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Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.

For a finitely generated algebra over $k$, choose a finite-dimensional generating subspace $V$ containing one. Its Gelfand–Kirillov dimension isChanging generators bounds the two filtrations by constant rescalings of the index, so the value is independent of $V$.

For the noncommutative Krull dimension, set the zero module's dimension to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Recursively, dimension at most $d$ means that in every descending chain of submodules the successive factors eventually have dimension less than $d$; use the ordinal version of this definition if necessary. The left Krull dimension of $R$ is that of its left regular module. For commutative Noetherian rings this agrees with the usual prime-chain Krull dimension, but prime-chain counting alone is not its definition for a noncommutative ring.

Write $U=U({\mathfrak{sl}}_2(\mathbb{C}))$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. By the Poincaré-Birkhoff-Witt theorem, $f^i{h}^j{e}^k$ form a basis. In the total-degree filtration there are $\left(\genfrac{}{}{0ex}{}{n+3}{3}\right)$ such monomials of degree at most $n$, soThe commutators lower degree, giving $\mathrm{gr}U\cong \mathbb{C}[f,h,e]$, which also proves that $U$ is left and right Noetherian by finite leading-term cancellation.

To determine the module-theoretic Krull dimension, use the generalized Weyl algebra description and verify the dimension criterion, rather than substituting the three-dimensional graded ring. PutCommutation with $h$ is immediate; with $e$, the contributions are $4he-2(eh+he)-4e=0$, using $he-eh=2e$. The corresponding calculation with $f$ also vanishes. Thus $z$ is a Casimir element. The subalgebra $D=\mathbb{C}[h,z]$ is a polynomial ring in two variables: for a nonzero polynomial in $z$ with coefficients in $\mathbb{C}[h]$, its highest $z$ power contributes a nonzero PBW term with that many copies of both $f$ and $e$, and no lower power can cancel it.

Take $\sigma (h)=h-2$, $\sigma (z)=z$ and $t=(z-h^2-2h)/4$. The relations becomeThey identify $U$ with $D(\sigma ,t)$. Surjectivity follows from its generators; injectivity follows from PBW: the graded pieces $D$, $D{e}^r$ and $D{f}^r$ for $r>0$ are linearly independent, using the difference between the $e$ and $f$ exponents and then the highest paired exponent. Every PBW monomial lies in one of these pieces by replacing paired $f,e$ powers with polynomials in $h,z$.

We use the following general Krull dimension criterion for rank-one generalized Weyl algebras: for a commutative Noetherian base $D$ of dimension $d$, the generalized Weyl algebra has Krull dimension $d$ provided no height-$d$ maximal ideal is periodic under $\sigma$ and none contains infinitely many translates ${\sigma }^n(t)$. This is Theorem 5.3, specialized to the commutative base, in Krull dimension of Generalized Weyl Algebras with non-commutative coefficients.

Here every maximal ideal is ${\mathfrak{m}}_{\lambda ,\mu }=(h-\lambda ,z-\mu )$ and has height two. Its translates have $h$ coordinate $\lambda +2n$, so no nonzero power of $\sigma$ fixes it. Further,This is a nonzero quadratic polynomial in $n$, so it vanishes for at most two integers. Both exceptional mechanisms are excluded for every maximal ideal, and the base has dimension two. The criterion consequently givesThe orbit and root calculations are needed for this upper bound on the entire left-ideal lattice; merely exhibiting a length-two chain of prime ideals would only prove a lower bound.

An extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$, meaning thatfor some monic polynomial. The Krull dimension of a ring is the supremum of the lengths $r$ of strict chains ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_r$ of prime ideals; the number counted is the number of inclusions.

We prove the prime-chain facts needed for integral extensions preserve Krull dimension. Taking quotients preserves an integral equation. Localizing both rings by the same multiplicative subset of $A$ also preserves integrality: for $b/s$, divide the equation for $b$ by $s^n$, so its new coefficients are $a_i/s^{n-i}$. Localizing just $B$ at all elements outside one of its primes would not be the same operation.

