I need to prove the following:
"Prove that is irreducible in if and only if is irreducible in "
I want to prove "if"; that is, irreducible irreducible.
So, I think we have that , with one of .
The hint states that we should argue by contradiction. So, in this instance, we shall assume that is reducible, and hence
we say, , and so
with , by the division algorithm. But, this is a contradiction, and so . Thus,
which contradicts the fact that is a unique factorisation domain.
I think this is wrong. I'd be really grateful if anyone could pick out the flaws in this proof, and if possible, provide a better (or, correct) proof.
Thanks in advance,
First of all ! Now define by . Since it means . Say that where and . This means, but , , . It follows that and , therefore is not irreducible. We have proved that if is reducible then is reducible, taking the contrapositive we have proven that if is irreducible then is irreducible. To prove the converse I am going to give you a hint . Define by (again ). Notice that .
and I think you apply Eisensteins criterion to finish the proof.
That's just a sketch of what I'd do. If it's correct, then I'll fill in the details and post the other proof and this to this thread.