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

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

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

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

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

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

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

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

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

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

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

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

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

The points of the spectrum of a commutative ring $A$ are its prime ideals. Its Zariski topology has closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and basis $D(f)=\{\mathfrak{p}:f\notin \mathfrak{p}\}$. The structure sheaf of a scheme can be defined as follows: a section over $U$ assigns an element of $A_{\mathfrak{q}}$ to every $\mathfrak{q}\in U$, and locally that assignment is represented by one fraction $a/s$, with $s$ outside every prime in the neighbourhood. Restrictions forget values. Local representability and equality can be checked on an open cover, so these assignments satisfy the sheaf gluing axiom. On $D(f)$ the sections identify with the localization of a ring $A_f$.

Evaluation at $\mathfrak{p}$ induces the ring homomorphism ${\mathcal{O}}_{X,\mathfrak{p}}\to A_{\mathfrak{p}}$. It is surjective because $a/s\in A_{\mathfrak{p}}$ is represented by a section on $D(s)$. For injectivity, represent a germ by $a/s$ on a neighbourhood of $\mathfrak{p}$. If its value in $A_{\mathfrak{p}}$ vanishes, there is $t\notin \mathfrak{p}$ with $ta=0$. The section is then zero after restricting to $D(st)$ inside that neighbourhood. Its germ is zero. Thus

Let $S=\{1,x,x^2,\dots \}$. Since a principal ideal domain is an integral domain and $x\ne 0$, $S$ is a multiplicative subset avoiding zero. The stated equivalence relation and operations are precisely those of the localization of a ring $S^{-1}R$.

For completeness, reflexivity and symmetry are immediate, while transitivity follows by multiplying the two cross-multiplication identities and cancelling a nonzero power of $x$. If $(r,x^n)\sim (s,x^m)$ and $(r^{\prime },x^{n^{\prime }})\sim (s^{\prime },x^{m^{\prime }})$, cross multiplication shows that the proposed sums and products are equivalent. Thus the operations do not depend on representatives. The usual fraction calculations then prove associativity, commutativity and distributivity, with identities $[(0,1)]$ and $[(1,1)]$. Therefore $R^{\prime }$ is a well-defined commutative ring.

For the K-theory transfer of a finite covering, let $V\to Y$ be a vector bundle and defineOn an evenly covered neighborhood the $n$ sheets identify this with a direct sum of $n$ local vector bundles. On overlaps the identifications permute the sheets and apply the transition maps of $V$, so these local descriptions glue to a vector bundle on $X$. This operation respects direct sums, hence extends to a homomorphism of Grothendieck groupsIt obeys the projection formula for the K-theory transfer:For actual vector bundles, this follows fiberwise by distributing the tensor product of vector bundles over the direct sum; it then extends to their Grothendieck groups.

Let $P=p_{!}(1_Y)$ be the permutation vector bundle of the covering. It has rank $n$ on every connected component, but need not be a trivial vector bundle; in particular one cannot replace $p_{!}{p}^{\ast }$ by multiplication by $n$ integrally. Instead writeThe componentwise nilpotence result from the preceding solution gives ${\eta }^N=0$ for some $N$. In the localization of a ring $K^0(X)\otimes \mathbb{Z}[1/n]$, the finite geometric seriesis an inverse. The projection formula for the K-theory transfer givesThus $p^{\ast }(a)=0$ implies $a=0$ after inverting $n$. More explicitly, $(n+\eta {)}^{-1}{p}_{!}$ is a left inverse, proving

Choose an open cover of $X$ by affine opens $U_i=\mathrm{Spec}{B}_i$. Since $X$ is a quasi-compact topological space, finitely many suffice. Let $f_i,a_i\in B_i$ be the restrictions of $f,a$. Part (a) identifies $U_i\cap X_f$ with $D(f_i)$, whose ring of regular functions is the localization of a ring $(B_i{)}_{f_i}$. The vanishing of $a$ there meansso $f_i^{n_i}{a}_i=0$ for some $n_i\ge 0$. Taking $n>0$ at least as large as every $n_i$ gives $(f^{n}a){|}_{U_i}=0$ on every member of the finite cover. The local identity axiom for the sheaf of rings ${\mathcal{O}}_X$ therefore gives

Restriction and division by powers of $f$ define a natural ring homomorphismIf $\Phi (a/f^n)=0$, then $a{|}_{X_f}=0$, and part (b) gives $f^{m}a=0$ for some $m$; this is exactly the criterion that $a/f^n=0$ in the localization of a ring $A_f$. Hence $\Phi$ is injective. Given $b$ in the target, part (c) gives $f^{N}b=a{|}_{X_f}$ for some $a\in A$, so $b=\Phi (a/f^N)$. Hence $\Phi$ is surjective and

The set $S=1+I$ is a multiplicative subset, since $(1+a)(1+b)=1+(a+b+ab)$. The localization of a ring $S^{-1}R$ consists of fractions $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe Localization of a Noetherian ring is Noetherian. Since a finitely generated module over a Noetherian ring is a Noetherian module, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. But $R$ is an integral domain, so its localization map into $(1+I{)}^{-1}R$ is injective. This is the non-Noetherian failure of the intersection formula for localization.

The localization of a ring $S^{-1}R$ consists of pairs $(r,s)\in R\times S$ modulofor some $u\in S$. Addition and multiplication are

False. LetThe element $1/y$ does not belong to $A$: assigning degrees $\mathrm{deg}x=\mathrm{deg}y=1$ and $\mathrm{deg}(x/y)=0$ shows that every element of $A$ has nonnegative total degree, whereas $1/y$ has degree $-1$.

If $A=T^{-1}R$ for some multiplicative subset $T$, the membership $x/y\in T^{-1}R$ would give $x/y=a/t$ for some $a\in R$ and $t\in T$. Since $R$ is a unique factorization domain and $x,y$ are coprime, the equation $xt=ay$ implies $y\mid t$. Write $t=yu$. As $t$ is invertible in $T^{-1}R$, so iscontrary to $1/y\notin A$. Thus a subring of a localization of a ring need not itself be a localization of the original ring.

A multiplicative subset $S\subseteq R$ contains $1$ and satisfies $s,t\in S\Rightarrow st\in S$. The localization of a ring consists of fractions $r/s$ modulo the relationIts structure map $\iota :R\to S^{-1}R$ makes every $s\in S$ invertible. The universal property of localization says that if $f:R\to T$ is any ring homomorphism for which every $f(s)$ is a unit, there is one unique homomorphism $\tilde{f}:S^{-1}R\to T$ satisfying $\tilde{f}\iota =f$, namely

The K-theory transfer of a finite covering distributes over a pulled-back tensor product of vector bundles:For an $n$-sheeted covering map, $p_{!}(1)$ is the permutation vector bundle of rank $n$. Its difference from $n$ is a nilpotent element by nilpotence of rank-zero K-theory classes, so it is a unit after inverting $n$. Thus pullback on Topological K-theory becomes injective after this localization of a ring.

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.