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I need a help with this exercise.

(i) Find a primitive root $\beta$ of $\mathbb{F}_2[x]/(x^4+x^3+x^2+x+1)$.
(ii) Find the minimal polynomial $q(x)$ in $\mathbb{F}_2[x]$ of $\beta$.
(iii) Show that $\mathbb{F}_2[x]/(x^4 +x^3+x^2+x+1)$ is isomorphic to $\mathbb{F}_2[x]/(q(x))$.

Now, what I did for (i) was writing $\beta=x+1$ and calculating $\beta^3$ and $\beta^5$ knowing that $x^4+x^3+x^2+x+1=0$. Doing the calculation I found out that $\beta^3\neq\beta^5\neq 1$. So $\beta$ has to have order 15, hence it generates the multiplicative group of 15 elements of units of $\mathbb{F}_{16}$, and so $\beta$ is a primitive root.

The problems comes with (ii). Infact I thought that in $\mathbb{F}_2$ there is no polynomial that has $\beta=x+1$ as root, and so I can't find the asked minimal polynomial. Am I missing something? I thought that there could be a typo in the textbook and said minimal polynomial has to be found in $\frac{\mathbb{F}_2[x]}{(x^4+x^3+x^2+x+1)}$. In this case if I write $q(\alpha) = \alpha^4+\alpha^3+1$ and evaluating $q(\beta)$ (knowing that $x^4+x^3+x^2+x+1=0$) I find out that $q(\beta)=0$. Trying with polynomial of degree $<4$ doesn't give any risult. So I thought the minimal polynomial of $\beta$ is $q(\alpha) = \alpha^4+\alpha^3+1$.

Now since $q(\alpha) = \alpha^4+\alpha^3+1$ is irreducible in $\mathbb{F}_2$ (having no roots in that field), $\frac{\mathbb{F}_2}{q(\alpha)}$ is a field and has 16 elements such as the starting field. Knowing that finite field of the same cardinality are always isomorphic, (iii) is done.

Is how I resolved the exercise correct? Was I correct assuming the typo or it can be solved as the problem is stated? Can in such fields a polynomial has another polynomial as root?

Hope you can help me :)

(I don't have Galois Theory notions)

Edit: after all $q(\alpha)$ is a polynomial with coefficients in $\mathbb{F}_2$.

Shaun
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cento18
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    We don't usually use fraction notation (with long, horizontal fraction lines) for quotient rings like $\Bbb F_2[x]/(x^4+x^3+x^2+x+1)$. I have no real idea why. It just isn't conventional for some reason. – Arthur Aug 24 '22 at 10:45
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    I think the only problem (if you can call it that) is that you have $x$s and $\alpha$s mixed up. It is normal to denote by $\alpha$ the root $x+\langle x^4+x^3+x^2+x+1\rangle$ in the field $K=\Bbb{F}_2[x]/\langle x^4+x^3+x^2+x+1\rangle$. That is, $\alpha$ is the coset of $x$ in the quotient ring $K$ (that happens to be a field). Switching to $\alpha$ has the nice side-effect that $x$ is "freed up" to be used as an indeterminate again. – Jyrki Lahtonen Aug 24 '22 at 11:28
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    (cont'd) Your main calculations are then correct: $\beta=\alpha+1$ (the coset of $x+1$, not $x+1$ itself) has order fifteen, and it satisfies the relation $\beta^4+\beta^3+1=0$, implying that the minimal polynomial of $\beta$ over the prime field $\Bbb{F}_2$ is $q(x)=x^4+x^3+1$. To answer part (iii) my hint is that use the relation that a zero of $x^4+x^3+x^2+x+1$, namely $\alpha$ is related to a zero $\beta$ of $q(x)$ by the relation $\beta=\alpha+1$. I guess you maya want to also check that both $x^4+x^3+x^2+x+1$ as well as $q(x)$ are irreducible over the prime field. – Jyrki Lahtonen Aug 24 '22 at 11:31
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    You can use the uniqueness of fields of cardinality sixteen in parti (iii), but in light of the earlier parts I suspect that your teacher would like to see an explicit isomorphism. That is not clear though. – Jyrki Lahtonen Aug 24 '22 at 11:35
  • I agree with Arthur. That way of writing quotient rings makes my eyes bleed. – Jyrki Lahtonen Aug 24 '22 at 11:36
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    Here is an example run of a very elementary way of finding that minimal polynomial. There are other ways. For example, if you have a list of irreducible quartic polynomials in $\Bbb{F}_q[x]$ that is a rather short list to be tested :-) – Jyrki Lahtonen Aug 24 '22 at 11:47
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    Ok, so if I am gettin it, what I've done is correct (except for some notation), now, just to check if I understood the thing I wanted to find all the roots of $q(x)=x^4+x^3+1$ in $F = \mathbb{F}_2/(x^4+x^3+x^2+x+1)$ ;). Now I know that a root is $\beta$, what about the others? Doing this kind of things I thought that I had to reduce $q(x)$ with knowing that in F $x^4+x^3+x^2+x+1=0$, but that leads to $x^2+x=0$ with roots $0,1$, but not $\beta$, so it doesn't seem to be consistent what I am doing, what am I missing? – cento18 Aug 24 '22 at 13:23
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    It takes a while to get used to different roles of $x$. That is one reason why I use $\alpha$, but naturally that may have confused more at this point. Anyway, the roots of $q(x)$ are simply the roots of $r(x)=x^4+x^3+x^2+x+1$ plus one. Remember that $\alpha$ could have been any root of the latter, yet $\beta=\alpha+1$ is a root of $q(x)$. You can see this by calculating that $r(x+1)=q(x)$ (as we are in characteristic two, we also have $q(x+1)=r(x)$). Anyway, the other roots of $r(x)$ are $\alpha^2$, $\alpha^3$ and $\alpha^4=\alpha^3+\alpha^2+\alpha+1$. – Jyrki Lahtonen Aug 24 '22 at 18:19
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    (cont'd) That follows from $r(x)\mid x^5-1$. The roots of $q(x)$ are thus $\alpha+1$, $\alpha^2+1$, $\alpha^3+1$ and $\alpha^3+\alpha^2+\alpha$. If you want to write them in terms of $\beta$, you can substitute $\alpha=\beta+1$ everywhere. Your choice :-) – Jyrki Lahtonen Aug 24 '22 at 18:21

