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 \(\displaystyle V = K^n\) viewed as a module over the polynomial ring \(\displaystyle K[T]\).

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 \(\displaystyle V = U \oplus W\) expresses \(\displaystyle V\) as a direct sum of \(\displaystyle K[T]\)-submodules precisely when \(\displaystyle A = \left(\begin{array}{cc}B&0\\0&D\end{array}\right)\)

with \(\displaystyle B\) an \(\displaystyle r \times r\) matrix

and

\(\displaystyle D\) an \(\displaystyle (n - r) \times (n - r)\) matrix, \(\displaystyle B\) and \(\displaystyle D\) giving the action of \(\displaystyle T\) on \(\displaystyle U\) and \(\displaystyle W\) 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 \(\displaystyle K^n\) as a right module over \(\displaystyle K[T|\) 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 \(\displaystyle U\) and \(\displaystyle V\) ...

I can see that \(\displaystyle U\) and \(\displaystyle V\) 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 \(\displaystyle U\) and \(\displaystyle V\) ... hope you can help ...
I will explain my difficulties by focusing on \(\displaystyle U = \mathcal{K}^r\) ... the same considerations apply to \(\displaystyle V = \mathcal{K}^{n-r} \) ... ...

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

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

Therefore, by the definition of the action we have:

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

Now consider the term \(\displaystyle uf_0\) in the above expression ...

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

... so how do we calculate/form \(\displaystyle uf_0\)?

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

so then how do we calculate/form \(\displaystyle Auf_1 \) ... 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 \(\displaystyle U\) and \(\displaystyle V\) ...

I can see that \(\displaystyle U\) and \(\displaystyle V\) 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 \(\displaystyle U\) and \(\displaystyle V\) ... hope you can help ...
I will explain my difficulties by focusing on \(\displaystyle U = \mathcal{K}^r\) ... the same considerations apply to \(\displaystyle V = \mathcal{K}^{n-r} \) ... ...

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

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

Therefore, by the definition of the action we have:

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

Now consider the term \(\displaystyle uf_0\) in the above expression ...

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

... so how do we calculate/form \(\displaystyle uf_0\)?

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

so then how do we calculate/form \(\displaystyle Auf_1 \) ... 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).

 

[blue_raver22]

In Example 2.1.2 (ii), B&K are discussing the module structure of the vector space V = K^n over the polynomial ring K[T]. This means that we are considering the action of polynomials in K[T] on the elements of V, where the coefficients of the polynomials act as scalars on the elements of V.

To understand this example, it is helpful to first consider the definition of a module. B&K define a module as an abelian group with an external scalar multiplication operation. In this case, the abelian group is the vector space V = K^n and the scalar multiplication operation is given by the polynomial ring K[T].

Now, in Example 2.1.2 (ii), B&K are showing that V can be decomposed into a direct sum of K[T]-submodules, where the action of T on the submodule U is given by the r x r matrix B, and the action of T on the submodule W is given by the (n-r) x (n-r) matrix D. In other words, the elements of U and W are invariant under the action of T, and the action of T on V can be understood by considering the actions of T on U and W separately.

To verify this statement, we need to show that the decomposition V = U \oplus W satisfies the definition of a direct sum of K[T]-submodules. This means that we need to show that U and W are both K[T]-submodules of V, and that their intersection is trivial (i.e. the only element in both U and W is the zero element).

To show that U and W are K[T]-submodules, we need to show that they are closed under the action of polynomials in K[T]. In other words, if u \in U and p \in K[T], then pu \in U. This follows from the fact that B gives the action of T on U, and since T is a polynomial in K[T], we can see that B also gives the action of polynomials in K[T] on U.

Similarly, we can show that W is also a K[T]-submodule by using the matrix D to show that it is closed under the action of polynomials. And since U and W are both K[T]-submodules, their intersection must also be a K[T]-submodule. However, since the only element in both U and W