Show that $\deg(fg) = m+n$

Question

Let $R$, a ring with a $1$ and $f,g$ two polynomials, where $\deg(f)=n, \deg(g)=m$. Also, there's a $c\in R$ such that $b_mc = 1$. Show that $\deg(fg)=m+n$.

I'd be glad for a guidance.

Thanks

Answer 1

One always has $\deg(fg)\leq\deg(f)+\deg(g)=n+m$ since all products of terms contributing to the product $fg$ satisfy this degree inequality. Also there is only one product of terms contributing in degree $n+m$, namely the product $a_nx^n.b_mx^m=a_nb_mx^{n+m}$ of the leading terms, so no additive cancellation of contributions is possible: one has $\deg(fg)=n+m$ is and only if $a_nb_m\neq0$. But since $a_n\neq0$ (by the definition of degree) and $b_m$ is invertible (so not a zero divisor), the latter condition is always true.

Answer 2

Hint You only need to show that $(fg)_{m+n} \ne 0$ is the leading coefficient. To do that look at $$fg(x) = \left( \sum_{j=0}^n a_j x^j \right) \cdot \left( \sum_{k=0}^m b_k x^k \right)$$ Note that all coefficients of power $> n+m$ must be $0$, so this shows $\deg(fg) = \deg f + \deg g$.