0

I don't quite understand how I know if a polynomial is irreducible, and I would like to know if someone could help me understand that.

For example, if I have the following polynomial: $x^5+x^2+x^3+1$, is this polynomial irreducible in $\mathbb{Z}_{2}[x]$? I know that $\mathbb Z_{2}[x]=\{x^2, x^2+1, x^2+x, x^2+x+1\}$ but I will not continue from there to answer that question

Thomas Andrews
  • 177,126
Taylor Pearson
  • 3
  • 1
    If you can find a root in the field, it is not. As $x=1$ is a root of that polynomial, you can write it as $(x-1)$ times the quotient. – David P Mar 02 '21 at 00:32
  • 1
    $\mathbb Z[x]$ has infinitely many elements. – Thomas Andrews Mar 02 '21 at 00:32
  • @DavidP But what about the roots in the field does not only apply to polynomials of degree 2 and 3? – Taylor Pearson Mar 02 '21 at 00:34
  • $x^5+x^2+x^3+1 = (x^3+1)(x^2+1)$ so its roots are fairly easy to state – Henry Mar 02 '21 at 00:34
  • 1
    @TaylorPearson Reducible does not imply roots, but roots imply reducibility. – David P Mar 02 '21 at 00:35
  • First you try to find linear factors. If there are, you factor them out and obtain a simpler version. At this point, if you have only linear, irreducible quadratic and irreducible cubic factors you are done. If this is not the case you go hunting for quadratic factors. And so on.... – Bumblebee Mar 02 '21 at 00:35
  • It’s true that a polynomial can be reducible even if it has no root. But if it has a root, it is reducible. – Thomas Andrews Mar 02 '21 at 00:35
  • At the proposed level of generality, this is genuinely a difficult question. I don't think anybody can "see" that a randomly given polynomial of high degree is or is not reducible over $\mathbb{Z}_2$ or another finite field. Or if they can, I'd love to hear about it. – leslie townes Mar 02 '21 at 00:38
  • @ThomasAndrews $\mathbb{Z[x]}$ has infinite elements, but $\mathbb{Z_{2}}[X]$ are polynomials of the form $ax^2+bx+c$ such that $a, b, c$ are in $\mathbb{Z_{2}}=${$0,1$} so $\mathbb{Z_{2}}[X]$ is not it has infinite elements but the polynomials indicated in the statement – Taylor Pearson Mar 02 '21 at 00:38
  • @TaylorPearson If that is what your teacher or book taught you, it is very non-standard notation. Usuallly, $\mathbb Z_2$ is usually the integers modulo $2$, and $\mathbb Z_2[x]$ is all polynomials in $x$ with coefficients in \mathbb Z_2.$ The $2$ has nothing to do with degree in the usual notation. – Thomas Andrews Mar 02 '21 at 00:42
  • @ThomasAndrews ok, so I only have to find roots in the fields and so I get that a polynomial is reducible? – Taylor Pearson Mar 02 '21 at 00:43
  • @leslietownes true at the proposed level of generality, this is really a tough question. but it is not a high degree polynomial given at random it is an exercise that they put me and I try to understand the issue to solve it – Taylor Pearson Mar 02 '21 at 00:45
  • @ThomasAndrews This is how they taught me but what you say are the polynomials indicated in the statement – Taylor Pearson Mar 02 '21 at 00:46
  • @leslietownes There is a decidable way to determine if a polynomial is prime in $\mathbb Z_p[x].$ If $q(x)$ has degree $n$ the $q(x)$ is irreducible iff $q(x)\mid x^{p^n}-x$ and $\gcd(x^{p^k}_x,q(x))=1$ for all $k<n.$ You can restrict further to $k\mid n.$ – Thomas Andrews Mar 02 '21 at 00:49
  • @ThomasAndrews cool! I did not know that. Many thanks – leslie townes Mar 02 '21 at 00:53
  • 1
    @leslietownes See https://math.stackexchange.com/a/1343849/207316 for more info on Rabin's Irreducibility Test. – PM 2Ring Mar 02 '21 at 12:01

2 Answers2

0

In general, showing that a polynomial is irreducible is hard. There are a few irreducibility tests that can be used, but there are few general methods that don't require lots of computation or case-checking. One special case that you might have seen is that a cubic is irreducible over a field $k$ if and only if it has no roots in $k$; this is because any nontrivial factorization of a cubic must involve a term of degree at most $1$, and thus give a root. You can generalize this to higher degrees -- for example, for a degree $5$ polynomial, you only need to check all possible linear and quadratic factors -- but if your polynomial or field is very large this can get tricky.

However, showing that a polynomial is reducible is "easy" (easy in the sense that the data needed is short, not that it's necessarily easy to find). All you need to do is find a nontrivial factorization of your polynomial, and then you're done.

In particular, for the degree-$5$ polynomial you've written down, $x=1$ is a root in $\mathbb Z/2\mathbb Z$, so you can write it as $(x-1)$ times a degree-$4$ polynomial. This is a nontrivial factorization, and so the polynomial is reducible.

Carl Schildkraut
  • 32,931
0

It is easy to see that the set of irreducible polynomials of degree $2$ over $\Bbb F_2$ also contains $x+1$ and $x^2+x+1$. Dividing $x^5+x^3+x^2+1$ by $x+1$, or $x^2+x+1$ we see that $$ x^5+x^3+x^2+1=(x^2 + x + 1)(x + 1)^3. $$

Dietrich Burde
  • 130,978