# Determine all the prime ideals.

I need to find all prime ideals of $$\mathbb{Z}[x]/(x^2-3x+2)$$.

This is what I know so far: Since $$\mathbb{Z}$$ is not a field, so $$\mathbb{Z}[x]$$ is not a PID, hence many useful tools cannot be used here. I also observe that $$x^2-3x+2 = (x-2)(x-1)$$. Since $$(x-2)$$ and $$(x-1)$$ are irreducible polynomial, so the ideals are maximal, hence these two ideals are co-maximal, so we are free to use the Chinese remainder theorem, i.e., $$\mathbb{Z}[x]/(x^2-3x+2) \cong \mathbb{Z}[x]/(x-1) \times \mathbb{Z}[x]/(x-2)$$.

What else do I need to do to get it done? Thanks in advance.

• No principal ideal of $\Bbb Z[x]$ is maximal.
– user239203
Commented Oct 12, 2019 at 19:28
• $(x-2)$ isn't maximal, as $(x-2,5)$ is larger. Commented Oct 12, 2019 at 19:28

The prime ideals of $$\mathbb{Z}[X]/(X-1)(X-3)$$ are the images of the prime ideals of $$\mathbb{Z}[X]$$ containing $$(X-1)(X-3)$$ by the canonical projection.
By definition of a prime ideal, those either contain $$(X-1)$$ or $$(X-3)$$. Hence, they corespond to prime ideals of $$\mathbb{Z}$$ by one of the following isomorphisms :
$$\mathbb{Z}[X]/(X-1) \rightarrow \mathbb{Z}$$ $$\mathbb{Z}[X]/(X-3) \rightarrow \mathbb{Z}$$
Hence the primes $$(X-a,p)$$ for $$a=1,3$$ and $$p$$ a prime number or $$0$$.