Consider a prime ideal $\mathfrak{p}\in\mathrm{Spec} \ \mathbf{Z}[x]$; the residue field at $\mathfrak{p}$ is the fraction field of $\mathbf{Z}[x]/\mathfrak{p}$. Can we classify the residue fields? I know that some of them are finite fields, and telling their characteristic is easy, but can we say something about them in general?
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It is not quite clear what you need help with. You get some different types of fields, depending on which "type" of prime ideal you take. – Tobias Kildetoft Aug 14 '14 at 18:59
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Maybe we're using terminology differently somehow, but I wouldn't call $R/P$ a residue field unless $P$ were maximal. It will just be a domain, in general. – rschwieb Aug 14 '14 at 19:16
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@rschwieb: "the fraction field of" – Martin Brandenburg Aug 14 '14 at 19:16
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Somehow missed that composition by reading it as "is the quotient" – rschwieb Aug 14 '14 at 19:24
1 Answers
These fields are precisely those which can be generated by a single element over their prime field. In fact, $Q(\mathbb{Z}[x]/\mathfrak{p})$ is generated by the class of $x$, and conversely if a field $F$ can be generated by a single element $a$ over its prime field $P$, then let $\mathfrak{p}$ be the kernel of $\mathbb{Z} [x] \to F, x \mapsto a$ and observe that this induces a homomorphism $Q(\mathbb{Z}[x]/\mathfrak{p}) \to F$ which is surjective since $a$ generates $F$. As you see, we don't need the classification of prime ideals of $\mathbb{Z}[x]$.
If $a$ is algebraic, then $F$ is a simple algebraic extension of $P$. If $P=\mathbb{F}_p$, this just means that $F$ is a finite field. If $P=\mathbb{Q}$, this means that $F$ is a number field. (Since every finite separable extension is simple.) If $a$ is trancendental, then $F \cong P(x)$.

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