1
$\begingroup$

I have the polynomials $p(X)=X^{5}+P_4X^{4}+P_3X^{3}+P_2X^{2}+P_1X+P_0 \in\mathbb{Z_2[x]}$
I need to determine all $p(x)$ that can be factored into irreducible polynomials of degree three and two.

The catch is that I cannot use most of the techniques usually described to solve this kind of problem. What would be the steps I need to take in order to solve this problem?

For the record I am using Gallian: contemporary abstract algebra as a book and can only use theorems 17.1 and 17.5(plus corollary 1 and 2) from that specific chapter. For reference I wrote them down below:

17.1: Let $F$ be a field. If $f(x) [ F[x]$ and deg $f(x)$ is 2 or 3, then $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in $F$.

17.5: Let $F$ be a field and let $p(x) \in F[x]$. Then $<p(x)>$ is a maximal ideal in $F[x]$ if and only if $p(x)$ is irreducible over $F$.

Corollary 1: Let $F$ be a field and $p(x)$ be an irreducible polynomial over $F$. Then $F[x]/<p(x)>$ is a field.

Corollary 2: Let F be a field and let $p(x), a(x), b(x) \in F[x]$. If $p(x)$ is irreducible over $F$ and $p(x) | a(x)b(x)$, then $p(x) | > a(x)$ or $p(x) | b(x)$.

$\endgroup$
  • $\begingroup$ If it has an irreducible factor of degree 4, then it also has a linear factor and so has a zero. $\endgroup$ – Morgan Rodgers Mar 24 '16 at 18:03
  • $\begingroup$ I am not sure I follow. $\endgroup$ – MSB Mar 24 '16 at 18:07
  • $\begingroup$ A polynomial of degree 5 is either irreducible, splits into irreducible factors of degree 2 and 3, or else has a linear factor (irreducible factors of degree 4 and 1, 3 and 1 and 1, or 2 and 2 and 1). It is easy to check for linear factors by testing if $p(1) = 0$ or $p(0)=0$. $\endgroup$ – Morgan Rodgers Mar 24 '16 at 19:58
3
$\begingroup$

Use the no roots criterion to determine all irreducible quadratics and cubics. Quadratics are very simple, there is only $x^2+x+1$. Cubics are a little more work. But not much. The cubic has to have shape $x^3+ax^2+bx+1$ where there is an odd number of $1$'s. Once you have your list of cubics, multiply each by $x^2+x+1$.

$\endgroup$
  • $\begingroup$ Thanks, this set me on the right path, will check it as an answer when I have solved the problem. $\endgroup$ – MSB Mar 24 '16 at 18:26
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Mar 24 '16 at 18:33
  • $\begingroup$ so for quadratics I have: $p(x)= x^{2}+x+1$ and $p(x) = 1$, removing the duplicates at the cubics I also have: $p(x)= x^{3}+x^{2}+1$ and $p(x)= x^{3}+x+1$ So all I need to do is multiply the cubic with the quadratic(that isn't 1) to get all of the polynomials of degree 5 that factor back in to them right? $\endgroup$ – MSB Mar 24 '16 at 18:38
  • $\begingroup$ $p(x)=1$ is not an irreducible quadratic, indeed not a quadratic at all. The two answers to your problem are $(x^3+x^2+1)(x^2+x+1)$ and $(x^3+x+1)(x^2+x+1)$. The first simplifies to $x^5+x+1$. You may be able to simplify the second without multiplying. $\endgroup$ – André Nicolas Mar 24 '16 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.