Show irreducible polynomial by factorization in $\mathbb{F}_2[X]$ and $\mathbb{F}_3[X]$

Question

I found a problem, I don't really know how to solve, although it should be something very easy, since it is stuff of Algebra I.

Let $f= 29X^5−13X^4−44X^3+ 18X^2+ 35X+ 10\in\mathbb{Z}[X]$.

1) Decompose $f$ into irreducible factors in $\mathbb{F}_2[X]$ and $\mathbb{F}_3[X]$.

2) Conclude: $f$ is irreducible over $\mathbb{Z}$ and $\mathbb{Q}$.

1) Is clear: $\bar{f}=X^5+X^4+ X\in\mathbb{F}_2[X]$ and $\bar{f}= 2X^5−X^4−2X^3+ 2X+ 1\in\mathbb{F}_3[X]$.

So $\bar{f}=X^5+X^4+ X=X(X^4+X^3+1)\in\mathbb{F}_2[X]$ and $X^4+X^3+1$ is irreducible, because $0+0+1=1=1+1+1$ in $\mathbb{F}_2$.

Edit: because no irreducible polynominal with degree $\le 2$ divides the latter polynominal.

Furthermore, $\bar{f}=2X^5−X^4−2X^3+ 2X+ 1=2X^5+2X^4+X^3+ 2X+ 1\in\mathbb{F}_3[X]$ is irreducible, because $f(0)=1$, $f(1)=8=2,f(2)=109=1$.

Edit: $\bar{f}=2X^5−X^4−2X^3+ 2X+ 1=2X^5+2X^4+X^3+ 2X+ 1=2(X^2+1)(X^3+X^2+X+2)\in\mathbb{F}_3[X]$ with irreducible factors. Thank you Nicolas!

2) It's clear: If $f$ is irreducible over $\mathbb{Z}$, then it is irreducible over $\mathbb{Q}$ by a basic theorem of algebra, which gives this result (in deed equivalence) for factorial rings (UFD) and their field of fractions.

But I don't know how to conclude irreducibility by using 1). Is it something about the degree of the irreducible polynomials?

Thank you for your answers!

Answer 1

We have $\bar{f}=X(X^4+X^3+1)\in\mathbb{F}_2[X]$ and $\bar f=2(X^2+1)(X^3+X^2+X+2)\in\mathbb F_3[X]$ (credit to Macavity) factorizations into irreducibles.
If $f$ is reducible over $\mathbb Z$ then $f=gh$ with $g,h\in\mathbb Z[X]$. We have $\deg g+\deg h=5$, $1\le\deg g<5$ and $1\le\deg h<5$. Moreover, the possible leading coefficients of $g$ and $h$ are $1$ and $29$, respectively $-1$ and $-29$, and all of them are non-zero modulo $2$, respectively $3$.
Now reduce $f=gh$ modulo $2$ and get $\bar f=\bar g\bar h$. Furthermore, $\deg\bar g=\deg g$ and $\deg\bar h=\deg h$, so $\deg g=1$ and $\deg h=4$ (or vice versa).
Then reduce $f=gh$ modulo $3$ and similarly get $\deg g=2$ and $\deg h=3$ (or vice versa).
Thus we reached a contradiction.

Answer 2

$X^4+X^3+1$ is irreducible, because $0+0+1=1=1+1+1$ in $\mathbb{F}_2$

Not true: or, at least, the part about being irreducible because it doesn't contain any linear factors. A polynomial can have no linear factors and still be reducible. Consider \begin{align} p = \left( X^2 + X + 1 \right) \in \mathbb{F}_2[X] \end{align}

Then certainly $p^2$ has no linear factors and is reducible.

To prove reducibility in a finite field, the only way that (I) know of is to find all reducible polynomials of degree (in this case) 3 or less. If none of those polynomials divide your polynomial, then its reducible.

The reasoning being that, if your polynomial was reducible in $\mathbb{Z}$, then $p=fg$ and thus, can be factored in $\mathbb{F}_p$, $p$ a prime. Since no such factorization exists in $\mathbb{F}_2$ (assuming that your polynomial over $\mathbb{F}_2$ is actually irreducible), there is no way that it can factor over $\mathbb{Z}$.