2 Answers2

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Recall that $\Phi_5(x)=\dfrac {x^5-1}{x-1}=x^4+x^3+x^2+x+1=p(x)$.

So it seems you will want a primitive $5$-th root of unity, $\zeta_5=e^{2\pi i/5}$.

Right, we need to prove that the polynomial is minimal over $\Bbb F_2$. Now it has no root. So the question is can we factor it into two quadratics.

So write $x^4+x^3+x^2+x+1=(x^2+bx+c)(x^2+dx+f)$. We get $(b+d)=1,(f+c+db)=1, fb+dc=1$ and $fc=1$. Thus $f=c=1$ and $db=1\implies d=b=1$, so a contradiction, since $b+d=1$.

You get $\Bbb F_{2^4}$ for the quotient, because it's all polynomials in $\alpha=\zeta_5 $ of degree $\lt4$, with coefficients in $\Bbb F_2$.

Now (following you) let's set $\beta=\alpha+1$. Then let $q(x)=p(x-1)$. Then $q(\beta)=p(\alpha) =0$.

Note $\beta^5=(\alpha +1)^5=\alpha^4+\alpha =\alpha (\alpha ^3+1)\ne0$. Remember we're in characteristic $2$.

Similarly $\beta^3=(\alpha +1)^3=\alpha ^3+\alpha ^2+\alpha +1=\alpha^4\ne0$.

Thus $\beta$ is a primitive element (order $15$).

Let's compute $p(x-1)=(x-1)^4+(x-1)^3+(x-1)^2+(x-1)+1=(x^4+1)+(x^3-x^2+x-1)+x^2+1+x=x^4+x^3+1 \pmod 2$..

This is likely the minimal polynomial of $\beta $. There's no root. Say $q(x)=(x^2+bx+c)(x^2+dx+f)$, then $b+d=1,f+c+bd=0,fb+dc=0, fc=1\implies f=c=1\implies 1=b+d=0$, a contradiction.

So $q(x)$ is the minimal polynomial of $\beta$.

Finally we get, by uniqueness of the field of order $16$, that $\Bbb F_2[x]/(p(x))\cong \Bbb F_2[x]/(q(x))$.

calc ll
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$x^4+x^3+x^2+x+1=\varphi_5(x)$ is a cyclotomic polynomial, hence the splitting field of such polynomial over $\mathbb{F}_2$ is $\mathbb{F}_{2^k}$ where $k$ is the least natural number such that $5\mid (2^k-1)$, i.e. the multiplicative order of $2\pmod{5}$, which is $4$ since $2$ is a generator of $\mathbb{F}_5^*$. You have checked that the multiplicative order of $x+1$ in $\mathbb{F}_2/(\varphi_5(x))$ is $15$. In this context $x$ can be regarded as a primitive fifth root of unity $\zeta$, so the minimal polynomial of $x+1$ over $\mathbb{F}_2$ is $\varphi_5(x-1)$ reduced $\pmod{2}$, i.e. $x^4+x^3+1$. Since they differ by one $\zeta$ and $\zeta+1$ are algebraic numbers with the same degree over $\mathbb{F}_2$, and $\mathbb{F}_2/(\varphi_5(x))$ and $\mathbb{F}_2/(\varphi_5(x-1))$ are both isomorphic to $\mathbb{F}_{16}$.

Jack D'Aurizio
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