Make first order exact

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

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

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

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 dual numbers.

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:

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

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

The two examples have the same repair equation. For a polynomial equation f(x,y)=0 at a point p, write J_p(r,s)=f_x(p)r+f_y(p)s for its linear part. Suppose a direction v passes the first-order test, so J_p(v)=0. Let Q_p(v) be the coefficient of ε² in f(p+εv). Substituting a correction w=(r,s) gives

A second-order lift exists exactly when J_p(w)=−Q_p(v) can be solved. At the node, J₀=0 and Q₀(a,b)=ab, so no correction can change the residual. At (1,1) on the hyperbola, J_p(r,s)=r+s, so every residual can be cancelled. Defining Q_p by coefficient extraction avoids division by two and works in every characteristic.

At each later order, the same method asks which residual has appeared and which still-free coefficients can change it. Passing the first-order test is only the first step of that calculation.

Next, make a hidden cubic appear: a direction passes both first and second order, but fails at third order. Then test a change of coordinates, where the same coefficient method asks whether two operations can be made to agree.

Continue with the universal first-order construction, the node and its smoothing, or the complex that organises directions and obstruction targets. The separate oriented-boundary calculation explains cancellation by signs; it is a different operation from discarding quadratic terms.

Sources. The dual-number description of tangent vectors is standard; see The Stacks Project, §33.16, “Tangent spaces”. The general lifting condition is developed in §10.138, “Formally smooth maps”.