The coordinate ring: functions you can actually write down
Guide 1 of this rung built affine varieties as zero-sets in A^n with their Zariski topology, and Guide 2 sealed the Nullstellensatz dictionary: radical ideals on the algebra side correspond exactly to closed sets on the geometry side. That dictionary was about subsets. Now we ask a sharper question — what are the functions on a variety V? The honest answer of algebraic geometry is: the only functions we trust are the ones we can write as polynomials. Restrict every polynomial in k[x_1, ..., x_n] to V, declare two polynomials the same when they agree at every point of V, and what survives is the coordinate ring of V.
Made precise, the coordinate ring is the quotient k[V] = k[x_1, ..., x_n] / I(V), where I(V) is the ideal of all polynomials vanishing on V — the very ideal whose radical-ness Guide 2 hammered home. Killing I(V) is exactly the act of forgetting the difference between two polynomials that agree on V. Because V is a variety it is irreducible, so I(V) is prime, so k[V] is an integral domain: no zero divisors, which is what will later let us divide. Concretely, an element of k[V] is a polynomial-shaped function f: V -> k, and you should always picture it as a function first, a coset second.
V = { x*y = 1 } in A^2 (a hyperbola)
k[V] = k[x, y] / (x*y - 1)
In k[V] the relation x*y = 1 holds, so y = 1/x.
Hence k[V] = k[x, 1/x] = k[x, x^{-1}] (Laurent polynomials).
The function 'y' became the genuine inverse of 'x'.Morphisms are exactly ring maps, backwards
A morphism of affine varieties phi: V -> W is a map given in coordinates by polynomials: phi(p) = (g_1(p), ..., g_m(p)) with each g_i in k[V]. The point is not the formula but what it does to functions. Any function h on the target W can be pulled back to a function on the source V by composing: h composed with phi. Since h is polynomial and phi is polynomial, the composite is polynomial — so pullback sends k[W] into k[V]. This pullback, written phi*, is a homomorphism of k-algebras, and it runs from W's ring to V's ring: the arrow flipped.
The miracle is that this loses nothing: every k-algebra homomorphism k[W] -> k[V] arises as phi* for a unique morphism phi: V -> W. So studying maps of varieties is literally the same as studying maps of their coordinate rings, with all arrows reversed. This is the contravariant equivalence the callout promised, and it is what makes regular functions the right notion: a morphism V -> A^1 is the same thing as one element of k[V], because a k-algebra map k[x] -> k[V] is determined by where it sends x. Functions on V are the maps to the affine line.
A clean example: the parametrized cubic t -> (t^2, t^3) is a morphism A^1 -> V where V = { y^2 = x^3 } is the cuspidal cubic. Its pullback sends x -> t^2 and y -> t^3, embedding k[V] = k[x,y]/(y^2 - x^3) into k[t]. The map is a bijection of points, yet it is not an isomorphism, because the image misses t itself — the ring k[t^2, t^3] is a strict subring of k[t]. The cusp at the origin is exactly the algebraic fingerprint of that missing generator. We will name that defect precisely in Guide 4 when we meet singular points.
The function field: letting yourself divide
Polynomials are honest but stingy: they are defined everywhere, yet there are far too few of them. On the affine line A^1 the only regular functions are the polynomials in one variable — you cannot even write 1/x, because it blows up at 0. To gain expressive power we pay a price: we allow ratios f/g of elements of k[V], and we accept that such a function is undefined wherever the denominator g vanishes. Because k[V] is an integral domain, these ratios form a field, the function field k(V), built as the field of fractions of k[V].
An element of k(V) is a rational function: a fraction f/g that is a genuine function only on the open set where g is nonzero. Two fractions are equal when they agree there, exactly as in elementary fractions. The subtle and beautiful fact is that a single rational function can have several fractional representations, and one representation may be defined at a point where another is not. So the domain of a rational function is the union over all its representations — the largest open set where it can be salvaged. We call a rational function regular at p if some representation f/g has g(p) nonzero.
