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Find an irreducible polynomial defining the extension Q(3^1/2, 5^1/2).

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Solution Summary

An irreducible polynomial defining an extension is found.

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Let K=Q(sqrt(3),sqrt(5))
Degree of K:Q=4
Let a=sqrt(3)+sqrt(5) (1),

THEN we'll prove that K=Q(a)

We have:a^2= 8+2*sqrt(15)
a^4= 124+ 32*sqrt(15)=16*a^2-4 (from ...

Solution provided by:
Chris Busu, PhD (IP)
Education
  • BSc, University of Bucharest
  • MSc, Ovidius
  • MSc, Stony Brook
  • PhD (IP), Stony Brook
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