# Finding an irreducible quartic in $\mathbb{Z}_2[x]$

Find an irreducible polynomial $p(x) \in \mathbb{Z}_2[x]$ with degree 4.

Thank you!

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What have you tried? –  Chris Eagle Jun 27 '12 at 16:48
I have tried to test several polynomials (with degree 4) if they were irreducible. But i hoped that there is an other (structured) way to solve this question. :) –  user1255553 Jun 27 '12 at 16:49
I'm assuming you mean the integers modulo $2$, rather than the $2$-adics. Is this correct? –  Arturo Magidin Jun 27 '12 at 16:54

An irreducible biquadratic cannot have a root, so the constant term must be $1$, and the number of nonzero coefficients must be odd. So we would have $$x^4 + ax^3 + bx^2 + cx + 1$$ with $a+b+c=1$ (addition is modulo $2$, of course, since $a,b,c\in\mathbb{Z}_2$).

Now, we also must avoid the case in which it is a product of two irreducible quadratics. What are the irreducible quadratics in $\mathbb{Z}_2[x]$? We have $x^2+x+1$... and that's it, because the other three quadratics, $x^2$, $x^2+1$, and $x^2+x$, all have roots.

So you need to make sure that $x^4+ax^3+bx^2+cx+1\neq (x^2+x+1)^2$.

So... can you find a polynomial that satisfies those two conditions ($a+b+c=1$ and $x^4+ax^3+bx^2+cx+1\neq (x^2+x+1)^2$) ?

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A quadratic has degree $2$. A polynomial of degree $4$ is called a quartic. –  Chris Eagle Jun 27 '12 at 16:51
@Chris: Actually, I meant to write "biquadratic" (which is another valid name for degree 4...) and messed up twice. Sigh; thanks. –  Arturo Magidin Jun 27 '12 at 16:52
I would guess, x^4 + 1. But I do not understand why a+b+c can only be 1. In the other quadratic you use (x^2 + x + 1), a+b+c=3? –  user1255553 Jun 27 '12 at 16:57
@user1255553: I'm doing addition modulo $2$, of course, since you are working in $\mathbb{Z}_2$. So $3$ is the same thing s $1$. –  Arturo Magidin Jun 27 '12 at 16:57
@user1255553: Don't guess. Check! –  Arturo Magidin Jun 27 '12 at 16:58

Hint $\rm\:f(x) = (x^5\!-\!1)/(x\!-\!1) = x^4\! +\! x^3\! +\! x^2\! +\! x\! +\! 1\:$ has no linear factors by $\rm\:f(0) = 1 = f(1).\:$ And $\rm\:f\:$ has no quadratic factors: the only irreducible quadratic is $\rm\:g(x) = x^2\!+\!x\!+\!1,\:$ but $\rm\:g\:|\:f\:$ $\Rightarrow$ $\rm\:(x\!-\!1)g\:|\:(x\!-\!1)f,\:$ i.e. $\rm\:x^3\!-\!1\:|\:x^5\!-\!1\:$ $\Rightarrow$ $\rm\:x^3\!-\!1\:|\:x^2\!-\!1 = (x^5\!-\!1)-x^2(x^3\!-\!1),\,$ contra degree.

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Can we also argue that since $5$ is prime we know that this polynimial is irreducible over $\mathbb{Q}$ and since it is fixed it is also irreducible over $\mathbb{Z}$, if it was reducible in $\mathbb{Z}/2\mathbb{Z}$ that would of been a contradiction ? –  Belgi Jun 27 '12 at 19:25
@Belgi Note that $\rm\:x^2+2\:$ is irreducible over $\rm\:\mathbb Q\:$ but reducible over $\:\mathbb Z/2.$ –  Bill Dubuque Jun 27 '12 at 19:30
Thanks, I was unsure about which direction is the correct one. –  Belgi Jun 27 '12 at 19:33