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The categorical commutation of iterated categorical limits theorem is the following. For small categories $I,J$ and a two-variable functor $F:I\times J\to \mathcal{D}$, if the two collections of inner limits and their outer limits exist, then there are canonical isomorphismsIn particular all these limits exist if $\mathcal{D}$ has all $I$-shaped and all $J$-shaped limits. The isomorphisms are natural in $F$ and uniquely characterized by preserving every projection to $F(i,j)$.

Part (iii) proves that each iterated limit represents the functor of cones on the product diagram, with the same flattened projections. Two such representations have unique inverse cone-compatible arrows, proving the isomorphisms. For a natural transformation $F\to F^{\prime }$, the two composites in any naturality square have the same product-diagram projections, hence agree by limit uniqueness. This proves naturality as well. Completeness of $\mathcal{D}$ is a convenient sufficient hypothesis, rather than a necessary one for this individual diagram.

First prove strict creation for the canonical forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$. Let a diagram of algebras $(A_i,a_i)$ have a chosen underlying limit cone $p_i:L\to A_i$. The maps $a_{i}T(p_i):TL\to A_i$ form a cone: if $h:A_i\to A_j$ is a diagram arrow, then $h{a}_i=a_{j}T(h)$ and $h{p}_i=p_j$. Thus the limit property gives a unique action $a:TL\to L$ withThe projections are jointly monic. By naturality of $\eta$ and the algebra unit laws, $p_{i}a{\eta }_L=a_i{\eta }_{A_i}{p}_i=p_i$, hence $a{\eta }_L=1_L$. Likewiseso $a{\mu }_L=aT(a)$. Therefore $(L,a)$ is an algebra and all $p_i$ are algebra morphisms.

For any algebra cone $h_i:(B,b)\to (A_i,a_i)$, let $h:B\to L$ be its unique underlying factor. Then $p_{i}hb=h_{i}b=a_{i}T(h_i)=p_{i}aT(h)$, so joint monicity gives $hb=aT(h)$. Thus the factor is an algebra morphism, unique as such. The action on the prescribed $L$ was itself forced by the cone equations. This proves monad algebra forgetful functor creates limits, without assuming $T$ preserves them.

For a monadic $G$ with comparison an equivalence, transport $(L,a)$ and its cone across that equivalence. This yields a limiting cone in $\mathcal{D}$ whose image is identified with the prescribed base cone by an isomorphism, uniquely up to the corresponding cone-compatible isomorphism. Thus a monadic functor creates limits in the equivalence-invariant sense. If monadicity uses a comparison isomorphism over the base, the lift is literally on the prescribed underlying cone, proving strict creation as well.

A qualification is necessary if equivalence is used to define monadicity but literal unique lifting is demanded. Let $\mathcal{C}$ be the indiscrete category on two objects $0,1$, let $\mathcal{D}$ be the terminal category, and let $G$ select $0$. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to $G$, and the comparison to the induced algebra category is an equivalence: the two algebra objects have exactly one map between each pair. Yet the chosen terminal cone with vertex $1$ in $\mathcal{C}$ has no literal lift, since the only object in the image of $G$ is $0$. The creation of limits up to isomorphism conclusion holds, because $0$ and $1$ are uniquely isomorphic. This distinguishes the two conventions rather than claiming an invalid strict conclusion from equivalence alone.

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\Rightarrow T$, $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$, $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism $(A,a)\to (B,b)$ is an arrow $h:A\to B$ with $ha=bT(h)$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$.

A functor $G:\mathcal{D}\to \mathcal{C}$ is monadic when it has a left adjoint $F$ and its comparison $K:\mathcal{D}\to {\mathcal{C}}^{GF}$ is an equivalence; this is the monadic adjunction convention. Some strict formulations require $K$ to be an isomorphism over the base. The distinction affects the meaning of literal limit creation, discussed in part (iii), but not the algebra construction.

