# Probably a simple explanation

• Mar 31st 2010, 09:45 AM
Ryaη
Probably a simple explanation
Could someone explain why $2x-10$ is not an irreducible in $Z[x]$ but is an irreducible in $Q[x]$.
• Mar 31st 2010, 12:02 PM
Opalg
Quote:

Originally Posted by Ryaη
Could someone explain why $2x-10$ is not an irreducible in $Z[x]$ but is an irreducible in $Q[x]$.

It factorises as 2(x–5) in both cases. However, 2 is a unit in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$.
• Mar 31st 2010, 08:55 PM
Ryaη
Quote:

Originally Posted by Opalg
It factorises as 2(x–5) in both cases. However, 2 is a unit in $\mathbb{Q}[x]$ but not in $\mathbb{Z}[x]$.

Why would that not be the other way around? (since it is irreducible in $Q[x]$ and reducible in $Z[x]$)
• Apr 1st 2010, 12:11 AM
Opalg
Quote:

Originally Posted by Ryaη
Why would that not be the other way around? (since it is irreducible in $Q[x]$ and reducible in $Z[x]$)

Reducible means having a nontrivial factorisation. Irreducible means having no nontrivial factorisation. Nontrivial means that neither of the factors is a unit. If one of the factors is a unit then the factorisation doesn't count as a nontrivial factorisation. Since 2 is a unit in $\mathbb{Q}$, the factorisation 2x – 10 = 2(x – 5) doesn't count as nontrivial. That is why 2x – 10 is irreducible in $\mathbb{Q}[x]$.