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A Noetherian ring is a ring whose ideals satisfy the ascending chain condition; equivalently every ideal is finitely generated. For the surjective ring endomorphism, put $K_j=\ker f^j$. The ascending chain of ideals $K_1\subseteq K_2\subseteq \cdots$ stabilizes, say $K_m=K_{m+1}$. If $a\in \ker f$, surjectivity of $f^m$ gives $b$ with $f^m(b)=a$. Then $f^{m+1}(b)=0$, so $b\in K_{m+1}=K_m$ and $a=0$. The endomorphism is therefore injective, hence a ring automorphism.

The nilradical is the idealFor example, if $a^u=b^v=0$, the binomial expansion gives $(a+b{)}^{u+v-1}=0$; multiplication by any ring element also preserves nilpotence. Every prime ideal contains each nilpotent element, since $a^j\in P$ implies $a\in P$. Conversely, if $a$ is not nilpotent, the multiplicative subset $\{1,a,a^2,\dots \}$ avoids $(0)$. By Zorn's lemma, choose an ideal maximal among those disjoint from this subset. It is a prime ideal: if $xy\in J$ but neither factor is in $J$, then $J+(x)$ and $J+(y)$ both meet the subset, and multiplying their two witnesses puts a power of $a$ in $J$, a contradiction. This prime ideal excludes $a$. HenceFor the zero ring the empty intersection is the whole ring, as required.

Here is a finite prime-intersection representation of a radical argument that does not assume the result. In a Noetherian ring, suppose some ideal $J$ has ideal radical not expressible as a finite intersection of prime ideals, and choose such a $J$ maximal by the ascending chain condition. It is proper and cannot be prime. Choose $a,b\notin J$ with $ab\in J$. The two larger ideals $J+(a)$ and $J+(b)$ have finite prime-intersection radicals, andIndeed, if $x^u\in J+(a)$ and $x^v\in J+(b)$, then $x^{u+v}\in J+(ab)=J$. This contradicts the choice of $J$. Applying the result to $(0)$ proves that $N(R)$ is the intersection of finitely many prime ideals. Empty intersections handle the unit ideal.

For reducedness of a formal power series ring, the inclusion of constants gives one implication: a nonzero nilpotent element of $R$ remains nonzero and nilpotent in $R[[X]]$. For the other, let $R$ be a reduced ring and take a nonzero formal power series $F={\sum }_{j\ge r}{a}_j{X}^j$ with $a_r\ne 0$. The coefficient of $X^{mr}$ in $F^m$ is $a_r^m\ne 0$, because $R$ is reduced. Thus $F$ is not nilpotent. ThereforeNo Noetherian ring hypothesis is needed for this argument.

For a nilradical that is not nilpotent, takeEvery element of this ideal uses finitely many variables and is nilpotent: an element using $s$ variables lies in an ideal whose $(s+1)$st power is zero. The quotient by this ideal is the field $k$, so it is exactly the nilradical. Nevertheless ${\bar{t}}_1\cdots {\bar{t}}_m\ne 0$ for every $m$, because the square-free monomials form a vector space basis. Thus $N(R{)}^m\ne 0$ for all $m$.

Suppose $S$ is a saturated multiplicative subset, and take $a\notin S$. The principal ideal $(a)$ is disjoint from $S$: otherwise $ra\in S$ would force $a\in S$. The prime ideal avoiding a multiplicative subset argument from Question 1 extends $(a)$ to a prime ideal $P_a$ disjoint from $S$. Thus every point of the complement belongs to such a prime ideal, andIf $S=R$, this is the empty union.

Conversely suppose $R\setminus S$ is a union of prime ideals. If $xy\in S$ and $x\notin S$, some prime ideal in that union contains $x$, and hence $xy$, a contradiction. The same holds for $y$. Thus The saturated complement as a union of prime ideals characterization holds.

For units with no finite prime-union complement, takeProducts of units are units, and a product is a unit only if both factors are units, so $S$ is a saturated multiplicative subset. Every nonunit integer lies in some $(p)$ with $p$ a positive prime, giving $\mathbb{Z}\setminus S={\bigcup }_p(p)$. If finitely many prime ideals sufficed, discard any $(0)$ and write the others as $(p_1),\dots ,(p_r)$. A prime $\ell$ outside this finite list belongs to none of them but is a nonunit, a contradiction. There is always such an $\ell$: a prime divisor of $p_1\cdots p_r+1$ is outside the list.