For finite rooted trees, use rooted-tree homeomorphic embedding: an injective vertex map preserves ancestors and lowest common ancestors, so edges may become paths and distinct branches remain distinct. The root can map to a vertex below the host root. More generally label vertices in a well-quasi-ordering $Q$ and require each source label to precede its image label. We prove this stronger labelled version of Kruskal's tree theorem, from which the unlabelled result follows by taking one label.

We first need the finite-word closure, and prove it rather than merely invoking Higman lemma. If finite words over $Q$ were not a well-quasi-ordering under order-preserving subsequence embedding with increased labels, choose a minimal bad sequence $w_0,w_1,\dots$, minimizing the length of $w_n$ among words admitting a bad continuation of the already chosen prefix. No $w_n$ is empty. Write $w_n=v_n{a}_n$ by removing its last letter. The perfect subsequence lemma gives $n_0<n_1<\cdots$ with $a_{n_r}\le a_{n_s}$ for $r<s$.

The sequence $w_0,\dots ,w_{n_0-1},v_{n_0},v_{n_1},\dots$ is still bad. An earlier $w_i$ embedding into a later $v_{n_r}$ would embed into $w_{n_r}$, contradicting the original sequence. An embedding of $v_{n_r}$ into $v_{n_s}$ extends by the last-letter comparison to an embedding of $w_{n_r}$ into $w_{n_s}$. Both possibilities are excluded. This contradicts minimality at position $n_0$, since $v_{n_0}$ is shorter. Thus the finite-word closure holds. The perfect subsequence lemma also proves finite product closure: first pass to a nondecreasing subsequence in one coordinate, then in the next, preserving all previous comparisons.

Suppose now that the labelled trees admit a bad sequence $T_0,T_1,\dots$. Choose a minimal bad sequence by vertex number. Let $\mathcal{U}$ be all the proper descendant-rooted subtrees of these trees. We claim that $\mathcal{U}$ is a well-quasi-ordering. If it had a bad sequence of subtrees $U_r$, its parent indices can be made strictly increasing, say $U_r$ comes from $T_{n_r}$. Indeed each finite initial collection of parent trees has only finitely many subtrees, and a bad sequence cannot contain infinitely many terms from a finite collection. Passing to a subsequence therefore makes $n_0<n_1<\cdots$.

Then $T_0,\dots ,T_{n_0-1},U_0,U_1,\dots$ is bad. A prefix tree embedding into $U_r$ would embed into its parent $T_{n_r}$, contrary to the original badness, and the tail is bad by construction. But $U_0$ has fewer vertices than $T_{n_0}$, contradicting minimality. This proves the claim.

Give the children of each root any fixed order. Its forest of child subtrees is a finite word in $\mathcal{U}$. The word theorem and finite product closure show that the pairs consisting of the root label and the child-subtree word are a well-quasi-ordering. Thus some $i<j$ have comparable root labels and an embedding of the child word of $T_i$ into that of $T_j$. Map the source root to the target root and use the obtained embeddings inside distinct target child subtrees. The paths from the target root into those distinct branches are disjoint except at the root, so this is a label-monotone tree embedding. It contradicts badness. Hence

For Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. There is $N(k)$ such that any $N(k)$ trees with $|T_i|\le k+i$ contain $i<j$ with $T_i⪯T_j$. Use indices starting at zero; shifting them only changes $k$. To prove this uniform finite bound, form the finite bad-sequence tree of finite bad prefixes obeying the size bounds. At position $i$ there are only finitely many rooted-tree isomorphism types with at most $k+i$ vertices, so every node has finitely many successors.

If no bound existed, this tree would have nodes of arbitrarily large depth. A node with arbitrarily long extensions has a child with arbitrarily long extensions, since it has only finitely many children. Choosing such a child repeatedly produces an infinite branch, as in the König infinity lemma. The branch is an infinite bad tree sequence, contradicting the theorem just proved. Therefore the required $N(k)$ exists. The same proof works with any fixed finite label alphabet and any specified size bound finite at each position.

