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Let $\mathfrak{m}$ be the maximal ideal and $k=A/\mathfrak{m}$ the finite residue field. The maximal ideal of an Artinian local ring is nilpotent, so ${\mathfrak{m}}^r=0$ for some $r$. Every quotientis both an Artinian $A$-module and a vector space over $k$. An Artinian vector space is finite-dimensional, hence each quotient is a finite set. The finite filtrationtherefore proves that the underlying set of $A$ is finite. This is the Finiteness criterion for an Artinian local ring.

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.

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.

For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

True. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

The maximum isThe product of two fields $K\times L$ attains it: its only nonzero proper ideals are $K\times 0$ and $0\times L$, and both are maximal ideals.

For the upper bound, suppose ${\mathfrak{m}}_1$ and ${\mathfrak{m}}_2$ are distinct maximal ideals of $R$. Their intersection cannot be nonzero, since a nonzero proper ideal is maximal and cannot be properly contained in either of two distinct maximal ideals. Hence ${\mathfrak{m}}_1\cap {\mathfrak{m}}_2=0$. Also ${\mathfrak{m}}_1+{\mathfrak{m}}_2=R$, so the Chinese remainder theorem givesBoth factors are fields, and this product has exactly two maximal ideals. Thus a third maximal ideal is impossible.

Assume tensor products of two nonzero modules never vanish. If $\mathfrak{m}\ne \mathfrak{n}$ were distinct maximal ideals, then $\mathfrak{m}+\mathfrak{n}=R$ andalthough both residue fields are nonzero. Hence $R$ has one maximal ideal $\mathfrak{m}$ and is a local ring.

For every module $M$,If $\mathfrak{m}M=M$, this tensor product vanishes. Since $R/\mathfrak{m}\ne 0$, the assumed property forces $M=0$. Thus condition (a) implies condition (b).

A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$, and the character space of an algebra is the set ${\Phi }_A$ of all such characters. Since $A$ is unital, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. The spectral radius estimate therefore yieldsThus every character is continuous and has norm one.

Let $M$ be a maximal ideal. Its norm closure is again an ideal. It cannot equal $A$, because then some $m\in M$ would satisfy $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Hence $M$ is closed. The quotient $A/M$ is a complex unital Banach division algebra, so the Gelfand-Mazur theorem identifies it with $\mathbb{C}$. Composing the quotient map with this isomorphism gives a character with kernel $M$. Conversely, a character kernel is maximal because its quotient is $\mathbb{C}$.

Now $\lambda \in {\sigma }_A(x)$ exactly when $x-\lambda 1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker \phi$, giving $\phi (x)=\lambda$. The reverse implication follows from the first paragraph, so

The Gelfand topology is the weak-star topology on ${\Phi }_A\subseteq A^{\ast }$. The Gelfand transform isIts values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finallyso it is continuous.

The valuation ring and its maximal ideal areIf ${\mathcal{O}}_K$ is Noetherian, its maximal ideal is finitely generated. Ideals in a valuation ring are totally ordered, so every finitely generated ideal is generated by one of its generators; write $\mathfrak{m}=(\pi )$. Then $v(\pi )$ is the smallest positive element of the value group. Subtracting integral multiples of this value shows that every value is an integral multiple of $v(\pi )$, so $v$ is discrete.

Now assume $K$ is complete and discretely valued. Parts i and ii, applied to $v$ and to the other discrete valuation $v^{\prime }$, giveIf $v(a)=0$ and $\pi$ is a uniformizer for $v$, then $\pi a^r\in {\mathfrak{m}}_v$ for every $r\in \mathbb{Z}$. Hencefor every integer $r$, forcing $v^{\prime }(a)=0$ and in particular $v^{\prime }(a)\ge 0$. Every nonzero $x$ has the form $x={\pi }^{m}u$ with $v(u)=0$, soThe proportionality constant is positive because $v^{\prime }$ is nontrivial. Thus the valuations, and their associated absolute values, are equivalent.

A commutative Artinian ring has finitely many prime ideals, and every one is a maximal ideal. If $\mathfrak{p}$ is prime, the Artinian domain $R/\mathfrak{p}$ is a field: stabilization of $(x)\supseteq (x^2)\supseteq \cdots$ gives $x^n=x^{n+1}a$, and cancellation gives $xa=1$. Infinitely many distinct maximal ideals ${\mathfrak{m}}_i$ would give a strictly descending chainbecause distinct maximal ideals are comaximal.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

The points of the real affine plane as a scheme are its prime ideals: the zero ideal, the principal ideals of irreducible real polynomials, and its maximal ideals. The latter have residue field $\mathbb{R}$ or $\mathbb{C}$, by the Zariski lemma and the real closed field property. They correspond respectively to real coordinate pairs and conjugate pairs of nonreal complex coordinate pairs. Thus the scheme contains both nonclosed generic points and closed points absent from the real locus. Its topology is the Zariski topology, and its structure sheaf has sections $\mathbb{R}[x,y{]}_h$ on each principal open subscheme $D(h)$.

A Noetherian local ring $(R,\mathfrak{m})$ is regular when the minimal number of generators of its maximal ideal equals its Krull dimension. Every regular local ring is an integral domain and a unique factorization domain.

If an elliptic curve has good reduction over a local field $K$, its integral projective model is smooth. Every point over the residue field lifts by the Hensel lemma, applied in a chart with a unit partial derivative. Thus reduction gives an exact sequence $0\to E_1(K)\to E(K)\to \tilde{E}(k)\to 0$. The kernel of reduction of an elliptic curve is the maximal ideal of the valuation ring, with the formal group of an elliptic curve as its group operation.

If a Weil divisor $D$ on a normal variety is not Cartier at $x$, every linearly equivalent divisor has $x$ in its support of a Weil divisor. Otherwise that representative would vanish near $x$, expressing $D$ as a principal Weil divisor there. On $V(ac-b^2)$ the prime divisor $V(a,b)$ is not Cartier at the origin: its ideal has two independent generators modulo the maximal ideal times that ideal. This gives an unavoidable support point even for representatives with negative coefficients.

A uniformizer of a discrete valuation ring $R$ is a generator $\pi$ of its unique nonzero maximal ideal. Every nonzero $x$ in the fraction field has a unique expression $x=u{\pi }^n$ with $u$ a unit and $n\in \mathbb{Z}$.