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Q: Find an irreducible quadratic polynomial with coefficients in GF(16), with at least one non-binary coefficient.

A: We can’t because all quadratic polynomial with coefficients in GF(16) has a coefficient of either 0 or 1.

My prof said there is an equation for that so there exists an irreducible quadratic polynomial with at lest one non-binary coefficient. Can someone please give me a hint?

hanamontana
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    There are only 256 monic quadratic polynomials in GF(16). There are only 136 distinct products of two linear factors. Eliminate these and the rest of the 256 are irreducible. – Somos Apr 12 '23 at 12:17
  • @Somos thank you, but does any of those remaining polynomials have non-binary coefficients (not 0 and 1)? – hanamontana Apr 12 '23 at 13:22
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    You seem to be confused about what $GF(16)$ is like. Polynomials over $GF(2)$ always have binary coefficients. Polynomials over $GF(16)$ might not. – MJD Apr 12 '23 at 17:14
  • If you have covered the basic properties of the trace function, then my answer gives you a roadmap. If you haven't then it may be kinda cruel. But you can, instead of using the trace, simply list the eight values the polynomial $x^2+x$ takes in the field $GF(16)$. If $\gamma$ is any element of $GF(16)$ not among those values, then $x^2+x+\gamma$ has no zeros in $GF(16)$ and is thus irreducible. Both $0$ and $1$ are values of $x^2+x$ in $GF(16)$, so the polynomial you end up with is non-binary. – Jyrki Lahtonen Apr 12 '23 at 17:14
  • I agree with MJD. There is some confusion. All the binary quadratic polynomials have their zeros in $GF(4)$, a subfield of $GF(16)$. There the binary quadratic polynomials always factor completely over $GF(16)$. – Jyrki Lahtonen Apr 12 '23 at 17:16
  • @Jyrki Lahtonen:- Be kind, please. dear sir, and explain to me what does mean $GF(16)$. – Piquito May 09 '23 at 15:27
  • @Piquito It is just an alternative notation for $\Bbb{F}{16}$. The (up to isomorphism) unique field with sixteen elements. This notation can be seen in older texts. Many authors in coding theory/crypto still use it. I switched to $\Bbb{F}{16}$ myself only at relatively old age on this site. Somewhere between 2012-14 :-) – Jyrki Lahtonen May 09 '23 at 16:38
  • @Jyrki Lahtonen.- Thanks you very much. So we have place to abstract monsters (If i am not wrong, then nobody can explain in understandable practical way what an irrational of degree $\ge2$ over $\mathbb F_2$ is). – Piquito May 09 '23 at 17:20
  • @Piquito A third root of unity $\omega$ will surely do. Its minimal polynomial is $x^2+x+1$, and that is irreducible over $\Bbb{F}_2$ (or $GF(2)$). It's not clear whether you want to call it "irrational" though. – Jyrki Lahtonen May 09 '23 at 18:42
  • @Jyrki Lahtonen: Maybe it is what french mathematicians call "un abus de langage" but it is all element not in $\mathbb F_p$ but in an extension of it. In your case of $\omega$ certainly it is not an element of an extension of any $\mathbb F_p$ but in an extension of $\mathbb Q$ – Piquito May 09 '23 at 20:35
  • @Piquito Abuse of language? Guilty as charged! But still (one of many) useful way(s) of thinking about $\Bbb{F}_4$. It spawns the constructions $\Bbb{F}_4=\Bbb{F}_2[x]/\langle x^2+x+1\rangle$ with $\omega$ identified with the coset $x+\langle x^2+x+1\rangle$. As well as the construction as a quotient ring of Eisensteinian integers reduced modulo two: $$\Bbb{F}_4=\Bbb{Z}[\omega]/\langle 2\rangle.$$ – Jyrki Lahtonen May 10 '23 at 04:25
  • (cont'd) The trouble with the use of algebraic integers here is that there is no one-size-fits-all construction of all the finite fields relying on such trick. For example, to get $\Bbb{F}_8$ this way you may want to do $\Bbb{Z}[\zeta_7]/\mathfrak{p}$, where $\mathfrak{p}$ is either one of the two prime ideals of $\Bbb{Z}[\zeta_7]$ lying above the rational prime $p=2$. Or any of the other ways. I don't remember how strong your background in algebraic number theory is, but you can do it that way, too. – Jyrki Lahtonen May 10 '23 at 04:29
  • @Jyrki Lahtonen.- The only thing I want to say is that the elements of $\mathbb F_p$ are not numbers and that the "nature" of elements of extensions of this field are almost unimaginable (Si yo pudiera escribirle en Español sería mucho más convincente, mi Inglés es muy primario). Regards and very respect for you really. – Piquito May 11 '23 at 00:51
  • @Jyrki Lahtonen. I know enough of algebraic number theory (my french doctoral tesis was on elliptic curves and I have written the book "Invitación al estudio de la aritmética de curvas elípticas"). I am an old man aged $84$ and I have known several very great mathematicians , L.Schwartz, A, Grothendieck , among others. By today i can do few in maths but i did enough in the past. – Piquito May 11 '23 at 01:03