Make this vivid on the circle V = { x^2 + y^2 = 1 }. The fraction f = (1 - y)/x looks undefined wherever x = 0. But on V the relation x^2 = 1 - y^2 = (1 - y)(1 + y) holds, which lets us rewrite (1 - y)/x = x/(1 + y) — the very same element of k(V). The second form is perfectly regular at (0, 1), exactly where the first one blew up, while the first is fine at (0, -1) where the second now fails. Between the two representations the only point genuinely outside the domain is (0, -1), a single bad point salvaged down from what looked like a whole missing line.
The set of points where a rational function is regular is always open and nonempty, and a function regular at every point of V turns out to be exactly an element of k[V] again — provided V is normal, a hypothesis it is dishonest to drop. On a smooth affine variety the slogan holds cleanly: regular-everywhere equals polynomial. The function field k(V) itself is a single invariant attached to V, and remarkably it depends only on a dense open piece of V, not on the whole thing — which is the doorway to the next idea.
Rational maps, dominance, and birational equivalence
If a rational function is a fraction allowed to skip its bad points, a rational map phi: V ⇢ W is the same idea with a vector of fractions: it is a morphism defined merely on some dense open subset U of V, with two such agreeing on the overlap counted as equal. The dashed arrow ⇢ is the standard warning that phi is not defined everywhere. Where phi is defined we call its domain, and the points it cannot reach are its locus of indeterminacy — a closed set of lower dimension, never the whole variety.
You cannot freely compose rational maps — the image of phi might land entirely inside the bad set of the next map. The fix is the notion of a dominant map: a dominant rational map is one whose image is dense in W. Only dominant maps can be safely composed, and only dominant maps pull functions back: a dominant phi: V ⇢ W induces a field homomorphism phi*: k(W) -> k(V), and conversely every k-algebra map of function fields comes from a unique dominant rational map. The contravariant dictionary survives the move to fields, now between fields instead of rings.
Now the payoff. A birational map is a rational map phi: V ⇢ W with a rational inverse — equivalently, by the dictionary, an isomorphism of function fields k(V) and k(W). Two varieties are birationally equivalent when such a map exists, which means they share a common dense open subset: they are the same variety away from lower-dimensional closed pieces. This is far weaker than isomorphism and far stronger than nothing, and it is the central equivalence relation of classical algebraic geometry. The blow-ups of Guide 4 are precisely birational surgeries that leave k(V) untouched while reshaping the singular locus.
Worked example: the circle is rational
The cleanest way to feel all four notions at once is to prove that the circle V = { x^2 + y^2 = 1 } is birationally equivalent to the line A^1. The construction is the stereographic projection you may have met for sketching: fix the north pole N = (0, 1), and to any point p on the circle assign the place where the line through N and p crosses the x-axis. This is a rational map both ways, an isomorphism of function fields, and it produces the famous rational parametrization of the circle — the same formulas that generate Pythagorean triples.
- From the circle to the line: send p = (x, y) on V to t = x / (1 - y), the x-intercept of line N-p. This is a rational function on V, regular everywhere except the north pole N = (0, 1) where 1 - y = 0.
- From the line to the circle: send t to ( 2t/(t^2 + 1), (t^2 - 1)/(t^2 + 1) ). Each coordinate is a rational function of t, regular everywhere on A^1 since t^2 + 1 never vanishes over a field where -1 is not a square; over an algebraically closed field it skips the two points t = ±i.
- Check they invert each other: plug the parametrization into x/(1-y) and simplify, getting (2t/(t^2+1)) / (1 - (t^2-1)/(t^2+1)) = (2t/(t^2+1)) / (2/(t^2+1)) = t. The composite is the identity wherever both are defined.
- Read off the conclusion: the two maps give an isomorphism k(V) ≅ k(t), so the circle is birationally equivalent to A^1 — it is a rational curve. The single bad point N on one side and the pair t = ±i on the other are exactly the lower-dimensional loci a birational map is allowed to ignore.