# Probably a simple explanation

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

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

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

Why would that not be the other way around? (since it is irreducible in \$\displaystyle Q[x]\$ and reducible in \$\displaystyle 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 \$\displaystyle Q[x]\$ and reducible in \$\displaystyle 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 \$\displaystyle \mathbb{Q}\$, the factorisation 2x – 10 = 2(x – 5) doesn't count as nontrivial. That is why 2x – 10 is irreducible in \$\displaystyle \mathbb{Q}[x]\$.