Prove if $D$ is an integral domain and $f,g\in D[X]$ are nonzero, then $fg$ does not equal $0$ and $\deg[f(x)g(x)]=\deg f(x) + \deg g(x)$.
I do not know much about this since I just learned about it. Here is what I can prove for you. Please tell me if I am right. If not, please provide a proof.
Proof: Write $f(x) = a_0 + \dots + a_nx^{n}$ and $g(x) = b_0 + \dots + b_mx^m$, with $a_m$, $b_m$ nonzero. Then $f(x)g(x) = a_0b_0 + (a_0b_1 + a_1b_0)x + \dots + a_nb_mx^{n+m}$.
The largest power of $x$ that can occur is $x^{n+m}$ so $\deg [f(x)g(x)] \leq \deg f(x) + \deg g(x)$. If $D$ has no zero divisors, then $a_nb_m$ does not equal zero and $n+m$ is the degree giving us $\deg[f(x)g(x)]=\deg f(x) + \deg g(x)$.
Best Answer
(Just to leave an answer:) Your proof is correct and is the standard one. It's so short and straightforward that it's hard to see room for improvement.