# Make first order exact

Henry Shin · Refacte · Refactor companion · November 3, 2026

Choose an arithmetic in which first-order change is exact, then restore the next term to expose a lifting obstruction.

The product rule can be made into exact arithmetic. Replace a small displacement by a symbol whose square is zero. Then change that arithmetic to keep one more order, and the term that disappeared becomes a test of whether a motion can continue.

## 1. Start before taking a limit

A rectangle has sides x, y. Change them by u, v. Its area changes by the exact identity

(x+u)(y+v) − xy = xv + yu + uv.

The two strips contribute xv + yu; the corner contributes uv. When both displacements scale by a small parameter, the corner scales by its square. That is why it disappears from first-order variation.

To retain just constant and first-order terms, store the pair of coefficients (a,b) from a+bε. Multiplying two such expressions produces constant term ac and linear term ad+bc; the remaining term has degree two. Forgetting terms of degree two or more is consistent with subsequent addition and multiplication, since those operations cannot lower their degree. Over a field k, this defines the multiplication rule

(a+bε)(c+dε) = ac + (ad+bc)ε.

In this arithmetic the pair (0,1), abbreviated ε, is nonzero but has square zero. These are the dual numbers. The rule ε²=0 records exactly which coefficients we chose to retain.

## 2. Let the arithmetic produce the rule

For a polynomial f, expand every monomial after substituting x+uε. Terms with two or more copies of ε vanish. Among the factors of x^{m}, there are exactly m choices for the one displaced factor. Therefore

f(x+uε) = f(x) + u f′(x)ε.

Multiplying two such expressions gives (fg)′ = f′g + fg′. Substituting the expression for g(x+uε) into f gives (f∘g)′(x) = f′(g(x))g′(x). Product and chain rules are consequences of multiplication and substitution in the testing algebra.

This calculation works in every characteristic. For example, in characteristic three, (x+uε)³ = x³: the formal derivative of x³ vanishes. It does not follow that the polynomial is constant. The test retains only first-order information. For smooth real functions, Taylor's remainder connects this algebraic calculation with the usual limiting derivative.

## 3. Turn a tangent direction into a substitution

Now consider the equation xy = 0 at the origin. It describes two crossing axes. Try the displacement x = aε, y = bε. Its product is zero because ε² = 0. Every pair (a,b) passes the first-order test.

Even (1,1) passes, although it points along neither axis. The test has not made an error: the equation has no linear term at the origin, so first order imposes no restriction. To ask whether the direction can continue, we need a test that retains the next term.

## 4. Restore the corner and try to repair it

Use k[ε]/(ε³), where ε² survives but ε³ vanishes. Allow arbitrary second-order corrections:

x = aε + rε²,   y = bε + sε².

The product is xy = abε². Both correction coefficients disappear from this order: multiplying either correction by the other coordinate introduces at least ε³. A lift exists precisely when ab = 0. The mixed direction (1,1) cannot be repaired. Its corner is an obstruction.

Compare the smooth curve xy = 1 at (1,1). The first-order motion x = 1+tε, y = 1−tε has product 1−t²ε². Allow corrections rε² and sε². The new product is 1+(r+s−t²)ε², so the condition is r+s=t². There are many choices; taking r=0 and s=t² gives

(1+tε)(1−tε+t²ε²) = 1   modulo ε³.

The nonzero base coordinate lets the extra strip cancel the corner. At the node, both base coordinates are zero and that repair is unavailable.

## 5. Keep the next order when it matters

For any polynomial equation, substitute a general candidate through the desired order and compare coefficients. The equation supplies the residual; the coefficients still free at that order determine which corrections can affect it. Passing the first-order test is only the first step of that calculation.

Continue with [the universal first-order construction](https://refacte.com/notes/what-a-derivative-retains.html#kahler), [the node and its smoothing](https://refacte.com/notes/what-a-derivative-retains.html#node-and-smoothing), or [the complex that organises directions and obstruction targets](https://refacte.com/notes/what-a-derivative-retains.html#cotangent-complex). The separate [oriented-boundary calculation](https://refacte.com/notes/what-a-derivative-retains.html#boundary) explains cancellation by signs; it is a different operation from discarding quadratic terms.