Let $A=\begin{pmatrix} 1 &-2 & 1\\ 1& -2& 1\\ 1& -2& 1 \end{pmatrix}$.
Find the invariant elements and the degree of the $R$-module $M=\mathbb{C}^3$ where $R$ is the polynomial ring $R=\mathbb{C}[T]$ where $T(w)=Aw$.
A start of an attempt:
We know that $\mathbb{C}^3\cong (\mathbb{C}[x])^r\oplus \bigoplus_{i=1}^nR/\langle a_i\rangle $. There's a surjective homomorphism $\phi:R^3\to M$ which is $\phi:(\mathbb{C}[x])^3\to \mathbb{C}^3$ which I'm not sure how to define $\phi(f_1,f_2,f_3)$.
Then we wish to find the relations module $\ker(\phi)$.
To begin with notice that all elements in $ \mathbb{C} $ are torsion elements; if $ p_T(x) $ is the minimal polynomial for $ T $, then $ p_T(x).v = 0 $ for all vectors $ v\in \mathbb{C} $. Hence we even know that, as a $ \mathbb{C}[x] $-module, $$ \mathbb{C}^3 \cong \bigoplus_{i=1}^n \mathbb{C}[x]/\langle p_i(x) \rangle, $$
where $ p_i $ are the invariant elements. Not sure what you mean by degree though, is it the degree of the minimal polynomial $ p_T $? Anyways, we know that $ B = xI - A $ is the relation matrix so the relations of $ \ker \phi $ are generated by $ eB = 0 $, where $ e = (e_1 \; e_2 \; e_3) $. In order to find the invariant elements we need to get $ B $ on its smith normal form. After some row and column operations we get that $$ B \sim C = \begin{pmatrix} 1 & 0 & 0\\ 0 & x & 0\\ 0 & 0 & x^2 \end{pmatrix} \quad \text{wherefore} \quad \mathbb{C}^3 \cong \frac{\mathbb{C}[x]}{\langle x \rangle} \oplus \frac{\mathbb{C}[x]}{\langle x^2 \rangle}. $$ Hence the invariant elements are $ x $ and $ x^2 $. To be more precise we have that $ C = PBQ $, where $$ P = \begin{pmatrix} 1 & 0 & 0\\ -1 & 1 & 0\\ x+1 & -2 & 1 \end{pmatrix}. $$ Hence \begin{align*} 0 &= eB = eP^{-1}PB = eP^{-1}PBQ = eP^{-1}C \\ &= (e_1+e_2+(1-x)e_3\quad e_2+2e_3 \quad e_3) \begin{pmatrix} 1 & 0 & 0\\ 0 & x & 0\\ 0 & 0 & x^2 \end{pmatrix}. \end{align*} Thus the sought relations are $ x^2.e_3 = 0 $ and $ x.(e_2+2e_3) = 0 $ so that $ \ker \phi = \langle x^2.e_3, x.(e_2+2e_3) \rangle$.