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Note: this is from a paper on Galois Theory, so I believe the technique will come from Galois... Possibly.

I am going to reduce it to the monic case: $x^{21}+a_{20}x^{20}+... a_0$ where all $a_i$ belong to $\{0,1\}$ . I was thinking of trying to generate a polynomial that would satisfy Eisenstein's criterion but it cannot have any prime coefficients.

The easiest starting points are $x^{21}+x$, $x^{21}+x+1$, $x^{21}+x^2$ ...etc

I do not know how to check that these are irreducible without Eisenstein's criterion.. Is it true that since the highest power is odd, if it were irreducible, then it would have a linear factor. So can we just plug in values of $0$ and $1$ until we find a polynomial that gives us a non -zero answer?

So for example, can we say that $f(x)=x^{21}+x+1$ is irreducible, since $f(0)=f(1)=1 \neq 0$?

Is it possible to use Galois Theory to answer these questions?

user26857
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thinker
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  • Checking for roots only settles whether it has first degree factors. You still need to check whether there are some factors of degrees $2,3,4,\ldots,10$. At that point you have done enough. Do you see why? – Jyrki Lahtonen Apr 08 '16 at 12:37
  • In other words, this is NOT easy in general without a CAS with suitable algorithms in place. Too bad you didn't ask for degree 20. – Jyrki Lahtonen Apr 08 '16 at 12:38
  • @JyrkiLahtonen ah yes, for example it could be composed of $2$ irreducible polynomials any degree so long as their powers add to $21$, right? So a proof by exhaustion seems a little out of the question time-wise , so is their a clever way to do these? – thinker Apr 08 '16 at 12:40
  • Ok. It wasn't too difficult this time either. More or less the same trick works with $21$ and $20$. Only difference being that, unlike the fifth, the seventh cyclotomic polynomial is not irreducible modulo 2. – Jyrki Lahtonen Apr 08 '16 at 12:58

4 Answers4

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The following trick springs to mind.

We know that the minimal polynomials of seventh roots of unity are the cubic irreducibles $$ p_1(x)=x^3+x+1\qquad\text{and}\qquad p_2(x)=x^3+x^2+1. $$

Now, because $2$ is of order $21$ modulo $49$ (leaving it to you to check that), we can deduce (Galois theory is all we need here) that:

  • The 49th roots of unity generate the field $\Bbb{F}_{2^{21}}$.
  • Hence their minimal polynomials are irreducible of degree $21$.
  • Hence their minimal polynomials are (drums, please) $$ p_1(x^7)=x^{21}+x^7+1\qquad\text{and}\qquad p_2(x^7)=x^{21}+x^{14}+1. $$
Jyrki Lahtonen
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  • Very nice indeed ! – Dietrich Burde Apr 08 '16 at 12:45
  • I don't get the $2^{21} \pmod {49}$ thing. $p_1(x) = x^3+x+1$ is irreducible in $\mathbb{F}_2[x]$ hence $p_1(x^7)$ is irreducible too ? hence $\mathbb{F}_2[x] / (p_1(x^7))$ is the field with $2^{21}$ elements, hence what ? – reuns Apr 08 '16 at 12:51
  • @user1952009: The flow of the logic is: $2$ has order $21$ modulo $49$ => a $49$th root of unity $\alpha$ generates the extension field $\Bbb{F}_{2^{21}}$ => its minimal polynomial has degree $21$. BUT OTOH $\alpha^7$ is a seventh root of unity, so $\alpha^7$ is a zero of one of those cubics => $p_1(x^7)$ and $p_2(x^7)$ are irreducible. – Jyrki Lahtonen Apr 08 '16 at 12:53
  • I have covered the necessary theory so many times here that I have lost count. Here is one of the most recently upvoted occasions. Nothing deep. Just using the fact that the Galois group is generated by the Frobenius mapping. – Jyrki Lahtonen Apr 08 '16 at 19:33
  • @JyrkiLahtonen thanks for the answer. Can I just ask a question, about the general case: finding irred. polynomial of degree $d$ in $\mathbb{F_n}$. We need to find an $x$ such that $n$ is of order $d$ mod $x$.. since (I believe) there are multiple $x$ that satisfy this, should we choose a square , like $49$ in this case? – amiz9 Jun 02 '16 at 14:33
  • @amiz9: It depends. It is also essential that we can factor the (characteristic zero) cyclotomic polynomial easily enough. In the two cases handled here I used the square of a prime $p$. It then often happens that the order of $n$ modulo $p^2$ is $p$ times the order of $n$ modulo $p$. When that happens we can just plug in $x^p$ into the minimal polynomial of a root of unity of order $p$, and get the minimal polynomial of a root of unity of order $p^2$. In other words, this is a bit ad hoc, and anything but guaranteed to work in all cases. – Jyrki Lahtonen Jun 02 '16 at 15:43
  • @JyrkiLahtonen I tried an example: "Find the irreducible polynomial of degree $20$ in $\mathbb{F_2}[X]$." I found that $2$ is order $20$ mod $25$. So we need to consider the minimal polynomials of $\zeta_5$ and then plug in $x^5$. The minimal polynomial for $\zeta_5$ is $\phi_5[X]$, right? Why did you have two minimal polynomials in the case of $\zeta_7$? Thank you so much – amiz9 Jun 02 '16 at 17:46
  • @amiz9: The point of this answer is that the minimal polynomial of $\zeta_7$ is not $\phi_7(x)$. Modulo two $\phi_7(x)$ is a product of two cubics. The minimal polynomial of $\zeta_n$ over $\Bbb{F}_q$ has degree $r>0$, where $r$ is the smallest integer with the property $q^r\equiv1\pmod n$. With $n=7$ we have $2^3\equiv1\pmod 7$, so $\zeta_7$ has a cubic minimal polynomial. Or rather, three of the primitive roots of order $7$ share one cubic factor of $\phi_7(x)$ as a minimal polynomial, and the remaining primitive roots of order $7$ share the other. – Jyrki Lahtonen Jun 02 '16 at 20:03
  • oh so we take $\phi_n(X)$ and consider it modulo $n$? – amiz9 Jun 02 '16 at 20:19
  • @amiz9 We consider over the field we are interested in. This time I looked at $\phi_7(x)$ modulo two. In the case of degree 20, it was easier, because $\phi_5(x)$ is irreducible modulo two. – Jyrki Lahtonen Jun 02 '16 at 20:38
  • ok perfect. thanks, this helped a lot :) – amiz9 Jun 02 '16 at 22:06
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Using the Berlekamp algorithm it follows that $$ x^{21}+x+1=(x^{14} + x^{12} + x^7 + x^6 + x^4 + x^3 + 1)(x^7 + x^5 + x^3 + x + 1) $$ over $\mathbb{F}_2$, hence it is reducible. However, $$ g(x)=x^{21}+x^2+1 $$ is irreducible over $\mathbb{F}_2$.

