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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.

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

1 Answer 1

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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 $0$.

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