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The set $\tilde{P}$ contains zero. It is closed under negatives because $-f\in P$ whenever $f\in P$. If $f,g\in P$ and $n,m>0$, thensince $P$ is an ideal of $\mathbb{Z}[t]$. Also, any $q\in \mathbb{Q}[t]$ can be written $q=h/d$ with $h\in \mathbb{Z}[t]$ and $d>0$, givingThis proves that $\tilde{P}$ is an ideal of $\mathbb{Q}[t]$.

Clearly $P\subseteq \tilde{P}\cap \mathbb{Z}[t]$, by taking denominator one. Conversely, if $h\in \tilde{P}\cap \mathbb{Z}[t]$, then $nh\in P$ for some positive integer $n$. The condition $P\cap \mathbb{Z}=\{0\}$ implies $n\notin P$. Since $P$ is a prime ideal, $nh\in P$ forces $h\in P$. HenceIn particular, $\tilde{P}$ is nonzero because it contains the nonzero ideal $P$, and it is proper: otherwise its intersection with $\mathbb{Z}[t]$ would contain $1$. This is the concrete extension of an ideal and contraction of an ideal calculation for the localization that inverts nonzero integers.

Let $P$ be a prime ideal of $R$ disjoint from the multiplicatively closed set $Y$. In the localization $S=Y^{-1}R$, its extension of an ideal isIndeed, finite sums of such fractions have a common denominator in $Y$ and numerator in $P$, and every generator $p/1$ has this form.

We show that its contraction of an ideal recovers $P$:One inclusion is immediate. Conversely, if $r/1=p/y$ with $p\in P$ and $y\in Y$, equality of fractions in the field of fractions of the integral domain $R$ gives $yr=p\in P$. Since $y\notin P$ and $P$ is a prime ideal, $r\in P$.

If $S{P}_1=S{P}_2$, contracting would therefore give $P_1=P_2$, contrary to the hypothesis. HenceWe will also need the fact that $SP$ is a prime ideal, part of the prime ideal correspondence for localization. It is proper by the contraction identity, since otherwise $1\in P$. If $(a/u)(b/v)\in SP$, multiplication by the denominator gives $ab/1\in SP$, and contraction gives $ab\in P$. Thus $a\in P$ or $b\in P$, so one of the two fractions belongs to $SP$.

Write $N=\sqrt{(0)}$. Every element of the extension of an ideal $N{R}_{\mathfrak{p}}$ is a fraction $a/s$ with $a\in N$ and $s\notin \mathfrak{p}$: a finite sum of such fractions has a common denominator and numerator still in the ideal $N$. If $a^n=0$, then $(a/s{)}^n=0$, soConversely, let $a/s$ be a nilpotent element of the localization at a prime ideal $R_{\mathfrak{p}}$. For some $n\ge 1$, $a^n/s^n=0$. The vanishing criterion in a module localization supplies $t\notin \mathfrak{p}$ with $t{a}^n=0$. It follows thatso $ta\in N$. Since $ts\notin \mathfrak{p}$,This proves that the nilradical commutes with localization:The multiplication by $t$ is necessary: the numerator $a$ of a nilpotent element in a localization need not itself be nilpotent in $R$.

Let $S=A\setminus \mathfrak{p}$, let $S$ act on $B$ through $\phi$, and write $\mathfrak{p}B$ for the extension of an ideal generated by $\phi (\mathfrak{p})$. Since $\kappa (\mathfrak{p})=S^{-1}(A/\mathfrak{p})$, the fiber ring has the canonical presentationsThe first isomorphism sends $b\otimes (\bar{a}/s)$ to $\phi (a)b/s$ modulo $\mathfrak{p}B$. Its inverse sends the class of $b/s$ to $b\otimes (\bar{1}/s)$. The tensor balancing relation, reduction modulo $\mathfrak{p}$ and the fraction relations show that these maps are well-defined and inverse. Thus the description is canonical and does not depend on chosen representatives.

