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Irreducible polynomial on polynomial ring |
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| Nov8-05, 08:20 PM | #1 |
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Irreducible polynomial on polynomial ring
How would I prove that [itex]x^2+1[/itex] is irreducible in [itex] Z_p[x][/itex], where p is an odd prime of the form 3+4m.
I know that for it to be rreducible, it has to have roots in the ring. So [itex]x^2=-1 (mod p)[/itex]. Or [itex] x^2+1=k(3+4m) [/itex], for some k. I tried induction on m, but it does not work because [itex}x^2+1[/itex] is only reducible on [itex] Z_p[x][/itex] if p is prime, which is not the case for all m. Apperently, there exists a two-line solution. Any tips would be appreciated. |
| Nov8-05, 08:47 PM | #2 |
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Hint: Fermat's little theorem and this lemma: if R is a commutative ring with identity, and a in R is invertible, then a^n=1 and a^m=1 => a^gcd(n,m)=1.
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