Two elementary field observations are useful. If a field $C$ is integral over a subdomain $R$, then $R$ is a field, by the inverse argument given in Question 2. Conversely, if an integral domain $C$ is integral over a field $R$, every nonzero $c\in C$ has a monic equation with nonzero constant term: cancel a factor of $c$ from an equation if necessary. Dividing by $c$ and by that constant term expresses $c^{-1}$ as a polynomial in $c$. Thus $C$ is a field. In particular, the contraction of a maximal ideal under an integral extension is maximal.

For the Lying-over theorem, fix $\mathfrak{p}\in \mathrm{Spec}A$ and localize by $S=A\setminus \mathfrak{p}$. The nonzero domain $S^{-1}B$ is integral over the local ring $A_{\mathfrak{p}}$. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}B$. Its contraction is maximal in $A_{\mathfrak{p}}$, so it is the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. The prime ideal correspondence for localization now pulls $\mathfrak{n}$ back to a prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{q}\cap A=\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ and $\mathfrak{q}\cap A=\mathfrak{p}$. The injection $A/\mathfrak{p}\hookrightarrow B/\mathfrak{q}$ is still integral. Apply the just-proved Lying-over theorem to ${\mathfrak{p}}^{\prime }/\mathfrak{p}$. Pulling the resulting prime back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap A={\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ have the same contraction $\mathfrak{p}$. Write $R=A/\mathfrak{p}$ and $C=B/\mathfrak{q}$. The latter is a domain integral over $R$. Localizing both by $R\setminus \{0\}$ makes $S^{-1}C$ integral over the fraction field of $R$, so $S^{-1}C$ is a field. The prime ${\mathfrak{q}}^{\prime }/\mathfrak{q}$ is disjoint from $S$ and therefore localizes to a proper prime, which must be zero in this field. Since $C$ is a domain and all denominators are nonzero, an element of $C$ whose localization is zero was already zero. Hence ${\mathfrak{q}}^{\prime }/\mathfrak{q}=0$, proving ${\mathfrak{q}}^{\prime }=\mathfrak{q}$.

A strict prime chain in $B$ therefore contracts to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, start over the bottom of any finite prime chain in $A$ using Lying-over theorem, and successively use going-up theorem to lift the entire chain. Distinct contractions ensure that the lifted inclusions are strict, giving the reverse inequality. Taking suprema, including when arbitrarily long chains exist, proves

For an algebraic extension $K/\mathbb{Q}$, its ring $\mathcal{O}$ of algebraic integers is an integral extension of $\mathbb{Z}$. The primes of $\mathbb{Z}$ are $(0)$ and the maximal ideals $(p)$, so the longest strict chains have length one; for example $(0)⊊(2)$. The preceding argument yieldsThis Krull dimension of rings of algebraic integers argument also covers infinite algebraic extensions. It uses integrality, not an assumption that $K$ is a number field or that $\mathcal{O}$ is Noetherian.

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

Let $A=\mathbb{C}[x,y,z]$. This polynomial ring is a unique factorization domain of Krull dimension three. For a height-one prime in a unique factorization domain $\mathfrak{p}$, choose a nonzero element of it and an irreducible factor $q$ belonging to $\mathfrak{p}$. In a unique factorization domain, $q$ is prime, so $(q)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Strict containment would give a chain $(0)⊊(q)⊊\mathfrak{p}$ of length two. ThusIf $\mathfrak{p}$ has height three, it is maximal: any strictly larger proper prime would extend a length-three chain and contradict $\dim A=3$. The Weak Hilbert Nullstellensatz over the algebraically closed field $\mathbb{C}$ then givesfor some $a,b,c\in \mathbb{C}$. This proves both requested generator assertions.