A finite field is a field with finitely many elements. Its characteristic is the least positive integer $p$ for which $p\cdot 1=0$; such an integer exists because the additive group is finite. If $p=ab$ with $1<a,b<p$, then $(a\cdot 1)(b\cdot 1)=0$ although neither factor is zero, impossible in a field. Hence $p$ is prime, and the multiples of $1$ form its prime field ${\mathbb{F}}_p$. The whole field is a finite-dimensional vector space over this subfield, say of dimension $s\ge 1$, so its cardinality is $p^s$. Thus $q=p^s$ and its characteristic is $p$.

For existence, let $K$ be a splitting field over ${\mathbb{F}}_p$ of $P(T)=T^{p^s}-T$. Since $P^{\prime }(T)=-1$, it has exactly $p^s$ distinct roots. Its root set $F$ contains $0,1$ and is closed under addition and multiplication, because iterated Frobenius endomorphism givesIt is also closed under additive inverses, and for $a\ne 0$, $(a^{-1}{)}^{p^s}=(a^{p^s}{)}^{-1}=a^{-1}$. Thus $F$ is a field with exactly $p^s$ elements. Since it contains every root, it is the entire splitting field. This proves existence.

For uniqueness, let $L$ be any field with $p^s$ elements. Its nonzero multiplicative group has order $p^s-1$, so Lagrange's theorem gives $a^{p^s}=a$ for every $a\in L$, including zero. Hence $L$ is a splitting field of the same polynomial. Here is the embedding argument giving uniqueness, rather than merely invoking it by name. Extend the identity on ${\mathbb{F}}_p$ from one root at a time. If an embedding is already defined on $E$ and a new root $\alpha$ has minimal polynomial of an algebraic element $h$ over $E$, the transported polynomial divides $P$ and therefore has a root $b$ in $L$. Evaluation at $b$ induces an embedding $E[T]/(h)\to L$, extending the existing embedding to $E(\alpha )$. Repeating for the finitely many roots embeds the constructed field $F$ in $L$. They have the same finite cardinality, so this embedding is an isomorphism. Thus the field of order $p^s$ is unique up to isomorphism.

We prove cyclicity of the multiplicative group, also establishing the more general finite multiplicative subgroup of a field is cyclic result. Let $H$ be a finite subgroup of a field's multiplicative group, and let $e$ be the least common multiple of all element orders. For every prime $\ell$ dividing $e$, choose an element whose order has the maximal $\ell$-power ${\ell }^{a_{\ell }}$; raising it to the factor of its order coprime to $\ell$ gives an element of order ${\ell }^{a_{\ell }}$. Since the group is abelian, the product of these elements has order ${\prod }_{\ell }{\ell }^{a_{\ell }}=e$: for commuting elements of coprime orders, an equation $(ab{)}^t=1$ forces $a^t=b^{-t}$ into the intersection of their cyclic subgroups, which is trivial, so both orders divide $t$. Every member of $H$ is a root of $T^e-1$. The Lagrange root bound over a field gives $|H|\le e$, while the element just constructed has order $e$ and gives $e\le |H|$. Thus equality holds and that element generates $H$. In particular

For the field with nine elements, use ${\mathbb{F}}_3[u]/(u^2+1)$. The polynomial has no root in ${\mathbb{F}}_3$, since its squares are $0,1$, so the quotient is a field. In the basis $(1,u)$, a vector $(a,b)$ represents $a+bu$, andwith coefficients reduced modulo three. Put $\beta =1+u$. Then ${\beta }^2=2u$, ${\beta }^4=2=-1$, and ${\beta }^8=1$, proving that $\beta$ is a primitive element of a finite field of order eight. Its complete power/vector table isFor completeness, the full addition and multiplication tables in these labels are displayed below. The subscripts are element labels; they are not the scalar residues of ${\mathbb{F}}_3$.

