# Derivative in an abstract polynomial ring

1. Apr 15, 2008

### PingPong

"Derivative" in an abstract polynomial ring

1. The problem statement, all variables and given/known data
Let R be any ring and define D:R[X]-->R[X] by setting $D[\sum a_nX^n]=\sum na_nX^{n-1}$.

a) Check that, if $f(X)=\sum a_nX^n$ and $g(X)=\sum b_nX^n$, then D[f+g]=D[f]+D[g]
b) Check that $D[a_ib_jX^{i+j}]=D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j]$
c) Now using the additivity in a), show that D[f(X)g(X)]=D[f(X)]g(X)+f(X)D[g(X)] for all polynomials f and g in R.

2. Relevant equations
None.

3. The attempt at a solution
I see the clear parallel to polynomial derivatives, but it's not defined in terms of epsilons and deltas - it's just a "black box" function that really has no meaning. But in any case, I've been able to get parts (a) and (b) without any problems. But I'm having trouble with (c).

I've chosen to write the product of $f(X)=\sum_i a_iX^i$ and $g(X)=\sum_j b_jX^j$ as just $\sum_{i,j}a_ib_jX^{i+j}$ so that it's similar to the result of part (b). Then
$$D[f(X)g(X)]=D[\sum_{i,j}a_ib_jX^{i+j}]=\sum_{i,j} D[a_ib_jX^{i+j}]=\sum_{i,j} \left(D[a_iX^i]b_jX^j+a_iX^iD[b_jX^j]\right)$$
$$=\sum_{i,j}D[a_iX^i]b_jX^j+\sum_{i,j}a_iX^iD[b_jX^j]=D[f(X)]g(X)+f(X)D[g(X)].$$

So does this work? I tried it another way by checking what's in the nth position on each side of the equation and showing them to be equal, but got bogged down in keeping track of indexes and stuff. My algebra professor said there was an easier way - is this it?