# Formula for Multiplying Two Elements in a Polynomial Ring

1. Dec 21, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
I would like to show that if $p(x) = \sum_{i=1}^m a_i x^i$ and $q(x) = \sum_{j=1}^n b_j x^j$, then $p(x)q(x) = \sum_{k=0}^{m+n} \left( \sum_{i+j=k} a_i b_j \right) x^k$, where the polynomial ring is assumed to be commutative.

2. Relevant equations

3. The attempt at a solution

The base case of $m=n=1$ is trivial; one just simply compares the formula to a "brute force" calculation. So, suppose that the formula holds for polynomials $p$ and $q$ where $p$ has length $1$ and $q$ has length $n$. Then

$$p(x)q(x) = p(x) \sum_{j=1}^{n+1} b_j x^j = p(x) \sum_{j=1}^n b_j x^j + p(x)b_{n+1}x^{n=1}$$

By the induction, hypothesis, $p(x) \sum_{j=1}^n b_j x^j = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k$, and so

$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$

The term $a_0 b_{n+1} x^{n+1}$ can be identified with $k=n+1$ and $a_1 b_{n+2} x^{n+2}$ will be the leading term. Hence,

$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$

I feel that this last point is a bit shaky, but I'll let you be the judge of that. By commutativity and symmetry we get the other induction. Now for the double induction. Assume the formula holds for polynomials of length $m$ and $n$. Then

$$p(x)q(x) = \sum_{i=1}^{m+1} a_i x^i \sum_{j=1}^{n+1} b_j x^j = \left( \sum_{i=1}^{m} a_i x^i + a_{m+1} x^{m+1} \right) \left(\sum_{j=1}^{n} b_j x^j + b_{n+1} x^{n+1} \right)$$

$$= \sum_{i=1}^{m} a_i x^i \sum_{j=1}^{n} b_j x^j + \sum_{i=1}^m a_i b_{n+1} x^{i+n+1} + \sum_{j=1} a_{m+1} b_j x^{j + m+1} + a_{m+1} b_{n+1} x^{m+n+2}$$

At this point, I can apply the induction hypothesis on the first term, but I am unsure how to properly combine this mess to get the desired formula...

Last edited: Dec 21, 2016
2. Dec 21, 2016

### Ssnow

A suggestion: don't confuse the length with the degree of a polynomial , a polynomial of length $1$ can have a degree $m$, so it is of the following monomial form $P(x)=a_{m}x^{m}$. For the rest the induction is the correct idea ...

3. Dec 23, 2016

### Ray Vickson

Fix up the lower summation limits: they should all start at either 0 or at 1 (and, preferably, all at 0).