# Why is the polynomial in AES MixColumns multiplied modulo a reducible polynomial?

I know that the AES MixColumn step is calculated as follows

$b(x) = (a(x)c(x)) \mod l(x)$

with $a(x)$ being the column to encode, $c(x)$ the fixed polynomial $c(x) = 3x^3 + x^2 + x + 2$ and $l(x) = x^4+1$ which is reducible over $GF(2^8)$ since $x^4 + 1 = (x+1)^4$.

I wondered about the purpose of the reducibility property of $l(x)$ and so I looked it up in my text book as well as on the internet but I could not find any further explanation on this. As it is my understanding only an irreducible polynomial $f(x)$ would be able to create a field $GF(2^8)[x]/(f(x))$. So my question is:

Is there a particular reason why l(x) must be reducible over $GF(2^8)$ in AES MixColumns?

For MixColumns, the result must be a polynomial of degree 3 or less, which requires a degree 4 reduction. We are not reducing the elements of the finite field here, but the coefficients of the polynomial generated by multiplication of the fixed and input polynomials. With reduction by $x^4+1$, the reduction can also be represented as replacing the coefficient of $x^0$ with that of $x^4$.
The choice of degree 4 reduction polynomial does not have to be irreducible in $GF(2^8)$ if the fixed polynomial in the matrix multiplication has an inverse, which it does. In the case of $c(x)=3x^3+x^2+x+2$, $c^{-1}(x)=11x^3+13x^2+9x+14$.