C[T] modules - Berrick and Keating Exercise 1.2.8 (c)

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with Exercise 1.2.8 (c) ...

Exercise 1.2.8 (c) reads as follows:View attachment 5097Now ... from this exercise we have that \(\displaystyle M \equiv \mathbb{C}^3\)

... and ...

\(\displaystyle A = \begin{pmatrix} 0&1&1 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}\)
Now ... given a vector \(\displaystyle m \in M \equiv \mathbb{C}^3\) and the matrix \(\displaystyle A\) we have that the products \(\displaystyle Am, A^2 M , \ ... \ ...\) are all elements of \(\displaystyle \mathbb{C}^3\) ...

So ... following B&K Example 1.2.2 (iv) ... ... see below ... ... consider \(\displaystyle M \equiv \mathbb{C}^3\) as a right module over the polynomial ring \(\displaystyle \mathbb{C} [T]\) where

\(\displaystyle m f(T) = mf_0 + Am f_1 + \ ... \ ... \ A^r f_r \)

where

\(\displaystyle f(T) = f_0 + f_1 T + \ ... \ ... \ f_r T^r \in \mathbb{C} [T]
\)
Now, we are given:

\(\displaystyle L(v) \equiv\) submodule of M generated by vThat is \(\displaystyle L(v) \equiv v f(T)\) ...

and

\(\displaystyle \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) ...
BUT ... now how do we show \(\displaystyle \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \in v f(T)\) for some choice of \(\displaystyle f\) ...
Can someone please help ... ?
Further ... can someone show me how to find all \(\displaystyle v\) with dim\(\displaystyle (L(v)) = 2\)?Hope someone can help ...

Peter

 

[Deveno]

Some preliminaries: let $z_1,z_2,z_3$ be any three arbitrary complex numbers.

If $v = \begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}$, then:

$Av = \begin{bmatrix}z_2 + z_3\\z_3\\0\end{bmatrix}$, and

$A^2v = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix} = \begin{bmatrix}z_3\\0\\0\end{bmatrix}$

and $A^kv = 0$ for all $k \geq 3$.

So if $f(T) = \sum\limits_{i = 0}^n f_iT^i$, then

$v \cdot f(T) = vf_0 + Avf_1 + A^2vf_2$ (all the higher terms are 0).

Now if $v = \begin{bmatrix}1\\0\\0\end{bmatrix}$, we see that $Av$ and $A^2v$ are both $0$, so that for this special $v$:

$v\cdot f(T) = vf_0$, in other words:

$L_0 = \left\{ \begin{bmatrix}a\\0\\0\end{bmatrix}: a \in \Bbb C\right\}$

In this case, the ring that is "actually" acting on $\Bbb C^n$ (which due to vector addition is an abelian group) is:

$\Bbb C[T]/(T^3)$, because $A$ is *nilpotent*, of nilpotency $3$ (the minimal polynomial of $A$ is $T^3$, since $A^3 = 0$, and no factor of $T^3$ annihilates $A$).

So we may as well only consider (to calculate $L(v)$) $f$ of the form: $f(T) = a + bT + cT^2$ with $a,b,c \in \Bbb C$.

Going back to our original $v$ (with arbitrary entries) we see that for such an $f$:

$v\cdot f(T) = va + Avb + A^2vc = \begin{bmatrix}az_1\\az_2\\az_3\end{bmatrix} + \begin{bmatrix}b(z_2+z_3)\\bz_3\\0\end{bmatrix} + \begin{bmatrix}cz_3\\0\\0\end{bmatrix}$

(since $\Bbb C$ is a field, I am using commutivity of the field multiplication to put the coefficients of the polynomial in front of the $z$'s)

$= \begin{bmatrix}az_1 + bz_2 + (b+c)z_3\\az_2+bz_3\\az_3\end{bmatrix}$

Can you take it from here?

 

[Math Amateur]

Some preliminaries: let $z_1,z_2,z_3$ be any three arbitrary complex numbers.

If $v = \begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}$, then:

$Av = \begin{bmatrix}z_2 + z_3\\z_3\\0\end{bmatrix}$, and

$A^2v = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix} = \begin{bmatrix}z_3\\0\\0\end{bmatrix}$

and $A^kv = 0$ for all $k \geq 3$.

So if $f(T) = \sum\limits_{i = 0}^n f_iT^i$, then

$v \cdot f(T) = vf_0 + Avf_1 + A^2vf_2$ (all the higher terms are 0).