2 Answers2

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A quick solution using the properties of the trace map is the following. Let $\alpha\in GF(16)$ be a zero of the polynomial $p(x):=x^4+x+1\in GF(2)[x]$ (hopefully known to be irreducible. Then $1/\alpha$ is a zero of the reciprocal polynomial $x^4p(1/x)=x^4+x^3+1$. It follows that the trace of $1/\alpha$ is equal to $1$.

Therefore the quadratic polynomial $$ m(x)=x^2+x+1/\alpha\in GF(16)[x] $$ fits the bill:

  • $1/\alpha$ is not an element of $GF(2)$, so $m(x)$ is not binary.
  • If $\beta\in GF(16)$ were a zero of $m(x)$, then we run into the following contradiction: $tr(\beta)=tr(\beta^2)$, so $tr(\beta+\beta^2)=0$ irrespective of whether $tr(\beta)$ is $0$ or $1$. Therefore $$tr(m(\beta))=tr(\beta^2)+tr(\beta)+tr(\frac1{\alpha})=tr(1/\alpha)=1,$$ implying that $m(\beta)$ cannot be zero. It follows that $m(x)$ is irreducible over $GF(16)$.
Jyrki Lahtonen
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Let's try to find a polynomial without roots of the form $x^2 + c x + 1$ ( note that $x^2 + c x + d$ has roots if and only if $x^2 + \frac{c}{\sqrt{d}} + 1$ does). So now we need some $c$ that is not of the form $t + \frac{1}{t}$ with $t$ in the field. We'll be using the generator of the field provided by Jyrki. So let $\alpha \in GF(2^4)$ be a root of $x^4 + x + 1 = 0$. We have

$\alpha^4 + \alpha = 1$ so multiplying by $\alpha^5$ we get $\alpha^9 + \alpha^6 = \alpha^5$. Now raise to the square and get $\alpha^3 + \alpha^{12} = \alpha^{10}$. (raising again to the square returns to the start).

Now multiply $\alpha + 1 = \alpha^4$ by $\alpha^7$ and get $\alpha^8 + \alpha^7 = \alpha^{11}$. We can now take some squares: $\alpha^1 + \alpha^{14} = \alpha^{7}$, $\alpha^{2} + \alpha^{13} = \alpha^{14}$, $\alpha^{4} + \alpha^{11} = \alpha^{13}$. We still have to check two values: $\alpha^{0} + \alpha^{0} = 0$ and $\alpha^{5} + \alpha^{10}$. The last one is invariant under raising to the square, so it must be $1 = \alpha^{0}$. So here is a little table \begin{eqnarray} \alpha^{0} + \alpha ^{0} & =& 0\\ \alpha^{1} + \alpha^{14} & = & \alpha^7\\ \alpha^{2} + \alpha^{13} &=& \alpha^{14}\\ \alpha^{3} + \alpha^{12} & = & \alpha^{10}\\ \alpha^{4} + \alpha^{11}& = & \alpha^{13} \\ \alpha^{5} + \alpha^{10} & = & \alpha^{0}\\ \alpha^{6} + \alpha^{9} & = & \alpha^{5} \\ \alpha^{7} + \alpha^{8} & = & \alpha^{11} \end{eqnarray}

We can choose for instance $\alpha^{1}$ which is not of the form $t + \frac{1}{t}$, $t \in GF(2^4)$, so $x^2 + \alpha x + 1$ is irreducible.

orangeskid
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