Mathematics 392 – Seminar in Computational Commutative Algebra Comment/Solution for Problem Set 2 September 18

1.5/13. This problem studies the “formal derivative” of a polynomial

(1)  f = a_{0}x^{n} + a_{1}x^{n−1} + ⋯ + a_{n−1}x + a_{n} ∈ k[x].

Without using limits or any other properties of the real number system, we can define

f′ = na_{0}x^{n−1} + (n − 1)a_{1}x^{n−2} + ⋯ + a_{n−1}

in a purely algebraic fashion. So this actually makes sense for polynomials with coefficients in any field. The surprise (?) is that all of the usual rules for derivatives still work.

a) Let a ∈ k. Then

(a · f) = (aa_{0})x^{n} + (aa_{1})x^{n−1} + ⋯ + (aa_{n−1})x + (aa_{n})

So by (1)

(a · f)′ = n(aa_{0})x^{n−1} + (n − 1)(aa_{1})x^{n−2} + ⋯ + (aa_{n−1})
= a(na_{0}x^{n−1} + (n − 1)a_{1}x^{n−2} + ⋯ + a_{n−1})
= a · f′

b) Let f and g be any two polynomials, say with deg f ≥ deg g. By adding terms with zero coefficients at the start of g, we can write both using the same range of exponents:

\begin{aligned}f & = a_{0}x^{n} + a_{1}x^{n−1} + ⋯ + a_{n−1}x + a_{n} \\ g & = b_{0}x^{n} + b_{1}x^{n−1} + ⋯ + b_{n−1}x + b_{n} \\ ⇒ (f + g) & = (a_{0} + b_{0})x^{n} + (a_{1} + b_{1})x^{n−1} + ⋯ + (a_{n−1} + b_{n−1}x + (a_{n} + b_{n}) \\ ⇒ (f + g)′ & = n(a_{0} + b_{0})x^{n−1} + (n − 1)(a_{1} + b_{1})x^{n−2} + ⋯ + (a_{n−1} + b_{n−1}) \end{aligned}

Using the distributive law for multiplication over addition (valid in all fields), and regrouping the terms, this is

na_{0}x^{n−1} + (n − 1)a_{1}x^{n−2} + ⋯ + a_{n−1} + nb_{0}x^{n−1} + (n − 1)b_{1}x^{n−2} + ⋯ + b_{n−1} = f′ + g′.

c) As some of you know, I’m a big beliver that 80% or so of doing mathematics is really being lazy in creative ways(!) To prove the product rule (fg)′ = f′g + g′f in a reasonable (i.e. creatively lazy(!)) fashion, let’s notice that parts a and b combined say that the formal derivative mapping

\begin{matrix}′ : k[x] & → k[x] \\ f & ↦ f′ \end{matrix}

is linear, under the usual vector space structure on k[x]. If f = a_{0}x^{n} + a_{1}x^{n−1} + ⋯ + a_{n−1}x + a_{n} and g = b_{0}x^{m} + b_{1}x^{m−1} + ⋯ + b_{m−1}x + b_{m}, then the product fg is the sum of all the monomial products

(2)  a_{i}x^{n−i} · b_{j}x^{m−j} = a_{i}b_{j}x^{n+m−i−j}

Apply the definition of the formal derivative to this one-term polynomial:

(3)  \begin{aligned}(a_{i}b_{j}x^{n+m−i−j})′ & = (n + m − i − j)a_{i}b_{j}x^{n+m−i−j−1} \\ & = (n − i)a_{i}b_{j}x^{n−i−1} · x^{m−j} + (m − j)a_{i}b_{j}x^{n−i} · x^{m−j−1} \\ & = (a_{i}x^{n−i−1})′ · (b_{j}x^{m−j}) + (a_{i}x^{n−i}) · (b_{j}x^{m−j})′ \end{aligned}

In other words, the product rule works for the single term (2) in the product.

Now we use the linearity of the formal derivative. For the entire polynomials f, g, (fg)′ is the derivative of the sum of the terms (2) over all 0 ≤ i ≤ n and 0 ≤ j ≤ m. By (2) and linearity of derivatives, (fg)′ is the sum of the terms in the last line of (3) for the same i and j. It is easy to see that we get f′g + gf′. This is what we wanted to show.