Coordinate rings and regular maps
Recover polynomial maps from homomorphisms in the opposite direction and check isomorphisms through explicit inverse formulas.
On this page
The question for today
What you’ll understand
- Interpret a coordinate ring as polynomial functions modulo equality on a set.
- Translate a regular map into its contravariant pullback.
- Verify a sheared parabola isomorphism and distinguish finite maps from isomorphisms.
Coordinates describe a chosen embedding, while the coordinate ring captures the polynomial functions that survive on the embedded set. A map can then be studied by asking what it does to functions rather than following individual points. Functions pull back in the opposite direction, and this reversal is the source of a powerful algebraic dictionary. Explicit inverse formulas keep the dictionary concrete and expose why a bijection alone may be insufficient.
Functions rather than formal expressions
For a closed algebraic set X in affine n-space, define its coordinate ring as the ambient polynomial ring modulo I(X). Two polynomials represent the same element precisely when they agree at every point of X. The ring is a finitely generated reduced k-algebra; it is a domain exactly when X is irreducible. This construction does not identify two expressions merely because they agree on a finite plotted sample. Equality means agreement on the complete algebraic set over the specified ground field.
A regular function on an open subset is locally a quotient of polynomial functions with a denominator nonvanishing near the point in question. A fundamental affine result identifies the globally regular functions on X with its coordinate ring, even though the local definition initially allows many denominators. Principal opens behave differently: on D(f), where f never vanishes, the regular-function ring is the localization k[X] at f. The hyperbola xy=1 illustrates this distinction: y is globally regular there because it is already the polynomial coordinate representing x inverse.
A map reverses the direction of functions
A regular map between affine algebraic sets is given by coordinate functions in the source coordinate ring, with the target relations satisfied after substitution. If F maps X to Y, composing any regular function on Y with F gives a regular function on X. The resulting pullback is a k-algebra homomorphism: it preserves addition, multiplication, the identity, and every scalar from k. Relations must descend through the quotient, which is why specifying images of generators without checking the target equations is incomplete.
Conversely, a k-algebra homomorphism from k[Y] to k[X] supplies images of the target coordinate functions. Evaluating those images at a point of X gives a point of Y because every target relation maps to zero. Over the algebraically closed field, these constructions are inverse. Thus affine algebraic sets and finitely generated reduced k-algebras are contravariantly equivalent. Restricting to irreducible varieties corresponds to domains. Nilpotent algebras belong to the larger scheme dictionary and should not be inserted into this reduced point-set statement unnoticed.
The strength of an inverse and the limits of a projection
An isomorphism requires a regular inverse, and equivalently an invertible pullback homomorphism. A set-theoretic inverse can fail this condition at a singular point even if the original map is bijective. Likewise a finite morphism is defined by a module-finite coordinate-ring extension, not by being one-to-one or by having only finitely many points in a typical fibre. Algebraically finite maps have finite fibres, but the converse needs additional hypotheses and does not follow from counting a generic fibre alone.
The squaring map of the affine line in characteristic zero gives a simple diagnostic. Its pullback sends the target variable u to t², so the image contains exactly the even polynomials. The source ring is nevertheless a free module of rank two over that image, with basis one and t. The map is finite but not an isomorphism: the function t cannot be recovered polynomially from t², and most nonzero target points have two preimages. In characteristic two it is bijective on an algebraically closed field but its pullback still is not surjective.
Work it through
An isomorphism after a shear
Let in characteristic zero. Show that F(t)=(t+t²,t²) identifies the affine line with Y, and describe its pullback.
Substituting u=t+t² and v=t² gives u−v=t, so the target relation becomes t²−t²=0. Thus the two polynomial coordinate formulas define a regular map into Y.
Define G(u,v)=u−v. Then G(F(t))=t. Conversely the equation of Y gives v=(u−v)², so F(G(u,v))=((u−v)+(u−v)²,(u−v)²)=(u,v). Both inverse identities hold on their entire domains.
The pullback sends the classes of u and v to t+t² and t². Its inverse sends t to the class of u−v. Checking these images on generators proves a k-algebra isomorphism in the direction opposite to F.
Y is a parabola in sheared coordinates, isomorphic to the affine line. The inverse is the globally regular polynomial u−v.
The living notebook
Explore the model
Interactive lab · matrix
Coordinate changes preserve the algebraic curve
Pause and Reset, then set scale a = 1 and shear b = 1. The final matrix sends (x,y) to (x+y,y), so predict the parabola point (1,1) becomes (2,1) and satisfies v=(u−v)². Its inverse is (u,v)↦(u−v,v). Reset and set both a = 2 and b = 0: the final map becomes (2x,y), also invertible. The instrument draws only the grid and square; apply the formulas to the parabola yourself, and distinguish playback interpolation from the selected final map.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forAn invertible linear coordinate change gives polynomial inverse formulas, not just a visually similar curve.
Pause & reason
Check your understanding
Your turn
Try it yourself
Over characteristic zero, analyze the map Q:A¹→A¹ given by t↦t². Find the pullback image, a module basis for the source ring over that image, and the fibres over zero and one. Is Q an isomorphism?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Separate every polynomial into its even and odd terms. A ring inverse would have to recover t from a polynomial in t².
Compare with the solution
The pullback is k[u]→k[t], u↦t², with image k[t²]. Every polynomial uniquely has the form a(t²)+t b(t²), so k[t] is free of rank two over k[t²]. The ordinary fibre over zero is {0}; the fibre over one is {1,−1}. Since t is not an even polynomial, the pullback is not surjective and Q is not an isomorphism. The zero fibre's scheme quotient is k[t]/(t²), which retains a nilpotent although its underlying ordinary set has only one point. Finiteness and the behaviour of a special reduced fibre are separate questions.
Keep the idea
Take it with you
- The coordinate ring identifies polynomials that agree on the entire algebraic set.
- Regular maps pull functions back in the opposite direction.
- A finite or bijective morphism need not possess a regular inverse.
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