Rational maps and function fields
Work on dense open subsets, recognize birational inverses, and distinguish a rational function from a globally regular one.
On this page
The question for today
What you’ll understand
- Define a function field and rational-map equivalence on an irreducible variety.
- Translate dominant rational maps into function-field embeddings.
- Check a birational map with an exceptional fibre and a cusp function that does not extend regularly.
Polynomial formulas are global, but useful inverses often contain denominators. Rational maps allow those formulas on a nonempty open subset and identify two descriptions when they agree after restricting further. Irreducibility makes that overlap dense and keeps a single fraction field available. Birational geometry then studies the shared generic information while deliberately leaving room for exceptional fibres, omitted points, and singular local behaviour.
One fraction field needs an irreducible source
For an irreducible affine variety X, its coordinate ring A is a domain, so define k(X) as its fraction field. A rational function is an element f/g with nonzero g, initially regular on the nonempty principal open D(g). Different fractions can describe the same element, and cancellation can enlarge the actual regularity domain. A fraction's displayed denominator therefore gives a valid domain, not necessarily the maximal domain. Regularity at a point means membership in the local ring, not merely a formal limit in a real picture.
An irreducible variety has intersecting nonempty opens, so rational functions can be compared on a shared dense open subset. On a reducible algebraic set, zero divisors prevent one ordinary fraction field from describing all components. One can instead study component fields or an appropriate total quotient ring, but that is a different construction. Every function-field statement in this lesson assumes irreducibility; replacing it with reducedness alone would make division by a nonzero zero divisor illegitimate.
Dominance allows pullback through denominators
A rational map from X to Y is represented by a regular map on a nonempty open subset of X, with two representatives equivalent if they agree on a nonempty open overlap. It is dominant when its image is dense in Y. Dominance ensures that a nonzero rational function on Y does not pull back identically to zero, so denominators remain meaningful and pullback extends to an embedding k(Y) into k(X). A map with image in a proper closed subset does not generally define this field embedding.
For irreducible varieties, an isomorphism of their function fields over k corresponds to a birational map: there are dense open subsets on which the varieties are isomorphic. One can see the dense opens by writing finitely many coordinate functions and inverse coordinate functions as fractions and excluding their denominator zero loci. This does not prove the entire varieties are isomorphic. The removed subsets can contain precisely the singularity or exceptional fibre of interest, and recovering those subsets requires local geometry that the common field does not remember.
Cancellation and extension are local questions
A rational function may have an apparent singularity that disappears after applying a coordinate-ring relation. On a parabola y=x², the fraction y/x agrees with x wherever x is nonzero and therefore extends regularly across the origin. On the cusp y²=x³, the fraction y/x behaves differently: in the parameter t it equals t, whereas the coordinate ring contains only polynomials in t² and t³. There is no degree-one term available to supply a global polynomial representative. The similar-looking fractions therefore have different extension behaviour.
Even a regular map that is birational need not be finite or injective. A curve may be contracted to one point, and the image can omit other points on a closed subset. Those phenomena are invisible on a dense open where an inverse exists. To decide a stronger property, inspect the exceptional fibres and the ring extension, rather than concluding from the existence of a rational inverse. The explicit plane map below provides all three distinctions with formulas that can be verified directly.
Work it through
A birational plane map with a contracted line
Over characteristic zero consider F(x,y)=(u,v)=(x,xy). Find a rational inverse, the point image, and the fibre over the origin. Is F finite or an isomorphism?
On the open u≠0, set x=u and y=v/u. Substituting in either direction gives the identity on the corresponding opens x≠0 and u≠0. Thus F is birational and induces the field isomorphism k(u,v)≅k(x,y).
If u is nonzero, every v has the unique preimage (u,v/u). If u=0, then x=0 forces v=xy=0. Therefore the image is D(u) together with the single origin; every point (0,v) with v≠0 is omitted.
The fibre over (0,0) is the whole line x=0, with y arbitrary. This rules out both finiteness and an isomorphism. The image is dense because it contains D(u), and is constructible, but it is not closed because it omits points in its closure.
F is birational but contracts an affine line. Its image is the dense nonclosed constructible set D(u)∪{(0,0)}, so a rational inverse is much weaker than a global regular inverse.
The living notebook
Explore the model
Interactive lab · function
A rational inverse is stronger than counting parameters
Pause and Reset, then set a = 1 and x = 1; Reset again and set both a = 1 and x = −1. The same height one has two preimages under t↦t², unlike the birational plane map's unique inverse on u≠0. Predict why no rational function of u can equal both signs of t: the extension k(t)/k(t²) has degree two in characteristic zero. The instrument shows a real quadratic projection, not the plane map or a function-field algorithm.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA finite map between equal-dimensional varieties need not be birational; the induced function-field extension detects the difference.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the characteristic-zero cusp with coordinate ring A=k[t²,t³], show t=y/x belongs to the function field but is not regular at the origin. Contrast with y/x on the parabola y=x².
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
For the cusp, suppose t=a/b with a,b in A and b(0)≠0. Compare the coefficient of t in tb=a. For the parabola use its defining relation before judging the denominator.
Compare with the solution
The cusp fraction field contains t=t³/t². If t were regular at the origin, write t=a/b with a,b in A and b(0) nonzero. Every element of A has no degree-one term, but tb has degree-one coefficient b(0), whereas a has coefficient zero. This is impossible, so t does not lie in the cusp local ring at the origin. On the parabola, y=x² gives y/x=x on D(x), and the polynomial x extends it everywhere. A denominator vanishing at a point neither proves nor disproves a removable singularity until the coordinate-ring relations and possible alternate denominators are considered.
Keep the idea
Take it with you
- Irreducibility supplies the domain and common dense opens required by function fields.
- Dominant rational maps pull back to embeddings of function fields.
- Birationality preserves generic information while permitting exceptional fibres and different singularities.
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