Dietrich Burde
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  • Thank you, nice example. Yet how do you know $;g;$ is irreducible? – DonAntonio Apr 08 '16 at 12:43
  • I have not learnt this theory in class... Is this the only short way to go around this question – thinker Apr 08 '16 at 12:43
  • @Joanpemo Berlekamp says so. I suppose we can find a proof, too. – Dietrich Burde Apr 08 '16 at 12:44
  • @thinker There is another shortcut. People in coding theory have exhaustive lists of all monic irreducible polynomials in $\mathbb{F}_2[x]$ up to some degree. – Dietrich Burde Apr 08 '16 at 12:48
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    You can ask Wolfram alpha to: factor x^21 + x^2 + 1 mod 2 .(https://www.wolframalpha.com/input/?i=factor+x^21+%2B+x^2+%2B+1+mod+2) So you could have answered your question by guessing until you succeeded. (Not very enlightening.) – Ethan Bolker Apr 08 '16 at 12:51
  • @EthanBolker if only wolfram was available in the exam ;) – thinker Apr 08 '16 at 13:59
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About your very last question (with fields extensions theory, in fact):

we know $\;\Bbb F_{2^{21}}\cong\Bbb F_2[x]/{p(x)}\;$ , with $\;p(x)\in\Bbb F_2[x]\;$ irreducible. But we also know that $\;\Bbb F_{2^{21}}\;$ is the

splitting field over its prime field of $\;x^{2^{21}}-x\in\Bbb F_2[x]\;$ , so it must be that $\;p(x)\;$ is among the

divisors of this last polynomial.

Anyway, it doesn't look like an enjoyable task.

DonAntonio
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    This is a small question in an old exam paper on Galois Theory... So my suspician is there is some Galois Theory way to solve it... Since oractically we can soe d no longer than 5 minutes on these questions in the exam – thinker Apr 08 '16 at 12:41
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We may take two irreducible polynomials over $\mathbb{F}_2$: $$ p_\alpha(x)=x^3+x+1,\qquad p_\beta(x)=x^7+x+1 $$ and by assuming that $\alpha$ is a root of $p_\alpha$, $\beta$ is a root of $p_\beta$, just compute the minimal polynomial, over $\mathbb{F}_2$, of $\alpha+\beta$. Since $[\mathbb{F}_2(\alpha):\mathbb{F}_2]=3,\, [\mathbb{F}_2(\beta):\mathbb{F}_2]=7$ and $\gcd(3,7)=1$, $[\mathbb{F}_2(\alpha+\beta):\mathbb{F}_2]=21$ as wanted.

We just have to represent $(\alpha+\beta)^k$, for $k\in\{0,1,2,\ldots,20,21\}$, as a linear combination of $\alpha^i \beta^j$, with $i\in\{0,1,2\}$ and $j\in\{0,1,\ldots,6\}$, then perform Gaussian elimination.

Jack D'Aurizio
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