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I am having trouble proving problem 10.1.18 in Dummit and Foote.

The problem is stated as follows:

Let $F = \mathbb{R}$, $V = \mathbb{R}^2$ and let $T$ be the linear transformation from $V$ to $V$ which is rotation clockwise about the origin by $\pi/2$ radians. Show that $V$ and $0$ are the only $F[x]$-submodules for this $T$.

Here is the route I tried to take:

We proceed with the goal of satisfying the submodule criteria, namely that a submodule is non-empty and closed under addition and multiplication by elements of $F[x]$.

With this in mind, let $S \subset V$ be an $F[x]$-submodule of $V$. Then the elements of $S$ are of the form $(x,y)$.

Therefore, to satisfy the second criterion of the Submodule criteria, we need to show that $(x_1,y_1) + p(x)(x_2,y_2) \in S$, for $p(x) \in F[x]$.

We also see that the linear transformation stated above is given by $T(x,y) = (y,-x)$.

Now, if I understand how the polynomials act on elements of $V$, we should have:

$$(x_1,y_1) + p(x)(x_2,y_2) = (x_1,y_1) + a_n T^n(x_2,y_2) + a_{n-1} T^{n-1}(x_2,y_2) + \dots + a_0(x_2,y_2)$$

where $a_n$ is the $n$-th coefficient in the polynomial $p(x)$.

From the definition of componentwise addition, and the fact that the reals are closed under addition and multiplication, it seems clear to me that $S$ is closed under addition and multiplication by elements of $F[x]$. Although aside from just claiming that, I'm not sure how to rigorously show this. Is this sufficient? Or am I missing something?

In addition, this only shows that $V$ is a submodule. I'm unclear how I can show that besides $0$, $V$ is the only other submodule.

2 Answers 2

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In the same chapter, note the correspondence between $F[x]$ submodules and those subspaces $W \subseteq V$ that are "invariant" under the action of $T$.

So, we can rephrase the question: what are the possible subspaces of $V$ so that $T(W) \subseteq W$? One can take $(x,y) \mapsto (-y,x)$ as what $T$ is really doing. In particular, notice that if there is some $x \in W$, then $\langle x,Tx \rangle=0$ , meaning that they are orthogonal, and so $W$ has at least dimension $2$, the full dimension of $V$.

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    Ok gotcha, I guess I should have read that section more carefully! So in other words, we're looking for $F $-submodules (vector subspace) that are $T$-stable subspaces. These are exactly the $F[x]$-submodules. – 2017-02-09
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    @FofX well, they can be identified with them. Formally, they are not the same, but you got it – 2017-02-10
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Hint: You're looking at $ \mathbb C $-submodules of $ \mathbb C $.

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    Hmm... ok I think I may understand the hint. The elements of $\mathbb{C}$ can be written as ordered pairs $(x,y)$ and have the particular addition and multiplication operations defined on them. I have to go to work, so I'll have to think about this more later, but am I on the right path? – 2017-02-09
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    @FofX You should remember that multiplying by $i$ in the complex plane is the same as rotating by $\frac{\pi}{2}$. More strictly speaking, note that $T^2+1=0$, hence the $F[x]$-action on $V$ induces an $F[x]/(x^2+1) \cong \mathbb C$-action on $V$ and this action is precisely given by the usual complex multiplication on $V=\mathbb R^2=\mathbb C$. – 2017-02-10