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
where
Now,
and so,
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,
HTale.
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 .
I think in this instance, we have that , and so if we set , we get
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.
HTale
Yes . The only issue I will take with your proof is that you wrote . This is not a geometric sequence, these are just polynomials (I mean we can add and multiply polynomials we cannot divide them). What I would do to be more formal is say that - this can be easily seen by multiplying out the LHS. Now if you let then you get . We can "cancel" the (I say "cancel" because we are not really dividing since there is no such thing as division of polynomials, however, it is as if we are canceling) to get . Now apply Eisenstein criterion by using the fact that divides if .
If the polynomials have integer coefficients then you can define to get a transformed polynomial. This transformed polynomial is irreducible if and only if the original polynomial was irreducible by following the exact same proof except is replaced by . We need to avoid the case because we cannot define an inverse transform because we are in the integers.