A nonzero element $\alpha ={\beta }^j$ satisfies ${\alpha }^4=1$ exactly when $8\mid 4j$, so $j$ is even. Thus, with $e=1$,

Broadly in favour, if “trivial” means transparent once the correct structure has been identified. The value of category theory is often to isolate a universal property that makes a repeated argument automatic.

For example, two objects with the same universal property have unique connecting maps in each direction. Their composites and the identities satisfy the same defining conditions, so uniqueness makes those composites identities. This one argument explains the uniqueness of categorical products, categorical limits, adjoint functors and many other constructions. It also specifies which isomorphism is canonical, a point that an arbitrary objectwise choice would miss.

The Yoneda lemma makes this perspective systematic. A natural transformation out of $\mathcal{C}(c,-)$ is determined by its value at $1_c$. Once that evaluation principle is recognized, questions about representable functors reduce to compositions in the original category. Similarly, the Beck monadicity theorem organizes many free-and-forgetful constructions around monad algebras, reflection of isomorphisms and specified coequalizers. The proof supplies a reusable explanation of why the algebraic structure can be reconstructed from the forgetful functor.

The substantial work lies in finding suitable definitions and verifying their hypotheses. The Freyd general adjoint functor theorem, for instance, needs completeness, local smallness and a solution set; none follows merely from calling a construction natural. The partial-operation tower below even has a monadic length that records how many distinct reconstruction steps are needed. These examples support the view that abstraction clarifies the source of an argument's simplicity. They also show why that simplicity is an achievement requiring proof.

An adjunction $L⊣U$ induces the monad $T=UL$ with unit $\eta$ and multiplication $U\epsilon L$. The Eilenberg-Moore comparison functor sends $B$ to $(UB,U{\epsilon }_B)$. A monadic adjunction is one for which that comparison is an equivalence of categories. Iterating comparisons, when their left adjoints exist, gives a tower; its monadic length is the number of steps until equivalence, with length zero for an equivalence and length one for a monadic adjunction that is not already an equivalence.

The precise monadicity theorem says that a right adjoint is monadic exactly when it reflects isomorphisms and creates coequalizers of pairs whose images admit split coequalizers.

Write $U_r:{\mathcal{C}}_r\to {\mathcal{C}}_{r-1}$ for one-step forgetting. For $r=1$, the free extension of nested partial unary operations is the familiar set $A\times \mathbb{N}$, with ${\alpha }_1(a,j)=(a,j+1)$ and unit $a\mapsto (a,0)$. A map into a set with endomorphism ${\beta }_1$ extends uniquely by $(a,j)\mapsto {\beta }_1^{j}f(a)$.

For $r\ge 2$, letKeep $A$ and all its old operations, and adjoin the disjoint set $S(A)\times \mathbb{N}$. On the new points define ${\alpha }_1(a,j)=(a,j+1)$; every old operation ${\alpha }_i$ with $i\ge 2$ is undefined there. Define the new top operation only at the old eligible points, by ${\alpha }_r(a)=(a,0)$ for $a\in S(A)$.

These domains satisfy the defining rules of the nested partial unary operation category: the new points have no ${\alpha }_1$-fixed points, so none has any higher operation defined. Given $f:A\to U_{r}B$, extend it byThe expression is defined because $f$ sends eligible points to eligible points and ${\beta }_1$ is total. It preserves every defined operation; preservation of ${\alpha }_r$ at old points determines $j=0$, and preservation of ${\alpha }_1$ determines the whole chain. Thus it is the unique extension, proving $L_r⊣U_r$.

For monadicity, $U_r$ reflects isomorphisms: an isomorphism of the lower structures takes $S(A)$ bijectively to $S(B)$, so the inverse of a map preserving the new operation preserves it too.

