A favorite problem from Linear Algebra Done Right

Solutions to Linear Algebra Done Right are all over the Internet, and I'm not really saying that's a good thing, but it means that I, in my independent study, am well able to verify my approaches.

That is, until I approach a problem differently than others have, and then I have to post my solution on math.stackexchange.com and hope that someone bothers to read it and confirm my approach. These are some of those problems, and incidentally, they were some of the more interesting ones I've enjoyed.

For some common notation: 

$\mathscr{P}(\textbf{F})$ : The vector space of (finite) polynomials on the field $\textbf{F}$
$\mathscr{P}_{n}(\textbf{F})$: The vector space of polynomials of degree $\leq n$
$\mathscr{L}(V, W)$: The vector space of linear maps from the vector space V to the vector space W (These maps can all be represented by matrices).

So, with that out of the way... 

Problem 4.11: 

Fix some $p \in \mathscr{P}(\textbf{F})$ with $p \neq 0$, where $p$ has degree $m$. Let $U = \lbrace pq | q \in \mathscr{P}(\textbf{F})\rbrace$.

Notice that U is a linear subspace of $\mathscr{P}(\textbf{F})$, composed of all polynomials with p as a factor. Since our vector space is commutative, we can talk about the quotient space $\mathscr{P}/U$, the space of cosets (or "affine subsets") of the form $r + U$ for some $r \in \mathscr{P}(\textbf{F})$ under typical coset addition.

The most important question we might want to ask is, what is the structure of this quotient space, and what is its dimension? It turns out that this quotient space is isomorphic to $\mathscr{P}_{m-1}(\textbf{F})$, and we will show this this by showing that U is the null space of a linear transformation, which we will set up in the next line.

Let $T \in \mathscr{L}(\mathscr{P}(\textbf{F}), \mathscr{P}_{m-1}(\textbf{F}))$ be the transformation defined as $Ts=r$  for $s \in \mathscr{P}(\textbf{F})$, where $r$ is the unique polynomial remainder of degree $<m$ defined via the division algorithm; the unique $r$ where $s = p q + r$ for some $q \in (\mathscr{P}(\textbf{F})$. (This ensures that T is well-defined and maps into $\mathscr{P}_{m-1}$)

Now, a natural question to ask would be, are you sure that T is actually a linear transformation? After all, all of the structural arguments that follow about kernels of maps rely on having a linear transformation.

First, we check additivity (homogeneity):

Let $s$ and $t$ be polynomials on $\textbf{F}$. 

By the division algorithm, $s=p q_1 + r_1$ and $t = p q_2 + r_2$ for $q_1, q_2$ polynomials, $r_1, r_2$ of degree $< m$. Evaluating T, we see that $Ts = r_1$ and $Tt = r_2$, and $Ts + Tt = r_1 + r_2$. 

Now we need to ensure that $Ts + Tt = T(s+t) $

Evaluating $s + t$, we see that: 

$s + t = p q_1 + p q_2 + r_1 + r_2 = p(q_1 + q_2) + (r_1 + r_2)$. 

Since $(r_1+r_2)$ also has degree $< m$, this is thus the unique representation of $s + t$ in this form, and $T(s+t) = r_1 + r_2$.

Now, we check scalar multiplication:

Let $\alpha \in \textbf{F}$, and consider $\alpha s$ for $s \in  \mathscr{P}(\textbf{F}$.
$\alpha s = \alpha(p q_1 + r_1) = p(\alpha q_1) + \alpha r_1$ 

Since $\alpha r_1$ is of degree $< m$ and of course $\alpha q_1$ is a polynomial over F, this again is the unique representation guaranteed by the division algorithm and $T(\alpha s)= \alpha T(s)$

Thus T is a well-defined and linear transformation.

Next, we show that U is the kernel of T: 

To show $U \subset null T$: Suppose $u \in U$. Then by the definition of U, $u = pq = pq + 0$ for some $q$, so $Tu = 0$ and $u \in null T $. 

To show $null T \subset U$: Suppose $u \in null T$. Then $u = pq + 0$ for some q, so $u = pq$ and $u \in U$.

And T is onto ... 

If $r$ is any polynomial in $\mathscr{P}_{m-1}(\textbf{F})$, then the inclusion $i(r) = r$ is the identical polynomial in $\mathscr{P}(\textbf{F})$ and $Tr = r$, since $ r = 0p + r$ and deg r $< m$. Thus all polynomials in $\mathscr{P}_{m-1}(\textbf{F})$ are in range T, and T is onto.

Now we can put it all together!

Using Theorem 3.91d, we see that $\mathscr{P}(\textbf{F})/null T \cong range T$.

Thus $\mathscr{P}(\textbf{F})/U \cong \mathscr{P}_{m-1}(\textbf{F})$, using the substitutions that U=null T and range T = $\mathscr{P}_{m-1}(\textbf{F})$.

This confirms the initial statement, and indicates that the dimension of the quotient space between these two infinite-dimensional vector spaces is $(m-1)$, where, again, $m$ is the degree of the polynomial $p$.