NOTE: The question is inspired by an operation in cryptography. But I try to recast it as a purely algebraic question.
Let $b(x)=b_{3}x^{3}+b_{2}x^{2}+b_{1}x+b_{0}$ be a polynomial whose coefficients are elements of $GF(2^8)$, and consider a fixed polynomial $a(x)=3x^{3}+x^{2}+x+2$. Define the multiplication between $a(x)$ and $b(x)$, denoted $d(x) = a(x) \otimes c(x)$ as their "ordinary" multiplication, modulo $x^4 + 1$.
EDIT: By "ordinary" multiplication, I mean $s(x) = a(x) \times c(x)$:
$$a(x)=a_3x^{3}+a_2x^{2}+a_1x+a_0\\ b(x)=b_{3}x^{3}+b_{2}x^{2}+b_{1}x+b_{0}\\ s(x) = s_6x^6 + \dots + s_0 \,, $$ where $s_i = \sum_{j=0}^i a_i \times b_{i-j}$ is computed over $GF(2^8)$: That is, the addition and multiplication of coefficients are the addition and multiplication of $GF(2^8)$. Finally, we set $d(x) = s(x) \bmod (x^4+1)$.
Can we say that this operation takes place over the quotient ring $R = GF(2^8)[x]/\langle x^4+1 \rangle$?