Next prove the split-coequalizer lifting for a fixed-point-domain operation. Let $f,g:A\rightrightarrows B$ be ${\mathcal{C}}_r$-morphisms and suppose their lower images have a split coequalizer $q:U_{r}B\to Q$, with $s:Q\to U_{r}B$, $t:U_{r}B\to U_{r}A$ satisfyingEquip $Q$ with ${\beta }_Q(z)=q{\beta }_B(sz)$ on its required eligible domain. The lower-structure maps $s,t$ preserve eligibility. For eligible $x\in B$,so $q$ preserves the top operation. If $h:B\to D$ coequalizes $f,g$, its lower factor $\overline{h}:Q\to U_{r}D$ satisfiesThus it is a morphism of the full structures. Uniqueness of the lifted operation follows from the eligible section $s$ and surjectivity of $q$. This creates the specified coequalizer, so the Beck monadicity theorem proves each one-step adjunction is monadic.

Finally, every free top operation just constructed has no fixed points: for $r=1$ it shifts the free chain, while for $r\ge 2$ it sends old eligible points to new distinct points and is undefined on the new points. Freely adjoining any further operation therefore changes nothing, since its required domain is empty. For $n>m$, the monad induced on ${\mathcal{C}}_m$ by the long free-and-forgetful adjunction is consequently the same monad, including unit and multiplication, as that for ${\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$.

The first comparison category is thus equivalent to ${\mathcal{C}}_{m+1}$. Its algebra action evaluates the new formal value by ${\alpha }_{m+1}$, so the comparison from ${\mathcal{C}}_n$ retains precisely that next operation and is the remaining forgetful functor. Iteration recovers one additional operation at each step. These remaining comparisons are not equivalences while an operation is still forgotten: take a two-element set with all retained operations equal to the identity, and choose the next operation to be the identity in one extension and the transposition in another. Higher operations can be identities in the first case and undefined in the second. The identity on the retained structure is not a morphism between these extensions, so the forgetful functor is not full. This includes ${\mathcal{C}}_1\to \mathbf{Set}$.

The monadic tower for nested partial unary operations therefore has exactly the claimed length:

Put $\mathcal{D}={\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, the quotient sheaf of abelian groups. At a closed point $P$, the discrete valuation gives an isomorphismIndeed, a local parameter on a smooth algebraic curve has valuation one, and the valuation kernel is precisely the unit group of the discrete valuation ring. Thus a local section represented by $f$ determines its local principal divisor, and changing $f$ by a regular function which is a unit does not change those coefficients.

These local valuations identify $\Gamma (X,\mathcal{D})$ with $\mathrm{Div}(X)$. To see that they give finite support, represent a section locally by finitely many rational functions on an algebraic variety, using Noetherian Zariski topology to take a finite cover. Each representative has only finitely many zeros and poles, so their combined support is finite. Conversely, for $D=\sum n_{P}P$, choose at each support point a local parameter $t_P$ and shrink its neighbourhood to exclude every other zero or pole of $t_P$ and every other support point. The local rational equation $t_P^{n_P}$ represents the desired valuations there; on the complement of the support use $1$. On overlaps these equations differ by units, so their classes glue in the quotient sheaf. Finally, a section with every valuation zero has zero germ everywhere and is the identity section. Hence the identification is both surjective and injective.

The map on global sections sends $f\in k(X{)}^{\ast }$ to $\mathrm{div}(f)$. Therefore its cokernel is

For a divisor on an algebraic curve $D=\sum n_{P}P$, define its line bundle associated to a divisor inside the sheaf of rational functions byLocally choose a rational equation $f_i$ for $D$, as in the construction of the quotient sheaf above. ThenConsequently this is an invertible sheaf. Multiplication gives an isomorphismas is checked on the displayed local generators. If $D=\mathrm{div}(f)$, multiplication by $f$ identifies ${\mathcal{O}}_X(D)$ with ${\mathcal{O}}_X$. Thus $D\mapsto [{\mathcal{O}}_X(D)]$ induces a group homomorphism from the divisor class group to the Picard group.

