# Prime ideals in $\mathbb{Z}[x]$ [duplicate]

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Am I right that all prime ideals in $\mathbb{Z}[x]$ has the form $p\mathbb{Z}[x]$ for some prime $p\in\mathbb{Z}$?

Thanks a lot!

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## marked as duplicate by YACP, 5PM, Hagen von Eitzen, Henry T. Horton, Davide GiraudoFeb 9 '13 at 22:09

Hint: they can't be all the primes since none of them are maximal (since modding out by them yields the non-field $\Bbb Z_p[x])$ – Bill Dubuque Sep 21 '12 at 1:24

No, this is not right. There are much more prime ideals. They come in two flavours:

1. Principal ideals $(f)$, where $f$ is either zero, a rational prime, or an irreducible polynomial.
2. Maximal ideals are of the form $(p,f)$, where $p$ is a rational prime, and $f$ is an irreducible polynomial which remains irreducible modulo $p$.

Graphically, here is a "picture" of $\mathrm{Spec}(\mathbb Z[X])$.

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What are the coordinate axes $0,1,2,...,\infty$ in the lower right corner supposed to mean ($\mathbb Z[X]$ has dimension 2) ? And what has $\overline {\mathbb Q}$ got to do here? – Georges Elencwajg Sep 21 '12 at 10:15
@GeorgesElencwajg I don't know. I was looking for the picture from Mumford's Red Book and thought the one above was similar enough. I didn't notice at first the axes in the corner. Maybe someone else will be able to answer. – M Turgeon Sep 21 '12 at 12:33
Dear M Turgeon: so apparently someone xeroxed Mumford's drawing and added a few strange notations. Weird. – Georges Elencwajg Sep 21 '12 at 13:51

You are missing some, for example: $\langle 0 \rangle$ and $\langle x \rangle$ since the quotient is an integral domain

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