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The key is a Laurent normalization by a monomial change of variables. We prove the slightly stronger assertion that every nonzero quotient of a Laurent polynomial ring in $r$ variables is finite as a module over an embedded Laurent polynomial ring in some number $m\le r$ of variables.

Induct on $r$. If the defining ideal is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomialin that ideal, with nonzero coefficients. For a sufficiently large positive integer $N$, the weightsare distinct for all $\mathbf{a}\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible monomial substitutionIts inverse is $Y_j=X_j{X}_r^{-N^j}$, so it is an automorphism of the Laurent polynomial ring. When $r=1$ there is no substitution and distinct exponents already give distinct weights.

As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero scalar multiples of monomials in $Y_1,\dots ,Y_{r-1}$, hence units inMultiplying by a power of $Y_r$ and the inverse leading coefficient givesHere $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0{Y}_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-module by $1,Y_r,\dots ,Y_r^{d-1}$.

The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable Laurent polynomial ring. By induction, it is finite over an embedded Laurent polynomial ring in $m\le r-1$ variables. Finiteness of modules is transitive, so the same is true of $T$; a finite ring extension is an integral extension, by the determinant trick applied to multiplication by an element.

Applying this to $r=3$ yields an embedded subringThe base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional vector space. The hypothesis excludes this, so $1\le m\le 3$, as required. No infinitude assumption on $k$ is used.

An integral extension $R\subseteq T$ means every $t\in T$ satisfies a monic polynomialTo prove the Lying-over theorem, localize at $S=R\setminus P$. The injection $R_P\hookrightarrow S^{-1}T$ remains injective, and $S^{-1}T$ is integral over $R_P$: a monic relation for $t$ divided by suitable powers of $s$ gives one for $t/s$. This is a nonzero ring, so choose a maximal ideal $M$ of it.

The contraction of $M$ to $R_P$ is maximal. To see this directly, a field integral over a subring forces that subring to be a field. For nonzero $a$ in the subring, $a^{-1}$ is integral; multiply its monic equation of degree $d$ by $a^{d-1}$ to express $a^{-1}$ as a polynomial in $a$ with coefficients in the subring. Apply this to the integral extensionSince $R_P$ is a local ring with unique maximal ideal $P{R}_P$, we obtain $M\cap R_P=P{R}_P$.

Contract $M$ to $T$, obtaining a prime ideal $Q$. Its contraction to $R$ is the contraction of $P{R}_P$, namely $P$. Thus

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

For a subring $S\subseteq R$ with the same identity, $R$ is an integral extension of $S$ if every $r\in R$ is an integral element over $S$: there exist $n\ge 1$ and $s_0,\dots ,s_{n-1}\in S$ such thatThe requirement that the polynomial be monic is essential. Merely satisfying some polynomial equation over $S$ is insufficient: for instance $1/2\in \mathbb{Q}$ satisfies $2T-1=0$ over $\mathbb{Z}$, but is not integral over $\mathbb{Z}$. To see this directly, a hypothetical monic equation for $1/2$, multiplied by $2^n$, would say that an odd integer $1$ plus a sum of even integers is zero.

The degree and coefficients may depend on $r$. Integrality does not by itself assert that the entire ring $R$ is a finite $S$-module; it does imply that the subalgebra $S[r]$ is generated as an $S$-module by $1,r,\dots ,r^{n-1}$.

We use three theorems for an integral extension $S\subseteq R$, none requiring Noetherian hypotheses. The Lying-over theorem says that each prime ideal $\mathfrak{p}$ of $S$ is $\mathfrak{q}\cap S$ for some prime ideal $\mathfrak{q}$ of $R$. The going-up theorem says that if $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ in $S$ and $\mathfrak{q}\cap S=\mathfrak{p}$, there is ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap S={\mathfrak{p}}^{\prime }$. The incomparability theorem for integral extensions says that comparable prime ideals of $R$ with the same contraction to $S$ are equal.

Take any strict chainin $R$. The contraction of an ideal operation gives ${\mathfrak{p}}_j={\mathfrak{q}}_j\cap S$, which are prime ideals of $S$, by the proof in Question 1(b), and remain strictly increasing by the incomparability theorem for integral extensions. Thus every chain length in $R$ occurs in $S$, giving $\dim R\le \dim S$.

Conversely, take any strict chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ in $S$. Use the Lying-over theorem to choose ${\mathfrak{q}}_0$ over ${\mathfrak{p}}_0$, then repeatedly use the going-up theorem to obtain ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\le \dim R$.

Taking suprema proves integral extensions preserve Krull dimension:The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.

No. In the Gaussian integers, chooseIndeed, substituting $i=-2$ gives $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, so $P$ is a prime ideal and its contraction is $(5)$. Put $\beta =2-i$. Modulo $P$, $\beta$ becomes $4\ne 0$ in ${\mathbb{F}}_5$, so $\beta \notin P$. ThereforeWe show that $\alpha$ is not an integral element over ${\mathbb{Z}}_{(5)}$, using a direct reduction argument. Write $B={\mathbb{Z}}_{(5)}[i]$, viewed inside $\mathbb{Q}(i)$. There is a ring homomorphismwhich reduces rational coefficients modulo $5$. It is well defined because denominators in ${\mathbb{Z}}_{(5)}$ are not divisible by $5$, and $2^2+1=0$ in ${\mathbb{F}}_5$. In particular, $\rho (\beta )=0$ and $\rho (1)=1$.

If $\alpha$ satisfied a monic equation over ${\mathbb{Z}}_{(5)}$, saymultiplication by ${\beta }^n$ would give an equality in $B$:Applying $\rho$ gives $1=0$ in ${\mathbb{F}}_5$, a contradiction. ConsequentlyThis exhibits why localization of an integral extension at a prime need not be integral. Localizing both rings by the same multiplicative subset preserves an integral extension, but $\mathbb{Z}[i{]}_P$ inverts additional elements, including $2-i$, whose image at the other prime ideal $(2-i)$ is zero. The original extension is integral because $i$ satisfies the monic equation $T^2+1=0$: the subalgebra generated by finitely many integral elements property applies to $\mathbb{Z}[i]$.

An element $t\in T$ is an integral element over $R$ when it satisfies a monic equationThe extension $R\subseteq T$ is an integral extension when every $t\in T$ is integral over $R$.

Let $P\in \mathrm{Spec}R$ and localize both rings at $S=R\setminus P$. The extension $R_P\subseteq S^{-1}T$ remains integral. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}T$. The contraction of a maximal ideal under an integral extension is maximal, and the local ring $R_P$ has unique maximal ideal $P{R}_P$, so $\mathfrak{n}\cap R_P=P{R}_P$. Contracting $\mathfrak{n}$ back to $T$ produces $Q\in \mathrm{Spec}T$ with $Q\cap R=P$. This proves the Lying-over theorem and hence the surjectivity of

Suppose $Q_1\subseteq Q_2$ and both contract to $P$. Quotient by $Q_1$ and localize the resulting integral domain at the nonzero elements of $R/P$. The localized ring is an integral domain integral over the field $\mathrm{Frac}(R/P)$, and is therefore itself a field. The localization of $Q_2/Q_1$ must consequently be zero; since the localized ring is a domain, $Q_2/Q_1=0$. Thus the incomparability theorem for integral extensions gives

The Krull dimension of a ring is the supremum of the lengths of its strict chains of prime ideals. Incomparability makes the contraction of every strict prime chain in $T$ strict, so $\dim T\le \dim R$. Conversely, start over the bottom prime of any chain in $R$ using lying over and lift each subsequent inclusion using the going-up theorem. Hence

For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and putThenand characteristic zero allows division by three. ThereforeDivision by the monic quadratic shows that $T$ is free over $k[u,v]$ with basis $1,z$. In particular, $u,v$ are algebraically independent and the requested Noether normalization of the quadratic surface xy plus yz plus zx is

A ring extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$: there is a monic polynomialwith all $a_i\in A$.

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

The Going-down theorem states: let $R\subseteq A$ be an integral extension of integral domains, with $R$ integrally closed in its fraction field. If ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ are prime ideals of $R$ and ${\mathfrak{q}}_1$ is a prime ideal of $A$ lying over ${\mathfrak{p}}_1$, then there is a prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ lying over ${\mathfrak{p}}_0$.

Put $K=\mathrm{Frac}R$, letbe the minimal polynomial of an algebraic element $y$ over $K$, and let $B$ be the integral closure of $R$ in a finite normal extension containing all roots of $f$. Since $y$ is integral over $R$ and $R$ is integrally closed domain, every $a_i$ belongs to $R$.

Write $y={\sum }_j{p}_j{z}_j$ with $p_j\in \mathfrak{p}$ and $z_j\in A$. Every $K$-embedding into the normal extension fixes the $p_j$ and sends each $z_j$ to an element integral over $R$. Thus every conjugate of $y$ lies in the extended ideal $\mathfrak{p}B$. Each nonleading coefficient of $f$ is, up to sign, an elementary symmetric polynomial in those conjugates, so it lies in $\mathfrak{p}B\cap R$.

For an integral extension, extension followed by contraction preserves a prime ideal:Indeed, the determinant trick gives $r^m\in \mathfrak{p}$ for $r\in \mathfrak{p}B\cap R$, and primality then gives $r\in \mathfrak{p}$. Hence $a_i\in \mathfrak{p}$ for every $i<d$.