For surjectivity, let $\mathcal{L}$ be an invertible sheaf. Trivialize its one-dimensional rational fibre over the function field $k(X)$; equivalently, choose a nonzero rational section and identify $\mathcal{L}$ with a subsheaf of the rational-function sheaf. A finite trivializing cover givesOn overlaps $a_i/a_j$ is a regular unit. The integers $-v_P(a_i)$ therefore agree wherever two descriptions apply and define a finite divisor on an algebraic curve $D$. Its local description gives $\mathcal{L}={\mathcal{O}}_X(D)$. A different rational trivialization multiplies all $a_i$ by the same rational function, changing $D$ by a principal divisor, so the resulting divisor class is intrinsic.

For injectivity, an isomorphism ${\mathcal{O}}_X(D)\to {\mathcal{O}}_X(E)$ becomes multiplication by some $g\in k(X{)}^{\ast }$ on the rational fibre. At each closed point its effect on the local free generators givesHence $D-E=\mathrm{div}(g)$. Conversely this equality gives the required isomorphism by multiplication by $g$. We have proved the group isomorphism

Since $X$ is an irreducible variety, every nonempty open subset is irreducible and the sheaf of nonzero rational functions on an irreducible variety has value $k(X{)}^{\ast }$ there. All its restriction maps between nonempty opens are identities, and restriction to the empty set is surjective. Thus ${\mathcal{K}}_X^{\ast }$ is a flabby sheaf, and the introductory argument gives $H^1(X,{\mathcal{K}}_X^{\ast })=0$.

Apply the long exact sequence in sheaf cohomology to the short exact sequence of sheavesIts relevant portion isThe connecting homomorphism is therefore surjective, with kernel the image of $k(X{)}^{\ast }$. The computation in part (a) now givesConcretely, local rational equations $f_i$ of a divisor on an algebraic curve differ by regular units on overlaps. Their ratios give the gluing data of an invertible sheaf; changing the local equations by units changes those data by a coboundary. This is the transition-function interpretation of the identification with the Picard group. Reversing all transition ratios replaces a class by its inverse; either consistent convention gives the same group isomorphism.

Let $X$ be an affine variety over an algebraically closed field $k$, with reduced coordinate ring $A=k[X]$. On an open subset $U$, define ${\mathcal{O}}_X(U)$ to be the functions $h:U\to k$ which, near every point, have an expression $a/b$ with $a,b\in A$ and $b$ nowhere zero on that neighbourhood. Restriction is ordinary restriction of functions. Compatible local functions glue uniquely and remain locally quotients, so this defines the structure sheaf. On the principal open subset $D(f)$ it gives $A_f$, and at $P$ it gives the local ring $A_{{\mathfrak{m}}_P}$. The principal-open identification can also be seen by identifying $D(f)$ with the affine algebraic set obtained by adjoining a coordinate $t$ and the equation $tf=1$; its coordinate ring is $A[t]/(tf-1)\cong A_f$. This construction permits several irreducible components; no division by a function vanishing identically on a component is assumed.

An abstract algebraic variety is a locally ringed space locally isomorphic to such affine models, with a finite affine cover; we use the usual separated convention. Reducible varieties are allowed here. The separation condition says that the diagonal morphism is a closed immersion. Two varieties are related by a birational map with an inverse when some dense open subset of one is isomorphic as a locally ringed space to a dense open subset of the other.

