K^n as a K[T]-module - Example 2.1.2

[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 understanding Example 2.1.2 (ii) (page 39) which concerns $$V = K^n$$ viewed as a module over the polynomial ring $$K[T]$$.

Example 2.1.2 (ii) (page 39) reads as follows:View attachment 2965In the above text by B&K we read:

" ... ... it is easy to verify that the decomposition $$V = U \oplus W$$ expresses $$V$$ as a direct sum of $$K[T]$$-submodules precisely when $$A = \left(\begin{array}{cc}B&0\\0&D\end{array}\right)$$

with $$B$$ an $$r \times r$$ matrix

and

$$D$$ an $$(n - r) \times (n - r)$$ matrix, $$B$$ and $$D$$ giving the action of $$T$$ on $$U$$ and $$W$$ respectively. ... ..."

I am trying to formally and rigorously verify this statement, but am unsure how to approach this task. Can someone please help me to get started on this verification ... ?

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Other relevant text in B&K that MHB members may need to interpret and understand the above example follows.

B&K's notation for polynomial rings is as follows:

View attachment 2966
B&K's definition of a module is as follows:
View attachment 2967
View attachment 2968
B&K's explanation and notation for $$K^n$$ as a right module over $$K[T|$$ is as follows:View attachment 2969

 

[Deveno]

Well, first let's look at what we need to happen for $\mathcal{K}^n$ to be the direct sum of $U$ and $V$ as $\mathcal{K}[T]$-modules.

First of all, we need $U$ and $V$ to act as $\mathcal{K}[T]$-submodules.

The closure under addition is clear: as vector subspaces, both $U$ and $V$ are abelian groups, and thereby closed under addition.

So what we need to do is verify that they are likewise closed under the $\mathcal{K}[T]$-action, that:

$u \cdot f(T) \in U$ for all $f(T) \in \mathcal{K}[T]$ (a similar consideration holds for $V$).

So we need $Au \in U$. This will ensure that $A^tu \in U$, and therefore that:

$A^tuf_j \in U$, and so (adding all the terms) $u \cdot f(T) \in U$.

If we write $A$ in block form, this ($Au \in U$) becomes:

$\begin{bmatrix}B&H\\K&D \end{bmatrix} \begin{bmatrix}u\\0 \end{bmatrix} = \begin{bmatrix}u'\\0 \end{bmatrix}$

To achieve this, we must have $Ku + D0 = Ku = 0$, for ALL $u \in U$. So $K$ is the 0-block.

A similar analysis with $V$ shows $H$ must be the 0-block.

Note that $U + W = V$ considered purely as abelian groups. Furthermore, note that:

$u \cdot 1_{\mathcal{K}[T]} = Iu\cdot 1 = u$, and similarly for $V$, so as $\mathcal{K}[T]$-modules these are non-zero (this is true even if the matrix $A$ is the 0-matrix, since the action of constant polynomials does not have any $A^tu$ terms).

Finally, since $U \cap W = \{0_V\}$ (since we have a direct sum of vector spaces), this is still true when we consider them as $\mathcal{K}[T]$-modules. So (DS1) and (DS2) are satisfied, we have a direct sum as modules.

(in my opinion this flows better with a left-action, but it's "essentially" the same).

 

[Math Amateur]

Well, first let's look at what we need to happen for $\mathcal{K}^n$ to be the direct sum of $U$ and $V$ as $\mathcal{K}[T]$-modules.

First of all, we need $U$ and $V$ to act as $\mathcal{K}[T]$-submodules.

The closure under addition is clear: as vector subspaces, both $U$ and $V$ are abelian groups, and thereby closed under addition.

So what we need to do is verify that they are likewise closed under the $\mathcal{K}[T]$-action, that:

$u \cdot f(T) \in U$ for all $f(T) \in \mathcal{K}[T]$ (a similar consideration holds for $V$).

So we need $Au \in U$. This will ensure that $A^tu \in U$, and therefore that:

$A^tuf_j \in U$, and so (adding all the terms) $u \cdot f(T) \in U$.

If we write $A$ in block form, this ($Au \in U$) becomes:

$\begin{bmatrix}B&H\\K&D \end{bmatrix} \begin{bmatrix}u\\0 \end{bmatrix} = \begin{bmatrix}u'\\0 \end{bmatrix}$

To achieve this, we must have $Ku + D0 = Ku = 0$, for ALL $u \in U$. So $K$ is the 0-block.

A similar analysis with $V$ shows $H$ must be the 0-block.

