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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!

Aspirin
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    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 01:24

2 Answers2

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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 $\mathrm{Spec}(\mathbb Z[X])$.

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

Belgi
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