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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.