Find kernel of homomorphism of rings

Question

$\phi: \mathbb Q[x] \to Mat_{2,2}(\mathbb Q): \phi(a) = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}, \phi(x)= \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$. Find kernel of $\phi$, and guess if $\phi(\mathbb Q[x])$ is a field. I am puzzled. The only way I see is to prove that $\phi$ is surjective and find ideal $(f):\mathbb Q[x]/(f) \cong Mat_{2,2}(\mathbb Q)$. That would be the kernel, but I can't even start.

Answer 1

The minimal polynomial of the matrix $A=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ equals its characteristic polynomial, that is, $X^2-2X+2$, so $\ker\phi=(X^2-2X+2)$.

From the fundamental theorem of isomorphism for rings we have $$\phi(\mathbb Q[X])\simeq \mathbb Q[X]/(X^2-2X+2)$$ which is a field since $X^2-2X+2$ is irreducible over $\mathbb Q$.

Answer 2

the characteristic equation of $\phi(x)$ is $$ \lambda^2 - 2 \lambda + 2 = 0 $$ so the element $x^2 - 2x + 2$ belongs to the kernel of $\phi$ and $\phi(\mathbb{Q}[X])$ is the field $\mathbb{Q}[X]/X^2 -2X +2$ which is isomorphic to the quadratic extension $\mathbb{Q}(1+i) \subset \mathbb{C}$