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I am trying to determine if the following polynomials are irreducible in $\mathbb{F_2}[X]$ are irreducible:

$f(X)=X^5+X^2+1$

$g(X)=X^5+X^3+1$

There are no linear factors since $f(0)=f(1)=g(0)=g(1) \neq 0$

However there is another possible factorization: $(X^3+aX^2+bX+c)(X^2+dX+e)$

I am unsure how to verify if $f$ and $g$ can be factored in this way in $\mathbb{F_2}[X]$

Would appreciate your help on this one

Community
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thinker
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  • Work out the product and impose the equality with $f$ or $g$ (the conditions are easily handled since you are on the field $\mathbb F_2$) – guestDiego Jun 02 '16 at 15:31
  • Ok so just expand in terms of $a$ and $b$ and see if they are equal to $f$ and $g$ using the fact that all coefficients are $0$ and $1$? – thinker Jun 02 '16 at 15:32
  • Well, your unknowns are $ a,b,c,d,e$. To start: note for example that you need $ce=1$ which implies $c=e=1$. Then work out the other conditions to have the equality of $(X^3+aX^2+bX+c)(X^2+dX+e)$ with $f$ or $g$. – guestDiego Jun 02 '16 at 15:35
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    Note that $g$ is reducible iff $f$ is reducible (as $g(X)=X^5,f(1/X)$). Thus, it suffices to check the reducibility of only one of them. – Batominovski Jun 02 '16 at 15:36
  • ok thanks. I am unsure about $f(1/X)$, how did you determine this? (what are the coefficents etc.? – thinker Jun 02 '16 at 15:39
  • Here is a relatively extensive discussion about various ways of finding all the irreducible quintics over $\Bbb{F}_2$. – Jyrki Lahtonen Jun 02 '16 at 16:03

2 Answers2

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Hint: find all irreducible polynomials of degree 2 (there are not many) and try to divide $f$ and $g$ by this (those) irreducible polynomial(s) of degree 2

H. Potter
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  • Oh ok, so in this case irreducible polynomials of degree $2$ will not have linear factors.So $f(x)=x^2+ax+b$. $f(0)=b \neq 0 \implies b=1$. $f(1)=1+a+1 \neq 0 \implies 1+a+1=1 \implies a=-1$. So are the two irreducible polynomials: $x^2+x+1$ and $x^2-x+1$.. are these equal ? – thinker Jun 02 '16 at 15:31
  • Those are the same polynomial because $-1 \mod 2 = 1$. – Steven Harding Jun 02 '16 at 15:35
  • Note: there are only an irreducible polynomial of degre 2 in $F_2[X]$ is the cyclotomic polynomial $ \Phi_3(X)$. – m.idaya Jun 02 '16 at 15:39
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From H. Potter's hint, it follows that $f(X)$ is reducible iff it is divisible by $X^2+X+1$. Now, in the same spirit, $g(X)$ is reducible iff it is divisible by $X^2+X+1$ as well. As $f(X)$ is reducible iff $g(X)$ is reducible, and $f(X)+g(X)=X^3+X^2=X^2(X+1)$ is not divisible by $X^2+X+1$, we conclude that $f(X)$ and $g(X)$ are irreducible over $\mathbb{F}_2$.

Batominovski
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