Use the convention that a multiplicative subset contains $1$ and is closed under multiplication. Usually $0$ is excluded; allowing it gives the harmless zero-ring localization. A prime ideal disjoint from $S$ is mapped toin the localization of a ring. This ideal is proper, andis an integral domain, since the denominators are nonzero in the integral domain $R/P$. Thus $S^{-1}P$ is a prime ideal.

Conversely, for a prime ideal $Q$ in $S^{-1}R$, its contractionis a prime ideal disjoint from $S$: each $s/1$ is a unit and cannot belong to a proper ideal. Membership in $Q$ is equivalent to numerator membership in $P$, since $a/s=(a/1)(s/1{)}^{-1}$. Hence $Q=S^{-1}P$.

To check the other composite, $a/1\in S^{-1}P$ implies $sa\in P$ for some $s\in S$, by the equality criterion for fractions in a localization of a ring. Since $s\notin P$, primality gives $a\in P$. Therefore extension and contraction are mutually inverse:If $0\in S$, both sets are empty. This is the prime ideal correspondence for localization.

A prime ideal $P$ is a proper ideal such that $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently $R/P$ is an integral domain.

For each $i$, and each $j\ne i$, choose $a_{ji}\in P_j\setminus P_i$, using incomparability. PutThen $b_i$ belongs to the other two prime ideals and does not belong to $P_i$, by primality. In particular each $b_i$ lies in the union. But $b_1+b_2+b_3\notin P_i$ for every $i$: modulo $P_i$, the sum equals $b_i\ne 0$. The union is not closed under addition and is not an ideal. This is a concrete instance of prime avoidance.

The union of three incomparable nonprime ideals example can retain pairwise incomparability. In the ringtakeThese distinct two-element ideals are pairwise incomparable, yetwhich is an ideal. None is prime: its quotient ring still contains a nonzero element of square zero. The three one-dimensional vector subspaces cover the two-dimensional vector space over ${\mathbb{F}}_2$.

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.

The height of a prime ideal is the supremum of the lengths of strict chains of prime ideals ending at $P$:The last equality follows from the prime ideal correspondence for localization. Here is a length-growth proof of the Krull height theorem that makes the bound explicit.

We first establish a prime-chain lower bound for local length. For a Noetherian local ring $(B,\mathfrak{m})$ admitting a prime chain of length $d$, put $H_B(t)=\mathrm{length}(B/{\mathfrak{m}}^t)$. Thenfor some $c>0$. For $d=0$, use $H_B(t)\ge 1$. For $d>0$, quotient by the first prime ideal of the chain; this only decreases $H_B$ and reduces us to a local integral domain $B$ withTake $0\ne a\in P_1$ and put $C=B/(a)$. It has a prime chain of length $d-1$, so induction bounds $H_C(t)$ below by a positive multiple of $t^{d-1}$.

The Artin-Rees lemma applied to $(a)\subseteq B$ gives, for a fixed integer $c_0\ge 1$,Because $a$ is a non-zero-divisor, the first term in the exact sequencehas length at least $H_B(t-c_0)$. HenceIterate this inequality about $t/(2c_0)$ times, while its arguments remain at least $t/2$. This gives the desired positive multiple of $t^d$. The particular Artin-Rees lemma inclusion used here follows directly from finite generation: the Rees algebra ${⨁}_{j\ge 0}{\mathfrak{m}}^j{t}^j$ is Noetherian, by the Hilbert basis theorem, and its graded submodule ${⨁}_j((a)\cap {\mathfrak{m}}^j)t^j$ has finitely many homogeneous generators. A bound on their degrees gives $c_0$.

Now localize at the given minimal prime ideal $P$. Write $B=R_P$, $\mathfrak{m}=P{R}_P$, and $J=I{R}_P$. The only prime ideal of $B$ containing $J$ is $\mathfrak{m}$, so $\sqrt{J}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, some ${\mathfrak{m}}^q\subseteq J$. Thus $B/J$ has finite module length, say $L$. The ideal $J$ is generated by at most $n$ elements.

