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\centerline{Mathematics 392 -- Seminar in Computational Commutative Algebra}
\centerline{Comment/Solution for Problem Set 2}
\centerline{September 18}
\bigskip
1.5/13.  This problem studies the ``formal derivative'' of 
a polynomial 
$$f = a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n \in k[x].\leqno(1)$$
Without using limits or any other properties of the real
number system, we can define
$$f' = na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \cdots + a_{n-1}$$
in a purely algebraic fashion.  So this actually makes sense
for polynomials with coefficients in {\it any field}.  The 
surprise (?) is that all of the usual rules for derivatives
still work.
\bigskip
\noindent
a)  Let $a \in k$.  Then 
$$(a\cdot f) = (a a_0) x^n + (a a_1) x^{n-1} + \cdots + (a a_{n-1}) x + (a a_n)$$
So by (1)
$$\eqalign{
(a\cdot f)' &= n(a a_0) x^{n-1} + (n-1)(a a_1) x^{n-2} + \cdots + (a a_{n-1})\cr
         &= a(na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \cdots + a_{n-1})\cr
         &= a\cdot f'\cr}$$
\bigskip
\noindent
b)  Let $f$ and $g$ be any two polynomials, say with $\deg f \ge \deg g$.  
By adding terms with zero coefficients at the start of $g$, we can write 
both using the same range of exponents:
$$\eqalign{
f &= a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n\cr
g &= b_0 x^n + b_1 x^{n-1} + \cdots + b_{n-1} x + b_n\cr
\Rightarrow (f+g) &= (a_0+b_0) x^n + (a_1+b_1) x^{n-1} + \cdots + (a_{n-1}+b_{n-1} x + (a_n+b_n)\cr
\Rightarrow (f+g)' &= n(a_0+b_0) x^{n-1} + (n-1)(a_1+b_1) x^{n-2} + \cdots + (a_{n-1}+b_{n-1})\cr}$$
Using the distributive law for multiplication over addition (valid in all fields),
and regrouping the terms, this is
$$ na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \cdots + a_{n-1} + 
 nb_0 x^{n-1} + (n-1)b_1 x^{n-2} + \cdots + b_{n-1} = f' + g'.$$
\bigskip
\noindent
c)  As some of you know, I'm a big beliver that {\it $80\%$ 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
$$\eqalign{
{}' : k[x] &\to k[x]\cr
      f  &\mapsto f'\cr}$$
is {\it linear}, under the usual vector space structure on $k[x]$.
If $f = a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1} x + a_n$ and 
$g = b_0 x^m + b_1 x^{m-1} + \cdots + b_{m-1} x + b_m$, then
the product $fg$ is the sum of all the monomial products
$$a_i x^{n-i} \cdot b_j x^{m-j} = a_i b_j x^{n+m - i - j}\leqno(2)$$
Apply the definition of the formal derivative to this one-term polynomial:
$$\eqalign{
(a_i b_j x^{n + m - i - j})' &= (n + m - i - j)a_i b_j x^{n + m - i - j - 1}\cr
&= (n - i) a_i b_j x^{n-i-1} \cdot x^{m - j} + (m - j) a_i b_j x^{n-i}\cdot x^{m - j - 1}\cr
&= \left(a_i x^{n - i - 1}\right)' \cdot \left(b_j x^{m-j}\right) + \left(a_i x^{n-i}\right) 
\cdot\left(b_j x^{m-j}\right)'\cr}\leqno(3)$$
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
$0 \le i \le n$ and $0\le j \le m$.  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.
\bye

