Then find the polynomial $ \ f \ $ for which $ \ H(f)=0 \ $

Question

Let $ \ R_1=\mathbb{Z_3}[x] \ $ be the ring of polynomials over $ \mathbb{Z}_3 \ $ and $ \ R_2 \ $ be the ring of $ \ 2 \times 2 \ $ entries in $ \ \mathbb{Z}_3 \ $. Let $ \ A=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \ $ .

Define a homomorphism $ \ H:R_1 \to R_2 \ $ by $ \ H(x)=f(A) \, f \in R_1 $ , where $ \ A^0=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \ $.

Then find the polynomial $ \ f \ $ for which $ \ H(f)=0 \ $ .

Does $ \ H \ $ have finite range ?

Is $ \ H \ $ a isomorphism ?

Answer:

Let $ f(x)=\sum_{k=0}^{n} a_k x^k \ $ , then $ \ H(f)=\sum_{k=0}^{n} a_k A^k \ $

Then,

$ f(x)=x \ \Rightarrow H(x)=A^1=A=\begin{bmatrix}1 & 0 \\ 0 & 2 \end{bmatrix} $

Similarly,

$ H(x^2+2)=A^2+2A^0=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}+ 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \ = \begin{bmatrix} 3 & 0 \\ 0 & 6 \end{bmatrix}=3 \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}=0 , \ \ (\because 3=0 \ \ in \ \mathbb{Z}_3 ) $

Now we have to find out $ \ f(x) \ $ for which $ \ H(f)=0 \ $

Now,

From above we see that , any polynomial $ \ f (x) \ $ which is a multiple of $ \ x^2+2 \ $ gives us $ \ H(f)=0 \ $

Thus $ \ H \ $ has no finite range .

Also $ Ker (H) \neq \phi \ $

Thus $ \ H \ $ can not be isomorphism .

I need confirmation of my work.

Answer 1

1. The range of $H$ is a subring of $R_2={\Bbb Z_3}^{2\times2}$, which itself is finite. So, $H$ is of finite range.
2. Correct, the kernel of $H$ is not trivial, hence $H$ cannot be isomorphism.
   Note that it also follows from 1., as $R_1$ is inifinite.