## Minimal polynomials

Dr Kim,

Could you tell me where i am going wrong here:

Homework Sheet 3 – Question 4 part 2

$a:=(1+\sqrt{8})/2$

$2a = 1 + \sqrt{8}$

$2a - 1 = \sqrt{8}$

$1/2(2a - 1) = \sqrt{2}$

$1/4(2a - 1)^2 = 2$

$a^2 - a - 7/4 = 0$

Irreducible since $Q(a)= Q(\sqrt{2})$ has degree 2. Not an Algebraic Integer.

$N=4$

This is not the same answer as in the solutions.

Also could you explain briefly why $Q(a)$ having degree 2 implies irreducibility. And that even if the answer is $X^2 - X - 31/4$ as in the solutions, N=2 rather than 4? (I would think that multiplying by 2 does not give a monic polynomial with integer coefficients or would 2 do this?)

—————————————————————————-

For the minimal polynomial, your answer is correct. The solutions contain a misprint.

For any algebraic number $\alpha$, we have

$Q(\alpha)\simeq Q[x]/(m(x))$

where $m(x)$ is the minimal polynomial of $\alpha$. So the degree of $m(x)$ is the same as the degree of the field extension. In the case of $a$, we see that $m(x)$ must have degree 2. Since it divides $x^2-x-7/4$ and both are monic, we then must have $m(x)=x^2-x-7/4$. In particular, the polynomial is irreducible. Of course in this case, it’s easy to see directly that it has no rational roots, and hence is irreducible. But if you think about it, even in general, if you know that the degree of the field extension $Q(\alpha)$ of $Q$ is $d$ and $f(x)\in Q[x]$ is a monic polynomial of degree $d$ such that $f(\alpha)=0$, then $f(x)$ must be the minimal polynomial and hence irreducible.

To get the $N$, you’ve already noted that $b=2a$ satisfies the monic integral polynomial equation

$(x-1)^2=8$

and hence, is an algebraic integer.