Factoring a polynomial over a GF I have the following question:


  What polynomial, when factored over the field $GF (2^8)$ based on the irreducible polynomial that is used in Rijndael, will factor into all the polynomials in the field?


As I understand it, the polynomial should be an irreducible polynomial and this polynomial must factor all the polynomials. How can I get this polynomial? What steps should I follow?
There are several questions hidden in your question.

You'll have to distinguish between the polynomial ring $\mathbb Z_2[X]$, and the factor field $GF(2^8) = \mathbb Z_2[X]/F$ (where $F$ is any irreductible polynomial of degree 8).

In $\mathbb Z_2[X]$, the only polynomial which is a factor of all others is the trivial polynomial $1$. (In general, in a (unitary) ring the set of elements with this property are called units, and this ring has just the trivial unit one.)

In $GF(2^8)$ (like in any field), every non-zero element is a unit and thus a factor of every other element, but we normally don't call them "polynomials".

But the original question actually asks something else:


  What polynomial when factored over the field $GF (2^8)$ based on the irreducible polynomial that is used in Rijndael will factor into all the polynomials in the field?


The expression "factors into" can be understood as "is a multiple of" ... and "when factored over $GF(2^8)$" could be understood as "when seen as a polynomial with coefficients in $GF(2^8)$ (i.e. an element of $GF(2^8)[X]$, and then factored (into its prime divisors)". But then the expression "into all the polynomials in the field" makes no sense, as a field element is either $0$ (which is not a factor other than for itself) or a unit (which is a factor for everything), and you likely don't want $0$ as the result.

Quite likely the original question was something else, but I have no idea, what.