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Can someone explain to me how i would go about doing a problem like this? I don't really know where to start. GF refers to a Galois field. I'm struggling to even understand exactly what they want me to do here.

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  • $\begingroup$ $GF(q)$, where $q$ is a prime power, refers to the finite field with $q$ elements. $\endgroup$ – Josué Tonelli-Cueto Apr 22 '14 at 16:17
  • $\begingroup$ I understand that, I'm just not sure what to do with that. I assume the prime power will be that of 2^n because of the field $\endgroup$ – Se yaJ Me Apr 22 '14 at 16:22
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The point here to realize is that all fields of size $p^n$ are isomorphic and that you construct a (and hence the) field of $p^n$ elements by taking ${\mathbb F}_p[x]/(f(x))$ for some irreducible polynomial $f(x) \in {\mathbb F}_p[x]$ of degree $n$.

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  • $\begingroup$ any way that you could explain this more or even give me an example of contructing that? I'm really having trouble with the notation. $\endgroup$ – Se yaJ Me Apr 22 '14 at 16:28
  • $\begingroup$ What is troubling? You need to give me some idea of your background. For instance, how would you construct a field of $p^n$ elements? $\endgroup$ – Magdiragdag Apr 22 '14 at 17:16
  • $\begingroup$ the way i see it, their is no integer q such that q will be a root for the polynomial in mod 2. thats why i don't get it. I can split polynomial fields with complex numbers, just not integers. $\endgroup$ – Se yaJ Me Apr 22 '14 at 21:13
  • $\begingroup$ If you mean, there is no $q \in {\mathbb F}_2$ (i.e., $q = 0$ or $q = 1$) such that that polynomial has $q$ as root: that is correct, but does not (yet) show that the polynomial is irreducible. It might factor as a product of two quadratic polynomials. (It doesn't, but it something you have to argue). If you mean something else (for instance, if the $q$ you mention in the comment has anything to do with the $q$ from the question), then I have no clue what you mean. $\endgroup$ – Magdiragdag Apr 22 '14 at 21:45
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    $\begingroup$ Yes, but there are still some technicalities you might want to understand. For someone who is familiar enough with this subject, the argument "$x^4 + x^3 + 1$ is irreducible over ${\mathbb F}_2$, therefore its splitting field is ${\mathbb F}_{2^4}$" is ok. $\endgroup$ – Magdiragdag Apr 22 '14 at 22:28

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