## 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 $V$.