For the monomial curve with exponents three, four and five, putAll three map to zero under the given parametrization, so $J\subseteq I=\ker \phi$. Reduce any monomial using $y^2\mapsto xz$, $yz\mapsto x^3$ and $z^2\mapsto x^{2}y$. Each reduction decreases the sum of the exponents of $y,z$, so it terminates. Every polynomial has, modulo $J$, the formIts image is $A_0(t^3)+t^4{A}_1(t^3)+t^5{A}_2(t^3)$. These three terms occupy different exponent classes modulo three, so they cannot cancel. A zero image forces each $A_i=0$. ConsequentlyThe quotient is the domain $\mathbb{C}[t^3,t^4,t^5]$, hence $I$ is a prime ideal. It is finite free over $\mathbb{C}[t^3]$, with basis of a module $1,t^4,t^5$, by the same normal-form argument. Its dimension is therefore one, using that integral extensions preserve Krull dimension. The polynomial-ring height and dimension formula givesFor the generator obstruction, let $\mathfrak{m}=(x,y,z)$. Since $I\subseteq {\mathfrak{m}}^2$, we have $\mathfrak{m}I\subseteq {\mathfrak{m}}^3$. The degree-two initial forms of $g_1,g_2,g_3$ areThey are linearly independent over $\mathbb{C}$ in ${\mathfrak{m}}^2/{\mathfrak{m}}^3$. Thus no nonzero constant linear combination of the $g_i$ belongs to $\mathfrak{m}I$, and their classes are independent in $I/\mathfrak{m}I$. Since the three generators span that quotient,Two global generators would span it by two vectors, which is impossible. This is the initial-form lower bound for ideal generators; it shows why height two does not force two generators, even though height one and height three behave as above.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

An integral extension $R\subseteq T$ means that every element $t\in T$ satisfies a monic polynomial with coefficients in $R$:The Krull dimension is the supremum of the lengths of strict chains of primes:There need not be a finite bound on these lengths.

Here are the prime-ideal facts behind dimension preservation, with their relevant proofs. If a domain $B$ is integral over a subdomain $A$ and $B$ is a field, then $A$ is a field: for $0\ne a\in A$, a monic equation for $a^{-1}\in B$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$. Conversely, an integral domain integral over a field is itself a field: a nonzero element has a polynomial equation with nonzero constant term after removing any factor of the indeterminate, and that equation expresses its inverse. Applied to quotients, these observations show that a prime in an integral extension is maximal if and only if its contraction is maximal.

For the Lying-over theorem, localize $R\subseteq T$ at $S=R\setminus \mathfrak{p}$. The inclusion remains injective and integral, and $S^{-1}T$ is nonzero. Any maximal ideal of $S^{-1}T$ contracts to the unique maximal ideal of $R_{\mathfrak{p}}$, by the field criterion just proved. The prime ideal correspondence for localization then gives a prime of $T$ contracting to $\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{q}$ lies over $\mathfrak{p}$ and $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$. The quotient inclusion $R/\mathfrak{p}\subseteq T/\mathfrak{q}$ is integral. Apply Lying-over theorem to the prime ${\mathfrak{p}}^{\prime }/\mathfrak{p}$; lifting back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ contracting to ${\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ contract to the same $\mathfrak{p}$. After localizing at $R\setminus \mathfrak{p}$, both are maximal ideals, because they lie over the maximal ideal of $R_{\mathfrak{p}}$. Their inclusion is therefore equality. The bijection between primes under localization gives $\mathfrak{q}={\mathfrak{q}}^{\prime }$.

Now contract a strict chain of primes in $T$. Incomparability theorem for integral extensions ensures that every contraction remains strict, so $\dim T\le \dim R$. Conversely, Lying-over theorem lifts the first member of any finite chain in $R$, and repeated going-up theorem lifts the remaining members; different contractions ensure a strict chain in $T$. Taking suprema, including the possibility of infinity, givesThis proves that integral extensions preserve Krull dimension.

