# is f irreducible

• May 25th 2009, 09:37 AM
georgel
is f irreducible
Is $\displaystyle f(x)=x^3+15x^2+8x+6$ irreducible in $\displaystyle \mathbb{Z}[x]$, $\displaystyle \mathbb{Z}[[x]]$, $\displaystyle \mathbb{Z}_5[x]$?

I've tried doing the $\displaystyle \mathbb{Z}[x]$, but we've only mentioned Eisenstein's criterion and I can't apply it here.

Thank you.
• May 25th 2009, 10:33 AM
NonCommAlg
Quote:

Originally Posted by georgel

Is $\displaystyle f(x)=x^3+15x^2+8x+6$ irreducible in $\displaystyle \mathbb{Z}[x]$, $\displaystyle \mathbb{Z}[[x]]$, $\displaystyle \mathbb{Z}_5[x]$?

I've tried doing the $\displaystyle \mathbb{Z}[x]$, but we've only mentioned Eisenstein's criterion and I can't apply it here.

Thank you.

the degree of $\displaystyle f$ is 3. so it's irreducible over $\displaystyle \mathbb{Z}$ iff it has no integer root. now suppose $\displaystyle x^3+15x^2+8x+6=0.$ then $\displaystyle x(x^2+15x+8)=-6$ and thus if $\displaystyle x \in \mathbb{Z},$ then we must have

$\displaystyle x=\pm 1, \pm 2, \pm 3, \pm 6.$ check these cases! the polynomial is reducible in $\displaystyle \mathbb{Z}[[x]]$ because for example: $\displaystyle f(x)=(1-x)(6+14x+29x^2 + 30x^3 + 30x^4 + 30x^5 + \cdots ).$

finally since $\displaystyle f(1) \equiv 0 \mod 5,$ we have $\displaystyle x-1 \mid f(x)$ in $\displaystyle \mathbb{Z}_5[x]$ and thus $\displaystyle f$ is reducible over $\displaystyle \mathbb{Z}_5.$
• May 25th 2009, 11:12 AM
georgel
Quote:

Originally Posted by NonCommAlg
the degree of $\displaystyle f$ is odd. so it's irreducible over $\displaystyle \mathbb{Z}$ iff it has no integer root.

I didn't know this. Could you possibly link the theorem and its proof, if it's available somewhere online?
Maybe it's obvious but I don't see it. :(

And thank you so much, again!
• May 25th 2009, 11:41 AM
georgel
Quote:

Originally Posted by NonCommAlg
the polynomial is reducible in $\displaystyle \mathbb{Z}[[x]]$ because for example: $\displaystyle f(x)=(1-x)(6+14x+29x^2 + 30x^3 + 30x^4 + 30x^5 + \cdots ).$

And I need further help with this as well. :( How did you get this above?

Thank you if you have the time to help, if not, I'm already grateful for all your help!
• May 25th 2009, 11:50 AM
NonCommAlg
Quote:

Originally Posted by georgel
I didn't know this. Could you possibly link the theorem and its proof, if it's available somewhere online?
Maybe it's obvious but I don't see it. :(

And thank you so much, again!

sorry, i meant 3 not "odd". i don't know why i wrote odd?? well, there are only two ways to factorize a polynomial of degree 3: all factors of degree 1 or one degree 1 and one degree 2.

so either way, there will be a factor of degree 1 and therefore a root.