Now if $v = \begin{bmatrix}1\\0\\0\end{bmatrix}$, we see that $Av$ and $A^2v$ are both $0$, so that for this special $v$:

$v\cdot f(T) = vf_0$, in other words:

$L_0 = \left\{ \begin{bmatrix}a\\0\\0\end{bmatrix}: a \in \Bbb C\right\}$

In this case, the ring that is "actually" acting on $\Bbb C^n$ (which due to vector addition is an abelian group) is:

$\Bbb C[T]/(T^3)$, because $A$ is *nilpotent*, of nilpotency $3$ (the minimal polynomial of $A$ is $T^3$, since $A^3 = 0$, and no factor of $T^3$ annihilates $A$).

So we may as well only consider (to calculate $L(v)$) $f$ of the form: $f(T) = a + bT + cT^2$ with $a,b,c \in \Bbb C$.

Going back to our original $v$ (with arbitrary entries) we see that for such an $f$:

$v\cdot f(T) = va + Avb + A^2vc = \begin{bmatrix}az_1\\az_2\\az_3\end{bmatrix} + \begin{bmatrix}b(z_2+z_3)\\bz_3\\0\end{bmatrix} + \begin{bmatrix}cz_3\\0\\0\end{bmatrix}$

(since $\Bbb C$ is a field, I am using commutivity of the field multiplication to put the coefficients of the polynomial in front of the $z$'s)

$= \begin{bmatrix}az_1 + bz_2 + (b+c)z_3\\az_2+bz_3\\az_3\end{bmatrix}$

Can you take it from here?

Thanks Deveno ... I think I understand your post ...

To outline the situation in the exercise ... ... we have the following:

$v = \begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}$ ... ... so \(\displaystyle v\) is a vector of complex constants where \(\displaystyle z_1, z_2\) and \(\displaystyle z_3\) are (arbitrary) complex constants ...

We also have $L_0 = \left\{ \begin{bmatrix}k\\0\\0\end{bmatrix}: k \in \Bbb C\right\}$ where \(\displaystyle k\) varies over all \(\displaystyle \mathbb{C}\) ... ...

... and we have ...$f(T) = a + bT + cT^2$ with $a,b,c \in \Bbb C$ ... ... so \(\displaystyle a, b\) and \(\displaystyle c\) vary over all \(\displaystyle \mathbb{C}\) Now we also have that

\(\displaystyle L(v) = v \cdot f(T) = \begin{bmatrix}az_1 + bz_2 + (b+c)z_3\\az_2+bz_3\\az_3\end{bmatrix} \)
Now, we have to show that \(\displaystyle L_0 \subseteq L(v)\)So consider \(\displaystyle \begin{pmatrix} m \\ 0 \\ 0 \end{pmatrix} \in L_0\) where \(\displaystyle m \in \mathbb{C}
\)We need to show that \(\displaystyle \begin{pmatrix} m \\ 0 \\ 0 \end{pmatrix} \in L_0 \in L(v) = \begin{bmatrix}az_1 + bz_2 + (b+c)z_3\\az_2+bz_3\\az_3\end{bmatrix}\)Now, if \(\displaystyle a = b = 0\) then

\(\displaystyle L(v) = \begin{pmatrix} (b + c) z_3 \\ 0 \\ 0 \end{pmatrix}\)so ... we see that if we set \(\displaystyle a = b = 0\) and set \(\displaystyle c\) such that \(\displaystyle m = c z_3\) then \(\displaystyle \begin{pmatrix} m \\ 0 \\ 0 \end{pmatrix} \in L_0\) ...Thus if we choose \(\displaystyle a = b = 0\) and choose \(\displaystyle c = m/z_3\) then \(\displaystyle \begin{pmatrix} m \\ 0 \\ 0 \end{pmatrix} \in L_0\) ... So we have shown that \(\displaystyle L_0 \subseteq L(v)\) ...
Is that correct? Indeed, does the above make sensse ... ?

Hope you can confirm it is correct ...

Peter

 

[Deveno]

One small detail...why must $v \neq 0$?

 

[Math Amateur]

One small detail...why must $v \neq 0$?

Hi Deveno,

Yes ... omitted consideration of \(\displaystyle v\) possibly being zero ...

so ... ... If \(\displaystyle v = \underline{0} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} \)Then we have \(\displaystyle L(v) = v \cdot f(T) = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)and then it follows that ...\(\displaystyle \begin{pmatrix} m \\ 0 \\ 0 \end{pmatrix} \notin L_0 \) ... since \(\displaystyle m\) can range over all \(\displaystyle \mathbb{C}\) ... ... and so \(\displaystyle L_0\) is not a subset of nor equal to \(\displaystyle L(v)\) ...
Hope the above analysis is correct ...

Peter