For the given quotient, put $B=k[Y][X]/(X^2+YX+Y^3)$. The relation is monic in $X$, so monic polynomial division gives a unique representative $a(Y)+b(Y)X$. Thus $k[Y]\hookrightarrow B$ is injective and $B$ is free of rank two as a $k[Y]$-module. In particular, $B$ is integral over $k[Y]$. The one-variable polynomial ring has dimension one: its zero prime is strictly below $(Y)$, and every nonzero prime is maximal because $k[Y]$ is a principal ideal domain. HenceThis is an instance of dimension of a monic plane hypersurface. No irreducibility or algebraic-closure assumption on $k$ is needed; monicity supplies the integral extension in every characteristic.

Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor a module whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t{)}^r$.

Here is an induction proof. An Artinian ring is Noetherian by question 1, so the Hilbert basis theorem makes $S$ Noetherian. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite module over the Artinian ring $A$, hence has finite length of a module, and $P_M$ is a polynomial.

For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded modules annihilated by $x$, hence modules over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis is the Hilbert series multiplication exact sequence. By induction, the right-hand side has denominator ${\prod }_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the Hilbert-Serre theorem. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and module generators; no analytic convergence of a series is involved.

For the local invariant, use the usual Noetherian local ring hypothesis of Hilbert–Samuel growth dimension. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the localization at a prime ideal of the polynomial ring $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, the vector space $\mathfrak{m}/{\mathfrak{m}}^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak{m}=P$ and $\kappa =R/\mathfrak{m}$. Its associated graded ringis generated over $\kappa$ in degree one, because $\mathfrak{m}$ is finitely generated. Thus its Hilbert series is rational with a denominator that is a power of $1-t$. After cancelling factors, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ is the degree of the cumulative Hilbert-Samuel polynomial, rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals Krull dimension by the local dimension theorem, although that theorem is not required to define it here.

For an example, take $R=k[x,y{]}_{(x,y)}$. Its associated graded ring is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.

The dimension of a topological space by irreducible chains is the supremum of the integers $d$ for which there is a chain $Z_0⊊Z_1⊊\cdots ⊊Z_d$ of nonempty irreducible closed subsets of the space. Closedness here is relative to the given locally closed space. For varieties this is their Krull dimension.

An algebraic group is a group whose underlying space is an algebraic variety and whose multiplication $G\times G\to G$ and inversion $G\to G$ are morphisms. An algebraic group action on a variety $V$ is a morphism $G\times V\to V$ satisfying the identity and associativity axioms of a group action.

For the homomorphism $\phi$, the kernel is ${\phi }^{-1}(1_H)$, so the kernel is closed. To prove closedness of the image, use the Chevalley constructibility theorem: the image of a morphism of varieties is constructible. Thus $S=\phi (G)$ is a constructible subset of a variety and an abstract subgroup. Its closure $C$ is also a subgroup: translation by elements of $S$ preserves $C$, and continuity then extends multiplication and inversion to $C$.

A dense constructible subset contains a dense open subset $U$ of its closure. For $c\in C$, both $U$ and $cU$ are dense open subsets of $C$, so their intersection is nonempty. If $u=cv$ with $u,v\in U\subseteq S$, then $c=u{v}^{-1}\in S$. Hence $S=C$, proving that the image is closed. This is the principle that a constructible subgroup is closed.

Every nonempty fiber of $G\to S$ is a translate of $\ker \phi$ and has that same dimension. The fiber dimension theorem therefore givesThis dimension formula for an algebraic group homomorphism is a dimension statement, so it does not require separability of $\phi$.

For a dimension vector of a quiver representation $\mathbf{n}$, setThe base change action on quiver representations isThe entries are regular functions on the product of the general linear groups and the quiver representation space, because inverse entries are cofactors divided by the invertible determinant. Thus this is an algebraic group action.

For nonzero $\mathbf{n}$, let $\Delta k^{\times }$ be the common scalar subgroup $(\lambda I_{n_i}{)}_i$ and define $\mathrm{PGL}(\mathbf{n})=\mathrm{GL}(\mathbf{n})/\Delta k^{\times }$. Common scalars act trivially, so the formula descends to an algebraic group action of this projective base change group of a quiver. This is a quotient by one common scalar, not a product of the individual projective groups. If every $n_i=0$, the representation space is a point and both actions are taken to be trivial.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis