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I am preparing for an exam and I'm looking for a systematic way to find all irreducible polynomials. I am given the following problem > Find all irreducible polynomials of degree 3 with coefficients over Z2. I know that a polynomial is reducible if it can be represented as a product of two or more polynomials, but I can't find an efficient algorithm which can help me to list all the irreducible polynomials. Any suggestions?

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    $\begingroup$ There are only eight polynomials of degree 3 over the field of two elements, and if a polynomial of degree 3 is reducible then it has a root, and the only possible roots are zero and one. So brute force is a pretty efficient algorithm in this case. And no one is going to ask you to find all irreducible polynomials of degree 100 with coefficients in the field of 65,537 elements, so don't worry about it. $\endgroup$ – Gerry Myerson Feb 6 '17 at 11:39
  • $\begingroup$ I have another question (maybe a stupid one) : How do you calculate that there are 8 polynomials? Is it because of the fact that we can represent the polynomial of degree 3 as : ax^3 + bx^2 +c*x + d and we have that a can be 1 , b can be {1,2},c {1,2} and d {1,2} so we would have 1*2*2*2 = 8 possible combinations. What confuses me is the "-" sign. Do we skip the polynomials with "-" coefficients because in the Z 2 field we calculate with mod 2 ? I mean for example -1 would be actually 1 (because -1 mod 2 = 1) . Did I grasp the idea? $\endgroup$ – Teo Feb 6 '17 at 12:00
  • $\begingroup$ Yes to all of your questions. $\endgroup$ – Gerry Myerson Feb 6 '17 at 12:01
  • $\begingroup$ You can search for irreducible polynomials $f(x) \in \mathbb{F}_q$ by checking all the finite possible factors, then $f(x)$ will be also irreducible over $\mathbb{Q}$. Also, there is an easy combinatorial proof that for every $k$, there exists $\sum_{d | k} \frac{q^d}{d}$ irreducible polynomials of degree $k$ over $\mathbb{F}_q$ $\endgroup$ – reuns Feb 6 '17 at 12:16
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If $f(x)\in \mathbb Z_2[x]$ and has degree $3$,then $f(x)=x^3+ax^2+bx+c$ and we have the possible states:

$a=0,b=0,c=0 \Rightarrow f(x)=x^3$

$a=1,b=0,c=0 \Rightarrow f(x)=x^3+x^2$

$a=1,b=1,c=0 \Rightarrow f(x)=x^3+x^2+x$

$a=1,b=1,c=1 \Rightarrow f(x)=x^3+x^2+x+1$

$a=1,b=0,c=1 \Rightarrow f(x)=x^3+x^2+1$

$a=0,b=1,c=0 \Rightarrow f(x)=x^3+x$

$a=0,b=1,c=1 \Rightarrow f(x)=x^3+x+1$

$a=0,b=0,c=1 \Rightarrow f(x)=x^3+1$

$f(x)$ is irreducible polynomial in $\mathbb Z_2[x]$ if $0$ or $1$ isnot roots, then $f(x)=x^3+x^2+1$ and $f(x)=x^3+x+1$ are irreducible polynomials in $\mathbb Z_2[x]$.

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