For $n>0$, products of these generators give a surjection from $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $B/J$ onto $J^j/J^{j+1}$. Summing the lengths gives the generator bound for primary-ideal lengthSince $J\subseteq \mathfrak{m}$, we have $J^t\subseteq {\mathfrak{m}}^t$, and thereforeThe prime-chain lower bound for local length now excludes any chain of length greater than $n$. ConsequentlyFor $n=0$, minimality over the zero ideal says that $P$ is a minimal prime ideal, so its height is zero. This also handles that boundary case.

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.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. A character of an algebra is a nonzero multiplicative complex linear functional $\chi :A\to \mathbb{C}$. It satisfies $\chi (1)=1$. Moreover $\chi (a)\in {\sigma }_A(a)$: otherwise $a-\chi (a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the spectrum of an element therefore gives $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

Every proper maximal ideal $M$ of $A$ is closed. Indeed its closure is an ideal; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann series, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The quotient Banach space $A/M$, with its quotient Banach algebra structure, is a complex normed division algebra. By the Gelfand-Mazur theorem, it is $\mathbb{C}$, so the quotient map gives a character of an algebra with kernel $M$. Conversely, the kernel of every character of an algebra is a maximal ideal, since the character is onto $\mathbb{C}$. The Zorn lemma supplies a maximal ideal containing every proper ideal, so the character space of an algebra $\Delta (A)$ is nonempty.

These facts give the exact relation between the character space and the spectrum of an element:One inclusion was proved above. For the other, if $a-\lambda 1$ is noninvertible, the principal ideal it generates is proper because $A$ is commutative. Contain it in a maximal ideal and use its corresponding character of an algebra to obtain $\chi (a)=\lambda$.

Give $\Delta (A)$ the Gelfand topology, namely its subspace topology from the weak-star topology on $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. This is a continuous function on $\Delta (A)$ by definition of the Gelfand topology. The Gelfand representation theorem gives a contractive unital algebra homomorphism over a fieldMultiplicativity and linearity follow by evaluating at each character of an algebra; the supremum norm equality follows from the preceding spectrum of an element identity. Its kernel isthe Jacobson radical. Equivalently, its elements have spectrum of an element $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a semisimple commutative Banach algebra, and it gives a faithful continuous representation of $A/\mathrm{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta (A)$, because distinct characters of an algebra differ on some $a$. An arbitrary Banach algebra need not have an isometric or surjective Gelfand transform, nor a uniformly dense range: those conclusions require further hypotheses.

For the Banach algebra $C(K)$ on a nonempty compact Hausdorff space $K$, all characters of an algebra are evaluation characters. To see this, let $M$ be a maximal ideal. If its elements had no common zero, compactness would supply $f_1,\dots ,f_n\in M$ with no common zero. The continuous function $h={\sum }_i\overline{f_i}{f}_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq \ker {\delta }_x$ and maximality gives equality. The associated character of an algebra must be ${\delta }_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.

The map $x\mapsto {\delta }_x$ is a continuous bijection $K\to \Delta (C(K))$, using separation of points by continuous functions. Compactness and the Hausdorff property make it a homeomorphism. Under this identification the Gelfand transform is $\boxed{\hat{f}({\delta }_x)=f(x)}$, so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty character space represents the zero function space; it was excluded by the nonzero unital convention above.

Now let $A$ be a commutative unital C-star algebra. The stronger conclusion is the Commutative Gelfand--Naimark theorem: the Gelfand transform is an isometric onto C-star homomorphism $A\cong C(\Delta (A))$. We prove the additional assertions without assuming this conclusion.

First every character of an algebra respects the C-star algebra involution. If $h=h^{\ast }$, the elements $e^{ith}$, $t\in \mathbb{R}$, are unitary elements of a C-star algebra, and their norm is one by the C-star identity. Continuity and multiplicativity give $\chi (e^{ith})=e^{it\chi (h)}$. Thus $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. Therefore the range of $\Gamma$ is closed under complex conjugation.

Every element of commutative $A$ is a Normal element of a C-star algebra. For a normal $b$, use the C-star identity, and then the same identity for the self-adjoint element $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. The Gelfand transform is therefore an isometry, and its range is complete and closed in the supremum norm. It contains constants, separates points and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence all of $C(\Delta (A))$.

The approximation step in Stone-Weierstrass theorem can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\epsilon >0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.