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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 $\mathbb{Z}[i]$.

Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.

LetIts minimal polynomial is $X^2-X+m$, and the algebraic norm givesSuppose that $N$ is composite and choose a prime divisor $q\le \sqrt{N}<m$. Since $X^2-X+m$ has the root $-n$ modulo $q$, the idealis a prime ideal of ideal norm $q$. The triviality of the ideal class group makes $\mathfrak{q}=(\beta )$ principal, so $|N_{L/\mathbb{Q}}(\beta )|=q$.

The ring of integers of a quadratic field consists of elementswhose norm is $(a^2+p{b}^2)/4$. If $b=0$, this norm is a square and cannot be the prime $q$. If $b\ne 0$, thenand integrality gives $q\ge m$, again a contradiction. Therefore the Euler prime-generating quadratic from class number one yields

Put $\alpha =\sqrt[n]{m}$. It is an algebraic integer because it satisfies the monic polynomial $X^n-m$. Choose a prime number $q$ dividing the square-free integer $m$, and choose a prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_L$ above $q$. Fromand prime ideal factorization, the $\mathfrak{p}$-adic exponents satisfybecause $q$ occurs to exponent one in $m$. Hence $n$ divides the ramification index of a prime ideal $e_{\mathfrak{p}}$. If $d=[L:\mathbb{Q}]$, then $e_{\mathfrak{p}}\le d$, while the equation ${\alpha }^n=m$ gives $d\le n$. Thereforeso

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and

Write the affine scheme as $U=\mathrm{Spec}B$. A point $x\in U$ corresponds to a prime ideal $\mathfrak{p}\subset B$, and its local ring is ${\mathcal{O}}_{X,x}\cong B_{\mathfrak{p}}$ with maximal ideal $\mathfrak{p}{B}_{\mathfrak{p}}$. HenceThe last condition defines the principal open subscheme $D(\bar{f})$, soEvery point of the scheme $X$ has such an affine neighbourhood. Thus $X_f$ is locally open, and hence is open in the Zariski topology.

On every open set $U$, define $F_U^{\#}(a)=a^p$. In characteristic of a ring $p$, the binomial theorem gives $(a+b{)}^p=a^p+b^p$, so these are ring homomorphisms; they commute with restrictions and hence define a morphism of sheaves of rings.

On an affine chart $\mathrm{Spec}A$, the inverse image of a prime ideal $\mathfrak{p}$ under the Frobenius endomorphism isby primality. The induced continuous map is therefore the identity. These local morphisms agree on overlaps, giving the Absolute Frobenius morphism $F_X:X\to X$. Its action on the underlying space and on every local section was prescribed, so the morphism is unique.

Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\mathrm{Spec}R$. The ideal $P=(x,z)$ is a height-one prime ideal, because $R/P\cong k[y]$ and $\dim R=2$. It therefore defines a prime Weil divisor $D=V(x,z)$.

Localizing at $x$ eliminates $y$, givinga unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(X)$ is generated by the height-one primes containing $x$. Sincethe only such prime is $P$, and hence $[D]$ generates. At the generic point of $D$, $y$ is a unit in a ring and $x=z^3/y$, so the order of vanishing is three:Thus $3[D]=0$.

This relation has exact order three. Indeed, if a principal divisor were supported on $D$, its defining rational function would be a unit on $D(x)$. The units of $R_x=k[z][x,x^{-1}]$ are precisely $c{x}^m$ with $c\in k^{\times }$ and $m\in \mathbb{Z}$, whose divisors are $3mD$. Consequentlygenerated by $[V(x,z)]$. This is the $n=3$ case of the Divisor class group of an A-type surface singularity.

Evaluation at $\alpha$ gives the ring isomorphismand reducing modulo $p$ givesFor each $i$,This quotient is a field because ${\overline{g}}_i$ is irreducible. Hence $P_i$ is a nonzero prime ideal; it is nonzero because it contains $p$. Its ideal norm is

The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both ${\overline{g}}_i$ and ${\overline{g}}_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a Bezout identity between them would make $1$ vanish there, a contradiction.

Because the chosen lifts satisfyfor some $H\in \mathbb{Z}[X]$, evaluation at $\alpha$ givesEvery generator of the product ideal ${\prod }_i{P}_i^{e_i}$ either contains a factor $p$ or is the displayed product, soOn the other hand, multiplicativity of ideal norms and ${\sum }_i{e}_i\mathrm{deg}{\overline{g}}_i=\mathrm{deg}f=[L:\mathbb{Q}]$ giveTwo finite-index ideals with one contained in the other and equal index are equal. Thus the Dedekind factorization theorem is proved in this monogenic case:

Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus

Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

The polynomial of $\alpha$ satisfies the Eisenstein criterion at $p$. Let $\mathfrak{p}$ be any prime ideal of ${\mathcal{O}}_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak{p}}(\mathfrak{p})=1$, and write $e=v_{\mathfrak{p}}(p)$ for its ramification index of a prime ideal. Put $t=v_{\mathfrak{p}}(\alpha )$. Inthe constant term has valuation $e$, while every other $a_j{\alpha }^j$ has valuation at least $e+jt$. The non-Archimedean valuation property then forcesIndeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.

The fundamental inequality ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}}{f}_{\mathfrak{p}}\le n$ now shows that $\mathfrak{p}$ is the unique prime above $p$ and has residue degree one. At this prime,while $(p,\alpha )$ has no factor away from $p$. Unique prime ideal factorization therefore givesFinally $v_P(\alpha )=1$, soThis is total ramification from an Eisenstein polynomial.

For every homogeneous $s\in S_{+}$,Henceso $U$ is open in the Zariski topology.

If $\mathfrak{p}\in U$, then ${\phi }^{-1}(\mathfrak{p})$ is a homogeneous prime ideal that does not contain the irrelevant ideal of a graded ring $S_{+}$. Thusdefines a map $U\to \mathrm{Proj}S$. On every standard affine open of Proj the graded homomorphism inducesand therefore an affine scheme morphismThese morphisms agree after localization on overlaps, so they glue to the required morphism of schemes $f:U\to \mathrm{Proj}S$.

PutSinceis an integral domain, $(x,z)$ is a prime ideal. Moreover $\dim R=3$ and $\dim R/(x,z)=2$, so its height of an ideal is one. Therefore $D=V(x,z)$ is a prime Weil divisor.

Localizing at $x$ eliminates $y$ and givesa unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(R)$ is generated by the height-one primes containing $x$. Sincethese are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the principal divisorThe units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c{x}^n$ with $c\in k^{\ast }$ and $n\in \mathbb{Z}$. Consequently the only relation supplied by localization is $[D]+[E]=0$, andThis is the divisor class group of the three-dimensional affine quadric cone.

Write $n=[L:K]$. The field trace gives a nondegenerate $K$-bilinear trace pairingThe inverse different, or codifferent, is the trace-dual latticeIt contains $S$, because the trace of an element integral over $R$ belongs to the integrally closed ring $R$. It is stable under multiplication by $S$: if $s,t\in S$ and $x\in {\mathfrak{D}}_{S/R}^{-1}$, then $\mathrm{Tr}(sxt)=\mathrm{Tr}(x(st))\in R$. Nondegeneracy of the trace pairing and finite generation of $S$ show that this trace dual is a finitely generated $R$-module spanning $L$. Therefore it is a fractional ideal of $S$.

Its inverseis the different ideal. Since $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$, every such $x$ lies in $S$; hence ${\mathfrak{D}}_{S/R}$ is an integral ideal of $S$.

The discriminant ideal $\mathrm{disc}(S/R)$ is locally generated bywhere ${\alpha }_1,\dots ,{\alpha }_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairingThis formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.

The asserted identity of ideals can be checked after localization at every nonzero prime ideal of $R$. We may therefore assume that $R$ is a discrete valuation ring and choose a basis ${\alpha }_1,\dots ,{\alpha }_n$ of $S$. Let ${\beta }_1,\dots ,{\beta }_n$ be its trace-dual basis, so $\mathrm{Tr}({\alpha }_i{\beta }_j)={\delta }_{ij}$; this is a basis of ${\mathfrak{D}}_{S/R}^{-1}$. If $T=(\mathrm{Tr}({\alpha }_i{\alpha }_j))$, thenThus the determinant measuring the inclusion $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$ is $\det T$. The determinant description of the norm of a fractional ideal consequently givesLocalization then proves the equality over the original Dedekind domain.

Finally, the determinant-of-pairing map identifies the invertible $R$-module $({\det }_{R}S{)}^{\otimes 2}$ with the discriminant ideal. Hence in the ideal class groupup to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.