Show that $\deg(PQ)=\deg(P)+\deg(Q)$ 
Let $R$ be an integral domain and $P,Q \in R[x]$ non-zero polynomials. Show that $PQ$ is non-zero and that $\deg(PQ)=\deg(P)+\deg(Q)$. Is $\Bbb Z_6[x]$ an integral domain?

Suppose that $PQ=0$, then $R$ being an integral domain so is $R[x]$ and so $PQ=0 \implies P=0$ or $Q=0$ which is a contradiction. Thus $PQ \ne 0$.
Let $P=a_nx^n+\dots+a_0 \in R[x]$ and $Q=b_mx^n+\dots +b_0 \in R[x]$. Then $PQ=a_nx^nb_mx^m+ \dots = a_nb_mx^{n+m}+ \dots$ so the degree of $PQ$ is $n+m$.
The second question is false I think since I can just take constant polynomials $2$ and $3$ in $\Bbb Z_6[x]$ and note that $2 \cdot 3=6 \equiv 0$, but $2 \ne0$ and $3\ne0$?
I'm not sure about the proof for $\deg(PQ)=\deg(P)+\deg(Q)$ since I think I'm assuming that $R[x]$ is commutative in order to conclude that $a_nx^nb_mx^m = a_nb_mx^{n+m}$?
 A: 
Let $P=a_nx^n+\dots+a_0 \in R[x]$ and $Q=b_mx^n+\dots +b_0 \in R[x]$. Then $PQ=a_nx^nb_mx^m+ \dots = a_nb_mx^{n+m}+ \dots$ so the degree of $PQ$ is $n+m$.

This isn't a completely satisfying argument because we need to collect all possible terms with $x^{m+n}$.  Of course there is just one, but you need to indicate why.  We also need an argument for why $a_nb_m\ne 0$.
Your concern about commutativity isn't terribly concerning to me, actually, because the $x^{m+n}$ term is obtained by $(a_nx^n)(b_mx^m) = a_nb_mx^{m+n}$.  We're not using commutativity, even though it looks like it.  Rather this is just the definition of formal products in the ring of polynomials.  The emphasis here is that we are computing formal products.  $x$ is not actually an element of the ring, it is a formal symbol that is effectively not much different from just a "placeholder".  For instance we can identify $2x^2-1$ with the sequence $(2,0,-1)$ and now there is no $x$ at all, and $x^2$ was just a placeholder for the coefficient, which we identify with the left-most coordinate of the tuple.
Long story short, no need to worry about commutativity.
Once you've supplied those two explanations, the proof will be satisfying.
Your argument showing that $\Bbb Z_6[x]$ is not an integral domain is fine.
A: ''Let P=anxn+⋯+a0∈R[x] and Q=bmxn+⋯+b0∈R[x]. Then PQ=anxnbmxm+⋯=anbmxn+m+… so the degree of PQ is n+m.''
Since $R$ is an integral domain and $a_n\ne0$ ($P$ has degree $n$) as well as $b_m\ne 0$ ($Q$ has degree $m$), $a_nb_m \ne 0$ and so $PQ$ has degree $m+n$.
