Example 10.27.2. In this example we describe $X = \mathop{\mathrm{Spec}}(\mathbf{Z}[x])$. Fix $\mathfrak p \in X$. Let $\phi : \mathbf{Z} \to \mathbf{Z}[x]$ and notice that $\phi ^{-1}(\mathfrak p) \in \mathop{\mathrm{Spec}}(\mathbf{Z})$. If $\phi ^{-1}(\mathfrak p) = (q)$ for $q$ a prime number $q > 0$, then $\mathfrak p$ corresponds to a prime in $(\mathbf{Z}/(q))[x]$, which must be generated by a polynomial that is irreducible in $(\mathbf{Z}/(q))[x]$. If we choose a representative of this polynomial with minimal degree, then it will also be irreducible in $\mathbf{Z}[x]$. Hence, in this case $\mathfrak p = (q, f_ q)$ where $f_ q$ is an irreducible polynomial in $\mathbf{Z}[x]$ that is irreducible when viewed in $(\mathbf{Z}/(q) [x])$. Now, assume that $\phi ^{-1}(\mathfrak p) = (0)$. In this case, $\mathfrak p$ must be generated by nonconstant polynomials which, since $\mathfrak p$ is prime, may be assumed to be irreducible in $\mathbf{Z}[x]$. By Gauss' lemma, these polynomials are also irreducible in $\mathbf{Q}[x]$. Since $\mathbf{Q}[x]$ is a Euclidean domain, if there are at least two distinct irreducibles $f, g$ generating $\mathfrak p$, then $1 = af + bg$ for $a, b \in \mathbf{Q}[x]$. Multiplying through by a common denominator, we see that $m = \bar{a}f + \bar{b} g$ for $\bar{a}, \bar{b} \in \mathbf{Z}[x]$ and nonzero $m \in \mathbf{Z}$. This is a contradiction. Hence, $\mathfrak p$ is generated by one irreducible polynomial in $\mathbf{Z}[x]$.

Comment #46 by Rankeya on

In the third line, "then it p corresponds to ..." should be "then p corresponds to".

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