How to prove this polynomial multiplication is associative and distributes over addition? The following is a construction from Emil Artin's Galois Theory Lecture Notes:
Let $f$ be a monic irreducible over $F$ with root $\alpha$ (i.e. $\min(\alpha,F)$).
Let $E_{\alpha}=\{g(\alpha)\in F[x]|\deg(g)<n\}$. Define a multiplication on $E_\alpha$ using the quotient remainder theorem: $g(\alpha)h(\alpha)=r(\alpha)$ where $g(x)h(x)=q(x)f(x)+r(x)$. Artin says

The product of $m$ terms $g_1(\alpha),g_2(\alpha),...g_m(\alpha)$ is again the remainder of the ordinary product $g_1(\alpha)g_2(\alpha)...g_m(\alpha)$. This is true by induction if we prove the easy lemma: The remainder of the product of of two remainders is the remainder of the product of these two polynomials. This shows our product is associative and commutative and will coincide with the ordinary product just as long as $\deg(g_1g_2...g_m)\leq n$. The fact that it is distributive is easily verified.

I think this is the "easy lemma" he is referring to:
Clearly if  $g(x)=q_1(x)f(x)+r_1(x)$ and $h(x)=q_2(x)f(x)+r_2(x)$, then 
$g(x)h(x)=q_1(x)q_2(x)f^2(x)+q_1(x)f(x)r_2(x)+q_2(x)f(x)r_1(x)+r_1(x)r_2(x)$, so $g(\alpha)h(\alpha)=r_1(\alpha)r_2(\alpha)$. I guess we can proceed by induction to see that $\displaystyle\prod_{i=1}^{m}g_i(\alpha)=\prod_{i=1}^mr_i(\alpha)$, where $g_i(x)=q_i(x)f(x)+r_i(x)$.
However, I don't see how this shows associativity or distributivity.
For associativity,  $(g(x)h(x))p(x)=q_1(x)f(x)p(x) + r_1(x)p(x)$ and $g(x)(h(x)p(x))=g(x)q_2(x)f(x) + g(x)r_2(x)$. But I don't know how we could show that $r_1(\alpha)p(\alpha) = g(\alpha)r_2(\alpha)$.
Similarly, I don't know how to show that $g(x)(h(x)+k(x))$ will give the same remainder at $\alpha$ as $g(x)h(x) + g(x)k(x)$.
Can someone clarify how these two properties follow from the remainder of a product being the product of remainders?
 A: I assume $n=\deg(f)$ in your question, and the root $\alpha$ lies in some field $K$ containing $F$.
I think the construction is easier to understand if we change viewpoint. We have the evaluation map $F[x]\to K$, $x\mapsto\alpha$. This is a homomorphism of rings which is the identity on $F$, so an $F$-algebra homomorphism. Its kernel clearly contains $f$, and since $f$ is irreducible we know (from the division algorithm) that the ideal $(f)$ is maximal. Thus the kernel is precisely $(f)$, $F[x]/(f)$ is a field, and we have an injective $F$-algebra homomorphism $F[x]/(f)\rightarrowtail K$.
Again, by the division algorithm, the elements of $F[x]/(f)$ are in bijection with the   polynomials in $F[x]$ of degree less than $n$, so the image of $F[x]/(f)$ is exactly your $E_\alpha$.
Thus, with regard to your original question, the ring structure of $E_\alpha$ is completely determined by the ring structure on $F[x]/(f)$, and so all questions about remainders/associativity/distributivity in $E_\alpha$ follow from their analogues in $F[x]/(f)$, which in turn all follow from the division algorithm: given any $g\in F[x]$, there exist unique $q,r\in F[x]$ with $\deg(r)<n$ such that $g=qf+r$ (and it is really the uniqueness that allows Artin to deduce all the relevant properties).
A more abstract take on this idea is the following: given a (monic) irreducible polynomial $f\in F[x]$, we can construct a field containing $F$ in which $f$ has a root, namely $F[x]/(f)$. If $K$ is any other field containing $F$ in which $f$ has a root, $\alpha$ say, then there is a unique map of $F$-algebras $F[x]/(f)\to K$ sending (the image of) $x$ to $\alpha$, and this yields a bijection between $F$-algebra homomorphisms $F[x]/(f)\to K$ and roots of $f$ in $K$.
A: To show associativity, all you need to show is that $f(x)\mid (\operatorname{rem}((g(x)h(x))p(x))-\operatorname{rem}(g(x)(h(x)p(x)))$, i.e. $f(x)\mid (r_1(x)p(x)-g(x)r_2(x))$.  
Given that $r_1(x)=g(x)h(x)-q_1(x)f(x)$ and $r_2(x)=h(x)p(x)-q_2(x)f(x)$, we have that: $$r_1(x)p(x)-g(x)r_2(x)$$$$=(g(x)h(x)-q_1(x)f(x))\cdot p(x)-g(x)\cdot(h(x)p(x)-q_2(x)f(x))$$$$=g(x)h(x)p(x)-g(x)h(x)p(x)-f(x)(q_1(x)f(x)-g(x)q_2(x))$$$$=f(x)(q_1(x)f(x)-g(x)q_2(x))$$ 
which is clearly divisible by $f(x)$.  
So $f(x)\mid (r_1(x)p(x)-g(x)r_2(x))$. Since $\alpha$ is a root of $f(x)$ it must then also be a root of $r_1(x)p(x)-g(x)r_2(x)$. 
So $$r_1(\alpha)p(\alpha)-g(\alpha)r_2(\alpha)=0$$ i.e. $$r_1(\alpha)p(\alpha)=g(\alpha)r_2(\alpha)$$ 
You can similarly show distributivity - you just need to show that $p(x)$ divides the difference of the remainder functions produced from the expressions $g(x)(h(x)+k(x))$ and $g(x)h(x)+g(x)k(x)$.
