I am going through this video 302.S2a: Field Extensions and Polynomial Roots by Matthew Salomone & there is a lot in this video which confuses me.

  1. I understand the construction by which he generates the field with 8 elements {$0, 1, t, 1 + t, t^2, 1 + t^2, t + t^2, 1 + t + t^2$}. He sets $p(t) = 0$ & constructs a set with those 8 elements. But I am unable to figure out why he is calling this field as $F_2[t]/\langle p(t) \rangle$ - I am assuming this is the same as calling it $F_2[t]/\langle t^3 + t + 1 \rangle$

$F_2[x]/\langle t^3 + t + 1 \rangle$ is the Quotient Ring generated by the ideal $t^3 + t + 1$ - it doesn't contain these 8 elements at all. Each element of the Quotient Ring isn't even a polynomial - it's an equivalence class of polynomials - for e.g. one such element of $F_2[x]/\langle t^3 + t + 1 \rangle$ is this equivalence class $[\bar t]$

$[\bar t]$ = {$...-t^3+1, t, t^3 + 1, t + 2, ...$}

Another element is this equivalence class below

$[\bar 5]$ = {$..., -t, 1, t^3 +t , 1, ...$}

So now why is he calling that field of 8 elements as $F_2[x]/\langle t^3 + t + 1 \rangle$ - isn't that a totally different ring/field?

I don't understand where the quotient construction using an ideal comes in here at all. We construct this new field by 2 things by just setting $t^3 + t + 1 = 0$ in $F_2[x]$. We don't need to use the ideal or the quotient ring generated by the ideal at all for this.

  1. A little further into the video, timestamp 7:30 - he says $t$ is a root of the polynomial. Again I don't understand what he means by $t$ is a root of a polynomial $p(t)$ - for a polynomial $p(t)$, we always say $t = something$ is a root - what exactly does saying $t$ is a root mean?
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    $\begingroup$ Can't be bother to check, but I rather suspect that he calls it $K=\Bbb{F}_2[x]/\langle x^3+x+1\rangle$. If we denote the coset $x+\langle x^3+x+1\rangle$ by $t$, then we have, indeed, $t^3+t+1=0$ and $$K=\{0,1,t,t+1,t^2,t^2+1,t^2+t, t^2+t+1\}.$$ Observe the different roles of $t$ and $x$. With that difference clear it makes perfect sens to call $t$ a zero of the polynomial $x^3+x+1$. Personally I use a Greek letter in place of $t$, $\alpha$ being the default. $\endgroup$ Apr 11, 2022 at 10:03
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    $\begingroup$ In this field $5=1$. At least after you have become familiar enough not to become confused you earn the privilege to drop the bar on top :-) $\endgroup$ Apr 11, 2022 at 10:07
  • $\begingroup$ @JyrkiLahtonen - your explanation kind of makes sense but a couple of things 1) He doesn't use $x$ at all in the video. He starts this part of the problem with this imgur.com/0LXGAEi.png at 5:04 $\endgroup$
    – user93353
    Apr 12, 2022 at 11:26
  • $\begingroup$ @JyrkiLahtonen 2) I feel a little confused about the concept of a root which is a variable - a root is always a constant in my mind! For e.g. the roots of $t^2-4$ are {2, -2} - I am unable to understand how either $[\bar x]$ or $t$ - can be a root of $t^3 + t + 1$. I am not sure if calling it $\alpha$ helps! If $\alpha$ is a constant, then it has only one value - so why use something like $\alpha$ $\endgroup$
    – user93353
    Apr 12, 2022 at 11:27
  • $\begingroup$ Think of it just like any other element. It is just a name. Consider the quotient ring $K=\Bbb{Q}[x]/\langle x^2-2\rangle$. This time let's denote the coset $x+\langle x^2-2\rangle$ by $\alpha$. We see that $\alpha^2-2$ is the coset $x^2-2+\langle x^2-2\rangle$, which is the same coset as $0+\langle x^2-2\rangle$. That is, the zero element of $K$. So $$\alpha^2-2=0\in K.$$ Consequently some might want to give $\alpha$ the name $\sqrt2$. Of course, it has nothing whatsoever to do with the real number $1.4142135\ldots$, but that's not the point. $\endgroup$ Apr 13, 2022 at 17:40

1 Answer 1


$x^3+x+1=p(x)$ is irreducible in $F_2[x]$ so $\langle p(x)\rangle$ is a maximal ideal of $F_2[x]$. It follows that $F_2[x]/\langle p(x) \rangle=:S$ is a field. Note that this is where $\langle p(x) \rangle$ being an ideal (maximal ideal in this case) is used.

