I have a question on a solution I was reading from the following link: https://brianbi.ca/artin/11.8
The question is 11.8.2 part (c) and reads:
Determine the maximal ideals of each of the following rings:
part (c): $\mathbb{R}[x]/(x^2 - 3x + 2)$.
The solution author writes:
Let R = $\mathbb{R}[x]/(x^2 - 3x + 2)$ ... (let $\alpha$ be the residue of x)
... "Now R/($\alpha$ - 1) $\cong$ $\mathbb{R}[x]/(x -1, x^2 - 3x + 2 )$ = $\mathbb{R}[x]/(x-1)$ $\cong$ $\mathbb{R}$.
Why is the above true? I'm not sure why the first statement holds in particular, namely R/($\alpha$ - 1) $\cong$ $\mathbb{R}[x]/(x -1, x^2 - 3x + 2 $)
Thanks!
We may see this as two morfisms. The first is $\phi : \mathbb{R}[x]\to R$ that sends $p(x) \mapsto p(x)+(x^2-3x+2)$. The second is $\varphi: R \to R/(\alpha + 1)$ that sends $p(x)+(x^2-3x+2) \mapsto (p(x)+(x^2-3x+2)) + (\alpha + 1)$ where $\alpha = x+(x^2-3x+2)$.
The composition $\varphi \circ \phi$ is a morfism that sends $\mathbb{R}[x]$ to $R/(\alpha + 1)$, let's take a look on its kernel:
$$(\varphi \circ \phi)^{-1}(0) = \phi^{-1}( \varphi^{-1}( \alpha + 1 ) ) = \varphi^{-1}( \alpha + 1 + (x^2-3x+2) ) = (x+1, x^2-3x+2)$$
Now, bringing the compositions back by the isomorphism theorem we got that: $$ \frac{R}{(\alpha + 1)} \approx \frac{\mathbb{R}[x]}{ker(\varphi\circ\phi)} = \frac{\mathbb{R}[x]}{ (x+1, x^2-3x+2) } $$