Let a element C. Prove: If a is an algebraic number of degree n over Q, then a +

Question

Let a element C. Prove: If a is an algebraic number of degree n over Q, then a + 5 is an algebraic number of degree n over Q. [Hind: Since a is an algebraic number of degree n, there is an irreducible Q-polynomial p(x) of degree n such that p(a) = 0. To show that a + 5 is an algebraic number, and has the degree n over Q, we must show that p(x - 5) is irreducible and has degree n (we already showed a + 5 is a zero of p(x - 5) above). Look over the Unit 14 for the result connecting irreducibility of polynomials p(x) and p(x - 5). Look over the Unit 13 for the results connecting degrees of polynomials p(x) and p(x - 5).]

Explanation / Answer

Write c for alpha throughout.

Given c is an algebraic number of degree n over Q.

To show that c+5 is also algebraic over Q of degree n over Q.

Let P(x) be an irreducible polynomial satisfied by c over Q.

Consider Q(x) = P(x-5)................................(1)

Clearly Q(c+5) =P(c+5-5) =P(c)=0.

So c+5 is an algebraic number , as Q(x) is a polynomial with rational coefficients., clear from (1)

Claim : Q(x) is irreducible

If Q(x) = R(x)S(x) with nontrivial factorization, then

P(x-5) =R(x)S(x)

hence

P(y) = R(y+5)S(y+5)

If we set U(y) =R(y+5) and V(y) =S(y+5), then

P(y) =U(y)V(y) would be a non-trivial factorization of P(y) (for example, consider the degrees of U and V ,which are the same as those of R and S).

This would contradict the assumption that P(y) is irreducible.

Hence Q(x) is irreducible of degree n, and we are done

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