Define the ring of rational functions on a reduced variety by pairs $(U,h)$, where $U\subseteq X$ is dense open and $h\in {\mathcal{O}}_X(U)$, identifying two pairs when their restrictions agree on a common dense open subset. Finite intersections of dense open subsets are dense, so addition and multiplication on these intersections are well defined. For an irreducible variety this gives the usual function field; for a reducible variety it need not be a field. Restriction to any dense open subset $V$ induces an isomorphismIndeed, intersecting domains with $V$ defines the restriction, and a dense open subset of $V$ is also dense open in $X$, which defines its inverse. An isomorphism between dense opens therefore induces an isomorphism of their rational rings. Thus birational varieties have isomorphic rings of rational functions, also in the reducible case.

For the finite-cover assertion, write the closed complement of $U$ as $V(I)$, where $I=(f_1,\dots ,f_r)$ is an ideal of the Noetherian ring $A$. ThenLet ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_t$ be the minimal prime ideals of $A$, corresponding to its irreducible components. Since $U$ is dense, it meets every component, so $I$ is contained in none of these primes. Prime avoidance supplies $s\in I$ outside their union. Consequently $D(s)\subseteq U$ meets a dense open subset of every component and is dense in $X$. Add $D(s)$ to the finite cover if necessary. This proves that a finite principal-open cover can be chosen with a dense member. It does not say that every principal-open cover already has such a member: for $X=V(xy)$, the dense open $X\setminus \{(0,0)\}=D(x)\cup D(y)$ has neither displayed member dense, although adding $D(x+y)$ supplies a dense member.

For clarity, the zero divisors in this reduced Noetherian ring are exactly ${\bigcup }_i{\mathfrak{p}}_i$. If $ab=0$ and $a$ avoids every ${\mathfrak{p}}_i$, reduction in the domain $A/{\mathfrak{p}}_i$ forces $b\in {\mathfrak{p}}_i$ for all $i$, hence $b=0$. Conversely, if $a\in {\mathfrak{p}}_i$, choose for every $j\ne i$ an element $b_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$, possible because distinct minimal primes are incomparable. Their product $b$ is not in ${\mathfrak{p}}_i$ but belongs to all the other primes. Then $b\ne 0$ and $ab\in {\bigcap }_j{\mathfrak{p}}_j=0$. For a single minimal prime, reducedness makes that prime zero and this argument uses $b=1$. It follows that

The total quotient ring $S^{-1}A$, where $S$ consists of these non-zero-divisors, maps to $\mathrm{Rat}(X)$ byThe equivalence relation for localization makes this a well-defined $k$-algebra homomorphism. For surjectivity, a rational class represented on $U$ restricts to the dense principal open $D(s)\subseteq U$ constructed above. Its section there belongs to $A_s$, so it is $a/s^r$ and lies in the image. For injectivity, if such a fraction is zero on a dense open subset, its numerator vanishes on that dense subset. The numerator's zero set is closed, so it vanishes everywhere on $X$ and is zero in the reduced coordinate ring. ThusIn particular $A\to S^{-1}A$ is canonically injective: $a/1=0$ implies $sa=0$ for some non-zero-divisor $s$, hence $a=0$.

Finally let $Y$ be the reduced closed union of exactly those irreducible components of $X$ which contain $P$. Removing all the other components gives an open neighbourhood of $P$ on which $X$ and $Y$ coincide, including their structure sheaves. Hence ${\mathcal{O}}_{X,P}\cong {\mathcal{O}}_{Y,P}$. Write $B=k[Y]$ and let ${\mathfrak{n}}_P$ be its maximal ideal at $P$. Every minimal prime of $B$ is contained in ${\mathfrak{n}}_P$, because every component of $Y$ contains $P$. If $b\notin {\mathfrak{n}}_P$, it avoids each such prime and is a non-zero-divisor. Therefore the map of localizationsis defined and injective: if its value is zero, a non-zero-divisor annihilates $a$, so $a=0$. We obtain the desired local ring embeds in rational functions on incident components:At an intersection point one must retain all incident components. For example, the local ring of $V(xy)$ at the origin has two nonzero germs $x,y$ with $xy=0$; it cannot embed into the function field of just one of its axes.