Probably a simple explanation

**Ryaη** · March 31st 2010, 09:45 AM

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

**Opalg** · March 31st 2010, 12:02 PM

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 $\mathbb{Z}[x]$ .

**Ryaη** · March 31st 2010, 08:55 PM

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 $Z[x]$ )

**Opalg** · April 1st 2010, 12:11 AM

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 $\mathbb{Q}[x]$ .

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