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On pg. 587 (in the finite fields chapter) of Abstract Algebra, 3rd ed. by Dummit and Foote, the following statement is made:

'If $f_1(x)=x^4+x^3+1$, $f_2(x)=x^4+x+1$ are two of the irreducible quartics over $\mathbb{F}_2$, then a simple computation verifies that $\alpha(x)=x^3+x^2$ is a root of $f_2(x)$ in $\mathbb{F}_{16}=\mathbb{F_2}/(x^4+x^3+1)$.'

I understand why $\alpha(x)$ is a root, but I just can't grasp how they came up with it. Its like it was pulled out of thin air, and they make no mention of it. I know that you could find all the elements of the quotient group and try each one individually, but that seems like an extremely weird thing not to mention. How did they come up with $\alpha(x)$?

pretzelman
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  • You just have to search for it, more or less. There is no magical method. – KCd Apr 26 '15 at 02:03
  • That is remarkably tedious. If this is true, I guess I'm just going to stick with arbitrary roots like $\theta$, $\alpha(x)$, etc. – pretzelman Apr 26 '15 at 02:14
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    The point is simply to give an example. They are just showing that the theory works: in the field generated over $\mathbf F_p$ by the root of one irreducible of degree $n$ in $\mathbf F_p[x]$ you really will find roots of every other irreducible of degree $n$ and more generally of degree dividing $n$. This gives a grip on reality a lot more than avoiding working with any examples at all. – KCd Apr 26 '15 at 02:29
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    Actually, for the specific choice of irreducibles in that example there is a simple way to figure out a root of $f_2(x)$ in the field generated over $\mathbf F_2$ by a root of $f_1(x)$. Notice the roots of $f_1(x)$ are reciprocals of the roots of $f_2(x)$ and vice versa: $f_2(\alpha) = 0$ if and only if $f_1(1/\alpha) = 0$. So when $f_1(x)$ has the root $x \bmod x^4 + x^3 + 1$ in $F := \mathbf F_2[x]/(x^4+x^3+1)$ we can use the inverse of $x$ as a root of $f_2(x)$. Since $1 \equiv x^4 + x^3 = x(x^3+x^2) \bmod x^4 + x^3 + 1$ we see that $x^3+x^2$ has to be a root of $f_2(x)$ in $F$. – KCd Apr 26 '15 at 02:32
  • In the statement of your question is a typo: $\mathbf F_2/(x^4+x^3+1)$ should be $\mathbf F_2[x]/(x^4+x^3+1)$. – KCd Apr 26 '15 at 02:33
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    What KCd said (after his 1st comment). Mostly this is just an example. If you play with these objects a lot, the fact that the two quartics are reciprocals of each other kinda screams at you :-) – Jyrki Lahtonen Apr 27 '15 at 06:08
  • If you have trouble verifying the results of calculations in small finite fields, you can first generate discrete log tables such as these. Unfortunately (for you) I used $f_2$ in defining the field there, so you cannot immediately reuse that table. – Jyrki Lahtonen Apr 27 '15 at 06:12
  • But @KCd, there is a magical method in this case. – Lubin Apr 28 '15 at 20:27
  • @Lubin, I agree and when I realized there is such a method later I forgot what I had first written above. – KCd Apr 29 '15 at 01:12

2 Answers2

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I think that @JyrkiLahtonen’s comment is most apposite.

The roots of $f_1$ are the reciprocals of the roots of $f_2$, as he points out. To clarify matters, let’s call $\xi$ the image of $x$ in $\Bbb F_2[x]/(f_1(x))$. The question asks to show that $\alpha=\xi^3+\xi^2$ is a root of $f_2$, in other words that $1/\alpha$ is a root of $f_1$.

But from $\xi\alpha=\xi^4+\xi^3=1$, we see that $1/\alpha$ is $\xi$ itself, certainly a root of $f_1$.

Lubin
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In a situation where the set of possible solutions is finite and checking whether a candidate solution is indeed a solution involves simple calculations, the most straightforward algorithmic method to determine all solutions consists of trying all candidates in turn. This is especially true if the "finite" is not infeasibly large.

Hagen von Eitzen
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