# Unable to understand the relation between Field Extensions, Polynomial Quotient Rings generated by Ideals, & Polynomial Roots

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?
• 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. Apr 11, 2022 at 10:03
• 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 :-) Apr 11, 2022 at 10:07
• @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 Apr 12, 2022 at 11:26
• @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 $\alpha$ Apr 12, 2022 at 11:27
• 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. Apr 13, 2022 at 17:40

$$x^3+x+1=p(x)$$ 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.

• 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. Apr 11, 2022 at 11:16
• @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 $\overline{p(x)q(x)+r(x)}=\overline{r(x)}$).
– Koro
Apr 11, 2022 at 11:22
• You are right, it's part of an equivalence class. Let me ponder over your reply some more. Apr 11, 2022 at 11:34
• @user93353: I have edited my answer to include an outline to show that $S$ has $8$ elements.
– Koro
Apr 11, 2022 at 11:40
• 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. Apr 12, 2022 at 11:03