Smooth complex equations and flat remainders

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by the writing AI. Text: CC0.

Prerequisite integration and source/proof self-check by GPT-6 Astra (OpenAI), Ultra, October 2026. Historical authorship and component terms are retained.

A complex equation in real variables can have a singular real zero set even when its complex differential is nonzero. Division nevertheless gives a useful normal form: its ideal can be generated by coordinate differences from a smooth complex graph. The graph need not be a graph of real zeros. Its imaginary part measures exactly the ambiguity of the parameter-only remainder.

We work near the origin, with real variables and , and complex-valued smooth functions. The notation means the local smooth ideal: its elements are finite sums with smooth complex coefficients. Local assertions can be restricted to one smaller neighborhood. When several input functions have different domains, that neighborhood is first chosen inside their actual common domain.

Our input is the order-one smooth complex division proved in Classical finite-order preparation and division and its complete AN04 proof component. If and , it gives The conclusion concerns smooth complex functions in real variables. It gives no holomorphic continuation of an arbitrary smooth dividend. For dividends supplied on a fixed larger domain, a fixed cutoff and the proved construction give a common smaller output neighborhood. An arbitrary germ must first be restricted to its own supplied domain.

The supplied scalar calculus foundation, §§12–13, proves all finite-coordinate derivative rules, compact smooth cutoffs and the exponential estimates used below. The finite algebra foundation, §§10.1–10.6, supplies the complex matrix, adjugate and determinant rules. The polynomial foundation, scalar division, proves the factor theorem used in the finite interpolation argument.

The complete integral Taylor proof supplies the finite-order remainders used here, including signed increments, complex-valued functions, directional derivatives and uniform bounds for all required derivatives on compact neighborhoods.

Several equations and the coefficient matrix

Theorem 1. Suppose , , and Every smooth then has, locally, a representation

Proof. Induct on ; the first case is (I1). Permuting real coordinates and equations places a nonzero entry of in position . Write . Divide by : At the mark, , and differentiation gives Row subtraction shows that the determinant of the latter matrix is . It is nonzero. Apply the induction hypothesis to and , treating as parameters: Substitution into (I4) yields There are only finitely many division steps. Choose their domains successively inside the common supplied domain. For a family supplied on one fixed larger domain, the fixed-neighborhood division input retains a common smaller patch. This proves the result.

Lemma 2. Let vanish at the mark. Suppose and the complex covectors are linearly independent. Then the two systems generate the same ideal near the mark.

Proof. Differentiating at the mark gives , since all vanish. If were singular, a nonzero row vector annihilating it would give a nontrivial linear relation between the . Thus . On a smaller neighborhood the determinant stays nonzero, and the adjugate formula gives a smooth inverse matrix. The identity proves the reverse ideal inclusion.

Theorem 3. Under (I2), there is a smooth complex map , with , such that

Proof. Apply Theorem 1 to each real coordinate : Evaluation at the mark gives . The functions have identity -Jacobian, so their differentials are independent. Lemma 2 proves (I7). No real inverse function theorem has been applied to the real zero set of a complex equation.

For example, already has this graph form. Its real zeros satisfy and , a crossing. The coordinate normal form of the ideal remains valid at that crossing.

Extending smooth coefficients to complex arguments

We give the auxiliary construction used below, so that a formal complex substitution does not stand in for a proof. Let be smooth on a real neighborhood. Multiply it by a fixed smooth cutoff, equal to one on a chosen smaller patch, and extend the product by zero. On a fixed compact parameter patch all derivatives are bounded. Put , with , and choose a smooth compactly supported equal to one near zero.

For positive radii , consider A derivative of total order at most of the -th term is bounded by . Each derivative in removes at most one power of , or differentiates the cutoff and costs ; the remaining derivatives act on bounded smooth coefficients. Only finitely many multiindices occur at each degree.

Choose the radii recursively so that and these bounds are at most for every . For each fixed derivative order the tail then converges uniformly, while its initial part is a finite sum. The fundamental theorem of calculus along coordinate segments identifies the successive uniform derivative limits with derivatives of the sum. Hence is smooth on a fixed complex neighborhood.

At , the cutoff is identically one near each individual term's origin, and only the term of degree contributes to the normal derivative of that order. Consequently These identities also hold after every derivative in . With all normal jets of vanish. Taylor's integral remainder, applied in , gives for all multiindices and integers Here and below denotes ordinary real derivatives in all displayed variables. The constants depend on the chosen compact patch and the function's smooth bounds. The neighborhood is fixed; the chosen radii and constants need not be uniform over all smooth functions. This is an almost-analytic extension, not an analytic continuation.

