is not an R-submodule of M.

Characterize the submodules of a given F[x]-module

Let F = \mathbb{R}, let V = \mathbb{R}^2, and let V be an F-module by componentwise multiplication. Let T : V \rightarrow V be the linear transformation given by T(x,y) = (-x,-y), and make V an F[x]-module in the usual way. Characterize the F[x]-submodules of V.


We begin with a lemma.

Lemma: Let F be a field and let V be an F-module. Fix an element \alpha \in F. Suppose T : V \rightarrow V is the linear transformation given by T(v) = \alpha \cdot v, and make V an F[x]-module in the usual way. Then every F-submodule of V is an F[x]-submodule and vice versa. Proof: Suppose first that W \subseteq V is an F-submodule of V. Let x,y \in W and let p(x) \in F[x]; note that p(x) \cdot v (thinking of V as an F[x]-module) is exactly the same as p(\alpha) \cdot v (thinking of V as an F-module and p(x) as a mapping F \rightarrow F). Then x + p(x) \cdot y = x + p(\alpha) \cdot y \in W. Certainly W is nonempty since 0 \in W, and so by the submodule criterion, W is an F[x]-submodule of V. Conversely, suppose W \subseteq V is an F[x]-submodule. Again W is nonempty. Now let x,y \in W and r \in F; thinking of r as a constant polynomial, we have x + r \cdot y \in W. Thus W is an F-submodule of V. \square

For this problem, we see that T(v) = -1 \cdot v. Thus the F[x]-submodules of V are precisely the F-submodules; hence, it suffices to characterize the F-submodules of V.

Let W \subseteq V be a submodule which is neither 0 nor all of V. In particular, there exists a nonzero element (x_0, y_0) \in W; either x_0 \neq 0 or y_0 \neq 0. Suppose now that there exists nonzero (x_1, y_1) \in W such that y_0x_1 \neq y_1x_0. Let r_1 = \dfrac{xy_1 - yx_1}{x_0y_1 - x_1y_0} and r_2 = \dfrac{xy_0 - yx_0}{x_0y_1 - x_1y_0}. Evidently, (x,y) = r_1 \cdot (x_0, y_0) + r_2 \cdot (x_1,y_1), and thus V \subseteq W, a contradiction. Thus, if (x,y) \in W, we have x_0y = y_0x. If x_0 \neq 0, then y = \dfrac{y_0}{x_0} x, and we have (x,y) = \dfrac{x}{x_0} \cdot (x_0, y_0). Similarly, if y_0 \neq 0 we have x = \dfrac{x_0}{y_0} y, and then (x,y) = \dfrac{y}{y_0} \cdot (x_0, y_0). In either case, we see that W = \{ r \cdot (x_0, y_0) \ |\ r \in F \}.

Characterize the submodules of a given F[x]-module

Let F = \mathbb{R}, let V = \mathbb{R}^2, and let T : V \rightarrow V be given by V(x,y) = (0,y). Let F act on V by componentwise multiplication and make V into an F[x]-module in the usual way. Characterize the F[x]-submodules of this module.


We claim that 0, V, X = \mathbb{R} \times 0, and Y = 0 \times \mathbb{R} are the only F[x] submodules of V.

Let W \subseteq V be an F[x]-submodule.

Suppose there exists an element (x_0, y_0) \in W such that x_0, y_0 \neq 0. Let (x,y) \in V. Note that (\frac{x}{x_0} + (\frac{y}{y_0} - \frac{x}{x_0})x) \cdot (x_0, y_0) = (x,y). Thus V \subseteq W, and in fact W = V.

Suppose now that for every element (x_0, y_0) \in W, either x_0 = 0 or y_0 = 0.

Suppose we have (x_0, 0) \in W with x_0 \neq 0. Suppose there exists (0,y_0) \in W with y_0 \neq 0; then (x_0,y_0) \in W with x_0,y_0 \neq 0, a contradiction. Thus we have W \subseteq X. Now let (x,0) \in X; since \frac{x}{x_0} \cdot (x_0,0) = (x,0), we have X \subseteq W. Thus W = X.

Suppose we have (0, y_0) \in W with y_0 \neq 0. As before, W \subseteq Y. If (0,y) \in Y, then \frac{y}{y_0} \cdot (0, y_0) = (0,y), and so Y \subseteq W. Thus W = Y.

Finally, if for all (x_0, y_0) \in W we have x_0 = y_0 = 0, then certainly W = 0.

Thus the only possible submodules of V are 0, X, Y, and V. 0 and V are trivially submodules; it remains to be shown that X and Y are in fact submodules; to accomplish this we use the submodule criterion.

Certainly (0,0) \in X, so that X is nonempty. Now suppose (a,0), (b,0) \in X and p(x) \in F[x]. Write p(x) = p_0 + xp^\prime(x). Note that x \cdot (b,0) = (0,0). Then (a,0) + p(x) \cdot (b,0) = (a,0) + p_0 \cdot (b,0) + p^\prime(x) \cdot (x \cdot (b,0)) = (a,0) + (p_0b, 0) + (0,0) = (a+p_0b,0) \in X. By the submodule criterion, X \subseteq V is a submodule.

Likewise (0,0) \in Y, so that Y is nonempty. Now suppose (0,a), (0,b) \in X and p(x) \in F[x]. Write p(x) = \sum p_ix^i. Note that the transformation T is idempotent; that is, that T^2(v) = T(v). In particular, x^k \cdot v = x \cdot v for all k \geq 1. Moreover, if v \in Y, then T(v) = v. Note that (0,a) + p(x) \cdot (0,b) = (0,a) + (\sum p_i x^i) \cdot (0,b) = (0,a) + \sum p_i \cdot (x^i \cdot (0,b)) = (0,a) + \sum p_i \cdot (0,b) = (0, a+ \sum p_ib) \in Y. Thus Y \subseteq V is a submodule.

Characterize the submodules of a given F[x]-module

Let F= \mathbb{R}, let V = \mathbb{R}^2, and let T : V \rightarrow V be given by V(x,y) = (y,-x). (Rotation about the origin by \frac{\pi}{2} radians.) Make V an F[x] module in the usual way. Show that 0 and V are the only F[x]-submodules of V.


Let W be an F[x]-submodule of V. Suppose now that there exists a nonzero element (x_0, y_0) \in W; then either x_0 \neq 0 or y_0 \neq 0, so that x_0^2 + y_0^2 > 0. Let (x,y) \in V. Now let r_1 = \dfrac{x_0x + y_0y}{x_0^2+y_0^2} and r_2 = \dfrac{y_0x - x_0y}{x_0^2+y_0^2}. Then (r_1 + r_2x) \cdot (x_0, y_0) = r_1 \cdot (x_0,y_0) + r_2 \cdot (x \cdot (x_0, y_0)) = (r_1x_0, r_1y_0) + r_2 \cdot (y_0, -x_0) = (r_1x_0, r_1y_0) + (r_2y_0, -r_2x_0) = (r_1x_0 + r_2y_0, r_1y_0 - r_2x_0) = (x,y). Thus (x,y) \in W, and we have W = V.

In particular, every nonzero F[x]-submodule of V is all of V. Hence 0 and V are the only F[x] submodules of V.