Suppose $\mathfrak{q}$ is a prime ideal of $B$ with ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$. Then $\mathfrak{p}B\subseteq \mathfrak{q}$ and $\mathfrak{q}$ is disjoint from $\phi (S)$. Its localization $S^{-1}\mathfrak{q}$ is a proper prime, since the quotient is the localization of the integral domain $B/\mathfrak{q}$. It contains $\mathfrak{p}{S}^{-1}B$, and therefore gives a primeConversely, for a prime $\mathfrak{Q}\subset C$, take its contraction $\mathfrak{q}$ under the ring homomorphism $\psi :B\to C$, $b\mapsto b\otimes 1$. Contraction preserves primality. Every $\phi (a)$ with $a\in \mathfrak{p}$ maps to zero, while every $\phi (s)$ with $s\in S$ maps to a unit and hence cannot belong to a proper prime. Thus ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$.

These constructions are inverse. For a prime disjoint from $\phi (S)$, the fraction $b/1$ belongs to its localized extension exactly when some $s\in S$ has $\phi (s)b\in \mathfrak{q}$; primality then implies $b\in \mathfrak{q}$. In the other direction, every element of $C$ has a representative $b/s$, and membership in $\mathfrak{Q}$ is equivalent to membership of $b/1$ because the denominator is a unit. Hence $\mathfrak{Q}$ is recovered by extending its contraction and quotienting. We have proved the canonical fiber of the map on spectra correspondence:If a denominator maps to zero, the fiber ring is the zero ring and both sets are empty; the same argument includes this case.

The induced map on the spectrum of a commutative ring is contraction of an ideal:To see that it is well defined, let $\mathfrak{q}$ be a prime ideal of $S$. Its inverse image is an ideal of $R$, and it is proper because $\phi (1)=1\notin \mathfrak{q}$. If $ab\in {\phi }^{-1}(\mathfrak{q})$, then $\phi (a)\phi (b)\in \mathfrak{q}$, whence $\phi (a)\in \mathfrak{q}$ or $\phi (b)\in \mathfrak{q}$. This proves that the inverse image is a prime ideal.

For any ideal $I\subseteq R$, its extension of an ideal to $S$ is $IS=\phi (I)S$. We haveThus inverse images of Zariski-closed sets are Zariski-closed sets, which proves that $f$ is a continuous map.

There is an exact sequence of $R$-modulesIf $A$ is a flat module over $R$, tensoring this sequence with $A$ preserves its left exactness. The image of each tensor product inside $A$ is the corresponding extension of an ideal, soBoth displayed inclusions therefore hold. This is the flat extension preserves finite ideal intersections property.

The prime ideals $P$ of $\mathbb{Z}[t]$ with $P\cap \mathbb{Z}=\{0\}$ are the zero ideal and the principal ideals $(f)$ generated by nonconstant primitive polynomials that are irreducible in $\mathbb{Q}[t]$.

To prove the description, invert the nonzero integers to obtain the localization $\mathbb{Q}[t]$. The extension of an ideal $P$ consists of the polynomials $f/n$ with $f\in P$ and $n>0$. Its contraction is $P$: if $nh\in P$ and $n\ne 0$, the defining condition puts $n\notin P$, so primeness gives $h\in P$. The prime ideal correspondence for localization makes this extension a proper prime ideal of $\mathbb{Q}[t]$.

If $P\ne 0$, its extension has a nonzero generator because $\mathbb{Q}[t]$ is a principal ideal domain. Dividing this generator by its rational polynomial content gives an integer-coefficient primitive polynomial $f$ generating the same extended ideal. Divisibility descent by Gauss lemma for polynomials then gives $P=(f)$ in $\mathbb{Z}[t]$. The proper prime ideal $(f)$ in $\mathbb{Q}[t]$ makes $f$ irreducible there. Conversely, any such polynomial generates a prime ideal in $\mathbb{Q}[t]$ whose contraction is $(f)\subset \mathbb{Z}[t]$ and whose intersection with $\mathbb{Z}$ is zero.

The intersection condition is essential: $(2,t)$ is a maximal ideal of $\mathbb{Z}[t]$, since its quotient ring is ${\mathbb{F}}_2$, but it is not principal. A common divisor of $2$ and $t$ must be a unit, which would generate the whole ring.