Let $F$ be a field and $P(x),Q(x)$ be distinct irreducible monic polynomials in $F$. Then how do i prove that $P(x),Q(x)$ are relatively prime?

I can prove this for an infinite field $F$ so that $P(x),Q(x)$ have a unique representation of polynomial respectively, but what if $F$ is finite?

  • $\begingroup$ What proof depends on the size of $F$? $\endgroup$ – Math Gems Feb 4 '13 at 13:50
  • $\begingroup$ @Math This is an exercise in my book, so I'm trying to prove it myself. There is no reference. $\endgroup$ – Katlus Feb 4 '13 at 13:55
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    $\begingroup$ But you say "I can prove ...". What proof is that? $\endgroup$ – Math Gems Feb 4 '13 at 14:06
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    $\begingroup$ It would really be good if you included the proof that Math Gems is asking for in the post. We're kind of surprised you discovered a split in the finite/infinite case, and we are (or at least I am) thinking there might be a misconception. If there is some misconception in the proof we would like to help you clear it up. If you remain silent then you and I won't ever know if there was a problem :) $\endgroup$ – rschwieb Feb 4 '13 at 14:10
  • $\begingroup$ @rschwieb I was confused with the notion "$=$". Is there a special notation to distinguish "Equivalence in $F[X]$" from "equivalence in $F$"? I don't understand why one does not use some equivalence notation such as $\equiv$ for polynomials.. $\endgroup$ – Katlus Feb 4 '13 at 20:46

Hint $\ $ Write $A\sim B$ for $A$ is associate to $B$, i.e. $A = u B$ for some unit $u$. If $P$ is irreducible then the gcd $(P,Q)\neq 1\:\Rightarrow (P,Q)\sim P,\:$ So if $Q$ is also irreducible then $(P,Q)\sim Q,\:$ hence $P\sim Q.$ Therefore $P = Q$ (since both are monic), contra hypothesis.


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