Which parameter-only functions belong to the ideal?

Fix the graph ideal

Lemma 4. If a smooth parameter-only function belongs to , then on a fixed smaller neighborhood

Proof. Write on its actual real domain. Apply (I9) to the finite collection of coefficients, choosing the cutoff to be one on a common smaller patch. Let be the resulting smooth extensions and define Every normal -jet of this expression vanishes at . Indeed, (I10) and the product rule identify that jet with . This also holds after all real -derivatives. Taylor remainder therefore gives on one fixed smaller patch, for every . Since , shrink the parameter patch once so that lies in this patch. At , all factors in the sum in (I14) vanish, so . Equation (I15) proves (I13), including points with . This uses genuine smooth extensions and a flat error; it does not evaluate an arbitrary real smooth coefficient at a complex point.

There is an equivalent algebraic mechanism in the classical proof. Repeated coefficient division produces, for each , a polynomial in of degrees through , plus terms of degree . On the real line , the lower-degree part is a polynomial with . Interpolating at fixed real values of and evaluating the interpolant at yields the same bound. Proof (I14) makes the common-neighborhood assertion explicit without an infinite sequence of neighborhood restrictions.

Flat values imply flat derivatives

The next result is useful beyond graph ideals.

Lemma 5. Let be locally Lipschitz and be smooth near zero in . Suppose, on a neighborhood independent of , Then on a fixed smaller neighborhood, for every multiindex and ,

Proof of the interpolation estimate. In dimension zero only the undifferentiated value occurs, so (I16) already proves the assertion. In positive dimension, fix integers and a ball inside the domain. The Taylor polynomial satisfies, for , Choose distinct fixed nodes in each coordinate in . Their tensor grid lies in the unit ball. Tensor Lagrange interpolation recovers every polynomial of degree at most in each coordinate, hence recovers . Its finite, fixed basis polynomials show that any derivative of at zero is bounded by a constant times the largest of its grid values. For completeness, for distinct scalar nodes define

The denominators are nonzero, and equals one for and zero otherwise. The difference between a polynomial of degree below and has degree below and vanishes at all nodes. Repeated scalar factorization makes that difference divisible by , so it is zero. Applying this identity successively in each of the coordinates yields

Differentiating this finite identity at zero bounds by the largest grid value times the finite constant . Thus the interpolation estimate has an explicit finite proof on the stated grid.

Since , (I18) gives For , the value estimate is immediate; alternatively use throughout. No assertion about an infinite-dimensional polynomial space is needed.

Completion of the proof. Choose a compact inner patch with positive distance from the boundary of a larger patch. Let be a Lipschitz constant and an upper bound for on that larger patch. Choose At a point with , put . The ball stays in the larger patch and Use exponent in (I16), and choose with in (I19). Smoothness bounds the finitely many derivatives of order on the larger compact patch. Substitution yields a constant times , which is bounded by a constant times , since is bounded. At a zero approached by positive values of , continuity gives (I17). At an interior point of , (I16) says vanishes on a neighborhood, so every derivative vanishes there. These two cases cover the zero set and prove the lemma.

For the graph ideal take , which is Lipschitz on compact patches because is smooth. Combining the two lemmas gives every-power bounds for every parameter derivative of .

All powers of the graph ideal

Theorem 6. For a smooth parameter-only function , the following are equivalent locally:

1. .
2. For every , on one neighborhood.
3. , with all the representations valid on one neighborhood.

More precisely, under condition 2 there is one real neighborhood such that for every there are globally smooth coefficient functions on satisfying These coefficients may depend on ; no uniform bound in is claimed.

Proof. Condition 1 implies 2 by Lemma 4. Put For real , . Lemma 5 therefore implies that every real derivative of , considered as a function of , is bounded by every prescribed positive power of . Derivatives in are zero. On , set The multinomial formula gives exactly (I21), since On a fixed compact smaller patch, all derivatives of are bounded. A derivative of order of is bounded by when , by repeated use of the chain rule. In each derivative of (I23), the derivative of supplies an arbitrarily large positive power of . Thus, for every , Define the coefficient to be zero on . To prove that it is smooth at this possibly singular set, choose a smooth which is zero on and one on , and multiply (I23) by . Each such function is smooth. In its transition region , derivatives of the cutoff cost at most a fixed negative power of , while (I25) supplies any positive power. For every fixed derivative order the cutoff functions therefore converge uniformly with their derivatives to the zero extension. Coordinate-segment integration again identifies these limits as derivatives. The extension is smooth, and all its derivatives vanish on .

