show that $x^p - x + 1$ is irreducible for p prime over Q(X)
#Irreducible polynomials
violet fox
tired gateBOT
tautau
jolly rainBOT
show that $x^p - x + 1$ is irreducible for p prime over Q(X)
- Do not ping the Moderators, unless someone is breaking the rules.
- Do not ping the Helper Moderators, unless there is a conflict between helpers.
- Do not ping other members randomly for help.
- Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
- Wait patiently for a helper to come along.
- If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:
+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:
exotic steppe
usually one asks whether this polynomial is irreducible over Z_p
what do you mean by "irreducible over Q(X)"?
@violet fox
violet fox
rational numbers
violet fox
yeah i know but i dont kno how
exotic steppe
assume f is reducible over Q, what does that mean?
violet fox
yeah there would be a epresentation of f= gq
exotic steppe
such that?
violet fox
degree of both are smaller than degree of f
exotic steppe
correct, now what if you took the same product of polynomials modulo p?
violet fox
degree smaller than p
exotic steppe
conclusion?
violet fox
contradiction becuase f hast degree p
exotic steppe
contradiction with what?
violet fox
that deg g + deg q = p
exotic steppe
why is that a problem?
violet fox
because deg f = deg g + deg q if f was reducible
exotic steppe
yes, if we assume for a contradiction f was reducible over Q, then it would follow f is also reducible over Z_p
now show that x^p - x + 1 is irreducible over Z_p and you're done
this is a known exercise
violet fox
so if i now take that polynomial mod p then it would be constant 1 because of langrange theorem
exotic steppe
so if i now take that polynomial mod p then it would be constant 1 because of la...
huh
polynomial modulo p means the coefficients are calculated mod p, that's all
violet fox
oh true
and how do i then show it is irreducible over Z_p
sorry my brain seems blocked
exotic steppe
no rush, we can continue some other time
violet fox
maybe because over Z_p it has no zero
violet fox
well its also primitive
exotic steppe
there are many paths to victory
violet fox
well the polynomial has no roots of it would be reducible i could maybe show that eather g or q has a root and that would contradict
exotic steppe
no roots just shows there are no linear factors
but that doesn't imply irreducible
Show by induction that if g(x) is a monic irreducible factor of f(x), then g(x+n) is also a factor where n in N
f(x) = x^p - x + 1 in this case
violet fox
okay thanks i will try
nimble ginkgoBOT
@violet fox