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Can someone please let me know if this looks ok? Thanks in advance!

An irreducible degree $5$ polynomial over $\mathbb{Z}_3$is one such that $f(0)\neq0,f(1)\neq0,f(2)\neq0$.

Take e.g. $p(x)=x^5+x^4+x^3+x^2+1$

$p(0)=1$

$p(1)=2$

$p(2)=1$

And, $$x^2,x^2+1, x^2+x+1\ \nmid\ x^5+x^4+x^3+x^2+1.$$

$\therefore x^5+x^4+x^3+x^2+1$ is an irreducible polynomial over $\mathbb{Z}_3$.

Rodrigo de Azevedo
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user551155
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    So you have shown that $p(x)$ has no linear factor, but the majority of the work is showing that it doesn't have any quadratic factor. If this is being graded, you would want to expand on the last part. – pancini Apr 02 '20 at 02:51
  • Maybe this post might help you. This is not same problem, but quite close. https://math.stackexchange.com/questions/2874241/x4x3x2x1-irreducible-over-mathbb-f-7 I would suggest you to have it a look. – InsideOut Apr 02 '20 at 02:52

3 Answers3

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Your problem is to find an irreducible polynomial of degree $5$ over $\mathbb Z_3$. You haven't found one yet, because $$x^5+x^4+x^3+x^2+1=(x^2+x+2)(x^3+2x+2).$$

Note that the irreducible (monic) polynomials of degree $2$ are $x^2+1$ and $x^2+x+2$ and $x^2+2x+2$. You need to check all three of them as possible factors.

Hint. Try $x^5+2x+1$.

bof
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Since it has degree $5$, it would need to have a root ( but it doesn't), or be divisible by an irreducible quadratic.

But, did you check $x^2+2x+2$?

There may be others.

  • $x^5+2x+2$ is irreducible – user551155 Apr 02 '20 at 07:18
  • $p(x)=x^5+2x+2$ is irreducible

    $p(0)=p(1)=p(2)=2$ $x^2, x^2+1, x^2+x+1, x^2+2, x^2+2x+2, x^2+x+2, x^2+2x+1$ do not divide p(x) Since p(x) does not have any roots or quadratic factors, and hence no cubic factors, p(x) is irreducible

     – user551155 Apr 02 '20 at 07:41
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    @user551155 Why bother with the reducible quadratics $x^2$ and $x^2+x+1=(x+2)^2$ and $x^2+2=(x+1)(x+2)$ and $x^2+2x+1=(x+1)^2$ when you already know that there are no factors of degree one? – bof Apr 03 '20 at 23:30
  • good point!!!!! – user551155 Apr 05 '20 at 13:57
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You can also use this proposition from Dummit and Foote, and Sage to get these polynomials.

Proposition 18. The polynomial $x^{p^{n}}-x$ is precisely the product of all the distinct irreducible polynomials in $\mathbb{F}_{p}[x]$ of degree $d$ where $d$ runs through all divisors of $n$.

So $x^{5^{3}}-x=x^{243}-x$ is the product of all the distinct irreducible polynomials in $\mathbb{F}_{3}[x]$ of degree $1$ or $5$. Using Sage: 

F=GF(3) 
x=F['x'].0 
factor(x^243-x)

We get

x * (x + 1) * (x + 2) * (x^5 + 2*x + 1) * (x^5 + 2*x + 2) * (x^5 + x^2 + x + 2) * (x^5 + 2*x^2 + x + 1) * (x^5 + x^3 + x + 1) * (x^5 + x^3 + x + 2) * (x^5 + x^3 + x^2 + 2) * (x^5 + x^3 + x^2 + 2*x + 2) * (x^5 + x^3 + 2*x^2 + 1) * (x^5 + x^3 + 2*x^2 + 2*x + 1) * (x^5 + 2*x^3 + x^2 + 1) * (x^5 + 2*x^3 + x^2 + x + 2) * (x^5 + 2*x^3 + x^2 + 2*x + 2) * (x^5 + 2*x^3 + 2*x^2 + 2) * (x^5 + 2*x^3 + 2*x^2 + x + 1) * (x^5 + 2*x^3 + 2*x^2 + 2*x + 1) * (x^5 + x^4 + 2) * (x^5 + x^4 + x + 2) * (x^5 + x^4 + 2*x + 1) * (x^5 + x^4 + x^2 + 1) * (x^5 + x^4 + x^2 + x + 1) * (x^5 + x^4 + x^2 + 2*x + 2) * (x^5 + x^4 + x^3 + x + 1) * (x^5 + x^4 + x^3 + x^2 + 2*x + 1) * (x^5 + x^4 + x^3 + 2*x^2 + x + 1) * (x^5 + x^4 + x^3 + 2*x^2 + x + 2) * (x^5 + x^4 + 2*x^3 + 1) * (x^5 + x^4 + 2*x^3 + 2*x + 2) * (x^5 + x^4 + 2*x^3 + x^2 + 2) * (x^5 + x^4 + 2*x^3 + x^2 + x + 1) * (x^5 + x^4 + 2*x^3 + 2*x^2 + 1) * (x^5 + x^4 + 2*x^3 + 2*x^2 + 2) * (x^5 + 2*x^4 + 1) * (x^5 + 2*x^4 + x + 1) * (x^5 + 2*x^4 + 2*x + 2) * (x^5 + 2*x^4 + 2*x^2 + 2) * (x^5 + 2*x^4 + 2*x^2 + x + 2) * (x^5 + 2*x^4 + 2*x^2 + 2*x + 1) * (x^5 + 2*x^4 + x^3 + x + 2) * (x^5 + 2*x^4 + x^3 + x^2 + x + 1) * (x^5 + 2*x^4 + x^3 + x^2 + x + 2) * (x^5 + 2*x^4 + x^3 + 2*x^2 + 2*x + 2) * (x^5 + 2*x^4 + 2*x^3 + 2) * (x^5 + 2*x^4 + 2*x^3 + 2*x + 1) * (x^5 + 2*x^4 + 2*x^3 + x^2 + 1) * (x^5 + 2*x^4 + 2*x^3 + x^2 + 2) * (x^5 + 2*x^4 + 2*x^3 + 2*x^2 + 1) * (x^5 + 2*x^4 + 2*x^3 + 2*x^2 + x + 2)
Delong
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