there's a surjective homomorphism between $\mathbb{R}[x]$ and $\begin{equation}\begin{pmatrix} a & b \\-3b & a\end{pmatrix}\end{equation}$ $a,b$ reals

Question

Let $F=\left\{\begin{pmatrix} a & b \\-3b & a\end{pmatrix} \mid a,b\in\mathbb{R}\right\}$. I want to show $F\simeq \mathbb{R}[x]/(x^2+3)$. By first isomorphism theorem I just need a surjective homomorphism between $\mathbb{R}[x]$ and $F$. But I don't have idea which funtion works. Can someone give me a suggestion?

Answer 1

Hint: Note that $$\begin{pmatrix} a & b \\ -3b & a \end{pmatrix} = a\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ -3 & 0 \end{pmatrix}.\tag{1}$$ Any ring homomorphism from $\mathbb R[x]$ that maps the multiplicative identity to the multiplicative identity is uniquely determined by its action on $x$. So, if $1$ is mapped to the identity matrix and having in mind $(1)$, where would you map $x$, then? If you make the obvious choice, surjectivity will be trivial. Check where $x^2+3$ is mapped.