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?
Best Answer
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 $g(x)h(x)+g(x)k(x)$.