I'm solving the following problem:
Identify all maximal ideals in the ring $\mathbb R[x]/(x^2-3x+2)$.
I want to solve it in two ways.
First method. Consider the surjective quotient homomorphism $\mathbb R[x]\rightarrow \mathbb R[x]/(x^2-3x+2)$ with kernel $I=(x^2-3x+2)$. The ideals of the image correspond bijectively to the ideals of $\mathbb R[x]$ containing $I$. Since $\mathbb R[x]$ is a PID, the only ideals in that ring containing $I$ are $(1),(x-1),(x-2),(x^2-3x+2)$. The image of $(1)$ is the whole ring and therefore cannot be maximal. The image of $(x^2-3x+2)$ is the zero ideal and also cannot be maximal. The images of $(x-1)$ and $(x-2)$ are $(x-1)+I$ and $(x-2)+I$ respectively. How do I show rigorously that the quotient of $\mathbb R[x]/I$ by each of these ideals is a field? (I think this is the most reasonable way to show that $(x-1)+I$ and $(x-2)+I$ are maximal; if not, please let me know if there is an easier way).
Second method. I want to use the fact that the maximal ideals of $\mathbb R\times \mathbb R$ are $(0)\times \mathbb R$ and $\mathbb R\times (0)$. To this end define a ring homomorphism $\phi: \mathbb R[x]\rightarrow \mathbb R \times \mathbb R$ by $f(x)\mapsto (f(1),f(2))$. This is an empmorphism with kernel $I$. By the first isomorphism theorem it induces an isomorphism $\mathbb R[x]/I\simeq \mathbb R \times \mathbb R$. Now we need to find the inverse images of $(0)\times \mathbb R$ and $\mathbb R\times (0)$ under $\phi$ and take their images under the quotient map $\mathbb R[x]\rightarrow \mathbb R[x]/I$. We have $\phi^{-1}(1,0)=-x+2; \phi^{-1}(0,1)=x-1$. This implies $\phi^{-1}(\mathbb R\times (0))=(x-2)$ and $\phi^{-1}((0)\times \mathbb R)=(x-1)$ because for example $-x+2\in \phi^{-1}(\mathbb R\times (0)) $ and since the inverse image of any ideal is an ideal, $(x-2)$ lies in $\phi^{-1}(\mathbb R\times (0))$. But why there is nothing else in $\phi^{-1}(\mathbb R\times (0))$ except $(x-2)$?
For the first method, it is the usual proof that for any field $K$ and $a \in K$ we have $K[X]/(X-a) \cong K$, as by the third isomorphism theorem $(K[X]/I)/(\mathfrak a/I) \cong K[X]/\mathfrak a$ for $I \subset \mathfrak a \subset K[X]$. For this define the ring morphism:
$$K[X] \to K, f \mapsto f(a).$$
This is clearly surjective as $K \subseteq K[X]$ and the ideal $(X-a)$ lies in the kernel. Hence get surjective morphism
$$\varphi \colon K[X]/(X-a) \to K, [f] \mapsto [f(a)],$$
where $[f]$ denotes the equivalence class of some $f \in K[X]$. To show that this is injective, let $\varphi([f]) = 0$. As $[X] = [a]$, we have $[f] = [f(a)]$, so $f(a) = 0$ and $(X-a) | f \implies f \in (X-a)$ and $\varphi$ injective. So $(X-a)$ is maximal as the fraction ring is a field.
The answer to your second question follows again from the maximality of $(X-2)$.