Finding an isomorphism between rings Find an isomorphism between $T: \mathbb{F_3}[x]/(x^2+x+2) \to \mathbb{F_3}[x]/ (x^2 + 1)$. We have that both of these are fields. The answer states the map $T$ is given by $T(a+bx) = (a+b) + bx$ so $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Could someone explain how they got this?
 A: There is a mechanical way to find isomorphisms like this. We know that such an isomorphism exists by the uniqueness of finite fields of a given order. Furthermore, such an isomorphism is completely determined by the image of $x$. Finally, we know that the image of $x$ can be represented by $a + bx$ such that $(a + bx)^2 + (a + bx) + 2 \in (x^2 + 1)$. Expanding, we get
\begin{align*}
(a + bx)^2 + (a + bx) + 2 &= (a^2 + a + 2) + (2a + 1)bx + b^2 x^2 \\
&= (a^2 + a + 2 - b^2) + (2a + 1)bx + b^2 (x^2 + 1).
\end{align*}
This gives the equations $a^2 + a + 2 - b^2 = 0$ and $(2a + 1)b = 0$. By solving for $a$ and $b$ in $\mathbb F_3$ we get $a = b = 1$ or $a = 1, b = 2$.
A: The additive rings are both vector spaces over the prime field, hence a ring isomorphism must be a vector space isomorphsm as well, which can be represented by a $2\times 2$ matrix. Also, the prime field must be left as is, hence the left columnt must be 1 0. Remains to "guess" the right column. But observe how $T$ has to act on $x^2+x+2$.
