MATH 113 Final: MATH113 Final Exam 2015 Fall
MATH 113: SAMPLE FINAL SOLUTIONS AND HINTS
1. Note ϕ(21) =12. Thus Euler theorem says 312 ≡1 mod 21. Then 375 ≡
(312)6×33≡6 mod 21. Note gcd(15, 21)|3. Thus the equation has exactly 3
solutions.
2. (a) Direct calculation (b) No. Suppose such γexists, then σand τare in the
same conjugate class, thus have the same order. Check: ord(τ)=10, ord(σ)=12.
Contradiction!
3. FFTTFTF
4. Prove by definition.
5. α2=2+√2, (α2−2)2=2, thus α4−4α2+2=0. By Eisenstein criteria,
this is irreducible over Q. One basis is {1, α, α2, α3}. For the last question, we can
use the same argument as in the solution of HW 10, question 6.
6. Note the equation has no integral solutions. If it is reducible, then it should
be a product of two polynomials with degree 2(integral coefficients). Expand it
and compare the coefficients. One can show that this is not possible, by direct
argument.
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Document Summary
Math 113: sample final solutions and hints: note (21) = 12. Thus euler theorem says 312 1 mod 21. Then 375 (312)6 33 6 mod 21. Thus the equation has exactly 3 solutions: (a) direct calculation (b) no. Suppose such exists, then and are in the same conjugate class, thus have the same order. Contradiction: ffttftf, prove by de nition, 2 = 2 + By eisenstein criteria, this is irreducible over q. For the last question, we can use the same argument as in the solution of hw 10, question 6: note the equation has no integral solutions. If it is reducible, then it should be a product of two polynomials with degree 2(integral coe cients). One can show that this is not possible, by direct argument.
