Find all ideals of $\mathbb R[x] / \langle x^2-3x+2\rangle$.
I know that $\langle f(x)\rangle \subseteq \langle x^2-3x+2\rangle$ iff $\langle f(x)\rangle$ divides $\langle x^2-3x+2\rangle$. But $\Bbb R[x] / \langle x^2-3x+2\rangle$ is all the combinations $(x^2-3x+2)q(x), q(x)\in \mathbb R[x]$, so isn't the ideals all of the polynomials in $\Bbb R[x]$? Probably not, but what am I missing here?
In general, the ideals of a quotient ring $R/I$ are the ideals of $R$ containing $I$. So your question is for ideals of $\mathbb{R}[x]$ containing $(x^2-3x+2)$ now $\mathbb{R}[x]$ is a principal ideal domain, so we seek $f(x)$ with $(x^2-3x+2) \subseteq (f(x))$ and this means that $f(x)|x^2-3x+2$. Now since $x^2-3x+2=(x-2)(x-1)$ there are only two possibilities for $f$ and the only ideals are $(x-2)$ and $(x-1)$.
Hint: Consider the map $\varphi\colon\mathbb{R}[X]\to\mathbb{R}\times\mathbb{R}$ defined by $\varphi(f)=(f(1),f(2))$. Prove that it is a ring homomorphism (with the obvious operations on $\mathbb{R}\times\mathbb{R}$. What's the kernel of $\varphi$?
The ideals of $R[x]/⟨x^2−3x+2⟩$ are $(x-2),(x-1),(x^2-3x+2)$ only because every ideal must contain $(x^2-3x+2).$
Also you can use Chinese Remainder Theorem. Since $(x-1)$ and $(x-2)$ are comaxal so, $\mathbb{R}[x]/(x^2-3x+2)$ isomorphic to $\mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x-2)=\mathbb{R} \times \mathbb{R}$. Now ideals of $\mathbb{R} \times \mathbb{ R}$ are only $\mathbb{R}\times {0},{0}\times \mathbb{R}$ and itself. Whose pre-image are respectively $(x-1),(x-2),0$
