An easy question (given as an image) of showing $\ker θ = \langle X^2-1\rangle$ where $ θ : ℝ[X] \to M_2(ℝ)$ is a homomorphism.

Question

Is my solution correct?

Let $f(X)\in ℝ[X]$. Then $f(X)= (X^2-1)q(X) + r + sX$ $(*)$ for some $q(X), (r + sX)\in ℝ[X]$ by the division theorem for polynomials.

So $$θ(f(X)) = \begin{pmatrix} r&s\\ s&r\\ \end{pmatrix} $$

So now suppose $f(X)\in \ker θ$ which means $θ(f(X))=0$ which means $r,s = 0$. From $(*)$ this means $f(X) = (X^2-1)q(X)\in \langle X^2-1\rangle$. Therefore, $\ker θ ⊆ \langle X^2-1\rangle$.

Now suppose $f(X)\in \langle X^2-1\rangle$ which means $f(X) = (X^2-1)q(X)$ for some $q(X)\in ℝ[X]$. Thus $$θ(f(X)) = \begin{pmatrix} 0&0\\ 0&0\\ \end{pmatrix} $$

So $f(X)\in \ker θ$ which means $\langle X^2-1\rangle ⊆ \ker θ$. Hence, $\ker θ = \langle X^2-1\rangle$.