$S$ actually has $8$ elements in it. The elements of $S$ look like $ax^2+bx+c+\langle p(x) \rangle$, where $a,b$ and $c$ are in $F_2$. The elements of $S$ are often written with the bar notation (in such notation, $\langle p(x) \rangle$ is dropped) as $\overline {ax^2+bx+c}$, i.e., $\overline {ax^2+bx+c}=ax^2+bx+c+\langle p(x) \rangle$.

Now, regard every element in $F_2[x]$ as an element of $S$ (using the bar notation).

With this view point, $\overline x\in S$ is a zero of $p(t)\in S$ as the following shows:

\begin{align} p(\overline x)=(\overline x)^3+\overline x+1&=(x+\langle p(x)\rangle)^3+(x+\langle p(x)\rangle)+1\\ &=x^3+x+1+\langle p(x)\rangle\\&=p(x)+\langle p(x)\rangle=\langle p(x)\rangle \end{align} Note that RHS is a 'zero' in $S$.

To see that $S$ has $8$ elements, note that if $\overline{q(x)}$, where $q(x)$ is of degree$>2$ lies in $S$ then $q(x)$ can be reduced modulo $p(x)$ to get $q(x)=p(x) a(x)+b(x)$, where $b(x)=0$ or degree b(x)<degree $p(x)=3$. It follows that $\overline {q(x)}= q(x)+\langle p(x) \rangle= b(x)+p(x)a(x)+\langle p(x)\rangle= b(x)+\langle p(x)\rangle=\overline{b(x)}$.

Since $b(x)$ is of degree at most $2$, it follows that $b(x)$ is of the form $dx^2+ex+f$. Note that each of $d,e$ or $f$ has two choices (as they are in $F_2$) so maximum number of elements in $S$ is $2^3=8$.

The only way when the number of elements in $S$ is less than $8$ is when any two elements in $S$ are equal (i.e., there is some sort of 'collapsing'). Now, show that there is no collapsing and you are done.

  • $\begingroup$ Why do the elements of $S$ have to look like $ax^2+bx+c+\langle p(x) \rangle$. Why can't $x^4 + x^3 + x + 1$ be one of the equivalence classes of $S$? $x^4$ belongs to $Z_2[x]$ & and can be added to $\langle x^3+x+1 \rangle$ to get one of the cosets. Considering that, I don't see how $S$ has only 8 elements - it would have far, far more elements. $\endgroup$
    – user93353
    Apr 11, 2022 at 11:16
  • $\begingroup$ @user93353: The point is as you have mentioned in OP also that the elements of S are equivalence classes. $\overline{x^4+x^3+x+1}$ is indeed in $S$ but it can be shown to be equal to some element of the form $\overline{ax^2+bx+c}$. To see this, divide $x^4+x^3+x+1$ by $p(x)$ and obtain by Euclid algorithm that $x^4+x^3+x+1=p(x) q(x)+r(x)$, where degree r(x)< degree $p(x)=3$. So in bar notation, one has $\overline {x^4+x^3+x+1}=\overline {r(x)}\quad $ (note that $\overline{p(x)q(x)+r(x)}=\overline{r(x)}$). $\endgroup$
    – Koro
    Apr 11, 2022 at 11:22
  • $\begingroup$ You are right, it's part of an equivalence class. Let me ponder over your reply some more. $\endgroup$
    – user93353
    Apr 11, 2022 at 11:34
  • $\begingroup$ @user93353: I have edited my answer to include an outline to show that $S$ has $8$ elements. $\endgroup$
    – Koro
    Apr 11, 2022 at 11:40
  • $\begingroup$ I have pondered :-). Your answer is really good - thank you!!! This is non-trivial stuff which should have been included in the video. I have a couple of questions however about your answer. $\endgroup$
    – user93353
    Apr 12, 2022 at 11:03

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