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# Help with Abstract Linear Algebra?

Please find [Q((√7 , √5) : Q] by finding f(x) such that Q (√7 , √5) ≅ Q[x]/(f(x)),

### 1 Answer

- kbLv 77 years agoFavorite Answer
Claim: Q(√5,√7) = Q(√5+√7).

Proof:

Clearly, Q(√5+√7) is contained in Q(√5, √7).

For the other containment, note that (√5+√7)^2 = 12 + 2√35 is in Q(√5, √7).

==> √35 is in Q(√5, √7).

==> √35 (√5+√7) = 5√7 + 7√5 is in Q(√5, √7)

==> (5√7 + 7√5) - 5(√5 + √7) = 2√5 and thus √5 is in Q(√5, √7)

==> (√5+√7) - √5 = √7 is in Q(√5, √7).

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To find f:

x = √5 + √7

==> x^2 = 12 + 2√35, by squaring both sides

==> (x^2 - 12)^2 = (2√35)^2 = 140

==> x^4 - 24x^2 + 4 = 0.

So, we let f(x) = x^4 - 24x^2 + 4.

This is irreducible over Q, because it clearly has no linear factors (the factors of 4, the only possible Q zeros by the Rational Root Theorem are checked to not be zeros). As for irreducible quadratic factors, we only need to check polynomials with Z-coefficients (by Gauss' Lemma).

x^4 - 24x^2 + 4 = (x^2 + ax + b)(x^2 + cx + d) for some a,b,c,d in Z

==> x^4 - 24x + 4 = x^4 + (a+c)x^3 + (b+ac+d)x^2 + (ad+bc)x + bd

Check that equating like coefficients yields no integer solutions.

Hence, f is irreducible over Q and is thus the minimal polynomial for √5+√7.

==> [Q(√5,√7):Q] = [Q(√5+√7):Q] = deg f = 4.

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I hope this helps!