Let $M$ be an $F_pC_q$- module represented by the matrix
$$\left( \begin{matrix} a & b\\ c & d \end{matrix}\right)$$ i.e., $m_1 g=am_1 + bm_2$ and $m_2g=cm_1 + dm_2$
where g is the generator of $C_q$, {$m_1, m_2$} is a basis for $M$ and $a,b,c,d \in F_p$ . If we consider $F_p[\xi]$ (the extension of the field $F_p$, where $\xi$ is the primitive $p$th root of unity), how can we represent $M$ in this case? I mean how can we change the matrix when we replace $F_p$ by $F_p[\xi]$?
Thanks for help.
Suppose $V$ is a vector space over a field $K$ and $L/K$ is a field extension. Let $\{v_1,\cdots,v_n\}$ be a basis for the space (assume it's finite-dimensional for convenience). Then every element looks like
$$a_1v_1+\cdots+a_nv_n $$
for some scalars $a_1,\cdots,a_n\in K$. To add two vectors, we merely add them coefficientwise, and to multiply by a scalar $c$ we multiply each scalar by $c$. Intuitively then to extend scalars to $L$, we are constructing a new vector space $V_L$ comprised of formal sums $a_1v_1+\cdots+a_nv_n$ where now we have scalars $a_1,\cdots,a_n\in L$ (not just in $K$). It should be clear how $V_L$ is a vector space over $L$ and how the underlying set of $V_L$ is bigger than $V$ (there are sum in $V_L$ that are not in $V$; just pick a sum of the $v_i$s with coefficients not in $K$).
Suppose $T:V\to V$ is a linear transformation. What is the matrix representing it, with respect to the basis we're using? For $i=1,\cdots,n$ we can apply $T$ to $v_i$ and then write the result as a sum:
$$T(v_1)=a_{1,1}v_1+a_{2,1}v_2+\cdots+a_{n,1}v_n $$
$$T(v_2)=a_{1,2}v_1+a_{2,2}v_2+\cdots+a_{n,2}v_n $$
$$\cdots\cdots$$
$$T(v_n)=a_{1,n}v_1+a_{2,n}v_2+\cdots+a_{n,n}v_n $$
for some scalars $a_{i,j}\in K$. These numbers determine $T$; to apply $T$ to any other vector, simply use the distributive property as $T(c_1v_1+\cdots+c_nv_n)=c_1T(v_1)+\cdots c_nT(v_n)$. Using this rule we can define a linear map $T_L:V_L\to V_L$ which does the same thing to basis vectors $v_1,\cdots,v_n$ as written above, and to apply $T_L$ to any other vector in $V_L$ just use the distributive property.
If we rewrite sums $a_1v_1+\cdots+a_nv_n$ as column vectors $(a_1,\cdots,a_n)^t$ then $T$'s matrix is
$$\begin{pmatrix}a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}\end{pmatrix}.$$
For example, applying the matrix to $(1,0,\cdots,0)^t$ yields $(a_{1,1},a_{2,1},\cdots,a_{n,1})^t$. By construction, the matrix of $T_L$ is the same matrix, because we are using the same basis and $T_L$ acts the exact same on the basis vectors.
Notice our construction of $V_L$ depended on using a basis for $V$. What about if we have a module that doesn't have a basis though? We need a basis-free way of doing extension of scalars. The idea is that we must be able to "pretend multiply" vectors in $V$ by scalars in $L$: the way to achieve "pretend multiplication" in a rigorous, formal manner is to use the tensor product.
We define $L\otimes_K V$ to be the collection of all sums of the symbols $l\otimes v$, and then impose some conditions: distributivity on the left and the right of $\otimes$ (so it works like multiplication) and the associativity rule $lk\otimes v=l\otimes kv$ for any $l\in L,k\in K,v\in V$.
Clearly this is a vector space over $L$ (to multiply by a scalar in $L$, simply multiply the scalars in front of all of the $\otimes$ symbols). If $\{v_1,\cdots,v_n\}$ is a basis for $V$, then $\{1\otimes v_1,\cdots,1\otimes v_n\}$ is a basis for $L\otimes_KV$, and by thinking of $1\otimes v_i$ as $v_i$ we see that $L\otimes_KV$ is the same as $V_L$. Now suppose $R$ is a ring and $M$ is an $R$-module. (Recall vector spaces are simply modules over fields.) If $S$ is a bigger ring containing $R$, we can turn $M$ into an $S$-module via the tensor product as $S\otimes_R M$ in the same exact way.