Note that $U + W = V$ considered purely as abelian groups. Furthermore, note that:

$u \cdot 1_{\mathcal{K}[T]} = Iu\cdot 1 = u$, and similarly for $V$, so as $\mathcal{K}[T]$-modules these are non-zero (this is true even if the matrix $A$ is the 0-matrix, since the action of constant polynomials does not have any $A^tu$ terms).

Finally, since $U \cap W = \{0_V\}$ (since we have a direct sum of vector spaces), this is still true when we consider them as $\mathcal{K}[T]$-modules. So (DS1) and (DS2) are satisfied, we have a direct sum as modules.

(in my opinion this flows better with a left-action, but it's "essentially" the same).

Thanks Deveno ... but I need your help in order to clarify some of the mechanics of the $\mathcal{K}[T]$-actions for $$U$$ and $$V$$ ...

I can see that $$U$$ and $$V$$ are both abelian groups under addition and are therefore closed under addition, but as I have indicated above I am having trouble understanding the mechanics of the $\mathcal{K}[T]$-actions for $$U$$ and $$V$$ ... hope you can help ...
I will explain my difficulties by focusing on $$ U = \mathcal{K}^r$$ ... the same considerations apply to $$ V = \mathcal{K}^{n-r} $$ ... ...

Now, consider the action $$u \bullet f(T)$$ ... ...

$$ u \bullet f(T) = u \bullet (f_0 + f_1T + f_2T^2 + ... \ ... + f_rT^r ) $$

Therefore, by the definition of the action we have:

$$ u \bullet f(T) = uf_0 + Auf_1 + A^2uf_2 + ... \ ... + A^ruf_r $$

Now consider the term $$uf_0$$ in the above expression ...

Let $$u = \begin{pmatrix} u_1 \\ . \\ . \\ . \\ u_r \end{pmatrix}$$, $$f_0 = \begin{pmatrix} f_{10} \\ f_{20} \\ . \\ . \\ . \\ f_{n0} \end{pmatrix}$$

... so how do we calculate/form $$uf_0$$?

Similarly $$A$$ is $$(n \times n)$$ , $$u$$ is $$(r \times 1)$$, and $$f$$ is $$(n \times 1)$$ ...

so then how do we calculate/form $$ Auf_1 $$ ... and so on?

Hope you can help ...

Peter

 

[Deveno]

Thanks Deveno ... but I need your help in order to clarify some of the mechanics of the $\mathcal{K}[T]$-actions for $$U$$ and $$V$$ ...

I can see that $$U$$ and $$V$$ are both abelian groups under addition and are therefore closed under addition, but as I have indicated above I am having trouble understanding the mechanics of the $\mathcal{K}[T]$-actions for $$U$$ and $$V$$ ... hope you can help ...
I will explain my difficulties by focusing on $$ U = \mathcal{K}^r$$ ... the same considerations apply to $$ V = \mathcal{K}^{n-r} $$ ... ...

Now, consider the action $$u \bullet f(T)$$ ... ...

$$ u \bullet f(T) = u \bullet (f_0 + f_1T + f_2T^2 + ... \ ... + f_rT^r ) $$

Therefore, by the definition of the action we have:

$$ u \bullet f(T) = uf_0 + Auf_1 + A^2uf_2 + ... \ ... + A^ruf_r $$

Now consider the term $$uf_0$$ in the above expression ...

Let $$u = \begin{pmatrix} u_1 \\ . \\ . \\ . \\ u_r \end{pmatrix}$$, $$f_0 = \begin{pmatrix} f_{10} \\ f_{20} \\ . \\ . \\ . \\ f_{n0} \end{pmatrix}$$

... so how do we calculate/form $$uf_0$$?

Similarly $$A$$ is $$(n \times n)$$ , $$u$$ is $$(r \times 1)$$, and $$f$$ is $$(n \times 1)$$ ...

so then how do we calculate/form $$ Auf_1 $$ ... and so on?

Hope you can help ...

Peter

$f \in \mathcal{K}[T]$, so when we write:

$f(T) = f_0 + f_1T + \cdots + f_nT^n$, each of the $f_j \in \mathcal{K}$, these are just field elements.

Now in our given basis for $\mathcal{K}^n$, a typical $u \in U$ looks like:

$u = \begin{pmatrix}u_1\\u_2\\ \vdots\\u_r\\0\\0\\ \vdots\\0 \end{pmatrix}$

This is an $n \times 1$ matrix, and $A$ is an $n \times n$ matrix, so $Au$ is an $n \times 1$ matrix.

$Auf_1$ is just the $n \times 1$ matrix where every entry of $Au$ is multiplied by the coefficient $f_1$ of $T$ in the polynomial $f(T)$ (we're only writing it on the right so we get a right-action).