Question

find the degree of √2+√5 over Q

Answer 1

(a): Find the degree of α =
√
2 + √
3 over Q, and also find its minimal
polynomial.
Answer: Note that
α
4−10α
2+1 = √
2 + √
3
4
−10 √
2 + √
3
2
+1 = (49−20√
6)−10(5+2√
6)+1 = 0,
so α satisfies the polynomial f(x) = x
4 − 10x
2 + 1 ∈ Q[x]. Now, by
Math 602 PS7#5(d), the only possible rational roots of f are ±1,
and f(1) = f(−1) = −8, so f is irreducible over Q. Hence, f is the
minimal polynomial of √
2 + √
3 over Q and so deg α = deg f = 4.
♣

Answer 2

this is just a hint I think you can do it////Note that
α
4−10α
2+1 = √
2 + √
3
4
−10 √
2 + √
3
2
+1 = (49−20√
6)−10(5+2√
6)+1 = 0,
so α satisfies the polynomial f(x) = x
4 − 10x
2 + 1 ∈ Q[x]. Now, by
Math 602 PS7#5(d), the only possible rational roots of f are ±1,
and f(1) = f(−1) = −8, so f is irreducible over Q. Hence, f is the
minimal polynomial of √
2 + √
3 over Q and so deg α = deg f = 4.