1

I'm struggling with the following construction. So, we took the finite field $F_{16}$ as $F_2[\alpha]/(\alpha^4+\alpha+1)$. We know that $F_4$ = $\{0,1,\alpha,\alpha+1\}$, and there are three irreducible polynomials of degree 4 over $F_2$, each with 4 roots inside $F_{16}$.

So that all seems fine. But then for roots of $\alpha^4+\alpha+1$ we said they were $\{\alpha,\alpha+1,\alpha^2,\alpha^2+1\}$. But this has an intersection of $\{\alpha,\alpha+1\}$ with $F_4$; I thought the elements need to be independent? In this case, we'll end up with at most 14 elements. Are these two $\alpha$'s somehow different? I'm pretty confused. It seems like it should be the same $\alpha$, because $\alpha$ surely is a root of $\alpha^4+\alpha+1$ over $F_2[\alpha]/(\alpha^4+\alpha+1)$ and an element of $F_4$. Can someone explain what's going on?

Chaotic Good
  • 524
lightnesscaster
  • 336
  • 3
    Sorry, so you're saying that the (unique) subfield of $F_{16}$ of order 4 is given by the set ${0, 1, \alpha, \alpha+1}\subset F_{16}$? That's not quite right, as (for example) this set is not multiplicatively closed. (Note, for instance, that it does not contain $\alpha^2$.) It also does not contain inverses – for instance, the inverse of $\alpha$ in $F_{16}$ is $-1-\alpha^3$ – Atticus Stonestrom Nov 09 '20 at 03:29
  • 2
    What gave you the impression that those four elements comprise the subfield of order 4? – Atticus Stonestrom Nov 09 '20 at 03:29
  • @AtticusStonestrom That's what my professor said. But yeah I see your point; that can't be true. – lightnesscaster Nov 09 '20 at 03:38
  • 3
    Perhaps are you getting confused by having a seen a different construction of $F_4$ as the quotient $F_2[\alpha]/(\alpha^2+\alpha+1)$? – Atticus Stonestrom Nov 09 '20 at 03:39
  • @AtticusStonestrom Here are the notes where he says it: https://ibb.co/y8hy6FB – lightnesscaster Nov 09 '20 at 03:39
  • 2
    @AtticusStonestrom Ah but yeah maybe he's using $\alpha$ to mean something totally different there. – lightnesscaster Nov 09 '20 at 03:40
  • 2
    Yes exactly – what "$\alpha$" means in this representation of $F_4$ is different than what "$\alpha$" means in the representation you give of $F_{16}$. I definitely agree though that it's somewhat confusing notation! Maybe would have been better to write $F_{16}$ as $F_2[\beta]/(\beta^4+\beta+1)$ – Atticus Stonestrom Nov 09 '20 at 03:42
  • 1
    @AtticusStonestrom Ok thank you that makes a lot more sense! – lightnesscaster Nov 09 '20 at 03:44
  • 2
    Happy it helped! For future reference: the subfield of order 4 in this representation of $F_{16}$ is given by ${0, 1, \alpha^2+\alpha+1, \alpha^2+\alpha}$. (It's worth checking this manually, by making sure that this set is closed under products, sums, and inverse!) – Atticus Stonestrom Nov 09 '20 at 03:49
  • 1
    @AtticusStonestrom nailed it. If you want to get a bit more experience you can take a look at these tables I prepared for referrals like this. The element I call $\gamma$ is a zero of $x^4+x+1$ whereas $\beta$ is a zero of $x^2+x+1$. You see that $\gamma^5=\gamma^2+\gamma+1$ and $\gamma^{10}=\gamma^2+\gamma$ are roots of unity of order three. Just what we need for them to belong to the subfield of four elements. – Jyrki Lahtonen Nov 09 '20 at 07:04
  • (@JyrkiLahtonen that reference is awesome, thanks for making/sharing) – Atticus Stonestrom Nov 10 '20 at 00:57

0 Answers0