The number of ideals in the quotient ring $\mathbb R[x]/\langle x^2-3x+2 \rangle$

Question

Finding the number of ideals in the quotient ring $\mathbb R[x]/\langle x^2-3x+2 \rangle$.

Attempt: $R[x]/\langle x^2-3x+2 \rangle = \{f(x)+\langle x^2-3x+2 \rangle~~|~~f(x) \in R[x]\}$.

Since $(x^2-3x+2)=(x-1)(x-2)$ is a reducible polynomials, the members of $R[x]/\langle x^2-3x+2 \rangle) = f(1)$ or $f(2) $ are essentially real constants.

out of which only $\{0\}$ is an ideal as $a \cdot 0 = 0 \in \{0\}~~\forall~~a \in \mathbb R$

Is my attempt correct?

Thank you for your help.

Answer 1

I didn't understand the argument, but I'm afraid the answer is wrong. The polynomial $x-1$ does not have any inverse in this quotient, so $(x-1)$ is another ideal, for example.

By the way, another way to approach this problem is by the Chinese remainder theorem. Since $(x^2-3x+2)=(x-1)\cdot(x-2)=(x-1)\cap(x-2)$, and $(x-1),(x-2)$ are co-prime, we have

$$\mathbb{R}[x]/(x^2-3x+2)\simeq\mathbb{R}[x]/(x-1)\times\mathbb{R}[x]/(x-2)\simeq\mathbb{R}\times\mathbb{R}.$$

That should simplify the task of finding all ideals.

Answer 2

I'm not sure what you mean by "$\mathbb R [x]/\langle x^2-3x +2 \rangle = f(1)$ or $f(2)$ are essentially real constants." Note that there are nonzero ideals in the quotient:

Hint. For any ring $R$ and any ideal $I$, there is a bijection between ideals of $R$ containing $I$ and ideals of $R/I$.

Answer 3

Since the polynomial factors you have

$$\mathbb{R}[x]/(x^2-3x+2)\cong \mathbb{R}[x]/(x-2)\oplus \mathbb{R}[x]/(x-1) \cong \mathbb{R}\oplus \mathbb{R}$$ So what are the ideals of $\mathbb{R}\oplus \mathbb{R}$ ?

Another way of thinking about it is you have a (reducible) variety defined by $x^2-3x+2=0$, this variety has two points.