Choose one compactly supported smooth cutoff in the original patch, equal to one on . Multiply every extended coefficient by this same cutoff and extend by zero outside the patch. This yields globally smooth coefficients without changing (I21) on , for every . At , condition 2 gives , so the identity holds there too. Thus 2 implies 3. Finally 3 implies 1 by taking .

The ideals in this theorem are ideals of smooth functions. Vanishing merely on their real zero set gives a much weaker condition. If is real valued, condition 2 forces everywhere on the parameter patch; the parameter-only remainder is then unique. If has a nonzero imaginary part away from a smaller zero set, nonzero flat ambiguities can occur.

Examples and graded problems

For the two equations the change of generators is explicit: Thus . A remainder for is , with the full identity The fact that is a remainder does not make it an element of the ideal: ideal membership would require all-power flatness in .

Exercise 1 (foundation). For the equations in (I26), verify the complex Jacobian hypothesis, write an explicit invertible generator-change matrix, and check (I27) by multiplication.

Exercise 2 (foundation). Suppose and is real valued. Prove that every representation has . Explain why this argument fails for complex .

Exercise 3 (intermediate). Let . Show that , for any integer , is not in . Let for and . Prove that for every , with a coefficient smooth across the origin.

Exercise 4 (intermediate). In one parameter dimension, let be Lipschitz and for every . Derive the first-derivative version of Lemma 5 directly from the centered finite difference and Taylor's formula, including points with .

Exercise 5 (advanced). Let . Determine its real zero set. Give a smooth function vanishing on that real zero set which is not in the ideal. Give a nonzero parameter-only function which belongs to every power of the ideal.

Exercise 6 (intermediate). In Theorem 6 with , write the three coefficients explicitly and verify the factorial convention in (I21). State their values on .

Exercise 7 (advanced). Two representations of the same in a graph ideal have parameter-only remainders and . Prove that their difference lies in every ideal power. For , construct two distinct remainders for the zero dividend. Contrast this with real-valued .

Exercise 8 (advanced). In Lemma 2, identify exactly where the common marked zero and the independence hypothesis are used. Give one counterexample when the independence hypothesis is removed, and another when the original generators do not vanish at the mark.

Complete solutions

Solution 1. At , the -Jacobian is the identity: the derivative of is zero at the mark. The matrix has determinant one and sends to . Expanding the right side of (I27) gives Every term and sign is retained; in particular the remainder is .

Solution 2. The point is an actual real point of the supplied domain after shrinking. Substitution there annihilates and gives the stated value of . Thus two parameter-only remainders agree. When is not real, it need not be an admissible real argument of or . A general smooth function has no specified analytic continuation. Lemma 4 constructs an almost-analytic extension with a controlled error; it does not license direct complex evaluation of the original dividend.

Solution 3. Membership would imply near zero. For nonzero , this requires , a contradiction as . This includes , whose value already fails at zero. Differentiating repeatedly gives a finite sum of powers of times the same exponential. Each tends to zero faster than every power of : for , the bound follows from , which follows, for example, from . The zero extension is smooth and satisfies every-power value and derivative bounds.

For , on set Every derivative is bounded by an arbitrarily high power of , using the exponential's derivative bounds and . The cutoff proof in Theorem 6 makes this zero extension smooth. The exact identity holds away from the origin and also at it. Thus for every .

Solution 4. Smoothness bounds and on a fixed compact larger interval. At points where , that bound already gives . At a point with , choose instead , with the same small constant as in (I20). Since , the interval stays in the larger patch and there. Taylor's formula at the two endpoints yields Indeed, subtracting the two Taylor formulas and dividing by bounds the error in the centered difference by . Use in the value bound. The first term is at most ; the second is at most a constant times . At a zero in the closure of , take a limit. In the interior of , the original function vanishes identically, so its derivative does too. The radius chosen for this elementary first-derivative proof depends on the desired exponent. Lemma 5 uses a fixed radius proportional to and a higher-order interpolation formula to handle every multiindex directly.

Solution 5. The real zero set is , the union of the two coordinate axes in the parameter plane with . The parameter-only function vanishes there, but would have to satisfy if it belonged to . Taking contradicts this as .

Let off zero, with , and put . The scalar function is smooth by Solution 3, hence its composition is smooth. It is nonzero whenever , and satisfies for every . Theorem 6 puts it in every ideal power on one fixed neighborhood, including the crossing. No smooth-manifold hypothesis about that zero set is used.

Solution 6. Write , , and . On , the three coefficients are Dividing the first and last by , and the middle by , gives All three coefficients are defined to be zero on , where also . Their smoothness follows from the all-derivative estimates in (I25).